BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Class-XII IIT-JEE Advanced Physics Study Package Session: 2014-15 Office: Rajopatti, Dumra Road, Sitamarhi (Bihar), Pin-843301 Ph.06226-252314 , Mobile:9431636758, 9931610902 Website: www.brilliantpublicschool.com; E-mail:
[email protected] STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS-XII Chapters: 1. Electrostatics 2. Capacitance 3. Current Electricity 4. Thermal and Chemical Effects of Electric Current 5. Magnetic Effects of Electric Current 6. Electromagnetic Induction and Alternating Current 7. Optics 8. Optical Instruments 9. Wave Optics 10. Modern Physics 11. Semiconductors Electronics STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P1. Electrostatics Index: 1. Key Concepts 2. Exercise I 3. Exercise II 4. Exercise III 5. Exercise IV 6. Answer Key 7. 34 Yrs. Que. from IIT-JEE 8. 10 Yrs. Que. from AIEEE 1 1 dq ˆ Continuous charge distribution E = r = d E . ELECTRIC CHARGE Charge of a material body is that possesion (acquired or natural) due to which it strongly interacts with other material body. Due to q →0 charge induction on the source of electric field.. S. surface and volume charge densities respectively. unit is coulomb. It can be postive or negative.. ε0εr = Absolute permittivity of the medium εr = 1 for air (vacuum) = ∞ for metals COULOMB’S LAW : F = NOTE : The Law is applicable only for static and point charges. NOTE : The force due to one charge is not affected by the presence of other charges.I.. Charge is quantized. conserved. 2. 4.. 1 q1q 2 1 q1q 2 F = r where . Ex = & Ey = at a point above the end of wire at r r r an angle 45º . Inductive Capacity = Dielectric Const. ELECTRIC INTENSITY OR ELECTRIC FIELD STRENGTH (VECTOR QUANTITY) “The physical field where a charged particle.. and additive. unit is the force experienced by the particle carrying unit charge E = q → 0 q V/m here Lim represents that this charge does not alter the magnitude of electric field. r (iv) Semi ∞ line of charge E = kλ kλ 2 kλ as ... S.. σ & ρ are linear.1.. experiences force is called an electric field”. 3. induction may change the charge distribution. 2 Page 2 of 16 ELECTROSTATICS KEY CONCEPTS . irrespective of the fact whether it is in motion or at rest. ELECTRIC FIELD. In vector form 4πε 0ε r r 3 4πε 0ε r r 2 ε0 = permittivity of free space = 8. Note E ≠ ∫ dE because E is a vector quantity .I. dE = electric field due to an elementry charge 4πε 0 ∫ r 2 ∫ 5. .85 × 10−12 N−1 m−2 c2 or F/m and εr = Relative permittivity of the medium = Spec. (i) ELECTRIC FIELD DUE TO 1 q 1 q r (vector form) rˆ = Point charge : E = 2 4π ∈0 r 3 4πε 0 r (ii) Where r = vector drawn from the source charge to the point . (iii) dq = λ dl (for line charge) = σ ds (for surface charge) = ρ dv (for volume charge) In general λ. The direction of the field is the direction of the force experienced by a positively charged particle & the magnitude of the field (electric intensity) is F Lim unit is NC–1 . 2k λ Infinite line of charge E = where r = perpendicular distance of the point from the line charge . PRINCIPLE OF SUPER POSITION Force on a point charge due to many charges is given by F=F1+F2 +F3 +. Only applicable to static charges as moving charges may result magnetic interaction also and only for point charges as if charges are extended. UNSTABLE EQUILIBRIUM : If charge is displaced by a small distance the charge does not return to the equilibrium position.Uniformly charged ring . If an ELF is originated. 4πε 0 r 2 Behaves as a point charge situated at the centre for these points Ein = Qr ρr = . Quantity of ELF originated or terminated from a charge or on a charge is proportional to the magnitude of charge. r ≥R. r ≥ R conducting sphere . (i) (ii) uniformly charged cylinder with a charge density ρ is -(radius of cylinder = R) for r < R ρr ρR 2 Em = 2 ∈ . ELECTROSTATIC EQUILIBRIUM Position where net force (or net torque) on a charge(or electric dipole) = 0 STABLE EQUILIBRIUM : If charge is displaced by a small distance the charge comes (or tries to come back) to the equilibrium . Eaxis = (vi) Electric field is maximum when (vii) kQx ( x + R 2 )3 / 2 2 dE = 0 for a point on the axis of the ring. it must require termination either at a negetive charge or at ∞ . where σ is surface charge density ∞ charged conductor sheet having surface charge density σ on both surfaces E = σ/ε0 . Ecentre = 0 . 3 Page 3 of 16 ELECTROSTATICS (v) . Eout = 4πε 0 r 2 Behaves as a point charge situated at the centre for these points E in = 0 . Q Uniformly charged solid sphere (Insulating material) E out = . Preference of termination is towards a negative charge . ELF originates from positive charge or ∞ and terminate on a negative charge of infinity . Here we get x = R/√2. n (viii) (ix) (x) Just outside a conducting surface charged with a surface charge density σ. dx σ Infinite non conducting sheet of charge E = nˆ where 2ε 0 = unit normal vector to the plane of sheet. for r > R E= 2 ∈0 r 0 Uniformly charged cylinderical shell with surface charge density σ is ρr for r < R Em = 0 . r < R (xii) (xiii) 6. for r > R E = ∈ r 0 ELECTRIC LINES OF FORCE (ELF) The line of force in an electric field is a hypothetical line. tangent to which at any point on it represents the direction of electric field at the given point. electric field is always given as E = σ/∈0. (i) (ii) (iii) (iv) (v) 7. 3 4πε 0 R 3ε 0 r ≤ R where ρ = volume charge density (xi) Uniformly charged spherical shell (conducting or non-donducting) or uniformly charged solid Q . Properties of (ELF) : Electric lines of forces never intersects . . (r ≤ R) 4 πε 0 r 4πε0 R non conducting uniformly charged solid sphere : 11. between point A & B .I.d. 10. Used when EF varies in three dimensional coordinate system. Vin = . 4πε 0 r many charges V = q1 + q2 + q3 4πε 0 r1 4 πε 0 r2 4 πε 0 r3 + . (r ≥ R) . q WBA = w. (r ≥ R) . (ii) Q . we use – B →→ − VA – VB = ∫ E . unit is volt ( = 1 J/C) q POTENTIAL DIFFERENCE VAB = VA − VB = ( WBA ) ext VAB = p. It is called as equipotential region like conducting bodies..ELECTRIC POTENTIAL (Scalar Quantity) “Work done by external agent to bring a unit positive charge(without accelaration) from infinity to a point in an electric field is called electric potential at that point” . Potential of the whole region must remain constant as no work is done in displacement of charge in it. ELECTRIC FIELD & ELECTRIC POINTENIAL ˆ∂ ˆ∂ ˆ∂ E = − grad V = − ∇ V {read as gradient of V} grad = i + j +k ∂x ∂y ∂z .. . The region where E = 0. 4 Page 4 of 16 ELECTROSTATICS 8. ( W∞r ) ext . (v) Vout = 12.. (r ≤ R) 2 4πε 0 R EQUIPOTENTIAL SURFACE AND EQUIPOTENTIAL REGION In an electricfield the locus of points of equal potential is called an equipotential surface. (i) POTENTIAL DUE TO Q a point charge V = 4πε 0 r (iii) continuous charge distribution V = (iv) spherical shell (conducting or non conducting) or solid conducting sphere Q Q Vout = . by external source to transfer a point charge q from B to A (Without acceleration).. An equipotential surface and the electric field meet at right angles. 1 dq ∫ 4πε 0 r Vin = 1 Q(3R 2 −r 2 ) . S. For finding potential difference between two points in electric field. If W∞ r is the work done to bring a charge q (very small) from infinity to a point then potential at that point is V = 9.d.dt if E is varying with distance A if E is constant & here d is the distance between points A and B. (ii) If E is not uniform throughout the area A . dE d ˆ Force on a dipole when placed in a non uniform electric field is F=− − P. (i) ELECTRIC FLUX For uniform electric field. dx dx ( ) 16. F = 0 . A = EA cos θ where θ = angle between E & area vector ( A ). (e) (f) (g) Electric Dipole in uniform electric field : torque τ=px E . (a) ELECTRIC DIPOLE. φ = E .E.dA 5 Page 5 of 16 ELECTROSTATICS 13. Then sinθ and cosθ will be interchanged. of charge q in potential field U = qV. Interaction energy of a system of two charges U = q 1 V 2 = q 2 V1 . then φ = ∫ E.MUTUAL POTENTIAL ENERGY OR INTERACTION ENERGY “The work to be done to integrate the charge system .E . of an electric dipole in electric field U = − p. Flux is contributed only due to the component of electric field which is perpendicular to the plane. (b) (c) (d) 0 r = distance of the point from the centre of dipole p p ≈ − on the equitorial .” qq For 2 particle system Umutual = 1 2 4πε 0 r For 3 particle system Umutual = For n particles there will be q q q q q1q 2 + 2 3 + 3 1 4πε 0 r12 4πε 0 r23 4πε 0 r31 n (n −1) terms . . θ) in polar co-ordinate system is 2kp sin θ Radial electric field Er = r3 kp cosθ Tangentral electric field ET = r3 2 2 kp 2 Net electric field at P is Enet = E r + E T = 3 1 + 3 sin θ r kp sin θ Potential at point P is VP = r2 NOTE : If θ is measured from axis of dipole.E i = P. Total energy of a system = Uself + Umutual 2 14.E. Work done in rotation of dipole is w = PE (cos θ1 − cos θ2) P.r Dipole V = = p=qa electric dipole moment . Pθ p. 15. P. ˆi . E = 4πε 0 [ r 2 + (a 2 / 4)]3 / 2 4πε 0 r 3 At a general point P(r. If θ is angle between p and 2 3 4πε 0 r 4πε 0 r reaches vector of the point. O is mid point of line AB (centre of the dipole) on the axis (except points on line AB) pr p E= ≈ 2πε [r 2 − (a 2 / 4)]2 2πε r 3 ( if r < < a) 0 p = q a = Dipole moment . e. CONCEPT OF SOLID ANGLE : Flux of charge q having through the circle of radius R is q q / ∈0 x Ω = 2 ∈ (1 – cosθ) φ= 4π 0 ε E2 Energy stored p.u. 2ε 0 Electric pressure on a charged surface with charged density σ due to external electric field is Pele = σE1 IMPORTANT POINTS TO BE REMEMBERED (i) Electric field is always perpendicular to a conducting surface (or any equipotential surface) . in the direction of electric field) and negative charge from lower to higher (i. ε0 ε0 φ does not depend on the 18. When p||E the dipole is in stable equilibrium p||( − E ) the dipole is in unstable equilibrium (viii) (ix) σ ε0 (x) When a charged isolated conducting sphere is connected to an unchaged small conducting sphere then potential (and charge) remains almost same on the larger sphere while smaller is charged . . (vii) Positive charge flows from higher to lower (i.” φ = ∫ EdA = q = net charge enclosed by the closed surface . No tangential component on such surfaces . 19. opposite to the electric field) potential . between two points in an electric field does not depend on the path joining them . 20.e.GAUSS’S LAW (Applicable only to closed surface) “ Net flux emerging out of a closed surface is q q . the charge resides only on the surface. volume in an electric field = 0 2 σ2 Electric pressure due to its own charge on a surface having charged density σ is Pele = .d.e ρ depends only on r} behaves as if its charge is concentrated at its centre (for outside points). (iii) When a conductor is charged. 5R A spherically symmetric charge {i. (xi) Self potential energy of a charged shell = (xii) (xiii) (xiv) KQ 2 . (ii) Charge density at sharp points on a conductor is greater. (vi) Potential at a point due to positive charge is positive & due to negative charge is negative. (i) (ii) shape and size of the closed surface The charges located outside the closed surface. (iv) For a conductor of any shape E (just outside) = (v) p. 6 Page 6 of 16 ELECTROSTATICS 17. 2R 3k Q 2 . Self potential energy of an insulating uniformly charged sphere = Dielectric strength of material : The minimum electric field required to ionise the medium or the maximum electric field which the medium can bear without breaking down. Find the maximum angle through which the string is deflected from vertical. Q.6 The figure shows three infinite non-conducting plates of charge perpendicular to the plane of the paper with charge per unit area + σ. Q.4 10−9 C is located at the origin in free space & another charge Q at (2. 0). Find work done against electric forces in expanding it upto radius 2R.2 (a) Two particles A and B each carrying a charge Q are held fixed with a separation d between then A particle C having mass m ans charge q is kept at the midpoint of line AB. find the time period of the oscillations of C. Q. Find the velocity of this particle at the moment it passes through the centre of the ring. If the X−component of the electric field at (3. + 2σ and – σ.8 In the figure shown S is a large nonconducting sheet of uniform charge density σ.1 A negative point charge 2q and a positive charge q are fixed at a distance l apart. Where should a positive test charge Q be placed on the line connecting the charge for it to be in equilibrium? What is the nature of the equilibrium with respect to longitudinal motions? Q. of the particle if the amplitude does not exceed R. when (i) both are + ve (ii) both are – ve (iii) a is + ve and b is – ve (iv) a is – ve and b is + ve Q. If it is displaced through a small distance x (x << d) perpendicular to AB. Find the ratio of the net electric field at that point A to that at point B. 1.7 A thin circular wire of radius r has a charge Q. Q. Q. Find the electric field on the line passing through O and perpendicular to the plane of the figure as a function of distance x from point O. (assume x >> a) Q. If a point charge q is placed at the centre of the ring. 1) is zero. then find the increase in tension in the wire. Q. 7 Page 7 of 16 ELECTROSTATICS EXERCISE # I . Q. Find the angular acceleration of the rod just after it is released.12 A spherical balloon of radius R charged uniformly on its surface with surface density σ. 1)? A c h a r g e + Q. Is the Y−component zero at (3.11 A charge + Q is uniformly distributed over a thin ring with radius R. calculate the value of Q.3 Draw E – r graph for 0 < r < b.M. Find the frequency of S. If suddenly a charge +q is given to the bob & it is released from the position shown in figure.5 Six charges are placed at the vertices of a regular hexagon as shown in the figure. A negative point charge – Q and mass m starts from rest at a point far away from the centre of the ring and moves towards the centre. then find the time period of the oscillations of C.H.10 A particle of mass m and charge – q moves along a diameter of a uniformly charged sphere of radius R and carrying a total charge + Q. 0.9 A simple pendulum of length l and bob mass m is hanging in front of a large nonconducting sheet having surface charge density σ. if two point charges a & b are located r distance apart. The linear charge densities on the upper and lower half of the rod are shown in the figure.Q. 1. (b) If in the above question C is displaced along AB. A rod R of length l and mass ‘m’ is parallel to the sheet and hinged at its mid point. Find its potential. then find the relation between a & R.13 A point charge + q & mass 100 gm experiences a force of 100 N at a point at a distance 20 cm from a long infinite uniformly charged wire. If it is released find its speed when it is at a distance 40 cm from wire . Find the charge attained by the inner shell. 2m) respectively. pˆ k are located at (0. Find the final charge on the sphere C. Sphere C carries no charge.26 A charge Q is uniformly distributed over a rod of length l. They are then combined to form a bigger drop. Find the minimum possible flux of the electric field through the entire surface of the cube. Q. 0). y). Q. If charge q is situated on one of the vertices of the cube.19 Three charges 0. B and C. Q. If the energy is supplied to this system at the rate of 1 kW.24 The length of each side of a cubical closed surface is l. Q. Q. 0.14 Consider the configuration of a system of four charges each of value +q.20 Two thin conducting shells of radii R and 3R are shown in figure.22 A dipole is placed at origin of coordinate system as shown in figure.21 Consider three identical metal spheres A.23 Two point dipoles p kˆ and Q. Consider a hypothetical cube of edge l with the centre of the cube at one end of the rod. Q. Find the maximum distance from A on sheet where the particle can strike. The outer shell carries a charge +Q and the inner shell is neutral. Find the resultant 2 electric field due to the two dipoles at the point (1m.17 A particle of mass m and negative charge q is thrown in a gravity free space with speed u from the point A on the large non conducting charged sheet with surface charge density σ. Q. find the electric field at point P (0. Finally the sphere C is touched to sphere B and separated from it. Each has radius r and they are charged to a potential V0. Find the closest distance of approach. If one fourth (1/4th) of the flux from the charge passes through the disc.16 Two identical particles of mass m carry charge Q each.15 There are 27 drops of a conducting fluid. Q.18 Consider two concentric conducting spheres of radii a & b (b > a). Q. 0) and (1m. 0. The inner shell is earthed with the help of switch S. What charge should be given to the outer sphere so that potential of the inner sphere becomes zero? How does the potential varies between the two spheres & outside ? Q.25 A point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc. 8 Page 8 of 16 ELECTROSTATICS Q. Inside sphere has a positive charge q1. Find the work done by external agent in changing the configuration of the system from figure (i) to fig (ii).1 coulomb each are placed on the corners of an equilateral triangle of side 1 m. Sphere C is then touched to sphere A and separated from it. Spheres A and B are touched together and then separated. then find the flux passing through shaded face of the cube.Q. how much time would be required to move one of the charges onto the midpoint of the line joining the other two? Q. as shown in figure. 0. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards the first from a large distance with an initial speed V. Spheres A carries charge + 6q and sphere B carries charge – 3q. 7 A positive charge Q is uniformly distributed throughout the volume of a dielectric sphere of radius R . The clock hands do not disturb the net field due to point charges. Q. A particle of mass m and positive charge q is projected from the point P ( ) 3 R. Q.. At what time does the hour hand point in the same direction is electric field at the centre of the dial. charge = + 2e. Q. Make the following assumptions : the charge on the electrometer is equally distributed between the bar & the rod the charges are concentrated at point A on the rod & at point B on the bar. mass of α-particle = 4m. − 3q. Charge on the small mass remains constant throughout the motion. Considering the case when the beads are stationary. as shown.. (a) (b) 9 Page 9 of 16 ELECTROSTATICS EXERCISE # II . − 12q fixed at the position of the corresponding numerals on the dial.5 2 small balls having the same mass & charge & located on the same vertical at heights h1 & h2 are thrown in the same direction along the horizontal at the same velocity v . The 1st ball touches the ground at a distance l from the initial vertical . charge=+e. Neglect any resistance other than electric interaction.6 Two concentric rings of radii r and 2r are placed with centre at origin. The reading are taken on a quadrant graduated in degrees .4 A circular ring of radius R with uniform positive charge density λ per unit length is fixed in the Y−Z plane with its centre at the origin O..0 on the positive X-axis directly towards O. find the total work required to perform all three steps. Find the minimum velocity v so that it can penetrate R/2 distance of the sphere.. A proton and an α-particle are projected with velocity v0 = Q. The length of the rod is l and its mass is m . Smaller ring is now rotated by an angle 90° about Z-axis then it is again rotated by 90° about Y-axis. what are the values of the charges for which the beads continue to remain stationary.. The distance between their initial velocities is L.. Find the smallest value of the speed v such that the particle does not return to P. The angle α. Two bead of equal masses m each and carrying charges q1 & q2 are connected by a cord of length 1 & slide without friction on the wires.2 A rigid insulated wire frame in the form of a right angled triangle ABC. (b) The tension in the cord & The normal reaction on the beads. What will be the charge when the rod of such an electrometer is deflected through an angle α . ke 2 each.3 A clock face has negative charges − q. .0.. is set in a vertical plane as shown. with initial velocity v . Q. If finally larger ring is rotated by 90° about X-axis. At what height will the 2nd ball be at this instant ? The air drag & the charges induced should be neglected. If the cord is now cut.) .1 (a) (c) Q. Find the work done by electrostatic forces in each step.8 An electrometer consists of vertical metal bar at the top of which is attached a thin rod which gets deflected from the bar under the action of an electric charge (fig... when ml they are far away from each other. − 2q.Q. Find their closest approach distance. Two charges +q each are fixed at the diametrically opposite points of the rings as shown in figure. A point mass having charge + q and mass m is fired towards the centre of the sphere with velocity v from a point at distance r (r > R) from the centre of the sphere. mass of proton=m. Q. determine. 10 Page 10 of 16 ELECTROSTATICS Q. the direction of the velocity of the first ball changes by 60º and the magnitude is reduced by half . Show that: R (a) the total charge on the sphere is Q = π ρ0 R3 and 2 (b) the electric field inside the sphere has a magnitude given by.10 Two identical balls of charges q1 & q2 initially have equal velocity of the same magnitude and direction. Find the electric field at a distance r from the axis for (a) r < a (b) a < r < b (c) r > b Q. What is the potential difference between the 2 balls? Total charge of balls is constant. Find the ratio between the charges on the 2 balls at which electrostatic energy of the system is minimum.9 .11 Electrically charged drops of mercury fall from altitude h into a spherical metal vessel of radius R in the upper part of which there is a small opening. Determine the charge q of each part. An electron e is kept inside the cavity at an angle θ = 45° as shown . x = a. The kinetic energies of the released balls are found to differ by K at a sufficiently long distance from the polygon. The charged density i. In what proportion will the velocity of the second ball changes ? Q. charge per unit length is λ. E = KQr .gon with side a.e. ρ = ρ0 . E x Q. Given that the (n + 1)th drop just fails to enter the sphere. The cylinders have equal and opposite charges per unit length λ .14 2 small metallic balls of radii R1 & R2 are kept in vacuum at a large distance compared to the radii. At a certain instant. The distance between the centres of the sphere and the cavity is a .12 Small identical balls with equal charges are fixed at vertices of regular 2004 . It is then placed on a rough nonconducting horizontal surface plane. After a uniform electric field is applied for some time. Q.13 The electric field in a region is given by E = 0 i . What is the number 'n' of last drop that can still enter the sphere. At time t = 0. Q. How long will it take to touch the sphere again? Q. where ρ0 is a constant and r is the distance from the centre of the sphere. Take E0 = 5 × 103N/C. having a volume charge density of ρ. y = a. R4 Q. z = 0 and z = a. The direction of the velocity of the second ball changes there by 90º. The mass of each drop is m & charge is Q.15 Figure shows a section through two long thin concentric cylinders of radii a & b with a < b . one of the balls is released & a sufficiently long time interval later. Determine the friction force (magnitude and direction) acting on the ring. when it starts moving. y = 0. Find the charge contained inside a cubical volume l bounded by the surfaces x = 0. l = 2cm and a = 1cm.17 A nonconducting ring of mass m and radius R is charged as shown.A cavity of radius r is present inside a solid dielectric sphere of radius R. Q. the ball adjacent to the first released ball is freed.16 A solid non conducting sphere of radius R has a non-uniform charge distribution of volume charge r density. a uniform electric field E = E 0i is switched on and the ring start rolling without sliding. Calculate the energy required to take a test charge q from infinity to apex A of cone. An electron is released on the axis of the hole at a distance 3R from the centre.20 Find the electric field at centre of semicircular ring shown in figure. If the incident current is 100 µ A. The bobs are immersed in a liquid of relative permittivity εr & density ρ0. Find the ratio b/a if a charge particle placed on the axis at z = a is in equilibrium.Q. Find the density σ of the bob for which the angle of divergence of the strings to be the same in the air & in the liquid ? . At what angle will it terminate at − q2? 11 Page 11 of 16 ELECTROSTATICS Q. A line of force emerges from q1 at angle α with line AB. (e = 1.22 An infinite dielectric sheet having charge density σ has a hole of radius R in it.0×10−31 kg) Q. Q.19 An electron beam after being accelerated from rest through a potential difference of 500 V in vacuum is allowed to impinge normally on a fixed surface. (e = charge on electron and m = mass of electron) Q.24 Two charges + q1 & − q2 are placed at A and B respectively. one of radius 'a' and the other of radius 'b' have the charges +q and – (2 5)−3 / 2 q respectively as shown in the figure. determine the force exerted on the surface assuming that it brings the electrons to rest. m = 9.6×10−19 C . What will be the velocity which it crosses the plane of sheet. Q.21 A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. The slant length is L.18 Two spherical bobs of same mass & radius having equal charges are suspended from the same point by strings of same length.23 Two concentric rings. Q. 4 (i) (ii) (iii) Q. for 0 < r < ∞ (C) decreases as r increases. for all values of z0 satisfying 0 < z0 ≤ R (C) approximately simple harmonic. A – ly charged particle P is released from rest at the point (0..2 A metallic solid sphere is placed in a uniform electric field.5 m carries a total charge of 1. brought into contact with S2 & removed.. Then the potential at the origin potential at a point due to a charge Q at a distance r from it to be 4π∈0 r due to the above system of charges is : q n 2 q (A) 0 (B) (C) ∞ (D) 4π∈ x 8π∈0 x 0 n2 0 0 A non-conducting solid sphere of radius R is uniformly charged . 12 [ JEE '99. 1 ] Select the correct alternative : [JEE '98 2 + 2 + 2 = 6 ] A + ly charged thin metal ring of radius R is fixed in the xy−plane with its centre at the origin O . A conducting sphere S1 of radius r is attached to an insulating handle . S2 is initially uncharged . .. 7 + 1 ] Q. provided z0 << R (D) such that P crosses O & continues to move along the −ve z-axis towards x = −∞ A charge +q is fixed at each of the points x = x0. 2] Q. 0.. Then the motion of P is: (A) periodic. The value of the line integral =0 ∫ =∞ −E. The points A & B are on the cavity surface as shown in the figure. The magnitude of the electric field due to the sphere at a distance r from its centre : (A) increases as r increases.. Here x0 is a +ve constant . S1 is given a charge Q. x = 6x0. Find the electrostatic energy of S2 after n such contacts with S1. for R < r < ∞ (D) is discontinuous at r = R . . ∞ on the x-axis & a charge −q is fixed at each of the points x = 2x0.. 2] Q. 3 ] Page 12 of 16 ELECTROSTATICS E . What is the limiting value of this energy as n → ∞? [ JEE '98.1 The magnitude of electric field in the annular region of charged cylindrical capacitor (A) Is same throughout (B) Is higher near the outer cylinder than near the inner cylinder (C) Varies as (1/r) where r is the distance from the axis (D) Varies as (1/r2) where r is the distance from the axis [IIT '96. x = 4x0. The lines of force follow the path (s) shown in figure as : (A) 1 (B) 2 (C) 3 (D) 4 [IIT'96 . x = 3x0.5 (a) (b) (B) − 1 (C) − 2 (D) zero[JEE '97. for all values of z0 satisfying 0 < z0 < ∞ (B) simple harmonic.. z0) where z0 > 0 .3 A non-conducting ring of radius 0. ∞ . This procedure is repeated n times. for r < R (B) decreases as r increases. Take the electric Q .11 × 10−10 C distributed non-uniformly on its circumference producing an electric field E every where in space. x = 5x0. Then : (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is q/ε0 .. Another conducting sphere S2 of radius R is mounted on an insulating stand .6(i) An ellipsoidal cavity is carved within a perfect conductor.d (l = 0 being centre of the ring) in volts is : (A) + 2 Q.EXERCISE # III Q. A positive charge q is placed at the center of the cavity. S1 is recharged such that the charge on it is again Q & it is again brought into contact with S2 & removed. A non-conducting disc of radius a and uniform positive surface charge density σ is placed on the ground, with its axis vertical . A particle of mass m & positive charge q is dropped, along the axis of the disc, from q 4ε0 g a height H with zero initial velocity. The particle has . = σ m Find the value of H if the particle just reaches the disc . Sketch the potential energy of the particle as a function of its height and find its equilibrium position. [ JEE '99, 5 + 5 ] (ii) (a) (b) () Q.7(a) The dimension of 12 e0 E2 (e0 : permittivity of free space ; E : electric field) is : (A) M L T −1 (b) (B) M L2 T − 2 (C) M L T −2 (D) M L2 T − 1 (E) M L−1 T − 2 Three charges Q , + q and + q are placed at the vertices of a right-angled isosceles triangle as shown . The net electrostatic energy of the configuration is zero if Q is equal to : [ JEE 2000(Scr) 1 + 1] −q −2q (A) 1+ 2 (B) 2+ 2 (D) + q (C) − 2 q 27 m , − 3 m, 2 2 Four point charges + 8 µC , − 1 µC , − 1 µC and + 8 µC , are fixed at the points, − (c) + 23 m and + 27 m respectively on the y-axis . A particle of mass 6 × 10 −4 kg and of charge ge 2 + 0.1 µC moves along the − x direction . Its speed at x = + ∞ is v0 . Find the least value of v0 for which the particle will cross the origin . Find also the kinetic energy of the particle at the origin . Assume that space is gratity free. (Given : 1/(4 π ε0) = 9 × 109 Nm2/C2) [ JEE 2000, 10 ] Q.8 Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in [JEE 2001 (Scr)] (B) (A) Q.9 (C) (D) A small ball of mass 2 × 10–3 Kg having a charge of 1 µC is suspended by a string of length 0. 8m. Another identical ball having the same charge is kept at the point of suspension. Determine the minimum horizontal velocity which should be imparted to the lower ball so tht it can make complete revolution. [ J E Q . 1 0 T w a t x ( A Q.11 o e q u a l t h e o a l o n g ) x p o r i g i n . t h e T i n t h e c h a r g e s c h a n g e x - a x i s , i s a r e i n a p p r o ( B f i x e d t h e x i m ) x 2 a t x = e l e c t r i c a l a t e l y – a a n d x = p o t e n t i a l p r o p o r t i o n a l + a o n e n e r g y t h e o f Q x - a x i s . A , w h e n n o t h e r i t i s p o i n t d i s p l a c e d E c h a r g e Q b y a l l a s m 2 0 0 i s 1 ] p l a c e d d i s t a n c e t o (C) x3 (D) 1/x [JEE 2002 (Scr), 3] A point charge 'q' is placed at a point inside a hollow conducting sphere. Which of the following electric force pattern is correct ? [JEE’2003 (scr)] (A) (B) (C) (D) 13 Q.13 A charge +Q is fixed at the origin of the co-ordinate system while a small electric dipole of dipole-moment p pointing away from the charge along the x-axis is set free from a point far away from the origin. (a) calculate the K.E. of the dipole when it reaches to a point (d, 0) (b) calculate the force on the charge +Q at this moment. [JEE 2003] Q.14 Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to [JEE 2004 (SCR)] (A) q2 (B) only the positive charges (C) all the charges (D) +q1 and -q1 Q.15 Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively? [JEE 2004 (SCR)] (A) +, -, +, -, -, + (B) +, -, +, -, +, − (C) +, +, -, +, -, − (D) −, +, +, −, +, − Q.16 Two uniformly charged infinitely large planar sheet S1 and S2 are held in air parallel to each other with separation d between them. The sheets have charge distribution per unit area σ1 and σ2 (Cm–2), respectively, with σ1 > σ2. Find the work done by the electric field on a point charge Q that moves from from S1 towards S2 along a line of length a (a < d) making an angle π/4 with the normal to the sheets. Assume that the charge Q does not affect the charge distributions of the sheets. [JEE 2004] Q.17 Three large parallel plates have uniform surface charge densities as shown in the figure. What is the electric field at P. [JEE’ 2005 (Scr)] 4σ ˆ 2σ ˆ 4σ ˆ 2σ ˆ k k k (B) ∈ k (C) – (D) (A) – ∈ ∈ ∈ 0 0 0 0 Q.18 Which of the following groups do not have same dimensions (A) Young’s modulus, pressure, stress (B) work, heat, energy (C) electromotive force, potential difference, voltage (D) electric dipole, electric flux, electric field [JEE’ 2005 (Scr)] Q.19 A conducting liquid bubble of radius a and thickness t (t <<a) is charged to potential V. If the bubble collapses to a droplet, find the potential on the droplet. [JEE 2005] Q.20 The electrostatic potential (φr) of a spherical symmetric system, kept at origin, is shown in the adjacent figure, and given as q φr = ( r ≥ Ro) 4π ∈o r q ( r ≤ Ro) φr = 4π ∈o R o Which of the following option(s) is/are correct? (A) For spherical region r ≤ Ro, total electrostatic energy stored is zero. (B) Within r = 2Ro, total charge is q. (C) There will be no charge anywhere except at r = Ro. (D) Electric field is discontinuous at r = Ro. [JEE 2006] 14 Page 14 of 16 ELECTROSTATICS Q.12 Charges +q and –q are located at the corners of a cube of side a as shown in the figure. Find the work done to separate the charges to infinite distance. [JEE 2003] ANSWER KEY Q.1 a = l(1 + 2 ), the equilibrium will be stable Q.3 (i) Q.4 3 – 11 Q.8 3σ λ 2 m ∈0 Q.12 – Q.16 (ii) (a) mπ3ε 0d 3 πσ 2 R 3 ε0 Q2 mπ ∈0 V 2 0 Q.6 7 − kp kˆ 8 mπ3ε 0 d 3 2 Qq (iv) 3/ 2 3 × 10–9 C Q.5 (b) Qq (iii) qQ 8π 2ε0 r 2 0 Q.7 1 qQ 2π 4πε 0 mR 3 Q.9 σ q0 2 tan–1 2 ε mg 0 Q.10 Q.13 20 ln2 Q.14 – 2 ∈0 u 2 m qσ q1 1 1 − ; a ≤r ≤ b Vr = 4πε 0 r a b q 1 1 (i) q2 = − q1 ; (ii) Vb = 1 − ; r =b a 4πε 0 b a V = 1 q1 + q 2 ; r≥b r 4πε 0 r r Q.17 Q.18 Q.19 1.8 × 105 sec Q.20 – Q/3 Q.23 Q.2 Page 15 of 16 ELECTROSTATICS EXERCISE # I Q.24 q 24 ∈0 ( kq 2 3− 2 a Q.21 1.125 q Q.22 R 3 Q.26 Q.25 a = 2kQ 2 mR Q.11 ) Q.15 9V0 kP ( − i − 2 j) 2 y3 Q 2ε 0 EXERCISE # II kq1q 2 (c) 3 mg , mg . q1 & q2 should have unlike charges for the beads to remain 2 stationaly & q1q2 = − mg l2/k Q.1 (a) 60º (b) mg + 5 + 89 L Q.2 8 Q.3 9.30 λq 2ε 0 m Q.4 Q.6 8 4 Kq 2 Wfirst step = − , Wsecond step = 0, Wtotal = 0 5 r 3 Q.8 α α q = 4l 4πε 0 mgsin sin 2 2 Q.11 n= 4πε 0 mg( h−R )R q2 Q.12 Q.9 4πε 0 Ka 6 2mr ∈0 eρa Q.5 H2 = h1 + h2 − g V Q.7 2KQq r −R 3 mR r + 8 Q.10 v 3 Q.13 2.2 × 10–12C 15 Q.14 1/ 2 Q1 R1 = Q2 R 2 2 -.4 (i) A.C.86 m/s Q. (iii) A.8 C Q. K.19 V' = 3t . 3 Q.17 λ R E 0 ˆi Q.9 5. +. (b) QP along positive x-axis 2π ε 0 d 3 4 π ε0 d 2 -.21 Qq 2π ∈0 L Q. (b) U = mg 3 2 h 2 + a 2 − h equilibrium at h = a .19 7.2. C 2 R RQ 2 where a = .B.15 ] P Q .6 (i) C.14 C 1/ 3 Q.24 b = 2 sin -1 sin α 2 Q.E. .16 (σ1 − σ 2 )Qa 2 2 ε0 Q.7 (a) E. (ii) D. C.23 2 q1 q 2 EXERCISE # III Q.15 0.10 B Q. (b) U (n → ∞ ) = 2 r +R 8π∈0 r 2 4a . (b) B.20 A. (ii) (a) H = Q.V .3 A Q. Q.E = Q.11 A [ 1 q2 4 · 3 3 −3 6 − 2 Q.Q. at the origin = (27−10 6 ) × 10 −4 J approx. +.D 16 a Q.20 – 4kq ˆ i πR 2 Q.1 C Q. +.17 C Q.18 D Q.5 ×10 −4 J Q.2 Kλ .0 r Q.2 D Q.22 v= σeR mε 0 Q. (c) v0 = 3 m/s .5 × 10–9 N Page 16 of 16 ELECTROSTATICS Q.5 (a) U2 = a 2 Q 2 1−a n 8π∈0 R 1−a Q.13 (a) K.18 σ= ε r ρ0 εr − 1 Q.12 – 4 π ε0 a 6 Q. Capacitance Index: 1. Answer Key 7. Exercise II 4.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P2. 34 Yrs. from IIT-JEE 8. from AIEEE 1 . Exercise I 3. Exercise III 5. Que. Key Concepts 2. 10 Yrs. Que. Exercise IV 6. PARALLEL PLATE CAPACITOR : UNIFORM DI-ELECTRIC M EDIUM : (i) If two parallel plates each of area A & separated by a distance d are charged with equal & opposite charge Q. C1 = 3. the outer conductor is earthed . m n( ba ) 2 Page 2 of 12 CAPACITANCE KEY CONCEPTS . it may have different combinations of C1 and C2. C= ∈0 A with air as medium d This result is only valid when the electric field between plates of capacitor is constant.1. then the system is called a parallel plate capacitor & its capacitance is given by. COMPOSITE M EDIUM : C= ∈0 A t3 t1 t2 ∈r1 +∈r 2 +∈r 3 CYLINDRICAL CAPACITOR : It consist of two co-axial cylinders of radii a & b. The capacitance per unit length is C = . then the distance between the plates is effectively reduced by t − t irrespective of the position of ∈r the di-electric slab . The capacitance C = 4π ∈0 R exists between the surface of the sphere & earth . (ii) M EDIUM PARTLY AIR : C= ∈0 A d − t − ∈t r When a di-electric slab of thickness t & relative permittivity ∈r is introduced between the plates of an air capacitor. C= ∈0 ∈r A in a medium d . This sphere is at infinite distance from all the conductors . SPHERICAL CAPACITOR : It consists of two concentric spherical shells as shown in figure. We have 4 π ∈0 ab and C2 = 4π ∈0 b b−a Depending on connection. (iii) 4. Here capacitance of region between the two shells is C1 and that outside the shell is C2. CAPACITANCE OF AN ISOLATED SPHERICAL CONDUCTOR : C = 4π ∈0∈r R in a medium C = 4π ∈0 R in air * * 2. The di-electric constant of the medium filled in the space between the cylinder is 2π∈0∈r Farad ∈r . . If all dC's are in parallelCT = ∫ dC If all dC's are in series =∫ CT ∈0 k ( x ) A ( x ) COMBINATION OF CAPACITORS : (i) CAPACITORS IN SERIES : In this arrangement all the capacitors when uncharged get the same charge Q but the potential difference across each will differ (if the capacitance are unequal).. where as in a condenser it is stored within the condenser in its EF. charge Q & potential difference V . ENERGY STORED IN A CHARGED CAPACITOR : Capacitance C. (iii) The capacitance of a capacitor depends only on its size & geometry & the di-electric between the conducting surface . 1 1 1 1 1 = + + + .. 9. SHARING OF CHARGES : When two charged conductors of capacitance C1 & C2 at potential V1 & V2 respectively are connected by a conducting wire. CAPACITORS IN PARALLEL : When one plate of each capacitor is connected to the positive terminal of the battery & the other plate of each capacitor is connected to the negative terminals of the battery... The common potential (V) after sharing of charges. (ii) 7. V but the charge on each one is different (if the capacitors are unequal). 1 dx . then the capacitors are said to be in parallel connection... we choose a small dc in between the plates and for total capacitance of system. whether it is copper. HEAT PRODUCED IN SWITCHING IN CAPACITIVE CIRCUIT Due to charge flow always some amount of heat is produced when a switch is closed in a circuit which can be obtained by energy conservation as – Heat = Work done by battery – Energy absorbed by capacitor. gold etc) 3 Page 3 of 12 CAPACITANCE 5. (i.. Ceq. A or d varies in the region between d the plates. . = C1 + C2 + C3 + . C eq.e. independent of the conductor. This energy is stored in the electrostatic field set up in the di-electric 2 2 2 C medium between the conducting plates of the capacitor . + . (ii) The energy of an uncharged condenser = 0 . C1 C 2 (V1 − V2)2 .CONCEPT OF VARIATION OF PARAMETERS: ∈0 kA . In this process energy is lost in the connecting wire as heat . until the potential of the two condensers becomes equal.. the charge flows from higher potential conductor to lower potential conductor. The capacitors have the same potential difference.. 8. 2 (C1 + C 2 ) REMEMBER : (i) The energy of a charged conductor resides outside the conductor in its EF. This loss of energy is Uinitial − Ureal = 10. if either of k. silver. then energy stored is U= 1 1 1 Q2 CV2 = QV = .. Cn C1 C2 C3 As capacitance of a parallel plate capacitor isC = 6. V= C V + C 2V2 net ch arg e q +q = 1 2 = 1 1 .. like. net capaci tan ce C1 + C2 C1 + C 2 charges after sharing q1 = C1V & q2 = C2V. + Cn . 4 In the given network if potential difference between p and q is 2V and C2 = 3C1. 3µF and 5µF are independently charged with batteries of emf’s 5V.8 In the following circuit. 20V and 10V respectively.5 Find the equivalent capacitance of the circuit between point A and B. A charge q = 20µC is given to the inner sphere. Q. The capacitor is then connected across an uncharged capacitor of same capacitance as first one (= C). Q. Q. Now a battery of 20V and an uncharged capacitor of 4µF capacitance are connected to the junction A as shown with a switch S.3 The plates of a parallel plate capacitor are given charges +4Q and –2Q. If the plate area of either face of each plate is A and separation between plates is d.9 (a) (b) Three capacitors of 2µF.2 The capacitor each having capacitance C = 2µF are connected with a battery of emf 30 V as shown in figure. then find the amount of heat liberate after closing the switch. Find the final potential difference between the plates of the first capacitor. After disconnecting from the voltage sources. Then find the potential difference between a & b. the resultant capacitance between A and B is 1 µF.7 Find heat produced in the circuit shown in figure on closing the switch S. Q. When the switch S is closed. Q. These capacitors are connected as shown in figure with their positive polarity plates are connected to A and negative polarity is earthed. Find (a) the amount of charge flown through the battery (b) the heat generated in the circuit (c) the energy supplied by the battery (d) the amount of charge flown through the switch S Q. the inner sphere is connected to the shell by a conducting wire Q.Q. Q. find : the potential of the junction A. 4 Page 4 of 12 CAPACITANCE EXERCISE # I .1 A solid conducting sphere of radius 10 cm is enclosed by a thin metallic shell of radius 20 cm. Find the heat generated in the process. Find the value of C.6 The two identical parallel plates are given charges as shown in figure. final charges on all four capacitors. When switch is closed. 3. If the dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 440V is applied across the plates. 3. Q. Find the heat produced in (i) and work done by external agent in step (ii) & (iv).13 Five identical capacitor plates. 3.123. When the switch is closed. Q. Q.102. 3. 3.205 5 Page 5 of 12 CAPACITANCE Q.141. Q.177.16 Two parallel plate capacitors of capacitance C and 2C are connected in parallel then following steps are performed. The +ve plate of the first capacitor is connected to the –ve plate of the second capacitor.200.Q. (ii) A dielectric slab of relative permittivity k is slowly inserted in capacitor C. List of recommended questions from I.3 cm fill the space between the plates. 3. 3. Draw the graph which best describes the charge on the +ve plate of the 20 µF capacitor with increasing time. Irodov. 3.203. (iii) Battery is disconnected.124.103.15 In the circuit shown in figure. 3. each of area A. 3.204. find the amount of heat generated when switch s is closed.188. in series.11 Find the capacitance of the system shown in figure.132.10 Find the charge on the capacitor C = 1 µF in the circuit shown in the figure. 3. 3. Two parallel sided dielectric slabs of thickness 0.184. 3. 3. Q. the charge passing through the switch ____________ in the direction _________ to ________ . . 3. Find the effective capacitance between X and Y.18 A 10 µF and 20 µF capacitor are connected to a 10 V cell in parallel for some time after which the capacitors are disconnected from the cell and reconnected at t = 0 with each other . 3. are arranged such that adjacent plates are at a distance 'd' apart. (ii) the ratio of electric energies stored in the first to that in the second dielectric slab.121.14 In the circuit shown in the figure. 3.201.17 The plates of a parallel plate capacitor are separated by a distance d = 1 cm. Find : (i) the electric field intensities in each of the slabs. 3. through wires of finite resistance. the plates are connected to a source of emf V as shown in figure. 3. intially SW is open.12 The figure shows a circuit consisting of four capacitors. 3. (iv) Dielectric slab is slowly removed from capacitor. (i) A battery of voltage V is connected across points A and B. 3.117.113. Q. Q.142.122.7 cm and 0.199. 3.E. The charge on plate 1 is______________ and that on plate 4 is _________.101.133. 7 A capacitor of capacitance C0 is charged to a potential V0 and then isolated.8 When the switch S in the figure is thrown to the left. What are the final charges q1. 1µF be connected so as to produce a total capacitance of 3/7 µF. 2. Q. A battery charges the plates to a potential difference of V0. The junction of 1 & 3 and the plate 4 are connected to a source of constant e. discharged & charged again. Thw switch is now thrown to the right. the charges on plates 3 & 5. A plane parallel glass plate with a thickness of 0. At t =0.Q.01075 µF . Plates 2 & 5 are connected by a conductor while 1 & 3 are joined by another conductor . (b) How should 5 capacitors.2 The gap between the plates of a plane capacitor is filled with an isotropic insulator whose di-electric π constant varies in the direction perpendicular to the plates according to the law K = K1 1 + sin X . The surface charge density of the charge on capacitor the plates.1 (a) For the given circuit.1m apart. Find the potential difference across all the capacitors. Initially the capacitors C2C3 are uncharged. A small capacitor C is then charged from C0. The battery is then disconnected & a di-electric slab of constant K & thickness d is introduced. Given that : kglass = 6.3 (i) (ii) Q. V=35volt. the effective capacity of the system between the terminals of the source. The area of the plates is S. between the plates & K1 is a constant.5 cm are placed in the space between the capacitor plates find : Intensity of electric field in each layer. the plates of both the capacitors are 0. The drop of potential in each layer.2 µF & C = 0.5 A charge 200µC is imparted to each of the two identical parallel plate capacitors connected in parallel.6 A parallel plate capacitor has plates with area A & separation d . d where d is the separation. 4 & 5 are fixed parallel to and equdistant from each other (see figure). the process being repeated n times.f. kparaffin= 2 Q. Find . the plates of capacitors C1 acquire a potential difference V.001m/s and plates of second capacitor move apart with the same velocity. Determine the capacitance of the capacitor.4 (i) (ii) (iii) Five identical conducting plates 1. find the value of n for C0 = 0. If V0 = 100 volt.m. Given d = distance between any 2 successive plates & A = area of either face of each plate . A potential difference of 300 V is applied between the plates of a plane capacitor spaced 1 cm apart. Is it possible to remove charge on C0 this way? Q. Calculate the positive work done by the system (capacitor + slab) on the man who introduces the slab. Q. Q.5 cm and a plane parallel paraffin plate with a thickness of 0. 6 Page 6 of 12 CAPACITANCE EXERCISE # II . each of capacities. q2 & q3 on the corresponding capacitors. The plates of first capacitor move towards each other with relative velocity 0. Find the current in the circuit at the moment. 3. The potential of the large capacitor has now fallen to V. V0. Find the capacitance of the small capacitor. Q. 10 An insolated conductor initially free from charge is charged by repeated contacts with a plate which after each contact has a charge Q due to some mechanism . show that the inner cylinder should have radius a = b/e if the potential of the inner cylinder is to be maximum radius a = b e if the energy per unit length of the system is to be maximum. Find the charge on the rightmost capacitor as a function of time given that it was intially unchanged. where α = 1 per volt. 7 . If breakdown of air occurs at field strengths greater than Eb.15 The capacitors shown in figure has been charged to a potential difference of V volts. (take amplitude of oscillation << d1) Q. with the distance between the plates as d1.11 A parallel plate capacitor is filled by a di-electric whose relative permittivity varies with the applied voltage according to the law = αV.12 A capacitor consists of two air spaced concentric cylinders. Switch S1 is closed at t=0. Find the spring constant K. Q. When the capacitor is connected with an electric source with the voltage V. the distance between the plates is d0. The area of each plate is A. (i) (ii) Q. Q−q Q. Q. Find the charge on the capacitor at t=2R1C + R2C. What is the maximum voltage for a given K in which an equilibrium is possible ? What is the angular frequency of the oscillating system around the equilibrium value d1. Find the electrostatic energy of the system stored in the region I and II. prove that the maximum charge which can be given to the conductor in this way is Qq . In the steady position. a new equilibrium appears. If q is the charge on the conductor after the first operation. Now the switch is closed at t = 0.14 Figure shows three concentric conducting spherical shells with inner and outer shells earthed and the middle shell is given a charge q. The same (but containing no di-electric) capacitor charged to a voltage V = 156 volt is connected in parallel to the first "non-linear" uncharged capacitor.(i) (ii) (iii) Page 7 of 12 CAPACITANCE Q. The outer of radius b is fixed. Q. At t=R1C switch S1 is opened and S2 is closed.16 In the figure shown initially switch is open for a long time.13 Find the charge flown through the switch from A to B when it is closed. The lower plate is fixed and the other connected with a vertical spring.9 A parallel plate capacitor with air as a dielectric is arranged horizontally. Determine the final voltage Vf across the capacitors. Mass of the upper plates is m. and the inner is of radius a. Q. so that it carries a charge CV with both the switches S1 and S2 remaining open. Q.18 Find the charge which flows from point A to B, when switch is closed. EXERCISE # III Q.1 (i) (ii) (iii) Two parallel plate capacitors A & B have the same separation d = 8.85 × 10−4 m between the plates. The plate areas of A & B are 0.04 m2 & 0.02 m2 respectively. A slab of di-electric constant (relative permittivity) K=9 has dimensions such that it can exactly fill the space between the plates of capacitor B. the di-electric slab is placed inside A as shown in the figure (i) A is then charged to a potential difference of 110 volt. Calculate the capacitance of A and the energy stored in it. the battery is disconnected & then the di-electric slab is removed from A . Find the work done by the external agency in removing the slab from A . the same di-electric slab is now placed inside B, filling it completely. The two capacitors A & B are then connected as shown in figure (iii). Calculate the energy stored in the system. [ JEE '93, 7 ] Q.2 Two square metallic plates of 1 m side are kept 0.01 m apart, like a parallel plate capacitor, in air in such a way that one of their edges is perpendicular, to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of e.m.f. 500 volt . The plates are then lowered vertically into the oil at a speed of 0.001 m/s. Calculate the current drawn from the battery during the process. [ JEE '94, 6 ] [di-electric constant of oil = 11, ∈0 = 8.85 × 10−12 C2/N2 m2] Q.3 A parallel plate capacitor C is connected to a battery & is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V volt. The charging battery is now disconnected & the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of other. The final energy of the configuration is : 3 25 9 (A) zero (B) CV2 (C) CV2 (D) CV2 [ JEE '95, 1 ] 2 6 2 Q.4 The capacitance of a parallel plate capacitor with plate area 'A' & separation d is C. The space between the plates is filled with two wedges of di-electric constant K1 & K2 respectively. Find the capacitance of the resulting capacitor. [ JEE '96, 2 ] Q.5 Two capacitors A and B with capacities 3 µF and 2 µF are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in figure with one wire from each capacitor free. The upper plate of a is positive and that of B is negative. an uncharged 2 µF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate : the final charges on the three capacitors The amount of electrostatic energy stored in the system before and after the completion of the circuit. [ JEE '97 (cancelled)] (i) (ii) 8 Page 8 of 12 CAPACITANCE Q.17 In the given circuit, the switch is closed in the position 1 at t = 0 and then moved to 2 after 250 µs. Derive an expression for current as a function of time for t > 0. Also plot the variation of current with time. An electron enters the region between the plates of a parallel plate capacitor at a point equidistant from either plate. The capacitor plates are 2 × 10−2 m apart & 10−1 m long . A potential difference of 300 volt is kept across the plates. Assuming that the initial velocity of the electron is parallel to the capacitor plates, calculate the largest value of the velocity of the electron so that they do not fly out of the capacitor at the other end. [ JEE '97, 5 ] Q.7 For the circuit shown, which of the following statements is true ? (A) with S1 closed, V1 = 15 V, V2 = 20 V (B) with S3 closed, V1 = V2 = 25 V (C) with S1 & S2 closed, V1 = V2 = 0 (D) with S1 & S2 closed, V1 = 30 V, V2 = 20 V [ JEE '99, 2 ] Q.8 Calculate the capacitance of a parallel plate condenser, with plate area A and distance between plates d, when filled with a medium whose permittivity varies as ; ∈ (x) = ∈0 + β x 0 < x < d2 d <x<d. ∈ (x) = ∈0 + β (d − x) [ REE 2000, 6] 2 Q.9 Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is [ JEE 2002 (Scr), 3] (A) ( 1 C V12 − V22 4 ) ( 1 C V12 + V22 4 (B) ) (C) ( 1 C V1 − V2 4 ) 2 (D) ( 1 C V1 + V2 4 ) 2 Q.10 In the given circuit, the switch S is closed at time t = 0. The charge Q on ( 1–e–αt). Find the 0 value of Q0 and α in terms of given parameters shown in the circuit. [JEE 2005] t h e Q.11 c a p a c i t o r a t a n y i n s t a n t t i s g i v e n b y Q ( t ) = Q Given : R1 = 1Ω , R2 = 2Ω , C1 = 2µF, C2 = 4µF The time constants (in µS) for the circuits I, II, III are respectively (A) 18, 8/9, 4 (C) 4, 8/9, 18 (B) 18, 4, 8/9 (D) 8/9, 18, 4 9 [JEE 2006] Page 9 of 12 CAPACITANCE Q.6 Page 10 of 12 CAPACITANCE ANSWER KEY EXERCISE # I Q.1 9J Q.2 (a) 20 µC, (b) 0.3 mJ, (c) 0.6 mJ. (d) 60 µC Q.3 3Q/2C Q.4 30 V Q.7 0 Q.8 32 µF 23 Q.9 (a) Q.11 Q.5 Q.6 C 2 1 q d 2 ∈0 A 100 volts; (b) 28.56 µC, 42.84 µC, 71.4 µC, 22.88 µC Q.10 10 µC 7 8 A ∈0 V 2A ∈0 V 25 ε 0 A µF Q.12 Q.13 ,– 3 d d 24 d Q.15 150 µJ Q.16 (i) Q.14 60 µc , A to B 3 1 1 CV2; (ii) – CV2(K – 1); (K + 2) (K – 1)CV2 ; 2 2 6 Q.17 (i) 5 x 104 V/m , 3 x 104 V/m; (ii) 35/9 Q.18 EXERCISE # II Q.1 (a) 12 V, 9 V, 3 V, 13 V, 16 V , (b) Q.2 C = ∈SπK1 2d 5 3 Q.3 (i) 2 ∈0 AVa 4 ∈0 AVa ∈0 A ; (ii) Q3= d , Q5 = d 3 3 d Q.4 (i) 1.5 × 104 V/m, 4.5 × 104 V/m, (ii) 75 V, 225 V, (iii) 8 × 10–7 C/m2 V 1 / n 1 1 Q.5 2µA Q.6 W = C0 V02 1− Q.7 C = C0 0 −1 = 0.01078 µF, n = 20 2 K V 2 C1C 2 C 3 V C1 V(C 2 +C3 ) q = q3 Q.8 q1 = C1C 2 +C 2 C 3 +C 3 C1 C1C 2 +C 2 C3 +C1C3 2 Q.9 ε 0 AV 2 2d12 (d 0 − d1 ) Q.14 UI = Q.16 q = , K 2 d0 Aε 0 3 3kq12 10 r 1/ 2 3/ 2 Kd13 − ∈0 AV 2 md13 , where q1 = − Q.11 12 volt 4q ; UII = 2 K (q + q1 ) 2 35 r Q.15 25 CV 1 − t / RC 1 − e 2 2 10 Q.13 69 mC 1 CV q = CE 1 − + 2 e e 8 ×108 m/s 2 9. QB = 150 µC.18 – t (x10–4s) – 0.1 × 10−5 J Q.1 (i) 0.2 4.4 K CK1 K 2 ln 2 (K 2 −K1 ) K1 Q. I = – 0.015 −6 For t > 250 µs.84 × 10−5 J . Uf = 18 mJ Q.3 B Q.I(amp) Q.11 D 1 2 Q.11e −4000( t − 250) ×10 amp.8 D CVR 2 Q.5 QA = 90 µC.11 400 µC 7 EXERCISE # III Q. (iii) 1.2 × 10−8 F.17 For t ≤ 250 µs.6 Q. 0.2 × 10−5 J .1 C Q.10 Q0 = R + R 1 2 11 . 1. (ii) 4.7 2 ∈ +β d βA n 0 2 2 ∈0 R1 + R 2 and a = CR R Q. Q. I = 0.4 mJ.425 × 10−9 Ampere Q. QC = 210 µC.04 e–4000 t amp .04 0. Ui = 47.9 4. 10 Yrs.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P3. Key Concepts 2. from AIEEE 1 . Current Electricity Index: 1. Answer Key 7. 34 Yrs. Que. Exercise I 3. Exercise II 4. Exercise IV 6. from IIT-JEE 8. Exercise III 5. Que. of electrons is proportional to the electric field in side the conductor as – ν d = µE where µ is the mobility of electrons I current density is given as J = = ne ν d A = ne(µE) = σE 1 where σ = neµ is called conductivity of material and we can also write ρ = → resistivity σ of material. SOURCES OF POTENTIAL DIFFERENCE & ELECTROMOTIVE FORCE : Dry cells . They move with a constant drift velocity . aluminium etc. where dq is net charge transported at Ampere is the unit of current . Positive charge flows from higher to lower potential and negative charge flows from lower to higher. It is called as differential form of Ohm's Law. The potential difference between the two terminals of a source when no energy is drawn from it is called the " Electromotive force" or " EMF " of the source. i = nvdeA. The direction of current is along the flow of positive charge (or opposite to flow of negative charge). free to migrate is a conductor of electricity. The unit of potential difference is volt. 5. These electrons are not accelerated by electric field in the conductor produced by potential difference across the conductor. 1 volt = 1 Amphere × 1 Ohm. silver. 2. dq . 2 Page 2 of 16 CURRENT FLECTRYCITY CURRENT . → → For random J or S. secondary cells . Metals such as gold. E AND ν D : In conductors drift vol. In a current carrying conductor we can define a vector which gives the direction as current per unit normal. cross sectional area. CHARGE AND CURRENT : The strength of the current i is the rate at which the electric charges are flowing. 3. The electric charge flows from higher potential energy state to lower potential energy state. RELATION IN J. I Thus J = n or I = J · S S Where n is the unit vector in the direction of the flow of current. ELECTRIC CURRENT IN A CONDUCTOR : In absence of potential difference across a conductor no net current flows through a corss section. where Vd = drift velocity . ELECTRIC CURRENT : Electric charges in motion constitute an electric current.ELECTRICITY 1. we use I = ∫ J ⋅ ds 4. copper. generator and thermo couple are the devices used for producing potential difference in an electric circuit. Any medium having practically free electric charges . are good conductors. If a charge Q coulomb passes through a given cross section of the conductor in t second the current I through the conductor is given by I = Q t = Coulomb sec ond = Q t ampere . When a potential difference is applied across a conductor the charge carriers (electrons in case of metallic conductors) flow in a definite direction which constitutes a net current in it . Thus E = ρ J. If i is not constant then i = dt a section in time dt. The closed loop can be traversed in any direction . OHM'S LAW : Ohm's law is the most fundamental of all the laws in electricity . 10.ELECTRICAL RESISTANCE : The property of a substance which opposes the flow of electric current through it is termed as electrical resistance. unit of G is ohm. 9. put a + ve sign in expression or if lower potential point is entered put a negative sign . temperature and internal structure of the conductor. DEPENDENCE OF RESISTANCE ON TEMPERATURE : The resistance of most conductors and all pure metals increases with temperature . S. II . V = R I . Rα l (cross section area of the conductor) A l .I. geometery. It is also known as KCL (Kirchhoff's current law) . Where α is called the temperature co-efficient of resistance . Here we assume that the dimensions of resistance does not change with temperature if expansion coefficient of material is considerable. −V1 −V2 +V3 −V4 = 0. at a given temperature R = ρ 8.Law (Loop analysis) :The algebric sum of all the voltages in closed circuit is zero. 7. It is also known as specific resistance of the material . the temperature coefficient of resistance is negative. If Ro & R be the resistance of a conductor at 0º C and θº C .Law (Junction law or Nodal Analysis) :This law is based on law of conservation of charge . Electrical resistance depends on the size. LAW OF RESISTANCE : The resistance R offered by a conductor depends on the following factors : R α L (length of the conductor) . It says that the current through the cross section or the conductor is proportional to the applied potential difference under the given physical condition . While traversing a loop if higher potential point is entered. The unit of α is K− 1 of ºC −1 reciprocal of resistivity is called conductivity and reciprocal of resistance is called conductance (G) . It states that " The algebric sum of the currents meeting at a point is zero " or total currents entering a junction equals total current leaving the junction . Then instead of resistance we use same property for resistivity as ρ = ρ0 (1 + αθ) The materials for which resistance decreases with temperature. It is also known as KVL (Kirchhoff's voltage law) . 3 Page 3 of 16 CURRENT FLECTRYCITY 6. Ohm's law is applicable to only metalic conductors . KRICHHOFF'S LAW'S : I . but there are a few in which resistance decreases with temperature . Boxes may contain resistor or battery or any other element (linear or non-linear). Σ IR + Σ EMF = 0 in a closed loop . A Where ρ is the resistivity of the material of the conductor at the given temperature . Σ Iin = Σ Iout. then it is found that R = Ro (1 + α θ) . . COMBINATION OF RESISTANCES : A number of resistances can be connected and all the complecated combinations can be reduced to two different types, namely series and parallel . (i) RESISTANCE IN SERIES : When the resistances are connected end toend then they are said to be in series . The current through each resistor is same . The effective resistance appearing across the battery . R = R1 + R2 + R3 + ................ + Rn and V = V1 + V2 + V3 + ................ + Vn . The voltage across a resistor is proportional to the resistance V1 = (ii) R1 R2 V;V2 = V ; R1+R 2 +.........+R n R1+R 2 +.........+R n etc RESISTANCE IN PARALLEL : A parallel circuit of resistors is one in which the same voltage is applied across all the components in a parallel grouping of resistors R1, R2, R3, ........, Rn . CONCLUSIONS : Potential difference across each resistor is same . (a) (b) (c) (d) I = I1 + I2 + I3 + .......... In . 1 1 1 1 1 = + + +..........+ . Rn R R1 R 2 R 3 Current in different resistors is inversally proportional to the resistance . Effective resistance (R) then I1 : I2 : ........... In = I1 = 1 1 1 1 : : :..........: . R1 R 2 R 3 Rn G1 G2 I , I2 = I , etc . G1+G 2 +.........+G n G1+G 2 +.........+G n I = Conductance of a resistor . R EMF OF A CELL & ITS INTERNAL RESISTANCE : If a cell of emf E and internal resistance r be connected with a resistance R the total resistance of the circuit is (R + r) . E E I= ; VAB = where R +r R +r where G = 12. E = Terminal voltage of the battery .If r → 0, cell is Ideal & V → E . 13. GROUPING OF CELLS : (i) CELLS IN SERIES : Let there be n cells each of emf E , arranged in series.Let r be the internal resistance of each cell. nE The total emf = n E . Current in the circuit I = . R +nr If nr << R then I = nE R If nr >> R then I = E → r → Series combination should be used . Series combination should not be used . 4 Page 4 of 16 CURRENT FLECTRYCITY 11. CELLS IN PARALLEL : If m cells each of emf E & internal resistance r be connected in parallel and if this combination be connected to an external resistance then the emf of the circuit = E . Internal resistance of the circuit = I= (iii) r . m mE E = . r mR + r R+ m If mR << r ; I = mE r → Parallel combination should be used . If mR >> r ; I = E R → Parallel combination should not be used . CELLS IN MULTIPLE ARC : mn = number of identical cells . n = number of rows m = number of cells in each rows . The combination of cells is equivalent to single cell of : (a) emf = mE Current I = R= & mE . R + mr n (b) internal resistance = mr n For maximum current nR = mr or mr = internal resistance of battery . n nE mE . = 2r 2R WHEAT STONE NETWORK : When current through the galvanometer is zero (null point or balance Imax = R P . When PS > QR; VC < VD & PS <QR ; VC > VD or = S Q PS = QR ⇒ products of opposite arms are equal. Potential difference between C & D at null point is zero . The null point is not affected by resistance of G & E. It is not affected even if the positions of G & E are inter changed. ICD α (QR − PS) . point) 14. POTENTIOMETER : A potentiometer is a linear conductor of uniform cross-section with a steady current set up in it. This maintains a uniform potential gradient along the length of the wire . Any potential difference which is less then the potential difference maintained across the potentiometer wire can be measured using this . The potentiometer equation is E1 I1 = . E 2 I2 5 Page 5 of 16 CURRENT FLECTRYCITY (ii) AMMETER : It is a modified form of suspended coil galvanometer it is used to measure current . A shunt (small resistance) is connected in parallel with galvanometer to convert into ammeter . S = Ig R g I−I g ; An ideal ammeter has zero resistance . where Ig = Maximum current that can flow through the galvanometer . I = Maximum current that can be measured using the given ammeter . 16. VOLTMETER : A high resistance is put in series with galvanometer . It is used to measure potential difference . Ig = Vo R g +R . R → ∞ , Ideal voltmeter . 17. RELATIVE POTENTIAL : While solving an electric circuit it is convinient to chose a reference point and assigning its voltage as zero. Then all other potential are measured with respect to this point . This point is also called the common point . 18. ELECTRICAL POWER : The energy liberated per second in a device is called its power . The electrical power P delivered by an electrical device is given by P = VI , where V = potential difference across device & I = current. If the current enters the higher potential point of the device then power is consumed by it (i.e. acts as load) . If the current enters the lower potential point then the device supplies power (i.e. acts as source) . 19. V2 . Power consumed by a resistor P = VI = R HEATING EFFECT OF ELECTRIC CURRENT : When a current is passed through a resistor energy is wested in over coming the resistances of the wire . This energy is converted into heat . 20. V2 t Joule . R JOULES LAW OF ELECTRICAL HEATING : The heat generated (in joules) when a current of I ampere flows through a resistance of R ohm for T second is given by : = I2 R W = VIt Joule ; = I2 Rt Joule ;= I 2 RT Calories . 4.2 If current is variable passing through the conductor then we use for heat produced in resistance in time H = I2 RT Joules ; = t 0 to t is: 2 H = ∫ I Rdt 0 21. UNIT OF ELECTRICAL ENERGY CONSUMPTION : 1 unit of electrical energy = Kilowatt hour = 1 KWh = 3.6 × 106 Joules. 6 Page 6 of 16 CURRENT FLECTRYCITY 15. all wires have equal resistance r.Q. Q. D. B. Q.1 A network of nine conductors connects six points A. Then find the resistance R. find the internal resistance of the cell.8 In the circuit shown in figure. Q.11 In the circuit shown in figure the reading of ammeter is the same with both switches open as with both closed.10 For what value of R in circuit.M. Find the equivalent resistance between A and D. Where R is the resistance of each part.4 Find the equivalent resistance of the circuit between points A and B shown in figure is: (each branch is of resistance = 1Ω) Q.7 Find the effective resistance of the network (see figure) between the points A and B. Q. E and F as shown in figure. Q.9 Find the resistor in which maximum heat will be produced. C. Find the equivalent resistance between A and B. Find the current passing through 2Ω resistance. one after the another. Q. Q. produces the same amount of the heat during the same time in two independent resistors R1 and R2. (ammeter is ideal) 7 Page 7 of 16 CURRENT FLECTRYCITY EXERCISE # I .2 In the circuit shown in figure potential difference between point A and B is 16 V. Q. The figure denotes resistances in ohms.5 Find the current through 25V cell & power supplied by 20V cell in the figure shown. current through 4Ω resistance is zero.6 If a cell of constant E. when they are separately connected across the terminals of the cell.F. Q.3 Find the current I & voltage V in the circuit shown. If a galvanometer shows no deflection at the point P. Two cells of emf ε1 = 2V and ε2 = 4V having internal resistances 1Ω and 5Ω respectively are connected as shown in the figure.149. Calculate the resistance R1.14 A battery of emf ε0 = 10 V is connected across a 1 m long uniform wire having resistance 10Ω/m. 3.175. 3. The meter deflects full scale for a current of 10 mA.18 An accumulator of emf 2 Volt and negligible internal resistance is connected across a uniform wire of length 10m and resistance 30Ω. The range is 0–10 A. Q. 3. 3. .147. find R1 and R2. 3.150.176. then find the value of unknown resistance R.19 The resistance of the galvanometer G in the circuit is 25Ω.1 A between O and R. R2 and R3. 3.196. range is 0 – 0. The appropriate terminals of a cell of emf 1.179. 3. 3. When R2 is shunted by a resistance of 10 Ω. if the terminals O and P are taken.189.E. balance shifts to 50 cm. find the voltage across points A and B.13 The figure shows a network of resistor each heaving value 12Ω. Here reading of ammeter is 5 ampere and voltmeter is 96V & voltmeter resistance is 480 ohm. Then find the resistance R Q.207 8 Page 8 of 16 CURRENT FLECTRYCITY Q. List of recommended questions from I. 3.154.16 In the figure shown for gives values of R1 and R2 the balance point for Jockey is at 40 cm from A.5 Volt and internal resistance 1Ω is connected to one end of the wire. 3. If the galvanometer shows zero deflection at the position C.17 A part of a circuit is shown in figure. find the distance of point P from the point a.15 A potentiometer wire AB is 100 cm long and has a total resistance of 10ohm. (AB = 1 m): Q. 3. range is 0 – 1 A between O and Q . What length of the wire will be required to produce zero deflection of the galvanometer ? How will the balancing change (a) when a coil of resistance 5Ω is placed in series with the accumulator.169. Q. The meter behaves as an ammeter of three different ranges.190. S2 and S3 in the figure are arranged such that current through the battery is minimum. 3.5 volt is shunted with 5Ω resistor ? Q. Find the equivalent resistance between points A and B.12 If the switches S1.194. 3.186. 3.Q. 3. Q. Irodov. (b) the cell of 1.155. and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. Find the maximum resistance of the network between points A and the point of sliding wire with BC. 3. N are V1..Q.3 What will be the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure. (i) (ii) Q. (b) Suppose that instead the current density is a maximum Jo at the surface and decreases linearly to zero at r the axis so that J = J0 . Q. Calculate R for which the heat generated in the circuit is maximum. Find the equivalent resistance across OP. R c o n d u c t o r ’ s c r o s s s e c t i o n a l a r e a i s A = Q. A battery of 2 volts of internal resistance 0.2 V5/2.44 the power dissipated in the rod is twice that dissipated in the resistance. A currant enters the rectangular system at one of the corners and leaves at the diagonally opposite corners.6 A network of resistance is constructed with R1 & R2 as shown in the figure.4 The current I through a rod of a certain metallic oxide is given by I = 0.2(a) The current density across a cylindrical conductor of radius R varies according to the equation r J = J 0 1 − . Calculate the current in terms of Jo and the πR2. Another wire of resistance α/3 from A can make a sliding contact with wire BC. Vn respectively each having a potential k time smaller than previous one. V3 .7 A hemisphere network of radius a is made by using a conducting wire of resistance per unit length r. What length of wire connected between input and output terminals would have an equivalent effect. where r is the distance from the axis. What value should the series resistance have so that : the current in the circuit is 0. k & R3. . Q.. where V is the potential difference across it. Calculate the current.1 ohm is connected across the circuit.. Find: R2 R1 and R 3 in terms of k R2 current that passes through the resistance R2 nearest to the V0 in terms V0.8 Three equal resistance each of R ohm are connected as shown in figure. Show that the current in the common side is 1/5th of the entering current. The rod is connected in series with a resistance to a 6V battery of negligible internal resistance... BC & CA of same material and of resistance α. 2. V2. Q. Thus the current density is a maximum Jo at the R axis r = 0 and decreases linearly to zero at the surface r = R. (i) (ii) Q.1 A triangle is constructed using the wires AB .5 A piece of resistive wire is made up into two squares with a common side of length 10 cm. Q. The potential at the points 1. 9 Page 9 of 16 CURRENT FLECTRYCITY EXERCISE # II . 2α & 3α respectively. 16 An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to I1. ABCD is a square.9 m. Q. PQ is a wire of uniform cross-section and of resistance R0. The material obeys Ohm’s law and its resistivity varies along the rod according to ρ (x) = ρ0 e–x/L. Show that relation between the current is 3 I3 I2 = 2 I1 (I2 + I3) Q.12 A rod of length L and cross-section area A lies along the x-axis between x = 0 and x = L. find the new position of the null point. When the cells are connected are in parallel. calculate the following : (i) Potential difference between points a and b when switch S is open. Q. (ii) Current through S in the circuit when S is closed. the current through the circuit is I3. find the reading on the ammeter.14 An ideal cell having a steady emf of 2 volt is connected across the potentiometer wire of length 10 m.10 In the circuit shown in figure. at what rate should the water drain from the bath tube to light bulb? How long can we keep the bulb on. The efficiency of generator is 90%.11 The circuit shown in figure is made of a homogeneous wire of uniform cross-section. Q. (a) Find the total resistance of the rod and the current in the wire. When he joins a second identical cell is series with the first.(g = 10m/s–2) Q. A is an ideal ammeter and the cells are of negligible resistance.A person decides to use his bath tub water to generate electric power to run a 40 watt bulb. (b) Find the electric potential in the rod as a function of x. Find the ratio of the amounts of heat liberated per unit time in wire A-B and C-D. The end of the rod at x = 0 is at a potential V0 and it is zero at x = L.15 Find the equivalent resistance of the following group of resistances between A and B. Q. Each resistance of the circuit is R (a) (b) Q.13 In the figure.9 . Find the value of ‘f’ for maximum and minimum reading on the ammeter. The jockey J can freely slide over the wire PQ making contact on it at S. 10 Page 10 of 16 CURRENT FLECTRYCITY Q. The bath tube is located at a height of 10 m from the ground & it holds 200 litres of water. if the bath tub was full initially. If a resistance of 5Ω is put in series with potentiometer wire. Q. If we install a water driven wheel generator on the ground. If the length of the wire PS is f = 1/nth of PQ.5 Ω/m. The potentiometer wire is of magnesium and having resistance of 11.17 Find the potential difference VA – VB for the circuit shown in the figure. the current becomes I2. An another cell gives a null point at 6. The ammeter reads 3A.22 A galvanometer (coil resistance 99 Ω) is converted into a ammeter using a shunt of 1Ω and connected as shown in the figure (i). What is the resistance of the galvanometer? Further if the full scale deflection of the galvanometer movement is 300 mA.21 In the primary circuit of potentiometer the rheostat can be varied from 0 to 10Ω. Find the internal resistance r of the 4. This voltmeter is connected as shown in figure(ii). 3 Q. (a) Find the length AP of the wire such that the galvanometer shows zero deflection.19 Find the current through 2 Ω resistance in the figure shown. .18 A resistance R of thermal coefficient of resistivity = α is connected in parallel with a resistance = 3R. The same galvanometer is converted into a voltmeter by connecting a resistance of 101 Ω in series. Find the value of αeff . Its reading is found to be 4/5 of the full scale reading. 50Ω. Q. New balancing length is found to 8m. Find (a) internal resistance r of the cell (b) range of the ammeter and voltmeter (c) full scale deflection current of the galvanometer 11 Page 11 of 16 CURRENT FLECTRYCITY Q.Q. find the emf of the cell. Initially it is at minimum resistance (zero).20 A galvanometer having 50 divisions provided with a variable shunt s is used to measure the current when connected in series with a resistance of 90 Ω and a battery of internal resistance 10 Ω. (b) Now the rheostat is put at maximum resistance (10Ω) and the switch S is closed.5V cell. having thermal coefficient of resistivity = 2α. It is observed that when the shunt resistance are 10Ω. Q. respectively the deflection are respectively 9 & 30 divisions. electric field and drift speed (B) drift speed only (C) current and drift speed (D) current only (ii) The dimension of electricity conductivity is _________.6 A 100 W bulb B1. P ≠ R. 6] Q. The quantity /quantities constant along the length of the conductor is / are : [JEE’97. Then (A) IR = IG (B) IP = IG (C) IQ = IG (D) IQ = IR [JEE’99. either by applying Kirchhoff’s rules or otherwise.Q.50 A (D) the 4Ω resistor is 0. the reading of the galvanometer is same with switch S open or closed. 3] 12 Page 12 of 16 CURRENT FLECTRYCITY EXERCISE # III . the current through : (A) the 3Ω resistor is 0.5 The effective resistance between the points P and Q of the electrical circuit shown in the figure is (A) 2 Rr / (R + r) (B) 8R(R + r)/(3R + r) (C) 2r + 4R (D) 5 R/2 + 2r [JEE 2002 (Scr). are connected to a 250 V source. [JEE’96. Calculate the potential difference across the resistance of 400 ohm.2(i) A steady current flows in a metallic conductor of nonuniform cross-section. as will be measured by the voltmeter V of resistance 400 ohm. 2] Q. B2 and B3 respectively. (iii) Find the emf (E) & internal resistance (r) of a single battery which is equivalent to a parallel combination 1 & V2 & internal resistances r1 & r2 respectively with their similar polarity connected to each other o f t w o b a t t e r ie s o f e m f s V Q. and two 60 W bulbs B2 and B3. as shown in the figure.1 An electrical circuit is shown in the figure.50 A (B) the 3Ω resistor is 0. 2] Q. Now W1. Then (A) W1 > W2 = W3 (B) W1 > W2 > W3 (C) W1 < W2 = W3 (D) W1 <W2 < W3 [JEE 2002 (Scr).3 In the circuit shown in the figure.1+2+5] (A) current. 3] Q. W2 and W3 are the output powers of the bulbs B1.4 In the circuit shown.25 A [JEE’98.25 A (C) 4 Ω resistor is 0. After appropriate connections are made. [JEE’ 2002. Are there positive and negative terminals on the galvanometer? Copy the figure in your answer book and show the battery and the galvanometer (with jockey) connected at appropriate points.12 In an RC circuit while charging. an unknown resistance X and a resistance of 12 Ω are connected by thick conducting strips. [JEE’ 2003] Q. A battery and a galvanometer (with a sliding jockey connected to it) are also available.10 How a battery is to be connected so that shown rheostat will behave like a potential divider? Also indicate the points about which output can be taken.11 Six equal resistances are connected between points P. if the same current is passing through all circuits and each resistor is 'r' [JEE’ 2003 (Scr)] (I) (II) (III) (IV) (A) P2 > P3 > P4 > P1 (B) P3 > P2 > P4 > P1 (C) P4 > P3 > P2 > P1 (D) P1 > P2 > P3 > P4 Q. Then the net resistance will be maximum between (A) P and Q (B) Q and R (C) P and R (D) any two points [JEE’ 2004 (Scr)] Q. as shown in figure. 1 + 2 + 2] Arrange the order of power dissipated in the given circuits. which of the solid curves best represents the variation of ln I versus time? [JEE’ 2004 (Scr)] (A) P (B) Q (C) R (D) S 13 Page 13 of 16 CURRENT FLECTRYCITY Q.(a) (b) (c) Q. Q and R as shown in the figure. it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A.7 . Connections are to be made to measure the unknown resistance X using the principle of Wheatstone bridge. no current is passing through the galvanometer. Obtain the value of resistance X. Answer the following question.9 In the given circuit. the graph of ln I versus time is as shown by the dotted line in the adjoining diagram where I is the current. When the value of the resistance is doubled. If the cross-sectional diameter of AB is doubled then for null point of galvanometer the value of AC would [JEE’ 2003 (Scr)] (A) x (B) x/2 (C) 2x (D) None Q.8 A thin uniform wire AB of length l m. (A) Power loss in second half is four times the power loss in first half. R2 or R3.1 mA (C) 10.1 mA [JEE’ 2005 (Scr)] Q. the unknown resistance should be connected between [JEE’ 2004 (Scr)] (A) B and C (B) C and D (C) A and D (D) B1 and C1 .13 For the post office box arrangement to determine the value of unknown resistance. [JEE’ 2004] Q.15 In the figure shown the current through 2Ω resistor is (A) 2 A (B) 0 A (C) 4 A (D) 6 A [JEE’ 2005 (Scr)] Q. The smallest current required in the circuit to produce the full scale deflection is (A) 1000.693) (A) 6.18 An unknown resistance X is to be determined using resistances R1. a battery of emf 12 volt and a resistor of 2.52 sec.Q. B and C.C.16 An uncharged capacitor of capacitance 4µF. (B) Voltage drop in first half is twice of voltage drop in second half. Choose the correct option out the following.86 sec. (C) 20.14 Draw the circuit for experimental verification of Ohm's law using a source of variable D.1 mA (D) 100. (C) Current density in both halves are equal.1 mA (B) 1. The time after which vc = 3vR is (take ln2 = 0. (D) none of these [JEE’ 2005 (Scr)] Q.19 Consider a cylindrical element as shown in the figure. (D) Electric field in both halves is equal. A resistor 0.17 A galvanometer has resistance 100Ω and it requires current 100µA for full scale deflection. two galvanometers and two resistances of values 106 Ω and 10–3 Ω respectively. (B) 13. Find which of the above will give the most accurate reading and why? [JEE 2005] Q. a main resistance of 100 Ω. Clearly show the positions of the voltmeter and the ammeter.93 sec. Current flowing the through element is I and resistivity of material of the cylinder is ρ. 14 [JEE 2006] Page 14 of 16 CURRENT FLECTRYCITY Q. voltage.5 MΩ are connected in series.1Ω is connected to make it an ammeter. Their corresponding null points are A. 8 A D (c) 8 Ω Q.5 Volts Q. Out put can be taken across the terminals A and C or B and C Q.18 This is true for r1= r2. (iii) Q. V = 3.13 C Voltmeter 106 Ω G1 100 Ω Ammeter G2 Q.3Ω.22 EXERCISE # III Q.0278 Ω.4 A Q.8 0. (b) 9R/14 r + R 0 (f − f 2 ) 22 5 V – Q.3 I = 2. 5 Ω 3 Q.16 EXERCISE # II Q.V= 0 1 − .67 cm Q.12 R = Q. (ii) M–1L–3T3A2.6 V1r2 + V2 r1 r1 r2 .9 4Ω Q.1 (3/11)α Q.3125Ω 2 2 (k − 1) k V0 k ( k − 1) Q.13 9Ω Q. (b) 1Ω 9 4 (a) 1. R2 = 0. (c) 0. 0-10V.6 (i) .5 m. 6. (ii) (k − 1) R3 k Q.10 Battery should be connected across A and B.05 A Q.19 1A Q. I = ρ0 L e − 1 A e 1 − e −1 ε .15 (a) 5/7 R.25 Ω.21 (a) 6 m.5 (i) 10. 8.18 αeff = α Q. (b) Q.10 1Ω R1R 2 Q.01 W.1 20/3 V Q.16 B Q.5 12A.17 D E Q.18 7.3 Q.15 4 ohm Q. r1 + r2 r1 + r2 Q.14 7.19 A Page 15 of 16 CURRENT FLECTRYCITY EXERCISE # I .9 A Q. (b) 0-5A.5 A Q.15 B Q.7 (a) No.ANSWER KEY Q. 450 sec Q.6 Q.5 A.2 m Q.7 8/7 R Q. (ii) 0.14 10-3 Ω Q.7 ( 2 + π) a r 8 Q.1 1Ω Q.17 20 ohm Q.25m Q. 1 .20 233. –20W Q.11 600Ω Q.9 4/9 kg/sec.4 ( ) Q.13 Q.75m.11 A Q.14 46.12 Q.17 Q.2 (a) J0A/3. Imin f = 1/2 Q.10 (i) Vab = – 12 V.2 (i) D.12 B Q. So R2 given most accurate value 15 Q.4 22 Ω 35 3r 5 1V 10 Ω. (ii) 3 amp from b to a R2 3 = R1 5 7/5 times the length of any side of the square Q.5 Ω Q.19 R1 = 0. R3 = 2.2 3.8 Q..5 A Q. for Imax f = 0. (b) 2J0A/3 Q.52Ω . 144V Q.11 11+ 6 2 V0 A e ρ0 L 1 V (e − x / L − e −1 ) .3 D Q.3Ω Q. Exercise I 3. from IIT-JEE 8. Exercise II 4. Answer Key 7. from AIEEE 1 . Exercise III 5. Thermal and Chemical Effects of Electric Current Index: 1. Que. 34 Yrs. Que. 10 Yrs.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P4. Key Concepts 2. Exercise IV 6. State Joule’s law of heating.3 Sol. (2) It should have low resistance. Which of the two has larger resistance? The resistance of the dimmer bulb is larger. Q. Cation: It is a positively charged ion formed by removal of electrons from atoms or molecules. State Faraday’s laws of electrolysis. bases and salts which ionize when dissolved in water.) of the substance. the free electrons are accelerated due to the electric field.C. What is electrolysis? The process of chemical decomposition of a conducting liquid (electrolyte) into its components when a current is passed through it is called electrolysis.6 Sol. The practical unit of electrical energy is kilo watt-hour.I. is given by H = I2Rt.I. Q. In electrolysis. Q. In electrolysis.4 Sol.6 × 106 J Q. Define S. when a current flows through it for time t. then the masses of the various substances deposited at the respective electrodes are proportional to their chemical equivalents (equivalent weights). which appears as heat energy.E. These electrons collide frequently with the ions in the conductor and lose their kinetic energy. S. unit of electrical power. if a potential difference of one volt applied across it produces a current of one ampere in that circuit. Q.I. Q.. Q. Electrochemical equivalent of an element is the mass of the element released from a solution of its ion when a current of one ampere flows for one second during electrolysis. cations are attracted to the negative electrode (cathode). unit of electrical energy is joule and that of electrical power is watt. Q.11 Sol. then m = E . What is an electrolyte? Substances which conduct electricity because a fraction of their molecules dissociate into positive and negative ions are called electrolytes.1 Sol. The power consumed by an electric circuit is said to be one watt. Thus if m1 and m2 are the masses of two substances deposited and E1 m1 Q. Q. Faraday’s Second Law: If the same quantity of charge is passed through several electrolytes.THERMAL AND CHEMICAL EFFECTS OF CURRENT Q. E1 and E2 are their respective chemical equivalents. Write two characteristics of the wire of an electric heater. 2 2 Define electrochemical equivalent of an element. What is the cause of heating effect of current? When a potential difference is applied across a conductor.10 Sol. (c) the time for which the currents is passed through the conductor. Q.12 Sol.e. unit of electric power is watt. Heat produced in a conductor of resistance R. mαQ or m = ZQ = zIt The constant z is called the electrochemical equivalent (E.7 Sol. Of the two bulbs in a house. This is called Joule’s law.8 Sol. i. What are anions and cations? Anion: It is a negatively charged ion formed by addition of electrons to atoms or molecules.2 Sol. What are the SI units of electrical energy and electric power? What is the practical unit of electrical energy? The S. Two characteristics of a heater wire are: (1) It should have high melting point.13 Sol. one glows brighter than the other. What are the various factors on which the heat produced in a conductor depends? The various factors on which the heat produced in a conductor depends are (a) the current through the conductor. Q.5 Sol. (b) resistance of the conductor. anions are attracted to the positive electrode (anode). 2 .9 Sol. Faraday’s First Law: The mass of a substance deposited or liberated at an electrode is proportional to the quantity of charge that passes through the electrolyte. I kWh =103W × 60 × 60s = 3. The most common examples are acids. Convert kWh into joules. Some electrolytes are ionised to a small extent in their solutions. electrolysis is not possible with a. The SI unit of electrochemical equivalent (E. Those electrolytes which are completely ionised in their solutions are called strong electrolytes. What is Faraday constant? Faraday constant is the quantity of charge required to liberate one equivalent weight of a substance. Why does molten sodium chloride conduct electricity? Explain. During the electrolysis of water. 3 . no deposition on the electrodes will take place. A secondary cell is also called an accumulator or a storage cell. Thus a primary cell cannot be recharged. Q. What is an accumulator or a storage cell? Write names of two storage cells. What are weak electrolytes? Give examples.Q. NaCI etc. Write the SI unit of electrochemical equivalent. Current can be measured with voltameter. In these cells the chemical reaction is reversible. (b) The mobility of free electrons in metal is much higher than the mobility of ions in electrolytes. Q. A secondary cell is one in which the chemical reaction is reversible. Why? Pure Water does not undergo electrolysis as it does not dissociate into constituent ions. In case current reverses its direction periodically. So NaCl in molten state conducts electricity. Two secondary cells are: (a) Lead-acid cell (b) Ni-Fe cell. Q. why electrolytes have lower electrical conductivity than metallic conductors in general. What is an electrochemical cell ? An electrochemical cell is a system in which a chemical reaction produces an e. So they can be recharged.29 Sol. Q. For example. no electrolysis takes place.25 Sol.22 Sol.23 Sol.27 Sol. Sodium chloride in an ionic compound. Even in solid state it has Na+ and Cl– ions.f.18 Sol. For example HCI.15 Sol. A primary cell is one in which the chemical reaction that produces the emf is not reversible. Distinguish between primary and secondary cells.14 Sol.19 Sol.17 Sol.C.m. How does a voltmeter differ from a voltameter? Voltmeter is a device which is used to measure potential difference between two points. If pure water is used. Q. At the cathode the reaction is: 4H2O + 4e– → 4H– +2H2 At the anode the reaction is’: 2H2O → 4H+ + 4e– + O2 Q. Is electrolysis possible with alternating current? No. while voltameter is a vessel containing an electrolyte with two electrodes immersed in it. Thus a cell converts chemical energy into electrical energy. (a) The ionic density in electrolytes is very small as compared to the free electron density in metals. Some of the important uses of electrolysis are: (a) Electroplating (b) Purification of metals (c) Electrotyping (d) Medical applications etc.) is kg per coulomb.E. What are strong electrolytes? Give examples. Therefore water is acidified before electrolysis. Thus a secondary cell can be recharged.24 Sol.c. Write some uses of electrolysis.28 Sol. What is the difference between a fuse wire and a heater wire? The melting point of a fuse wire is very low while that of a heater wire is very high. Q. It is equal to 96500 C. Q. State two reasons.21 Sol. For electrolysis current should pass continuously only in one direction. Q.26 Sol.20 Sol. Q. Such electrolytes are called weak electrolytes. Name the electrodes on which hydrogen and oxygen are liberated during the electrolysis of water. hydrogen is liberated at the cathode and oxygen is liberated at the anode. Q. What is electroplating? Electroplating is the process in which a layer of some metal is deposited on any other surface by the process of electrolysis. Q. Q. Q. NH4Cl. The anode is made of the metal to be deposited and the cathode is made of the substance which is to be electroplated. Q. ions are free to move and became charge carriers.16 Sol. When it is melted. Q. 4 . Figure shows the labelled diagram of a dry Laclanche cell. The hydrogen is neutralized by MnO2 at the anode. Describe with the help of a labelled diagram the construction and working of a Daniel cell. As a result the current decreases and finally stops. This phenomenon is called polarization.30 Sol. As a result the zinc electrode is left negatively charged and acts as cathode. with a membrane (porous pot) through which ions can pass from one solution to the other. where they get discharged by removing electrons from the copper atoms:2H3O+ + 2e– → 2H2O + H2 The copper electrode is thus left positively charged and acts as anode.Q.31 Sol.34 Sol.33 Sol. according to the following reactions : Zn → Zn++ + 2e– Z++ + CuSO4 → ZnSO4 + Cu++ and Cu++ + 2e– → Cu The emf of the cell is 1. losing two electrons per atom. Q. zinc atoms in contact with the electrolyte ionize. Q.F. Due to these minute cells some internal currents are set up in the zinc rod which results in wastage of zinc. A copper rod and a zinc rod are immersed in the acid. Sol. This phenomenon is called local action. The electrons flow into the metal wire circuit and Zn++ ions pass into the solution. What is local action in a voltaic cell? In a voltaic cell.M. Zinc rod is amalgamated to avoid local action. so that the overall chemical reaction is Zn + 2NH4Cl + MnO2 → ZnCl2 + 2NH3 + H2O + Mn2O3 + Q Q being the energy released in the reaction.32 Draw a labelled diagram to show the components of a Leclanche cell and write the reactions taking place inside the cell. One thus has 2NH4+ + 2e– → 2NH3 + H2 at the anode. hydrogen is produced which migrates to the anode and covers it in the form of bubbles. What is the emf of a dry cell. Q. and in the body of the electrolyte. Describe briefly a simple voltaic cell ? A simple voltaic cell consists of a glass vessel containing dilute sulphuric acid which acts as electrolyte. What do you mean by polarisation in cells? In a primary cell.35 Sol. At the former electrode zinc ions pass into the solution and at the other electrode copper ions are deposited.wet and dry. The zinc is thus the cathode: Zn → Zn++ + 2e– The ammonium ions of the electrolyte remove electrons from the carbon anode. This cell is not very useful due to polarization. lying on the surface of the zinc rod. to which electrons flow in from the external circuit. on coming in contact with the acid form minute cells. Sulphuric acid and water dissociate as H2SO4 + H2O → 2H3O+ + SO4– Due to high concentration of Zn++ ions near the cathode. Action: Zinc atoms in contact with sulphuric acid give up electrons : Zn → Zn++ + 2e– The Zn++ ions pass into the electrolyte.5V. Laclanche cell has two forms . the H3O+ ions are pushed towards the copper electrode. Deniel Cell: It consists of a zinc electrode immersed in dilute sulphuric acid (or acidulated zinc sulphate solution) and a copper electrode in copper sulphate solution.of a dry cell is 1. When an external circuit is connected across the cell. E. impurities like carbon etc. Q.1 V. the Zn++ ion combines with the Cl– ions to form ZnCl2. Q. Draw the graph showing the variation of thermo-emf of a thermocouple with the temperature difference of its junctions.47 Sol. Seebeck arranged a number of metals in the form of a series according to the following criteria: (i) Current flows through the cold junction from the metal which appears earlier in the series to the metal which appears later.38 Sol. This temperature is called neutral temperature (θn) If the temperature is further increased. is called thermoelectric power or Seebeck dθ coefficient.37 Sol.45 Sol.I. unit. dE . The emf of this cell falls due to partial polarisation. the thermo-emf first increases and attains a maximum value. Can you use the Leclanche cell for supplying steady current? Leclanche cell can not be used for supplying steady current for a long time.43 Sol. Q. What is thermo electric series? Mention the first and the last members of the series. The temperature at which the thermo-emf changes direction is called temperature of inversion (θ) Q. Q. (ii) Greater the separation of the two metals in the series. What is the order of thermo-emf? The order of thermo-emf is 10–6 V Q. It's S. the emf decreases to become zero again and then it changes direction.46 Thermo-emf is given by the expression E = αθ + (1/2)βθ2 Write the expression for thermoelectric power (Seebeck coefficient) E = αθ + (1/2)βθ2 Sol. From which cell will you be able to draw larger current and why? The secondary cell will provide larger current as the internal resistance of a secondary cell is less than that of a primary cell. State the transformation of energy in a photo cell.39 Sol. What is thermoelectric effect (Seebeck effect) ? If two wires of different metals are joined at the ends and the two junctions are maintained at different temperatures.Q. Manganese dioxide (MnO2) Q. 5 . If the temperature of the hot junction of a thermocouple is gradually increased. Thermo electric power S = Q.I.36 Sol. Q.42 Sol.44 What is thermoelectric power? Write its S. On reversing the direction of current. heat is either absorbed or evolved at the junction. This is called Seebeck effect. Sol. How does its neutral temperature vary with the temperature of the cold junction? Neutral temperature is independent of the temperature of the cold junction. Q. unit is volt/Kelvin Q. The emf developed in the circuit is called thermo-emf. This phenomenon is called Peltier effect. The rate of change of thermo-emf with temperature. First member of the series -Sb Last member of the series -Bi Q. greater is the thermo-emf generated. Name the depolariser in Laclanche cell. You are given a primary and a secondary cell of the same emf. In a photo cell light energy is transferred into electrical energy. Define neutral temperature and temperature of inversion.41 Sol. dE = α + βθ dθ What is Peltier effect? If a current is passed through a junction of two dissimilar metals. the heating effect is also reversed. then a current starts flowing through the wires.40 Sol. anode is made of the impure metal and cathode of pure metal. Q. What is the relation between temperature of cold junction. What is the direction of the thermoelectric current at the hot junction of an iron-copper thermocouple? From copper to iron. Q. During discharging: At Cathode: Pb + SO4– – → PbSO4 + 2e– At anode: PbO2 + 2H+ + 2e– → PbO + H2O PbO + H2SO4 → PbSO4 + H2O 6 . neutral temperature and inversion temperature? Sol. θc is temperature of cold junction. (b) Platinum electrodes in dilute sulphuric acid.48 Sol. Mention some applications of thermoelectric effect. (b) H+ and OH– ions. and θi is inversion temperature. The primary cells can not be recharged while in secondary cells reversible reactions take place so that they can be recharged. Reactions: H2SO4 dissociates into H+ and SO– – ions.57 Sol. Q. The chemical process that occurred at the electrodes are then reversed and the cell recovers its original state. Q.50 Sol. Q. Write one main difference between primary and secondary cells. Define Peltier coefficient. Q. Some of the important applications of thermoelectric effect are: (a) Power generation (b) Measurement of temperature (c) Remgeration. Measurement of temperature.58 Sol. Write the expression which gives the relation of the thermoelectric emf of a thermocouple with the temperature difference of its cold and hot junctions. They are immersed in an electrolyte of dilute sulphuric acid.49 Sol. The electrodes consist of alternating parallel plates of lead dioxide (positive electrode) and spongy lead (negative electrode) insulated from each other. What is Thomson effect? The production of an electric potential gradient along a conductor as a result of a temperature gradient along it is called Thomson effect. Q.60 Sol. The electrolyte used is any soluble salt of pure metal. Thus points at different temperatures in a conductor are at different potentials.54 Sol. Describe briefly a lead-acid accumulator.59 Give one practical application of thermoelectricity.61 Sol.55 Sol. Lead-acid accumulator is a secondary cell which can be recharged by passing a current through it in the reverse direction. E = αθ + (1/2)βθ2 where a and β are constants. θn = Q. giving its charging and discharging chemical equations.52 Sol. Q. It is used to detect and measure the intensity of heat radiation. Copper-Constatantan thermocouple.Q.56 Sol. How is the electrical conductivity of an electrolyte affected by increase of temperature? The electrical conductivity of an electrolyte increases with the increase in temperature.53 Sol. Q. Why is an electrolyte dissociated when dissolved in liquids ? The ionic bonds between the ions of the solute are made weak by polar molecules of liquids. Peltier coefficient is defined as the amount of heat absorbed or evolved per second at a junction when a current of I A is passed through it. pure metal gets deposited on the cathode. When current is passed through the electrolyte.62 Sol. What is a Thermopile? Thermopile is a series combination of thermocouples. θi + θ c 2 where θn is neutral temperature. Name the carriers of current in the following voltameters: (a) Copper electrodes in CuSO4 solution. Q. Q. How are metals purified by electrolysis process? For purification of metals by electrolysis. Name the thermocouple which is used to measure a temperature of 300 K. Q. Q.51 Sol. Therefore the ions of electrolyte (solute) get dissociated. (a) Cu++ and SO– – ions. mercury is a good conductor.70 Sol.64 Sol. we have m = zit. where z is the electrochemical equivalent (ece). the charge on each ion is pe. For example. where e is electronic charge. deposits 0. therefore Fp N Now. What are the units in which the thermoelectric coefficients α and β are generally expressed? α: µ V/°C β: µ V/°C2 At room temperature. So the amount of charge required to liberate 1 mole of the substance is Fp where p is valency of the substance. Derive the relation connecting chemical equivalent and electrochemical equivalent of an element. Q. So.73 Sol. From Faraday’s first law of electrolysis.72 Sol.71 Sol. Seebeck effect is reversible.74 Sol. Q. Now we consider two substances having chemical equivalents E1 and E2. The charge required to liberate one atom of substance is. Q.66 Sol.65 Sol. Define International Ampere. from Faraday’s second law E = m 2 2 But m1 = z1it and m2 = z2it where z1 and z2 are the ece’s of the two substances. Q. Mercury On what factors does the magnitude of thermo-emf depend? The magnitude of thermo-emf depends on two factors: (a) Nature of the metal~ forming the thermo couple. let the masses of the two substances liberated be m1 and m2 respectively. Thus Q. which when passed through a silver voltameter. During charging: PbSO4 + 2H+ + 2e– → Pb + H2SO4 At cathode: At anode: PbSO4 + SO4– – + 2H2O → PbO2 + 2H2SO4 + 2e– Plot a graph showing the variation of thermoelectric power with temperature difference between the hot and the cold junctions.67 Sol. Define chemical equivalent and electrochemical equivalent of a substance. (b) Temperature difference between the two junctions.001118 g of silver in one second on the cathode. Faraday constant (F) is the amount of charge required to liberate 1 equivalent weight of a substance by electrolysis. The chemical equivalent of a substance is defined as the ratio of atomic weight to the valency. Thermo electric power S is given by S= α + βθ The graph is shown in figure Which one has lower internal resistance–a secondary cell or a primary cell? Secondary cell. Q. What does it mean ? It means that if the hot and the cold junctions are interchanged. E1 m1 Then.Q. F = Ne N Are all pure liquids bad conductors of electricity ? No. E1 E2 = m1 m2 = z1u z 2u or E/z = Constant. Fp = pe . the emf changes sign and the circulating current reverses direction. Q. The electrochemical equivalent of a substance is defined as the mass of the substance deposited on anyone of the electrodes when one coulomb of charge passes through the electrolyte.68 Sol. Q. When the same quantity of charge is passed through the electrolytes containing them. what is the order of the ratio of the conductivity of an electrolyte to-that of a conductor? 10–5 to 10–6 Name a liquid which allows current through it but does not dissociate into ions. Derive the relation between Faraday constant and Avogadro number. The constant is denoted by F and is called Faraday constant: E/z = F 7 .63 Sol. International ampere is defined as the steady current. Q. Q. Q.69 Sol. Inside the cell the current flows from anode to cathode. Q. Alkaline cell.76 Why do some covalent salts (which are not ionic in solid state) become conducting when dissolved in water? The dielectric constant of water is large (81). The anode is amalgamated zinc powder with gelatinised KOH electrolyte. Reactions taking place inside the cell are: At Cathode: Ag2O + H2O + 2e– → 2Ag+ 2OH– At anode: Zn + 2OH– → ZnO + H2O + 2e– The emf of the cell is 1. Lithium cell.75 Sol. 8 . the salts ionise and conduct electricity. What is the relation between Peltier coefficient and Seebeck coefficient? π = TS where π is the peltier coefficient. Alkali Accumulator: This accumulator consists of a steel vessel containing a 20% solution of KOH with 1% LiOH. the current flows from cathode to anode inside the cell and reaction is given by Ni(OH)4 + Fe→ Ni(OH)2 + Fe(OH)2 Sol. anode is connected to the positive terminal and cathode to the negative terminal of a d. It has a high energy output per unit weight and a constant voltage level. Which has higher internal resistance–lead accumulator or alkali accumulator? Alkali accumulator has higher internal resistance. Working:Potassium hydroxide dissociates as (a) 2KOH → 2K+ + 2OH– During charging. when no current flows through the cell. Another perforated steel grid stuffed with finely divided iron hydroxide is used as cathode. The anode is stuffed with nickel hydroxide. It is also known as silver oxide zinc cell.60 V. It weakens the attraction between the atoms of covalent salts. explain the construction and working of an alkali accumulator.Q.79 Sol.81 Sol.78 Sol. cameras etc.m. i.77 Sol. Q. Hydroxyl ions are attracted towards anode where they lose their charge and form nickel peroxide: (b) Ni(OH)2 + 2OH —→ Ni(OH)4 The positive ions move towards cathode and then react with Fe(OH)2 to form iron: (c) Fe(OH)2 + 2K → Fe + 2KOH The complete reaction during charging is given by adding (a) (b) & (c): Ni(OH)2 + Fe(OH)2 → Ni(OH)4 + Fe During discharging. State the condition in which terminal voltage across a secondary cell is equal to its e.80 Sol. Q. LiOH makes it conducting. Q. Main parts of a button cell are (a) Anode can (b) Cathode can(c) Separator (d) Gasket. source.e. Perforated steel grid is used as anode. which is widely used in electronic watches. In open circuit. (ii) Excess charging or discharging do not damage it. In some cases. What are the advantages of alkali accumulator over lead accumulator? (i) Alkali accumulator is not damaged if it is not charged for a long time.c. What is a button cell? Write its main components and reactions taking place at anode and cathode? Button Cell:The button cell is a solid state miniature dry cell. S is the Seebeck coefficient and T is the temperature of the cold junction. Cathode is of silver oxide and it is separated by an absorbent cellulosic material.. To lower the internal resistance traces of mercury oxide are used in it. Q.f. Q. With the help of a suitable diagram. Some other button cells are mercury cell. Silver Oxide zinc cell is shown in figure. .. If the domains are oriented preferentially by applying a magnetic field.....(4) valid except at low temperatures and high fields..(3) Diamagnetism In diamagnetism materials.... the relation can be expressed as B = µH.. Magnetic intensity H The magnetic intensity H is defined by the relation B = µ0 (H + M) .. The relation between B and H for ferromagnetic materials is nonlinear.. 9 . The magnetization can persist in hard magnetic materials to form a permanent magnet. Paramagnetism The permanent magnetic moment of an unpaired electron in a paramagnetic substance tends to become aligned with the magnetic field.... magnetic dipole moments are induced in molecules by the magnetic field.........(1) and a singular contribution due to the spin angular momentum.. For a Rowland ring.(2) The magnetization in a material is the magnetic moment per unit volume: M= < ∑ mi > ∆V .... the sample has a large magnetization... an orbiting electron has a magnetic moment proportional to its orbital angular momentum. m= – e 2m e L . magnetic dipoles. Large changes in the field occur over geological time intervals.. The magnetic field of the earth Outside its surface........... H is due to the macroscopic current in the windings. Ferromagnetism Molecular magnetic dipoles in a magnetic domain tend to be aligned in a ferromagnetic material...(5) For a linear medium with permeability µ. and the vectors M and B have opposite directions............... and hysteresis effects are present... the earth’s magnetic field is approximately a dipole field.MAGNETIC PROPERTIES OF MATTER Summary with Applications Atomic currents.. and magnetization In a simple model.. The vectors M and B are parallel and are related by Curie’s law M= CB µ0T ... m= e me S ...... (7) µ is called the magnetic permeability of the material. however.. Finally.40 1.00 0.. (Note the different scales: B >> B0) Initially (point a)... and reaches 98 percent saturation only when B0 is increased by about a thousand fold above that at point b.. The resulting magnetic field is the sum of that due to the current and that due to the iron. characteristic of the material inside the coil: B = µnI . the domains do not become completely unaligned. the total field follows the path efgb. µ. The value of µ. point c in Figure – 3..80 0. its value is very close to µ0 #.. if the current is again reduced to zero and then increased in the original direction. As the reverse current is increased further..80 1.20 a 0 Figure – 1 Iron–core torus 0.20 b g 0. enough domains can be turned around so B = 0 (point d). the domains become more and more aligned until at point b. (Point b is typically 70 percent of full saturation. Then the current I is slowly increased... The total field inside a solenoid in such a case can also be written by replacing the constant µ0 by another constant.80 B0 (10– Figure – 2 3 T ) Total magnetic field of an iron-core torus as a function of the external field B0 . which is essentially a long solenoid bent into the shape of a circle (Figure –1).. no domains are aligned. is not constant for ferromagnetic materials.. The iron is said to be approaching saturation.. It is sometimes convenient to write the total field in this case as sum of two terms : B = B0 + BM .. Measurements on magnetic materials are generally done using a torus..20 a 0 0...(6) Here.20 b B(T) B(T) 1. If the current is then reversed in direction. As the current is reduced to zero.. but follows the curved line shown in the graph of Figure –2... Suppose the torus has an iron core that is initially unmagnetized and there is no current in the windings of the torus. Indeed the field B0 inside a solenoid is given by B0 = µ0nI This is valid if there is only air inside the coil. often by hundreds or thousands of times.20 1.60 R d 0...40 0. Now suppose the external field B0 is reduced by decreasing the current in the coils. the field will be greatly increased. For ferromagnetic materials µ is much greater than µ0.40 1.. so that practically all the lines of B remain within the torus. 10 e Figure – 3 Hysteresis Curve . This occurs because the domains in the iron become preferentially aligned by the external field. the last few domains are very difficult to align). and B0 increases linearly with I. As B0 increases.Magnetic field in magnetic materials – Hysteresis The field of a long solenoid is directly proportional to the current.60 0.40 0. The total field B also increases. B0 refers to the field due only to the current in the wire (the “external field” ). Some permanent magnetism remains...00 0. it depends on the value of the external field B0. often BM >> B0..80 c 0. If we put a piece of iron or other ferromagnetic material inside the solenoid. the curve continues to rise very slowly.. it is equal to the field that would be present in the absence of a ferromagnetic material.20 B0 (10– 3 T) f 0. 1. as the following experiment shows... again approaching saturation at point b. the iron approaches saturation in the opposite direction (point e). nearly all are aligned. Φορ all other materials.. Then BM represents the additional field due to the ferromagnetic material itself. Materials for which this is true are said to have high retentivity. it is desired that ac and af be as large as possible. 11 . the result is a slight net magnetic effect which actually opposes the external field. heat treatment. in which µ is very slightly larger than µ0. Whether iron is “soft” or “hard” depends on how it is alloyed. a hysteresis curve such as that in Figure –4 occurs for so-called “soft iron” (it is soft only from a magnetic point of view). Atoms of diamagnetic materials have no net dipole moment. For a permanent magnet. made unmagnetized. whereas those revolving in the opposite direction are reduced in speed. The fact that the curves do not retrace themselves on the same path is called hysteresis. In such a cycle. the iron core is magnetized even though there is no current in the coils. The heads of a tape recorder are demagnetized in this way. On the other hand. This can be done by reversing the magnetizing current repeatedly while decreasing its magnitude. These points correspond to a permanent magnet. in the presence of an external field. Paramagnetic materials apparently contain atoms that have a net magnetic dipole moment due to orbiting electrons. electrons revolving in one direction are caused to increase in speed slightly. # B0 Figure – 4 Hysteresis Curve for soft iron B B0 Figure – 5 Successive hysteresis loops during demagnetization All materials are slightly magnetic. and may be referred to as “hard”. A ferromagnetic material can be demagnetized – that is. However. and diamagnetic. Non-ferromagnetic materials fall into two classes: paramagnetic. and the field can be reversed with less loss of energy. B At point c and f. it can be shown that the energy dissipated in this way is proportional to the area of the hysteresis loop. The curve bcdefgb is called a hyteresis loop. This results in the curve of Figure – 5. This is preferred for electromagnets since the field can be more readily switched off. in which µ is very slightly less than µ0. the alternating magnetic field acting at the heads due to a demagnetizer is strong when the demagnetizer is placed near the heads and decreases as it is moved slowly away. and other factors. and these become slightly aligned with an external field just as the galvanometer coil experiences a torque that tends to align it.Notice that the field did not pass through the origin (point a) in this cycle. much energy is transformed to thermal energy (friction) due to realigning of the domains. We should notice that. What about the electric field far from the source... undergoing f cycles per second........... Suppose. In the case of a radio station. the electric field reverses periodically as the wave passes.. for example – the field reverses in step with the charge reversal... The electric field wave a sent out by the antenna blankets the earth around Figure (3) The electric field wave from the antenna blankets an area it. The field shows the history of the charge on the dipole.. If the charges oscillate back and forth between the balls.. Propagation direction x ~ Figure (2) The alternating charges on the dipole antenna send an electric field disturbance out into space. the upward directed field were sent out one half cycle later.. as in the Figure(3). the field sent out along the x-axis is as shown in Figure(2).(2) a Figure (1) A portion of the instaneous electric field close to two charged balls.. What does this imply for the electric field outside the dipole? Close to the dipole – at point a.... Charges antenna are placed on the antenna by an ac voltage from a transformer system. the electric field at point a will alternately point up and down. you will see the antenna field Radio as a long wire stretched between two towers or station as a vertical wire held by a single tower. the electric field wave obeys the following relation between frequency f and wavelength λ λ= ν f . When the voltage source is a battery. that the battery is replaced by an ac voltage source.. when the top of the dipole was negative..(3) 12 .. At a certain instant. Thus the dipole continually reverses polarity... let us consider the situation shown in Figure... the transmitting antenna. If you visit Electr ic a radio transmitting site... Then the charges on the balls will vary sinusoidally. We see there a dipole consisting of two oppositely charged balls and the electric field the dipole generates... Its equation is Ey = Eoy cos 2πft .. The frequency of the oscillating electric field at a is the same as the frequency of the source. The charge on the top ball will vary as y .(1) q = q0 cos 2πft x The equal magnitude charge on the bottom ball is oppositely in sign and will vary in the same way.. however.... like all waves. the dipole (or antenna) is often simply a long wire. At a point such as a in the even quite distance from the station. The downward directed fields were sent out when the top of the dipole was positive...... the field is an electrostatic one and does not change. Hence the field at a oscillates in the y direction and varies in a sinusoidal fashion with the same frequency as the source.... Thus we see that an electric field wave is sent out by the oscillating dipole. however? How does it behave? We can think of the electric field as being the disturbance sent out by the dipole source much the same way on a string is the disturbance sent down the string by an oscillating source...ELECTROMAGNETIC WAVES The generation of em Waves To begin our study of electromagnetic waves (em waves). path of the wave. Types of Electromagnetic Waves As we discuss at greater length later. refer to Figure given below. the larger the acceleration (or deceleration) of the charge. This was an astonishing prediction because the speed he found was a well known speed. radio-type waves were foreseen by a 34 year old Scottish physicist.where ν is the speed at which the wave travels out through space. not of material particles. he was able to prove that these waves should have a speed of 3 × 108 m/s in vacuum. Notice. the magnetic field wave is perpendicular to both the electric field wave and the direction of propagation. the magnetic field travels out along the x-axis as a transverse wave. em waves consist of oscillating electric and magnetic fields. y An tenna x I Transformer z Magnetic field wave ~ Oscillator x Propagation direction Figure (4a) (a) As charge rushes up and down the antenna. y Electric field x Propagation direction z Magnetic Figure (5) field In an em wave. To see this. It is easy to see that a radio station’s antenna necessarily generates a magnetic field wave as it generates an electric field wave. Thus. James Clerk Maxwell. Because the direction of the current oscillates. Thus Maxwell was led to surmise that light waves are one form of em waves. if a fast-moving charged particle undergoes an impact. is always perpendicular to the direction of propagation. As such. as shown in the Figure (4b). Furthermore. as we shall see. we refer to this as the em wave spectrum. However. It turns out that whenever a charge undergoes acceleration. they reach maxima together). an oscillating magnetic field is produced. it will emit a burst of em radiation as it suddenly stops. Notice that the charges that oscillate up and down the antenna are accelerating. As shown in the Figure(5). Later we shall show that em waves travel through vacuum with the speed of light. You will recall that. At the radio station. c. Today we know that there are a variety of em waves that cover a wide range of wavelengths. That this is true is not obvious. we notice that the quantity that vibrates. As with the oscillating electric field. that the magnetic field is in the z direction.998 × 108 m/s. The two waves are drawn in phase (that is. many years before the first radio was invented. em waves are transverse waves and are much like waves on a string and other transversewaves. the magnetic field. charges are sent up and down the antenna in the Figure (4a) to produce the alternating charges we have been discussing. while the electric field is in the y-direction. however. Figure (4b) (b) A magnetic field wave is sent out as shown. the more energy it emits as em radiation. in the SI. they carry energy along their direction of propagation. it is the result of detailed computations. they can travel through empty space (vacuum). it emits em radiation. This charge movement constitutes an alternating current in the antenna. As we see. the magnetic field id perpendicular to both the electric field and the direction of propagation. which we designate by c. so too does that of the magnetic field. and because a magnetic field circles a current. Hence the electric field wave is a transverse wave . Further. And. the speed of light in vacuum. 13 . in 1865. Maxwell used the then known facts about electricity to show that em radiation should exist. There is one other feature of em wave generation that we should point out. namely the electric field vector. the speed of light in vacuum is defined to be c = 2. wvaelength range of each type of radiation. Since material particles and energy cannot travel faster than the speed of light. The various types of electromagnetic radiation are shown in Figure(11). At the shorter wavelengths. Let us now discuss briefly the nature of each type of radiation. of course. X-ray waves X-rays are electromagnetic radiation with λ < 10nm. The human eye is which shows the light spectrum in colour. and appears as thermal energy.03 cm can be charged during this short time. The shortest wavelength given in Figure for microwaves (10–3m) represents the lower limit of wavelengths that can be generated electronically at present. the relation 1022 1020 1018 1016 1014 1012 1010 10 8 106 λ = ν/f becomes λ = c/f for 10 4 electromagnetic radiation. Warm objects of all types radiate infrared rays. Ultraviolet waves Ultraviolet waves are radiation with wavelengths shorter than visible violet light but still stronger than about 10nm. Notice that. 14 .Wavelength The basic difference between (m) the various types of em waves –14 –12 –10 –8 10 10 10 10–6 10–4 10 –2 1 102 104 are the result of their different 10 Gamma rays Visible light Microwaves wavelengths. Even though these waves are all electromagnetic waves. 7 x 10–7 m. they are not distinct from x-rays.03 cm during one oscillation. The “antenna” that generates light waves is charge accelerating within an atom. The bars indicate the approximate frequency of the radiation. For this reason. how frequently should one reverse the charges on the antenna? Microwaves Microwaves are short-wavelength radio waves. light can travel only 0. infrared radiation is also called heat radiation. Gamma rays waves Gamma rays (γ-rays) are electromagnetic radiation given off primarily by nuclei and in nuclear reactions. We classify various 500 600 700 wavelength regions in this range by the names of colours. The energy contained in the waves is also absorbed. Sensitivity curve for the eye. Sensitivity Light waves Violet Green Red The wavelengths of the visible portion of electromagnetic –7 –7 Blue Yellow radiation extend only from about 4x10 to 7 ×10 m. only an antenna shorter than 0. Infrared waves Infrared waves have wavelengths between those of visible light. They differ from x-rays only in their manner of production. The earth receives from the sun a large amount of infrared radiation as well as light. See also Colour Plate II. You should learn most sensitive to greenish yellow light the approximate wavelengths of the various colours. Infrared radiation is readily absorbed by most materials. and microwaves. Examine this chart carefully to become familiar with the wavelength ranges involved. 40 0 Wavelength(nm) The sensitivity of the normal human eye to wavelengths in Figure (12) this region is shown in Figure(12). They are sometimes called radar waves. Since all electromagnetic radiation Ultraviolet travels through vacuum with X-rays light TV and radio waves Infrared light the speed of light. at a frequency of 1012Hz. which is the distance from the earth to the sun. this classification is reserved for the radiation given off by electrons in atoms that have been bombarded. If one wished to obtain a wave with λ = 108 km. Radiowaves We have already discussed radio waves in some detail. Usually. they differ considerably in their mode of interaction with matter. Notice that the spectrum of electromagnetic radiation encompasses waves with wavelengths extending from longer than 106m to shorter than 10–18m. Their wavelengths range from 1 m or so to more than about 3 × 106m for the waves sent out by ac power lines. This should indicate why very short wavelength waves are difficult to produce electronically. and we call this wavelength range light. Frequency (Hz) Hence a difference in λ Figure (11) implies a difference in the Types of electromagnetic radiation. The electric field vector is sinusoidal and perpendicular to the direction of propagation. the vertical component of the vibration will pass through and the horizontal component will be stopped. It derives its name from the fact that the electric field vector vibrates in a single plane when a beam of light is plane-polarized in only one plane. the transmitted light will be plane polarized and will consist of the sum of the electric field vector components parallel to the permitted direction.1 in which all vibrational planes are perpendicular to. the electric-field vibration will be as shown for three types of beams Unpolarized light can be conveniently plane . the x electric field vibrates up and down at a given point in space as the wave passes by.1. the approaching vectors would appear as shown in Figure – 2(b). the plane of the page in this case. in the plane of the page in Figure . The resulting sheet allows light to pass through it only if the electric field vector is vibrating in a specific direction. Unpolarized (a) Vertically Polarized Horizontally Polarized (b) (c) Figure – 2 If narrow beam of light is coming straight out of the page. Hence. the vectors would appear as in Figure – 2(c). Hence.polarized using a polarizing sheet. If the direction of propagation is to the right. many waves like the one in Figure . if the electric field is oriented as shown in Figure 3(a). If we pass light that is vibrating at the angle shown in Figure 3(a)(b) through a polarizing sheet whose transmission direction is vertical. with the wave travelling straight towards us. and actually most of them do not. We call a wave such as this a plane polarized The electric field vector vibrates wave. Let us stand at the end of the x-axis in Figure – 1 and look back along it towards the coordinate origin. if unpolarized light is incident upon the sheet. in other words.polarized wave. The great multitude of waves coming towards us give rise to many individual electric field vectors that are randomly oriented. as in Figure 2(a). Most light consists of many. v/c We know that light is em radiation. it can be thought of as consisting of a vertical and horizontal component. However. As we shall see. This is a sheet of transparent plastic in which special needle like crystals of iodoquinine sulfate have been embedded and oriented. It is fundamental to the behaviour of light that concerns us in this section.polarized vertically. Any vector can be thought of as consisting of two perpendicular components. If the wave is travelling along the x -axis. 15 . For a horizontally plane . this fact is important when light is transmitted through certain materials. V E H (a) (b) Figure – 3 The electric field vector can be split into x and y components. There is a magnetic field wave perpendicular to the page and in Figure step with the electric field. It consists of waves such as the one y E shown in Figure . It is also a factor when light is reflected. they need not all vibrate in the plane of the page. as shown in 3(b). the electric field vectors must all vibrate perpendicular to the x-axis.1. that is.POLARIZATION OF LIGHT Many optical devices make use of the fact that light is a transverse vibration. If the waves were plane . Apart from partially and plane (i.e. + = Plane polarised light Plane polarised light with with vibration in the vibration perpendicular to plane of paper the plane of paper · · · · Un polarised light The devices such as polaroids or Nichol prism are called polarizer when used to produce plane polarized light and analyzer when used to analyzed (i. light can also be circularly or elliptically polarized that too left handed or right handed. Vertical Polarizing sheet Vertical Polarizing sheet Horizontal Polarizing sheet (a) Vertical Polarizing sheet (b) Figure .. These are also transmitted by the analyzer (the second sheet) since it too is vertical. (b) The second polarizing sheet (the analyzer) and the polarizer are crossed. Therefore (almost) no light comes through the combination. In part (b). however. the polarizer (the first sheet) allows only the vertical vibrations to pass. Elliptically and circularly polarized lights result due to super position of two mutually perpendicular plane polarized lights differing in phase by (π/2) with unequal or equal amplitudes of vibrations respectively.e. These are completely stopped by the vertically oriented analyzer. the polarizer has been rotated through 900 and now allows only horizontal vibrations to pass. identify) the given light.Consider what happens when unpolarized light is passed through two polarizing sheets as shown in Figure 4. z E oz z E oz E ωt E ωt y y E oy Eoy Right handed Elliptically Polarised Light (B) Left handed Elliptically Polarised Light (A) 16 .4 (a) the unpolarized light is polarized by the first polarizing sheet (the polarizer). In this latter case we say that polarizer and analyzer are crossed. Light (transverse wave) Sound (longitudinal wave) (a) (b) I0 Unpolarized light Light source I0/2 Unpolarized light Polarizer Plane Polarized Light In case of interference of polarized lights the interfering waves must have same plane of polarization otherwise unpolarized (or partially polarized) light will result. and the beam is completely stopped by the analyzer. In part (a).. linearly) polarized. So if unpolarized light passes through proper thickness of these. 17 .. quartz and tourmaline etc..Methods of Obtaining Plane Polarized Light (A) By Reflection : In 1811. both reflected and refracted light becomes partially polarized. Era y O-ray Calcite crystal IE Era y IO O-ray Blackened surface Screen (B) (A) By using the phenomenon of double refraction and isolating one ray from the other we can obtain plane polarized light which actually happens in a Nichol prism. (D) By Dichroism : Some crystals such as tourmaline and sheets of Iodosulphate of quinone has the property of strongly absorbing the light with vibrations perpendicular to a specific direction (called transmission axis) transmitting the light with vibration parallel to it... Nichol prism is made up of calcite crystal and in it E-ray is isolated form O-ray through total internal reflection of O-ray at Canada balsam layer and then absorbing it at the blackened surface as shown in Figure.. reflected light is partially polarized. incident unpolarized light splits up into two light beams of equal intensities with perpendicular polarizations. And as in one reflection about 15% of the light with vibration perpendicular to plane of paper is reflected.ray) while the other does not obey laws of refraction and is called E-rays (extraordinary ray) this why when an object is seen through these crystals we usually see two images of an object and if the crystal is rotated one image (due to E-ray) rotates around the other (due Canada balsam layer to O-ray)... (B) By Refraction : In this method a pile of glass plates is formed by taking 20 to 30 microscope slides and light is made to be incident at polarizing angle (570).. (iv) For glass θp = tan–1(3/2) ~ 570 while for water θp = tan–1(4/3) ~ 530. This selective ... (iii) For i < or > θp.” In case of polarization by reflection – (i) For i = θp. reflected and refracted rays are perpendicular to each other. This specific angle of incidence is called polarizing angle θp and is related to the refractive index PP L µ of the material through the relation – UPL tan θp = µ . Polaroids work on this principle. r i = θp PL Partially 57 0 Reflected Light 57 0 (C) By Double Refraction : It was found that in certain crystals such as calcite.. the reflected light is completely plane polarized with vibrations in a plane perpendicular to the plane of incidence.. One of the ray behaves as ordinary light and is called O-ray (ordinary . (ii) For i = θp. Brewster discovered that when light is incident at a particular angle on a transparent substance. after passing through a number of plates as shown in the Figure emerging light will become plane polarized with vibrations in the plane of paper.. the transmitted light will be plane polarized with vibrations parallel to transmission axis.absorption of light is called dichroism...(1) known as “Brewster law... In accordance with Brewster law the reflected light will be plane polarized with vibrations perpendicular to the plane of incidence (which is here plane of paper) and the transmitted light will be partially polarized. . Transmission axis Pl perpendicular to the plane of paper Tourmaline crystal (A) θ A sin θ y I = K(A cos θ)2 = KA2 cos2 θ or I = I0 cos2 θ [as i0 = KA2] . 1 1 1 1 θ + sin 2θ = (cos )av = x 2 2π 2 2 0 2 so from eq...e.. intensity becomes half. From this it is clear that – If the incident light is unpolarized than as vibrations are equally probable in all directions (in a plane perpendicular to the direction of wave motion). So the intensity of emergent light will be A cos θ z Intensity of Light Emerging From a Polaroid (1) polarised an (E) By Scattering : When light is incident on atoms and molecules.. A cos θ. i.. (2). This process is called scattering. Parallel polaroids (A) Crossed polaroids (B) 18 .. the electrons absorb the incident light and reradiate it in all directions.. then component of vibrations parallel to transmission axis will be A cosθ while perpendicular to it A sin θ.. Now as polaroid will pass only those vibrations which are parallel to its transmission axis.. θ can have any value from 0 to 2π and hence 2π (cos2θ)av = 2π 1 1 1 cos 2 θ dθ = x (1 + cos 2θ) dθ ∫ 2π 0 2 2π ∫0 2π i. I= 1 I 2 0 i.e. If an unpolarized light is converted into plane polarized light (say by passing it through a polaroid or a Nichol – prism).e. we have. It is found that scattering light in directions perpendicular to the direction of incident light is completely plane polarized while transmitted light is unpolarized. (2) If light of intensity I1 emerging from one polaroid called polarizer is incident on a second polaroid (usually called analyzer) the intensity of the light emerging from the second polaroid in accordance with Malus law will be given by I2 = I1 cos2 θ ′ where θ ′ is the angle between the transmission axis of the two polaroids. Light in all other directions is partially polarized.(2) This law is called “Malus law”..Optic axis is Poloroid (B) y Plane Incident Light Unpolarised po la ri s ed x e z If plane polarized light of intensity I0 (=KA2) is incident on a polaroid and its vibrations of amplitude A makes an angle θ with the transmission axis. This phenomenon is called optical activity or optical rotation and substances optically active. (b) There is a variation in intensity of emergent light with minimum not equal to zero. it is said to be dextrorotatory or right handed. The optical activity of a substance is related to the asymmetry of the molecule or crystal as a whole. I2 = I1 cos2 θ = I1 And if the two polaroids are crossed.. 19 . polarimeter in terms of specific rotation which is defined as the rotation produced by a solution of length 10cm (1 dm) and unit concentration (i.polarized or plane polarized. However. if the substance rotates the plane of polarization anti clock wise it is called laevo–rotatory or left handed. θ ′ = 900. solution of cane sugar is dextrorotary due to asymmetrical molecular structure while crystals of quartz are dextro or levo rotatory due to structural asymmetry which vanishes when quartz is fused. Polariser Analyzer Substance Leavo Rota tory · Unpolarised Light Plane polarised Light Dextr o Rota tory Polarimeter If the optically active substance rotates the plane of polarization clockwise (looking against the direction of light). the incident light will be partially polarized as in partially polarized light vibrations exist in all directions but are more in some directions than in others.e. 1 g/cc) for given wavelength of light at a given temperature. Identification of given Light Polaroid or Nichol prism is used to examine whether a given light is unpolarized partially .. (c) There is variation in intensity of emergent light with minimum equal to zero. the incident light is plane (or linearly) polarized as in plane polarized light vibrations are confined along one direction only and so transmission axis of analyzer will become parallel and perpendicular to vibrations twice in its one complete rotation giving rise to maximum and zero intensity twice in each rotation. e.So if the two polaroids have their transmission axis parallel to each other. have their transmission axes perpendicular to each other.e. I2 = I1 cos2900 = 0 So if an analyzer is rotated from 0 to 900 with respect to polarizer. the incident light will be unpolarized as in case of unpolarized light the vibration of same amplitude are equally probable in all directions.. the plane of polarization of light is rotated about the directionof propagation of light through a certain angle. i. the intensity of emergent light changes from maximum value I1 to minimum value zero.g. the given light is passed through a polaroid (called analyzer) and the polaroid is rotated about the incident light and emergent light is seen. i. i. [α ] λ t0C = θ LC where θ is the rotation in length L at concentration C. θ ′ = 0. Then if – (a) There is no variation in intensity of emergent light in any position. Optical activity of a substance is measured with the help of.e. For this. Optical Activity When plane polarized light passes through certain substances.e. What are the various types of communication systems? There are two types of communication systems: (a) Analog communication system: It makes use of analog signals. What is an analog signal? A continuously varying signal (voltage or current) is called an analog signal. is called carrier wave. 20 . Very Short and Short-Answer questions What do you mean by Communication? Communication is the processing.5 Sol. for example. Modulated wave: The resultant wave produced by superimposing the audio signal on a high frequency carrier wave is called modulated wave. Q. Q. Modulating waves: The audio signal to be transmitted over long distances is called modulating wave. sending and receiving various information with the help of suitable devices and transmission medium.4 Sol. carrier wave and modulated wave. Q.PRINCIPLES OF COMMUNICATION Q. a square wave.9 Sol. Q. Q. Carrier wave: A high frequency wave.1 Sol. Distinguish between the terms modulating wave. (c) Phase modulations What is amplitude modulation? Amplitude modulation: When the amplitude of high frequency carrier waves is changed in accordance with the intensity of the modulating wave. Q. What is phase modulation? Phase modulation: The process in which the phase of the carrier wave is varied in accordance with the modulating wave is called phase modulation. Sol. it is called amplitude modulation.3 Sol.6 Sol. What is a digital signal? A voltage or current signal that can have only two discrete values is called a digital signal. over which audio signals are to be superimposed. What do you mean by radio communication? Radio communication involves transmitting and receiving a message in the form of a radio wave signal in between two stations without connecting them with wire.7 Sol.11 Sol. Frequency modulation: When the audio signal is superimposed on the high frequency carrier wave in a manner that the amplitude of the modulated wave is same as that of the carrier wave but its frequency is modified in accordance with the intensity of the modulating wave. Q.12 Show graphically amplitude modulation. (b) Digital communication system: It makes use of digital signals.10 What is frequency modulation? Sol. What is modulation? The process of superimposing electrical audio signals on high frequency carrier waves is called modulation. What are the various methods of modulation? There are three methods of modulation: (a) Amplitude modulation (b) Frequency modulation Q. Q. Q. it is called frequency modulation.2 Sol. Q.8 Sol. i.19 Write the advantages of frequency modulation. Yes. For strong and clear reception the modulation index must be high. Q.e.e. in case of frequency modulation the value of modulation index can be greater than unity. Q. Q. ma = Amplitude change of carrier wave Amplitude of normal carrier wave Q. i. In amplitude modulation.depth of modulation).23 Give expressions for bandwidth in (a) AM transmission (b) FM transmission Sol (a) AM transmission: Bandwidth = 2 × maximum frequency of modulating signal. Q. Q. Q. Q. (b) FM transmission: Bandwidth = 2n × frequency of modulating signal where n is the number of significant sidebands. The reception is generally noisy.Q. Sol. 21 .14 What is the value of bandwidth in amplitude modulation? Sol.22 What do you mean by bandwidth? Sol. The frequency range in which a transmitting system makes transmission is called bandwidth. Modulation factor is defined as the ratio of the change of amplitude of the carrier wave to the amplitude of the normal carrier wave. The modulation index determines the strength and quality of the transmitted signal. Sol. Sol. (ii) Frequency modulation can be used for the stereo sound transmission due to the presence of a large number of sidebands. the process of recovering the audio signal from the modulated wave is known as demodulation or detection. Modulation index in case of frequency modulation is given by δ mf = f s where δ is frequency deviation and fs is modulating frequency. Messages cannot be transmitted over long distances using amplitude modulation.20 Give two disadvantages of frequency modulation. Q. The reverse process of modulation.15 Define the term frequency deviation.(i) The frequency modulation transmitting and receiving equipments are very complex as compared to those used in amplitude modulation transmission. Q.. In frequency modulation the maximum variation of the frequency of modulated wave from the carrier frequency is called frequency deviation.16 Sol. bandwidth is twice the signal frequency.(i) Frequency modulation transmission is highly efficient.(ii) A.24 What is demodulation? Sol.13 Define modulation factor(modulation index.(i) (ii) (iii) What are the limitations of amplitude modulation? The efficiency of amplitude modulation is low. Sol.wider frequency channel is required in frequency modulation transmission.18 Can the value of modulation index be greater than unity? Sol.21 What is the importance of modulation index? Sol. Q. Sol.17 Write an expression for the modulation index for frequency modulation. 34 What is a passive satellite? Sol.25 What is pulse amplitude modulation? Sol. both functions are included in a modem.. Q. (a)Two wire lines (b) Coaxial cables (c) Radio link (d) Fibre link Q.27 What is sampling? Sol. Q.35 What is an active satellite? Sol. Q. 22 . Q. Q. Q. PVC insulation.33 What do you mean by remote sensing? Sol.(i) (ii) (iii) Write three merits of digital communication. As the name implies. 24 hours. The process of modulation in which the amplitude of the pulses is varied in accordance with the modulating signal is called pulse amplitude modulation. Satellite communication can be used for establishing mobile communication with greater ease.(i) (ii) (iii) Write three merits of satellite communication. Q. Q.37 Name various transmission media used in communication systems? Sol. A satellite which is used to reflect the signals back to the earth is called a passive satellite.38 What are the two types of two-wire line? Sol. for example. Q. (b) Twisted power line: It consists of two insulated copper wires twisted. An artificial object placed in an orbit around the earth or any other planet is called an artificial satellite.32 Sol. Digital signals do not get distorted by the noise. In remote and hilly areas it is most cost-effective. process it. The process of generating pulses of zero width and of amplitude equal to the instantaneous amplitude of the analog signal is called sampling. A satellite whose period of revolution around the earth is same as that of earth about its own axis. i.26 Draw a sketch to illustrate the basic elements required to transmit and receive an audio signal. Q.Q. Q. around each other. A satellite equipped with electronic devices to receive the signal from the earth.(a) Parallel wire line: In this two metallic wires run parallel to each other in an insulation coating. Digital signals are easy to receive. Satellite communication covers wide area. Q.29 What is modem? Sol.28 Sol. amplify it and then retransmit it back to the earth is called an active satellite. The name modem is a contraction of the terms modulator and demodulator. Digital signals can be stored as digital data.30 What is an artificial satellite? Sol. Remote sensing is the process of obtaining and recording information from a distance without physical contact.31 What is a geostationary satellite? Sol. is called a geostationary satellite and the orbit is called synchronous or geostationary orbit.e.36 What is transmission medium? Sol. It is a link which transfers information from the information source to the destination. Q.43 What is the use of optical fibres? Sol. The glass coating of relatively lower refractive index. It is a device which is based on the phenomenon of total internal reflection. the copper mesh shields it electrically from the external electrical disturbances. to distinguish it from spontaneous emission which occurs on its own. Q. the 2-wire line is used for transmission of signals over a small distance.47 Write an expression for the angle of acceptance in an optical fibre. If i is the angle of acceptance then sin i = µ12 − µ 22 Where µ1 is the refractive index of glass core and µ2 is the refractive index of cladding.41 What is the use of copper mesh in a coaxial cable? Sol. A laser beam can be sent to a far off place and be reflected back without any significant loss of intensity. Q. Q. surrounding the glass case in optical fibre is called cladding. Q. Sol.51 What is stimulated emission? Sol. the energy losses in a 2-wire line become very large. The term LASER stands for Light amplification by stimulated emission of radiation. Q.40 What is a coaxial cable? Sol. The phenomenon of transmission of information from one place to another using optical carrier waves is called optical communication.39 What is the main drawback of 2-wire line. it may make a transition to a lower energy state by the ‘impact’ of another photon of the required energy. The optical fibres are used to transmit light signals from one place to another without any appreciable loss in the intensity of light. highly coherent and perfectly parallel.Q. This process is known as stimulated emission.42 What is an optical fibre? Sol.45 What is cladding? Sol. when the central copper wire carries the signal.46 What do you mean by angle of acceptance in optical fibre? Sol. Therefore. Q. 23 .48 What is optical communication? Sol.49 What does LASER stands for? Sol. A coaxial cable consists of a central copper wire surrounded by a PVC insulation over which there is a sleeve of copper mesh. It consists of a very thin glass or quartz fibre which is coated with a material of lower refractive index. Q. At microwave frequencies.highly monochromatic. The main characteristics of a laser beam are . Finally it is covered with an outer thick PVC material.44 Sol. Q. The attenuation of signals increases with the length of the wire. Sol. (a) (b) What are the two types of optical fibres? The optical fibres are of the following two types: Monomode optical fibre Multimode optical fibre Q. Q. If an atom or a molecule is in an excited state. In a coaxial cable. The maximum angle of incidence in air for which light is totally reflected at the glass core-cladding interface is called the angle of acceptance. Q.50 Write the main characteristics of a laser beam? Sol. 58 What is FAX? Sol.59 Name the layer of earth’s atmosphere which is most useful in long distance radio communication. Amplitude modulation. Q. Q. Sol. The electronic transmission of a document to a distant place via telephone line is known as FAX or fascimile. Sol.56 Which modulation is used in TV broadcast and why? Sol.65 Is it necessary to use satellites for long distance TV transmission? Give one reason. So they can not be transmitted directly over large distances. Amplitude modulated band consists of radiowaves of frequencies less than 30MHz. Q. Q. TV broadcast requires larger bandwidth and hence frequency modulation is used. Q.57 What is the approximate dimension of an antenna? Sol. Short waves are used for long distance radio telecast as they can be easily reflected back to the earth by the ionosphere.64 What is amplitude modulated (AM) Band? Sol. The process of modulating a light beam from an optical source in accordance with the information signal is called light modulation. Q. Q. Audio signals are weak signals. TV signals cannot be sent directly to large distances. (i) If the waves travel along the surface of the earth from one place to another. 30MHz. it is called sky wave. Sol. (ii) If the wave is transmitted towards the sky and reaches another location after getting reflected from the ionosphere.66 Write an expression for the distance upto which the TV signals can be directly received from a TV tower of height h. Q. distance d = 2hR Where R is radius of the earth and h is the height of the antenna. it covers large distances on the earth.53 What is light modulation? Sol. Q.62 What is the frequency range of UHF band? Sol.55 Which modulation is used for commercial broadcasting of voice signal? Sol..52 What do you mean by optical pumping? Sol.60 What is the highest frequency that can be reflected back by the ionosphere? Sol. Q. it is called ground wave. Audio signals emitted from a certain location on the earth can be received at another location by two different ways. Q. Since the satellite is high above.54 Why do we need modulation. Hence antenna of large dimensions (–λ) is required.61 Distinguish between ground wave and sky wave? Sol. Of the order of the wavelength of the signal. Moreover the wavelength associated with audio signals is very large. Sky wave transmission is limited to a frequencyof 30MHz. Therefore the signals from the broadcasting station are beamed towards a geostationary satellite. Ground wave transmission is limited to 1500KHz. TV signals have very high frequencies which are not reflected by the ionosphere. which relays them back to the earth. It reflects the radio waves back to the earth. Why? Sol. Ionosphere.Q. It is the process of raising atoms from a lower energy state to a higher energy state by induced absorption. . modulation is needed. Q. 24 . Further. To overcome these difficulties. Ultra high frequency (UHF) band has frequency ranging from 300MHz to 3000MHz. Q.63 Which waves are used for long distance radio telecast? Sol. Q. 71 What is FM Band? Sol The electromagnetic waves in the frequency range 80 MHz to 200 MHz constitute frequency modulated (FM) band. The attenuation of ground waves increase with increase in frequency due to interaction with objects in its path.81 What is LED? Sol. Q. If a pn junction diode is forward biased.75 What is the bandwidth required to telecast picture through a TV channel? Sol.79 What do you mean by sensitivity and responsivity of a detector? Sol.78 What is an optical detector? Name three optical detectors. It is a device that generates electrical signals when light falls on it. (ii) Optical fibre. photo transistor.73 Name the electromagnetic wave used in satellite communication. TV signals have frequency much higher than 30 MHz.76 What is an optical source? Name any two optical sources. 25 . the energy released is in infrared region.Silicon photodiode.Q.69 Why is ground wave transmission limited to a frequency of 1500 kHz. Sol. Q. Q. Q.72 What is a geostationary satellite? Sol.(i) Low beam divergence (ii) Greater transmission distance (iii) High modulation rate Q. Light source used in optical communication is called an optical source. Sol. Q. Microwaves (109 – 1012 Hz) Q. Sensitivity is a measure of the ability of a detector and determines how weak a signal can be detected. Sol. It has the time period of revolution around the earth equal to the rotational period of the earth about its axis. the energy released is in visible range. if the diode is made of gallium arsenide or indium phosphide. Photonics is a subject that deals with generation and detection of photons. Q.e. The ability of the detector to respond quickly to the changing light pulses is called responsivity.LED and LASER. Q. i.70 Why sky waves are not used in the transmission of TV signals? Sol. Examples. sky waves are not used in the transmission of TV signals. Therefore.7MHz per channel. Q.74 Name two primary requirements for optical transmission. Microwave. However. An LED works on the process of spontaneous emission when a p-n junction is forward biased. MF stands for medium frequency band (300–3000 kHz). Sol. Sol. A satellite which appears to be fixed at a place above the earth is called a geostationary satellite. Sol.67 Which part of the electromagnetic spectrum is used in operating a radar? Sol. 4.80 What is photonics? Sol. Such high frequency signals cannot be reflected by the ionosphere. avalanche photodiode. Q. energy is released due to recombination of electrons and holes. In case of silicon and germanium diodes. So ground wave transmission is limited to a frequency of 1500 kHz. Q.. Examples .77 Write three advantages of LASER over LED as optical source.68 What do you understand by HF and MF bands? Sol. 24 hour Q. HF stands for high frequency band (3–30 MHz). (i) Optical source Q. etc. Why is modulation necessary in a communication system? Modulation is necessary in a communication system due to the following reasons: Antenna Length: It can be shown theoretically that in order to transmit a signal effectively. A carrier wave. This is due to their small wavelengths. The capacitor C1 acts as a bypass for carrier waves and the audio frequency voltage appears across the resistor R. The figure shows a TV transmitting antenna of height AB = h located at A on the surface of the earth of radius R. After modulation the transmission is at high carrier frequencies. superimposing them on high frequency carrier waves makes it possible to transmit them over long distances. signals can be directly received from a TV tower of height h. Q. At low audio frequencies the efficiency of radiation is not good.1 Sol. The audio signal frequencies are low and hence their energies are small.3(a) Explain the use of a geostationary satellite for long distance communication. This rectified output is fed to the parallel combination of a capacitor C1 and a resistor R. Sol. Therefore..82 Derive an expression for the distance upto which the TV. (a) (b) (c) Q. they cannot be transmitted over large distances directly. Draw a circuit diagram for demodulation (detection) and explain its working. (R+ h)2 = d2 + R2 d2 = h2 + 2hR Since h < < R. which can be conveniently installed. microwaves are used as carrier waves since they can pass through the atmosphere without significant loss of energy.The audio frequencies range from 20 Hz to 20 kHz: If they are transmitted directly. OB2 = OS2 + BS2 Now. the signal is radiated into space. the length of the required antenna would be very large. say 1000 kHz. So. for a frequency of 20 kHz. This would require an antenna of about 300 metres length. (b) Explain remote sensing. greater is the energy possessed by it. which is quite efficient. The output of the diode is a series of positive half cycles of radio frequency current pulses. However. The signal transmitted can be received within a circle of radius AS on the surface of the earth. Sol. on the other hand. Long distance transmission is carried out without wires. The peaks of these pulses vary in accordance with the modulating audio signal. BS ≈ AS = d. The energy of a wave depends on its frequency-greater the frequency of a wave. the antenna length would be (3 × 108)/(20 × 103) = 15. neglecting h2 we have d2=2hR or d = 2hR LONG-ANSWER QUESTIONS Q. i.(a) For long distance transmission.000 metres. the length of the transmitting antenna should be of the order of the wavelength of the signal. which is too large.Q. Now in the right angled triangle OBS. has a much higher frequency. Thus it is not practicable to transmit audio signals directly. This circuit is called tuning circuit which is used to select the desired modulated radio frequency.e. For example.2 Sol. Signals from the 26 . The basic circuit using a junction diode is shown: The input circuit consists of a parallel combination of an inductor L and a variable capacitor C. It can be shown that the height of such a satellite above the earths’ surface is about 36. it can send back the signals to a large part of the earth’s surface. nature. Q. Explain the experimental arrangement and theory of ruby laser with the help of energy level diagram. Figure B shows the path of two light waves of wavelength λ1 and λ2. These rods are about 5 cm in length and 0. At the centre.48 at the outer surface. 2. One major application is to gather information (e. The main component of it is a ruby rod.000 km. As such. which decreases to 1. Its time period is 24 hours. Ruby is a crystal of aluminium oxide doped with 0. satellite remote sensing has become very important. Such a satellite is equipped with various instruments (such as cameras.5 cm in diameter. It is the technique of obtaining information about an object or some region from a large distance. Different types of optical fibres areMonomode optical fibre . size etc. 1. Such a satellite is called a geostationary satellite.05% of chromium oxide. Communication links all around the earth have been established by putting many geostationary satellites in the equatorial plane at this height. State the principle of an optical fibre. State the principle of LASER. Since the satellite is at a large height. Since the refractive index of a material depends on the wavelength. In a graded index multimode fibre. there is no boundary between the core and cladding. The end faces of the rod are made parallel and coated with silver. the refractive index decreases smoothly from its centre to the outer surface of the fibre. broadcasting station are beamed towards an artificial earth satellite which reflects them back to the earth. etc. droughts. Constant lighting angles help the observations about that region to be more standard and easier to interpret. For this purpose a remote sensing satellites. one can cause population inversion and then the process of stimulated emission can be used to produce highly coherent.4 Sol.5 Sol. Optical fibres are based on the phenomenon of total internal reflection. arrive at other end of the fibre at the same time. As a result.52. the ray of light undergoes total internal reflection. such that one face becomes fully reflecting while the other face is partially reflecting so as to allow the laser beam to emerge out of the ruby 27 . This makes it possible to have similar lighting conditions every time it passes over the particular region of the earth. regardless of their wavelength. also called an active satellite.It has a narrow core of diameter about 5 mm surrounded by a relatively big cladding (125 mm in diameter). highly monochromatic and perfectly parallel beam of light.(b) Q.g. inaccessible regions of the earth and to estimate the damage caused by floods. Multimode optical fibre . As a result a faulty and distorted signal is obtained. The satellite is placed in sun-synchronous orbit around the earth. Explain in brief the various types of optical fibres. all the light waves. Ruber Laser is a solid state laser. temperature. its orbit must be such that it appears stationary relative to the earth’s surface.These are further classified as(i) Step index multimode fibre: In this type of optical fibre the diameter of the core is about 50 mm which is very large in comparison with monomode optical fibre. there is practically no loss in the intensity of the output light signal. (ii) Graded index multimode fibre: In this type of optical fibre. In recent years. The path of ray is shown in figure A. is used. LASER is based on the principle that in the atomic systems possessing metastable states. the refractive index is 1. For a satellite to be useful for sending signals to particular regions. microwaves scanners etc. Remote sensing.) to enable it to record and send the desired information.) about remote. In such optical fibres. Consider a ray BO incident on the glass core from air at an angle i such that the ray of light inside the core meets the core-cladding interface at an angle greater than the critical angle for it. the two light waves will reach the other end following different and unequal paths. modem converts digital data into analog signal for use in modulating a carrier by a signal.This type of modem provides transmission in only one direction. Now. The flash tube provides the necessary pumping energy to the ruby rod. Q.rod. The flow of data in one direction takes place at one time and in the opposite direction at a second time. A modem inside the Fax corverts these digital pulses into analog signals which are then transmitted via the telephone link. The modem at the other end converts back the analog signal into digital pulses which are printed by the printer of the Fax and produce Fax copy of the original document. It requires one transmission bidirectional channel. the amount of light that bounces back is detected by the sensor. The energy released during the process is lost to ruby rod and before it raises its temperature it is cooled by liquid nitrogen.This type of modem are able to transfer data in both directions. the scanner scans the document. (ii) Half duplex mode . state E1 (metastable) becomes more populated. it produces a flash intense light of wavelength 5500Å which acts as pumping radiation for chromium ions. (iii) Full duplex mode .In this mode of operation. phase and direction. once a photon is produced due to transition from the metastable state E1 to the ground state E0. Ruby laser is a three energy level system. When an intense beam of light falls on the scanned document.reflected which creates a pulse of low voltage while from the blank paper a pulse of high voltage is produced.. Operation of modem are classified into the following three modes: (i) Simplex mode . Thus modems are placed at both the ends of communication circuits.6 Write short notes on (a) Modem (b) Fax. Initially the atoms are distributed in various energy levels. only the printer is active to receive the incoming signal.At the receiver end reverse of it takes place.(a) MODEM:As the name implies. the sender dials in the phone number of Fax machine at the receiver’s end. In the transmitting mode. In fascimile transmission an exact reproduction of a document is produced at the receiving end. it stimulates the emission of another photon of the same energy. The rod is kept cool by circulating liquid nitrogen around it. In a short duration there is population inversion i. The process of amplification continues till an intense beam of laser emerges out from the partially coated face of the rod. transmission takes place in both directions at the same time. The emitted photons suffer multiple reflection between the two polished ends and stimulate further emissions. In this case two transmission channels are required. (b) FAX (FASCIMILE). In the receiving mode. Sol. The electronic transmission of a document to a distant place via telephone line is known as fascimile or Fax. When the document is fed into the machine. The atoms are raised from the ground state to the higher energy state E2. The three energy levels of chromium ion are the ground state E0 the metastable state E1 and a higher energy level E2. The ruby rod is placed along the axis of a Xenon flash tube. At the sender’s end Fax machine is in transmitting mode while at the receiver end it is in receiving mode. The electrons cannot stay in excited state for more than 10–8 s and they drop to either energy level E1 or directly to ground state E0. When the Xenon lamp is switched on. little light is. a modem can perform both the functions of modulation and demodulation. 28 . When ink is present. In order to send Fax copy of some document.e. from IIT-JEE 8.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P5. Magnetic Effects of Electric Current Index: 1. Que. 34 Yrs. Exercise IV 6. Answer Key 7. Key Concepts 2. Exercise I 3. Exercise II 4. from AIEEE 1 . 10 Yrs. Que. Exercise III 5. A moving charge produces both electric field ans magnetic field and both electric field and magnetic field can exert force on it. µ0 = 4 π × 10–7 Hm–1 ( 4 MAGNETIC INDUCTION dBP = 5. It is a vector quantity which may be defined in terms of the force it produces on electric currents . A static charge produces only electric field and only electric field can exert a force on it. current) . The number of lines of B crossing a given area is referred to as the MAGNETIC FLUX linked with that area. 4π r3 4π r2 here the quantity Idl is called as current element strength. Lines of magnetic induction may be drawn in the same way as lines of electric field . The vector quantity B known as MAGNETIC INDUCTION is introduced to characterise a magnetic field .KEY CONCEPTS 1. It can not produce electric field as net charge on a current carrying conductor is zero. µ 0qv sin θ 4πr 2 µ 0I 2πr MAGNETIC INDUCTION DUE TO SEMI INIFINITE ST. µ0 = permeability of free space µr = relative permeability of the medium (Dimensionless quantity). The number of lines per unit area crossing a small area perpendicular to the direction of the induction bring numerically equal to B . CONDUCTOR B= 7. produces a magnetic field . 2. µ = permeability of the medium = µ0 µr . CONDUCTOR B= 6. DUE TO A MOVING CHARGE → µ q( v xr ) dB = 0 In vector form it can be written as 4π r 3 MAGNETIC INDUCTION DUE TO AN INIFINITE ST. M AGNETIC INDUCTION PRODUCED B Y A CURRENT (B IOT-SAVART LAW): The magnetic induction dB produced by an element dl carrying a current I at a distance r is given by → µ µ I d x r µ 0 µ r Idsinθ 0 r : dB = or dB= . ) µ0I 4 πr M AGNETIC INDUCTION DUE TO A CURRENT CARRYING STRAIGHT CONDUCTOR B= µ0 I 4πR (cos θ1 + cos θ2) If the wire is very long θ1 ≅ θ2 ≅ 0º then . A magnetic field is detected by its action on current carrying conductors (or moving charges) and magnetic needles (compass) needles. Unit of µ0 & µ is NA–2 or Hm–1 . Magnetic charge (i. For this reason B is also called 3 MAGNETIC FLUX DENSITY . A current carrying conductor produces only magnetic field and only magnetic field can exert a force on it. B = µ0I 2πR 2 .e. 8. where µ B = magnetic induction at the point . MAGNETIC INDUCTION DUE TO SOLENOID B = µ0nI. R >> r MAGNETIC INDUCTION DUE TO CURRENT CARRYING SHEET 1 µI 2 0 where I = Linear current density (A/m) B= 13. where n → no. direction along axis. of turns per m) 2πR N = total turns 12. This plane is called the MAGNETIC MERIDIAN. at earth's north pole. 3 . M AGNETIC FIELD DUE TO A FLAT CIRCULAR COIL CARRYING A CURRENT : (i) At its centre B= µ0 NI 2R .e. of turns per m. MAGNETIC INDUCTION DUE TO FLAT CIRCULAR ARC B= µ0 Iθ 4πR 10. The Earth's Magnetic poles are opposite to the geometric poles i. MAGNETIC INDUCTION DUE TO THICK SHEET At point P2 At point P1 14. 15. 9. direction Where N = total number of turns in the coil I = current in the coil R = Radius of the coil (ii) On the axis B= ( µ 0 NIR 2 2 x 2 +R 2 ) 3/ 2 Where x = distance of the point from the centre . µ = permeability of the medium GILBERT'S M AGNETISM (EARTH' S MAGNETIC FIELD) : (a) The line of earth's magnetic induction lies in a vetical plane coinciding with the magnetic North South direction at that place. I → current 11. Earth's magnetic axis is slightly inclined to the geometric axis of earth and this angle varies from 10.50 to 200. It is maximum at the centre . MAGNETIC INDUCTION DUE TO TOROID B = µ0nI where n = N (no. its magnetic south pole is situated and vice versa. 1 µ Id 2 0 Bin = µ0Jx Bout = M AGNETIZATION INTENSITY (H) : B The magnetic intensity (H) at any point in a magnetic field is defined as H = . line and F = 0 MOTION OF mv (b) When v is | to B : Motion will be in circular path with radius R = and angular qB qB and F = qvB. NEUTRAL POINT IN SUPERPOSED M AGNETIC FIELDS : When more than one magnetic fields are suspended at a point and the vector sum of the magnetic inductions due to different fields . 16. (e) Lines drawn on earth at different places having same dip angle are called as "isoclinic lines" and line of zero dip is called as "aclinic lines". B v = the vertical component of B in the magnetic meridian plane = B sin θ . 18. equal to zero . 17 AMPERES LAW → ∫ B . 19. generally i n c l i n e d t o t h e h o r i z o n t a l a t a n a n g l e c a l l e d t h e MAGNETIC DIP at that place . The angle between them is called " DECLINATION AT THAT PLACE" . F = nett electromagnetic force on the charge = q E + q V × B This force is called the LORENTZ FORCE . (d) Lines drawn on earth at different places having same declination angle are called as "isogonic lines" and line of zero declination is called as "agonic lines". MAGNETIC FORCE ON A STRAIGHT CURRENT CARRYING WIRE : F = I ( L × B) I = current in the straight conductor L = length of the conductor in the direction of the current in it B = magnetic induction. LORENTZ FORCE : An electric charge 'q' moving with a velocity V through a magnetic field of magnetic induction B experiences a force F . given by F = qV x B . BH (c) At a given place on the surface of the earth . There fore. d = µ∑I Σ I = algebric sum of all the currents . BH = the horizontal component of B in the magnetic meridian plane = B cos θ . such that B = total magnetic induction of the earth at that point. A CHARGE IN UNIFORM MAGNETIC FIELD : (a) When v is || to B : Motion will be in a st. Bv = tan θ .(b) On the magnetic meridian plane . (Uniform throughout the length of conduction) Note : In general force is F = ∫ I (d × B) 4 . if the charge moves in a space where both electric and magnetic fields are superposed . the magnetic meridian and the geographic meridian may not coincide . m mv sin θ and pitch (c) When v is at ∠ θ to B : Motion will be helical with radius Rk = qB velocity ω = PH = 2πmv cos θ and F = qvBsinθ. qB 20. the magnetic induction vector of the earth at any point. the point is a magnetic neutral point. MOVING COIL GALVANOMETER : It consists of a plane coil of many turns suspended in a radial magnetic feild. they magnetically interact with each other . C= K = GALVANOMETER CONSTANT. when a current is passed in the coil it experiences a torque which produces a twist in the suspension.e. but experience a torque given by τ = NI A × B = M × B = BINA sin θ When A = area vector outward from the face of the circuit where the current is anticlockwise. This deflection is directly proportional to the torque ∴ NIAB = Kθ K I = NAB θ K = elastic torsional constant of the suspension I=C θ 24. in opposite direction) or Attraction if the currents are parallel (i. 1. in the same direction) This force per unit length on either conductor is given by F = µ 0 I 1I 2 2π r . 2 NOTE: The rate of magnetic moment to Angular momentum of a uniform rotating object which is charged uniformly is always a constant. NAB FORCE EXPERIENCED BY A MAGNETIC DIPOLE IN A NON-UNIFORM MAGNETIC FIELD : ∂B | F | = M ∂r where M = Magnetic dipole moment. 25. MAGNETIC MOMENT OF A ROTATING CHARGE: If a charge q is rotating at an angular velocity ω. 2. This force is of : Repulsion if the currents are anti-parallel (i. it experience a zero nett force . B = magnetic induction of the uniform magnetic feild. (i) (ii) MAGNETIC INTERACTION FORCE BETWEEN TWO PARALLEL LONG STRAIGHT CURRENTS : When two long straight linear conductors are parallel and carry a current in each . FORCE ON A RANDOM SHAPED CONDUCTOR IN MAGNETIC FIELD Magnetic force on a loop in a uniform B is zero Force experienced by a wire of any shape is equivalent to force on a wire joining points A & B in a uniform magnetic field . M = magnetic moment of the current circuit = IN A Note : This expression can be used only if B is uniform otherwise calculus will be used. Where r = perpendicular distance between the parallel conductors 22. its equivalent current is given as I = magnetic moment is M = IπR2 = qω & its 2π 1 qωR2.e.21. 26.. 23. one experiences a force. Irrespective of the shape of conductor 5 M q = L 2m . MAGNETIC TORQUE ON A CLOSED CURRENT CIRCUIT : When a plane closed current circuit of 'N' turns and of area 'A' per turn carrying a current I is placed in uniform magnetic field . Q. Six wires of current I1 = 1A.d l around the closed path. 4.1 Figure shows a straight wire of length l carrying a current i.9 5 5 cm and 5 cm respectively carry current 5 Amp and Amp 2 2 respectively. The separation between the rod and the wire is a.6 Find the magnetic induction at point O. I2 = 2A.EXERCISE # I Q. Find the magnitude of magnetic field produced by the current at point P. The rod is placed parallel to a long wire carrying a current i. Q. Q.4 What is the magnitude of magnetic field at the centre ‘O’ of loop of radius 2 m made of uniform wire when a current of 1 amp enters in the loop and taken out of it by two long wires as shown in the figure. Find the force needed to move the rod along its length with a uniform velocity v. Q. I4 = 1A.10 Electric charge q is uniformly distributed over a rod of length l.7 Find the magnitude of the magnetic induction B of a magnetic field generated by a system of thin conductors along which a current i is flowing at a point A (O. Find the value of the integral ∫ B. O). I5 = 5A and I6 = 4A cut the page perpendicularly at the points 1. R. The plane of B is perpendicular to plane of A and their centres coincide.5 Find the magnetic induction at the origin in the figure shown. If a current of 1 A passes through each coil in the opposite direction find the magnetic induction. I3 = 3A. Q. Find the magnitude and direction of a minimum uniform magnetic field in tesla that will cause the electron to move undeviated along its original path. midway between them. if the current carrying wire is in the shape shown in the figure. A n e l e c t r o n m o v i n g w i t h a v e l o c i t y 5 × 1 0 6 . The ring is in yz plane. 2.2 Two circular coils A and B of radius Q. Q. 3. that is the centre of a circular conductor of radius R.8 Two circular coils of wire each having a radius of 4 cm and 10 turns have a common axis and are 6 cm apart .11 6 ms–1 ˆi in the uniform electric field of 5 × 107 Vm–1 ˆj . (i) (ii) Q. Q. 5 and 6 respectively as shown in the figure. At a point on the axis. Q. Find the magnetic field at the centre. At the centre of either coil .3 Find the magnetic field at the centre P of square of side a shown in figure. Find an expression for the magnetic field B. one edge along X-axis and the other edge along Y-axis as shown in the figure. a) with initial velocity v i (a) (b) (c) and leaves the quadrant at a point (2a. where b is a constant. Find the magnetic force acting on the loop is Q.18 A system of long four parallel conductors whose sections with the plane of the drawing lie at the vertices of a square there flow four equal currents.15 Two coils each of 100 turns are held such that one lies in the vertical plane with their centres coinciding.49 x 10 -5 T and angle of dip = 30º. Q.Q. The loop is placed in a homogeneous magnetic field with an induction B = 10-1 T directed at right angles to the plane of the drawing. carrying a current 1 amp. It is placed in a uniform magnetic field B0 such that B0 is perpendicular to the plane of the loop. Q. How is the vector of magnetic induction directed at the centre of the square? Q. 7 . mesured from the axis Q.19 A cylindrical conductor of radius R carries a current along its length . In space there is a uniform magnetic field B in . 0) with velocity – 2v j . Find the period of small oscillations that the loop performs about its position of stable equilibrium. passing through the centre of the loop at right angles to two opposite sides of the loop.12 A charged particle (charge q.17 A particle of charge +q and mass m moving under the influence of a uniform electric field E i and a magnetic field B k enters in I quadrant of a coordinate system at a point (0.z direction. How would you neutralize the magnetic field of the earth at their common centre ? What is the current to be passed through each coil ? Horizontal component of earth's magnetic induction = 3.20 A square current carrying loop made of thin wire and having a mass m =10g can rotate without friction with respect to the vertical axis OO1. Q. it is not uniform over the cross section of the conductor but is a function of the radius according to J = br. mass m) has velocity v0 at origin in +x direction. Find the total magnetic force on the loop if it carries current i.14 A rectangular loop of wire is oriented with the left corner at the origin. 0). The radius of the vertical coil is 20 cm and that of the horizontal coil is 30 cm . A current I = 2A is flowing in the loop. Q. The directions of these currents are as follows : those marked ⊗ point away from the reader. The current density J.13 A conducting circular loop of radius r carries a constant current i.16 Find the ratio of magnetic field magnitudes at a distance 10 m along the axis and at 60° from the axis. A magnetic field is into the page and has a magnitude that is given by β = αy where α is contant. Find the y coordinate of particle when is crosses y axis. however. while those marked with a dot point towards the reader. a) Rate of work done by both the fields at (2a. from the centre of a coil of radius 1 cm. (a) at r1 < R & (b) at distance r2 > R. Find Magnitude of electric field Rate of work done by the electric field at point (0. Q. 253. Find the time interval for which it remains inside the magnetic field.373.372.22 A proton beam passes without deviation through a region of space where there are uniform transverse mutually perpendicular electric and magnetic field with E and B. 3. 3.E. If the arc carries current I then find the force on the arc. 3.243.245. Then the beam strikes a grounded target. 3. The rectangular loop carries a conventional current I' in the clockwise direction.389. 3.254.236.390. it flies through a uniform transverse magnetic field B. 3. Find the net force on the rectangular loop. 3. 3. 3.237. Q. 3. 3. 3. currents flow in the same direction? List of recommended questions from I. 3. 3.229.258. 3. 3.24 An arc of a circular loop of radius R is kept in the horizontal plane and a constant magnetic field B is applied in the vertical direction as shown in the figure.391.226.Q.234. 3. 3.244.269. 3. 3.252. Irodov.257.228. The field occupies a region of space d.384.225. 3.251. Q. 3.386.227. What work per unit length of a conductor must be done to increase the separation between the conductors to r2 = 10 cm if .383. 3.23 An infinitely long straight wire carries a conventional current I as shown in the figure.220.223. 3. 3. Find the force imparted by the beam on the target if the beam current is equal to I.25 Two long straight parallel conductors are separated by a distance of r1 = 5cm and carry currents i1 = 10 A & i2 = 20 A . 3. 3.230. 3.242 3.21 A charged particle having mass m and charge q is accelerated by a potential difference V. 3. Q. 3.224.396 8 . Q. 3. EXERCISE # II Q. B.7 Q charge is uniformly distributed over the same surface of a right circular cone of semi-vertical angle θ and height h. Find the magnetic field on the x = axis at x = 2R on the y = axis at y = 2R.3 An infinite wire. The cone is uniformly rotated about its axis at angular velocity ω. C and D. What will be the force OC if current in the wire B is reversed? Q.1 Three infinitely long conductors R. Find the minimum value of B so that the board will be able to rotate up to horizontal level. are fixed at y = – a metre and y = +a metre respectively. Q. has current I1 in positive z-direction. The rod is at a distance a from the origin. The currents in the respective conductors are IR = I0sin (ωt + 2π ) 3 IS = I0sin (ωt) 2π ) 3 Find the amplitude of the vertical component of the magnetic field at a point P. Aconducting rod placed in xy plane parallel to y-axis has current I2 in positive y-direction. Let the z-axis be the axis of the conductor. placed along z-axis. IT = I0sin (ωt − Q. 9 .2 Four long wires each carrying current I as shown in the figure are placed at the points A. At t = 0. a vertical downward magnetic field of induction B is switched on. Find net force on the rod. If the wires A and B each carry a current I amp into plane of the paper. as shown in the figure. distance 'a' away from the central conductor S. Q. Q. Find the magnitude and direction of (i) magnetic field at the centre of the square. (ii) force per metre acting on wire at point D. A single wire is wound along the periphery of board and carrying a clockwise current I. The ends of the rod subtend + 30° and – 60° at the origin with positive x-direction.5 A straight segment OC (of length L meter) of a circuit carrying a current I amp is placed along the x-axis.6 A very long straight conductor has a circular cross-section of radius R and carries a current density J. Calculated associated magnetic dipole moment. Two infinitely ling straight wires A and B . Obtain the expression for the force acting on the segment OC.each extending form z = – ∞ to + ∞. Inside the conductor there is a cylindrical hole of radius a whose axis is parallel to the axis of the conductor and a distance b from it.4 A square cardboard of side l and mass m is suspended from a horizontal axis XY as shown in figure. (a) (b) Q. S and T are lying in a horizontal plane as shown in the figure. and let the axis of the hole be at x = b. is 1. Q.Q. Make use of the fact that impulse of force equals ∫ F dt . uniform external magnetic field of 70 mT (but the clock still keeps perfect time) at exactly 1:00 pm. & B = 2 webers/meter2.which equals mv. Page 10 of 20 MAGNETIC EFFECTS OF CURRENT Q. z = 0 . the number of conduction electrons per unit volume. where there is a constant . m = 10gm.tekoclasses. that is. find the torque acting on the loop due to the field. the wire carries a current 2. z = − 0.1 Wb/m2. is established in a strip of copper of height h and width w. z = 2 m . What will be the magnitude and direction of the field EH? Assume that n. the hour hand of the clock points in the direction of the external magnetic field (a) After how many minutes will the minute hand point in the direction of the torque on the winding due to the magnetic field ? (b) What is the magnitude of this torque. the size of the charge or current pulse. If a charge. (b) What are the magnitude and dirction of the magnetic force F acting on the electrons? (c) What would the magnitude & direction of homogeneous electric field E have to be in order to counter balance the effect of the magnetic field ? (d) What is the voltage V necessary between two sides of the conductor in order to create this field E? Between which sides of the conductor would this voltage have to be applied ? (e) If no electric field is applied form the outside the electrons will be pushed somewhat to one side & thereforce will give rise to a uniform electric field EH across the conductor untill the force of this electrostatic field EH balanace the magnetic forces encountered in part (b) . a current pulse q = ∫ idt .1x1029/m3 & that h = 0.8 . (a) Calculate the drift speed vd for the electrons. indicated by the crosses in fig. from the height h that the wire reaches. assuming that the time of the current pulse is very small in comparision with the time of flight. is sent through the wire. Calculate. = 20cm & h = 3 meters. A uniform field of magnetic induction B is applied at right angles to the strip. 10 FREE Download Study Package from website: www. The clock is located. the wire will jump up.0 A in the clockwise direction. (b) x = 2 m.12 A current i. Evaluate q for B = 0. A uniform magnetic field B0 of magnitude 10−6 T is directed parallel to the x-axis.11 A U-shaped wire of mass m and length l is immersed with its two ends in mercury (see figure).5 m Q. circular wall clock has a face with a radius of 15cm. Find its instantaneous acceleration (b) If an external uniform magnetic induction field B = B i is applied. i = 50 amp .9 A long straight wire carries a current of 10 A directed along the negative y-axis as shown in figure.com A wire loop carrying current I is placed in the X-Y plane as shown in the figure (a) If a particle with charge +Q and mass m is placed at the centre P and given a velocity along NP (fig). Six turns of wire are wound around its perimeter. (c) x = 0 . w = 0. The wire is in a homogeneous field of magnetic induction B.1cm .02 meter .10 A stationary. What is the resultant magnetic field at the following points? (a) x = 0 .[g = 10 m/s2] Q. 19 The figure shows a conductor of weight 1. 11 . Such frequency shifts were actually observed by Zeeman in 1896. Q. Q. what is its dipole moment? Q. Q. (a) Show that B for a point on the axis of the loop and a distance x from its centre is given by.10 T.16 A square loop of wire of edge a carries a current i . (a) If the electron is circulating clockwise. In Bohr's theory of the hydrogen atom the electron can be thought of as moving in a circular orbit of radius r about the proton . The coefficient of static friction between the conductor and the plane is 0. Find the minimum value of B needed to make the electron hit S . as viewed by an observer sighting along B. will the angular frequency increase or decrease? (b) What if the electron is circulating counterclockwise? Assume that the orbit radius does not change.14 Zeeman effect .1.13(a) A rigid circular loop of radius r & mass m lies in the xy plane on a flat table and has a current I flowing in it. What is the force needed to be the applied parallel to the inclined plane to sustaining the conductor at rest? Q. approximately by ∆v = ± 4πm Q. At this particular place . Can the result of the above problem be reduced to give field at x = 0 ? Does the square loop behave like a dipole for points such that x >> a ? If so .1m & the line GS makes an angle of 60° with the x-axis.0 N and length L = 0.Q. The electron are required to hit the spot S where GS = 0. Find the force per unit lenght on the coductor. A current of I = 10 A flows through the conductor inside the plane of this paper as shown. A uniform magnetic field B parallel to GS exists in the region outsiees to electron gun. B = Bx i + Bz k . as shown in the figure. Suppose that such an atom is placed in a magnetic field. Q. as shown in the fig. The figure shows the surface traced by the wire AB.18 Find the work and power required to move the conductor of length l shown in the fig.15 In above problem show that the change in frequency of rotation caused by the magnete field is given Be . the earth's magnetic field is B = Bx i + By j .5 m placed on a rough inclined plane making an angle 300 with the horizontal so that conductor is perpendicular to a uniform horizontal magnetic field of induction B = 0. with the plane of the orbit at right angle to B. B= (b) (c) ( 4 µ 0 ia 2 π 4 x2 + a 2 ) (4x 2 + 2a 2 ) 1/ 2 . one full turn in the anticlockwise direction at a rotational frequency of n revolutions per second if the magnetic field is of magnitude B0 everywhere and points radially outwards from Z-axis.17 A conductor carrying a current i is placed parallel to a current per unit width j0 and width d. How large must I be before one edge of the loop will lift from table ? (b) Repeat if.20 An electron gun G emits electron of energy 2kev traveling in the (+)ve x-direction. 2] (A) zero . are suspended along their diameter as shown. The magnitude of the force due to the magnetic field acting on the charge at this instant is : (A) (ii) µ 0 Iqv 2 πd (B) µ 0 Iqv πd (C) 2 µ 0 Iqv πd (D) 0 Let [∈0] denote the dimensional formula of the permittivity of the vaccum and [µ0] that of the permeability of the vacuum . find the frequency of oscillation.3 An electron in the ground state of hydrogen atom is revolving in anticlock-wise direction in a circular orbit of radius R . Its instantaneous velocity v is perpendicular to this plane. [ JEE '98. a point charge q is at a point equidistant from the two wires. One of the arcs AB of the ring subtends an angle θ at the centre . [JEE '96. The central wire is along the y-axis while the other two are along x = ± d. Find the torque experienced by the orbiting electron. initially the planes of the rings are mutually perpendicular when a steady current is set up in each of them : [IIT '95. a deuteron and an α-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field . The value of the magnetic induction at the centre due to the current in the ring is : [ JEE '95.1 A battery is connected between two points A and B the circumference of a uniform conducting ring of radius r and resistance R . [JEE '97. show that it will execute simple harmonic motion .2 Two insulated rings. 1] particles then : (A) rα = rp < rd (B) rα > rd > rp (C) rα = rd > rp (D) rp = rd = rα Q. 5] (i) (ii) Q. Obtain an expression for the orbital magnetic dipole moment of the electron The atom is placed in a uniform magnetic.6 (i) Select the correct alternative(s) . 1] (A) The two rings rotate to come into a common plane (B) The inner ring oscillates about its initially position (C) The outer ring stays stationary while the inner one moves into the plane of the outer ring (D) The inner ring stays stationary while the outer one moves into the plane of the inner ring Q. 2 + 2 + 2 ] Two very long. (A) [∈0] = M−1 L−3 T2 I (B) [∈0] = M−1 L−3 T4 I2 (C) [µ0] = MLT−2 I−2 (D) [µ0] = ML2 T−1 I 12 . parallel wires carry steady currents I & − I respectively. The distance between the wires is d.5 3 infinitely long thin wires each carrying current i in the same direction . If the central wire is displaced along the z-direction by a small amount & released.θ) (D) inversely proportional to r Q. If the linear density of the wires is λ . one slightly smaller diameter than the other. are in the x-y plane of a gravity free space . Find the locus of the points for which the magnetic field B is zero . only if θ = 180º (B) zero for all values of θ (C) proportional to 2 (180º . Induction B such that the plane normal of the electron orbit makes an angle of 300 with the magnetic induction. straight. If M = mass. At a certain instant of time. rd & rα denote respectively the radii of the trajectories of these [JEE '97. L = length.4 A proton . T = time and I = electric current . in the plane of the wires . If rp . 5] (i) (ii) Q.EXERCISE # III Q. with its centre at the origin O. You must express your answer in terms of t. What is the torque τ about O acting on the frame due to the magnetic field ? Find the angle by which the frame rotates under the action of this torque in a short interval of time ∆t. constant magnetic field B is directed at an angle of 45º to the x-axis in the xy-plane. E & B and their magnitudes v0. the speed of light in vacuum).7 A particle of mass m & charge q is moving in a region where uniform. each of mass m & charge q. 2 + 6] (a) (b) Q9 A charged particle is released from rest in a region of steady and uniform electric and magnetic fields which are parallel to each other. (b) Find the final velocity of the particle and the time spent by it in the magnetic field. The ratio of the magnitudes of the magnetic moment of the system & its angular momentum about the centre of the rod is : (A) q 2m (B) q m (C) 2q m (D) q πm Q. 8] v 0 . The variation of the magnetic field B along the XX’ is given by (A) (B) (C) 13 (D) . q and m (C) q and m (D) ω and m (ii) Two long parallel wires are at a distance 2d apart. m. 6 + 4] Q. Each side of the frame is of mass M & length L. E & B. if the magnetic field now extends upto 2. the frame is at rest in the position shown in the figure. as shown. & the axis about which this rotation occurs (∆t is so short that any variation in the torque during this interval may be neglected) Given the moment of inertia of the frame about an axis through its centre perpendicular to its plane is 4/3 ML2. A particle of mass m.1L.8 A uniform. are attached to the two ends of a light rigid rod of length 2 R . 2] Q.11(i)A particle of charge q and mass m moves in a circular orbit of radius r with angular speed ω. (a) Find the value of L if the particle emerges from the region of magnetic field with its final velocity at an angle 30° to its initial velocity. steady magnetic field B0 k . (assume that its speed is always << c. The particle will move in a (A) straight line (B) circle (C) helix (D) cycloid [JEE’99. [JEE ’99.10 The region between x = 0 and x = L is filled with uniform. the vectors [JEE '98. constant electric and magnetic fields E & B are present. At time t = 0. Find the velocity v of the particle at time t. [JEE '98. E & B are parallel to each other. positive charge q and velocity v0 i travels along x-axis and enters the region of the magnetic field. PQRS is a rigid square wire frame carrying a steady current I0(clockwise). with its sides parallel to the x & y axes.(iii) Two particles. At time t = 0 the velocity v0 of the particle is perpendicular to E . q. Neglect the gravity throughout the question. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (A) ω and q (B) ω. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. They carry steady equal currents flowing out of the plane of the paper. Q. then (A) positive ions deflect towards +y direction and negative ions towards −y direction (B) all ions deflect towards +y direction. Then (A) mAvA < mBvB (B) mAvA > mBvB (C) mA < mB and vA < vB (D) mA = mB and vA = vB [JEE. (ii) If an external uniform magnetic field B j is applied. 5 + 5] Q.[JEE 2000 (Scr)] Q. Each of the straight sections of the loop is of length 2a. One of the semicircles (KNM) lies in the x − z plane and the other one (KLM) in the y − z plane with their centers at the origin . (C) all ions deflect towards −y direction (D) positive ions deflect towards −y direction and negative ions towards +y direction. determine the forces F1 and F2 on the semicircles KLM and KNM due to this field and the net force F on the loop . The magnetic field due to this current at the point M is H1.14 Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. Each arc subtends the same angle at the centre. Current I is flowing through each of the semicircles as shown in figure . Assume that space is gravity free. Find the instantaneous force f on the particle.12 m. The speeds of the particles are vA and vB respectively and the trajectories are as shown in the figure. The magnetic field due to this loop at the point P (a.12 A circular loop of radius R is bent along a diameter and given a shape as shown in the figure. The magnetic field at M is now H2. A uniform magnetic field exists perpendicular to this plane. A current I flows through PQR. 2001 (Scr)] 2 3 14 . An infinitely long straight wire carrying a current of 10A is passing through the centre of the above circuit vertically with the direction of the current being into the plane of the circuit. 4 + 6] Q.(iii) An infinitely long conductor PQR is bent to form a right angle as shown. What is the force acting on the wire at the centre due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the centre? [JEE 2001.08 m and r2 = 0. (i) A particle of charge q is released at the origin with a velocity v = − v0 i . The circuit consists of eight alternating arcs of radii r1 = 0. a) points in the direction 1 ˆ ˆ ˆ 1 (− j + k + i ) (−ˆj + kˆ ) (B) (A) 3 2 1 ˆ ˆ ˆ 1 ˆ ˆ (i + j + k ) (i + k ) (C) (D) [JEE. The ratio H1/H2 is given by (A) 1/2 (B) 1 (C) 2/3 (D) 2 (iv) An ionized gas contains both positive and negative ions. (a) (b) Find the magnetic field produced by this circuit at the centre. [JEE 2000 Mains.15 A non-planar loop of conducting wire carrying a current I is placed as shown in the figure. 0. If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction. 2001 (Scr)] Q. another infinitely long straight conductor QS is connected at Q so that the current in PQ remaining unchanged.13 A current of 10A flows around a closed path in a circuit which is in the horizontal plane as shown in the figure. Now. a. extending from x = a to x = b.22 Figure represents four positions of a current carrying coil is a magnetic field directed towards right.Q. 3] Q.16 A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. The loop is now released and is found to stay in the horizontal position in equilibrium. nˆ represent the direction of area of vector of the coil. which remains hinged along a horizontal line taken as the y-axis (see figure). Take the vertically upward direction as the z-axis. 2001 Screening] (A) µ 0 NI b (B) 2µ 0 NI a (C) µ 0 NI 2( b − a ) ln b a (D) µ0I N 2( b − a ) ln b a Q. A uniform magnetic field B = (3 i + 4 k ) B 0 (a) (b) (c) exists in the region. The minimum value of v required so that the particle can just enter the region x > b is (A) q b B/m (B) q( b – a) B/m (C) q a B/m (D) q(b + a) B/2m [JEE 2002 (screening). What is the direction of the current I in PQ? Find the magnetic force on the arm RS. b and m. [JEE 2002.20 A rectangular loop PQRS made from a uniform wire has length a.17 A particle of mass m and charge q moves with a constant velocity v along the positive x direction. 3] field B at a point having coordinates (x. Find the expression for I in terms of B0. 3] (D) Q. the magnetic field at the centre is [JEE. The correct order of potential energy is : [JEE 2003 (Scr)] (A) I > III > II > IV (B) I < III < II < IV (C) IV < I < II < II (D) II > II > IV > I 15 . When a current I passes through the coil. The magnetic vector [JEE 2002 (screening). The loop is held in the x-y plane and a current I is passed through it. It is free to rotate about the arm PQ. width b and mass m. Mark correct option [JEE 2003 (Scr)] (A) coil expands (B) coil contracts (C) coil moves left (D) coil moves right Q.19 The magnetic field lines due to a bar magnet are correctly shown in (A) (B) (C) [JEE 2002 (screening). It enters a region containing a uniform magnetic field B directed along the negative z direction.18 A long straight wire along the z-axis carries a current I in the negative z direction. 1+1+3] Q.21 A circular coil carrying current I is placed in a region of uniform magnetic field acting perpendicular to a coil as shown in the figure. y) in the z = 0 plane is µ 0 I (xi + yj ) µ 0 I (xj − yi ) µ 0 I (xi − yj ) µ 0 I (yi − xj) (A) (C) (B) (D) 2 2 2 2 2 2 2 π (x + y ) 2 π (x 2 + y 2 ) 2π (x + y ) 2π (x + y ) Q. 2 find the maximum angular velocity ω0 with which the wheel can be rotate. The rod is suspended by light inextensible stringe and a magnetic field B is applied as shown in the figure. I. Find the ratio of radii of the circular paths of the two particles. (A) Net force on the loop is zero. (C) As seen from O. [JEE 2003] Q. If the breaking tension of the strings are 0 . torque on the coil can be expressed as τ = ki. enter a uniform magnetic field the direction of which is perpendicular to their velocities. The 3T initial tensions in the strings are T0. the loop rotates clockwise. the torsional constant of the spring.Q. Choose the correct option (s). id charge Q is passed through the coil almost instantaneously. area A and moment of inertia I is placed in magnetic field B. A and B. after being accelerated through same potential difference. The rectangular coil of the galvanometer having numbers of turns N.23 A wheel of radius R having charge Q. Find k in terms of given parameters N. (B) Net torque on the loop is zero. the loop rotates anticlockwise 16 [JEE 2006] .24 A proton and an alpha particle. (D) As seen from O. (Ignore the damping in mechanical oscillations) [JEE 2005] Q. uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. [JEE 2004] Q. if a current i0 produces a deflection of π/2 in the coil in reaching equilibrium position. where i is current through the wire and k is constant.25 (a) (b) (c) In a moving coil galvanometer. the maximum angle through which coil is deflected.26 An infinite current carrying wire passes through point O and in perpendicular to the plane containing a current carrying loop ABCD as shown in the figure. 18 In the plane of the drawing from right to left dB q α .21 t= m Q.22 mEI Be 3 . where α = sin–1 2 mV qB 2 IRB µ0 weber.9 Q.2 Q.16 4 µ 0 bR 3 Q.23 µ 0 I I′ C 1 1 − to the left 2π a b Q.6 µ J/m 2π r1 EXERCISE # II Q. i2 = 0.m–1 (i) µ 0 4I along Y-axis.14 F = αa2i ˆj Q. (c) zero (a) 4qa 4a Q. (ii) zero Q.1 µ0I0 3b 2 2 π (a + b 2 ) Q. 4π a µ0 I2 1 (ii) 4π 2a 10 .1 2 µ 0i 8π l Q.5 µ0I 3 ˆ 1 ˆ k + j 4R 4 π Q.7 B= Q.2 7 Q. (b) . B2 = (2 2 − 1) µ 0i πa µ0 i 3 π + 1 4πr 2 Q.096 A 3mv 2 3mv 3 .4 zero Q.57 s 6IB Q.1110 A.13 zero Q.3 ×10–4T. tan 4 + π with positive axis 3 17 .8 (i) 1.24 5 4π × 10–5 T 2 2 2 2 π2 − 2 π + 1 Q.19 B1 = Q.6 ( ) µ 0 br12 Q.11 10 kˆ Q.3 Q.12 2mv 0 qB Q.25 W µ 0 I1I 2 r2 = n = 27.15 i1 = 0.20 T0 = 2 π 3r2 m = 0.ANSWER KEY EXERCISE # I Q.10 µ 0iqv 2πa µ0 i 4πR Q.17 Q. 5 µ0 I2 L2 + a 2 n −kˆ . (b) Bx = 0 4 .7 x 10−6 V (top + .direction] mg 2 Il Q.3 (i) m = 4πm .3 µ 0 I1I 2 ln (3) along – ve z direction 4π Q. (b) decrease Q.7 Qω 2 h tan 2 θ 4 Q.7 v = E m m Q. (b) τ=BI − a ˆj 3 4 Q.17 mg πr ( B2x + B2y ) 1/ 2 (b) I = mg π r Bx µ 0 iJ 0 d tan −1 (−kˆ ) π 2h Q.4 x 10−4 m/s (b) 4.x=± d I . (c) 5 × 10 –6 T .8 (a) τ = BI 0 L2 ˆi −ˆj 2 ( ) (b) θ = 3 BI 0 ∆t 2 4 M Q.8 x 10−4 V/m (down) (d) 5.12 (a) 1. + x .4 ( ) Q. bottom −) (e) same as (c) Q.19 0.7×10–3 T Q.1 B Q.11 15 C Q.18 − 2 π r B0 i l . By = 2 4R 2 + b 2 4R 2 + b 2 Q.13 (a) I = Q.9 (a) 0 (b) 1.plane.6 (i) D (ii) B.41 × 10 –6 T .9 A 18 . (b) 5. C (iii) A q qB t + v 0 cos ωt + [v0 sin ωt] k .Q.88 N EXERCISE # III Q.4 A Q.94 x 10–2 Nm Q.62 N < F < 0.8 QV µ 0 I 3 3 π 3 2 (a) m 6a π −1 .5 ehB eh Q.2 A z=0.14 (a) increase.5 x 10−23 N (down) (c) 2. τ = 8πm Q. kˆ = ( v 0 x E )/ v 0 x E Q. − 2 π r B0 i l n Q.6 1 µ 0 J a 2 b a2 µ0 J a 2 R − µ J R − − (a) B = 2 2 R − b 2 . (ii) 2πd 3 µ0 πλ Q. where ω = . zero F= 2 π a2 Q.20 Bmin = 4. 45º in xz .10 (a) 20 min. (c) Q × QR 2 B NABπ 2li 0 19 rp mp qα 1 Q.C .19 D mg Q..23 ω = Q. (b) C = d T0 Q. 8 × 10–6Nt Q.17 B Q.26 A. (b) 0. Net force = F1 + F2 = 4 I R B i Q. time= qB 0 0 Q.14 B Q.22 A 2i 0 NAB π . (b) F = BI0b (3kˆ−4ˆi) .6 × 10–5T.12 (i) Q.24 r = m .20 (a) current in loop PQRS is clockwise from P to QRS. (c) I = 6bB0 Q.16 C Q. 0.25 (a) k = NAB.15 D Q.21 A Q.18 A Q.mv0 πm Q.13 (a) 6.10 (a) 2qB (b)velocity=-v. q = 2 α α p Q.11 (i) C (ii) B (iii) C (iv) C µ0 I − 4R q v0 kˆ (ii) F1 = 2 I R B F2 =2 I R B . Key Concepts 2. Exercise III 5. Answer Key 7. from AIEEE 1 . 34 Yrs.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P6. Electromagnetic Induction and Alternating Current Index: 1. Que. Exercise II 4. Exercise I 3. 10 Yrs. from IIT-JEE 8. Que. Exercise IV 6. LAW OF EMI : dφ . This induced emf opposes the causes of Induction. φ φs = Li Coefficient of Self inductance L = s or i 2 Page 2 of 16 E. E0 = maximum induced emf . & A. It's unit is Henry . an electromotive force (emf) is produced in the conductor. d A for non uniform B . A = BA cos θ weber for uniform B . dt 3. 4. φ = ∫ B . (i) FARADAY'S LAWS OF ELECTROMAGNETIC INDUCTION : An induced emf is setup whenever the magnetic flux linking that circuit changes. The negative sign indicated that the induced emf opposes the change of the flux . KEY CONCEPTS . ε α . 7. Hence an emf (and thus a current) is induced in the conductor as a result of its movement across the magnetic field." 1. A = area of one turn . COIL ROTATION IN MAGNETIC FIELD SUCH THAT AXIS OF ROTATION IS PERPENDICULAR TO THE MAGNETIC FIELD : Instantaneous induced emf .M. L = length of the conductor (m) . ω = uniform angular velocity of the coil .When a conductor is moved across a magnetic field. This is known as "ELECTROMAGNETIC INDUCTION. LENZ'S LAWS : The direction of an induced emf is always such as to oppose the cause producing it . EMF INDUCED IN A STRAIGHT CONDUCTOR IN UNIFORM MAGNETIC FIELD : E = BLV sin θ volt where B = flux density in wb/m2 . e=− dt 5. MAGNETIC FLUX : φ = B .C. B = magnetic induction . E = NABω sin ωt = E0 sin ωt . where N = number of turns in the coil . 2.I. (ii) The magnitude of the induced emf in any circuit is proportional to the rate of change of the magnetic dφ flux linking the circuit. The property of the coil or the circuit due to which it opposes any change of the current coil or the circuit is known as SELF & INDUCTANCE . 6. θ = angle between direction of motion of conductor & B . SELF INDUCTION & SELF INDUCTANCE : When a current flowing through a coil is changed the flux linking with its own winding changes & due to the change in linking flux with the coil an emf is induced which is known as self induced emf & this phenomenon is known as self induction . If the conductors forms part of a closed circuit then the emf produced caused an electric current to flow round the circuit. V = velocity of the conductor (m/s) . Em = 9.I.(i) shape of the loop & (ii) medium i = current in the circuit . the two coils are said to be electromagnetically coupled circuits . 10. This phenomenon is called MUTUAL INDUCTION & the induced emf in the latter circuit due to a change of current in the former is called MUTUALLY INDUCED EMF . Ip current in the primary d dI dφm =− (MI) = − M (If M is constant) dt dt dt M depends on (1) geometery of loops (2) medium (3) orientation & distance of loops . dφ s di d self induced emf es = =− (Li) = − L (if L is constant) dt dt dt 8.C. Thus in a superconducting loop flux never changes. (i) (ii) ENERGY STORED IN AN INDUCTOR : 1 2 W= LI . where M is mutual inductance . ε = 0. Than any change of current in one produces a change of magnetic flux in the other & the latter opposes the change by inducing an emf within itself . n = number of turns in the solenoid per unit length . SOLENOID : There is a uniform magnetic field along the axis the solenoid (ideal : length >> diameter) B = µ ni where . φ m flux linked with sec ondary Mutual inductance = M = = mutually induced emf . φs = magnetic flux linked with the circuit due to the current i . Therefore φtotal = constant. 2 Energy of interation of two loops U = I1φ2 = I2φ1 = MI1I2 . MUTUAL INDUCTION : If two electric circuits are such that the magnetic field due to a current in one is partly or wholly linked with the other. i = current in the solenoid . (or it opposes 100%) 11. Self inductance of a solenoid L = µ0 n2Al . µ = magnetic permeability of the core material . is called the primary & the other circuit in which the emf is induced is called the secondary. . A = area of cross section of solenoid .M. L depends only on . & A. 3 Page 3 of 16 E. The co0efficient of mutual induction (mutual inductance) between two electromagnetically coupled circuit is the magnetic flux linked with the secondary per unit current in the primary. SUPER CONDUCTION LOOP IN MAGNETIC FIELD : R = 0 . The circuit in which the current is changed. . R E I0 = . & A. R (i) L behaves as open circuit at t = 0 [ If i = 0 ] L behaves as short circuit at t = ∞ always .I.M. L Large Curve (1) → R L Curve (2) → Small R DECAY OF CURRENT : Initial current through the inductor = I0 .GROWTH OF A CURRENT IN AN L − R CIRCUIT : E I= (1 − e−Rt/L) . Current at any instant i = I0e−Rt/L 4 Page 4 of 16 E. (ii) 13.C. [ If initial current = 0 ] R L = time constant of the circuit . 12. Q.m.M.7 A horizontal wire is free to slide on the vertical rails of a conducting frame as shown in figure. If the radius of the loop starts shrinking at a constant rate of 1.2 A wire forming one cycle of sine curve is moved in x-y plane with velocity V = Vx i + Vy j . Find the total heat produced in R2. Resistance of the inner coil is R. Q.1 The horizontal component of the earth’s magnetic field at a place is 3 × 10–4 T and the dip is tan–1(4/3).f.02 T.I.4 Find the dimension of the quantity Q.C. Q. 5 Page 5 of 16 E. There is a magnetic field B = (50T) kˆ . Find the e. Current in the outer coil is increased from 0 to i . then find the emf induced in the loop. EXERCISE–I . induced in the rod. find the ratio of i1 to i2 where i1 is the initial (at t = 0) current and i2 is steady state (at t = ∞) current through the battery. Find the current through external resistance. There exist a magnetic field B = − B 0 k . The wire has a mass m and length l and the resistance of the circuit is R.10 In the circuit shown. then find the terminal speed of the wire as it falls under the force of gravity.0 mm/s.8 A metal rod of resistance 20Ω is fixed along a diameter of a conducting ring of radius 0. & A. Then the switch is shifted to position 2 for a long time. Q.9 In the given current.3 A conducting circular loop is placed in a uniform magnetic field of 0. Resistance of connector is r = 2Ω. Q. initially the switch is in position 1 for a long time. where symbols have usual meaining.25 m placed in the north-south position is moved at a constant speed of 10cm/s towards the east.1 m and lies on x-y plane. L . Find the motional emf develop across the ends PQ of wire. An external resistance of 10Ω is connected across the centre of the ring and rim. Find the external force required to keep the connector moving with a constant velocity v = 2m/s. with its plane perpendicular to the field . at the instant when the radius is 4 cm. then find the total charge circulating the inner coil.Q. Two resistances of 6Ω and 3Ω are connected as shown in figure.0 m is situated in a uniform magnetic field B = 2T perpendicular to the plane of loop. RCV Q. A metal rod of length 0. Q.6 Two concentric and coplanar circular coils have radii a and b(>>a)as shown in figure. If a uniform magnetic field B is directed perpendicular to the frame. The ring rotates with an angular velocity ω = 20 rad/sec about its axis.5 A rectangular loop with a sliding connector of length l = 1. Q. Q. where K and C are constants and t is time. Find the charge which passes through the battery in one time constant. Q.15 In the circuit shown in figure switch S is closed at time t = 0. (0 ≤ t ≤ C/K). Find the magnitude and direction of the induced emf in the loop.14 A uniform magnetic field of 0. The key K is shorted at time t = 0. Q. A current flows in coil 1.11 .I.17 In a L–R decay circuit. then find the instantaneous charge q on the upper plate of capacitor. Calculate the phase difference between the voltage and current in the circuit. Q. a resistance of 10Ω and an ac source of 50 Hz so that the power factor of the circuit is unity. Find the initial (t = 0) and final (t → ∞) currents through battery. where K is a constant and 't' is time.08 T is directed into the plane of the page and perpendicular to it as shown in the figure.C. Find the total charge that has flown through the resistor till the energy in the inductor has reduced to one–fourth its initial value. If an electron is released from rest in this field at a distance of ‘r’ from the axis of cylinder. the switch is closed.18 A charged ring of mass m = 50 gm. Q.19 A capacitor C with a charge Q0 is connected across an inductor through a switch S. Q. Find the total charge that will pass around the loop. between t = 0 and t = T.22 Find the value of an inductance which should be connected in series with a capacitor of 5 µF. have a mutual inductance = M and resistances R each. is applied perpendicular to the plane of the circular loop of radius ‘a’ and resistance R. Find the angular speed attained in a time t1 = 10 sec.M. Q.16 Two coils.010 m2. which varies with time as: I1 = kt2 .2 t) Tesla/sec is applied on to the ring in a direction normal to the surface of ring.20 A uniform but time varying magnetic field B = Kt – C . A wire loop in the plane of the page has constant area 0. Find the total charge that has flown through coil 2. its acceleration.0 × 10–4 Ts–1.0 henry is connected across an alternating voltage of frequency 300/2π Hz. 1 & 2. If at t = 0. Q. Q.12 There exists a uniform cylindrically symmetric magnetic field directed along the axis of a cylinder but varying with time as B = kt. 6 Page 6 of 16 E. Q. just after it is released would be (e and m are the electronic charge and mass respectively) Q. & A.Two resistors of 10Ω and 20Ω and an ideal inductor of 10H are connected to a 2V battery as shown. The magnitude of magnetic field decrease at a constant rate of 3. A magnetic field varying with time at a rate of (0. Find the ratio of the currents at time t = ∞ and t = 1 second.21 A coil of resistance 300Ω and inductance 1. the initial current at t = 0 is I. Q. charge 2 coulomb and radius R = 2m is placed on a smooth horizontal surface.13 An emf of 15 volt is applied in a circuit containing 5 H inductance and 10 Ω resistance. 4. 4.135.C circuit the potential difference across an inductance and resistance joined in series are respectively 12 V and 16V.23 In an L-R series A.331.313. when 100 V ac of 50 Hz is applied to the same coil.98.124. only 0. 4.5 amp flows.299.25 A 50W. 4. 100V lamp is to be connected to an ac mains of 200V.315.316.309.100. 4.C. 3.C.329. 3. Q.137.Q. 4. 3.301 to 3. 50Hz.99.335.121. Calculate the resistance and inductance of the coil. 4. 4.126.311.136. Irodov. 4.288 to 3. Q. 3. a current of one ampere flows through it. 3.E. 4.141.326 to 3.M. is applied across a coil.125. 3. 3. 3. What capacitance is essential to be put in seirs with the lamp.134.I. 3. 4.333 to 3.144 7 Page 7 of 16 E. List of recommended questions from I. 4. Find the total potential difference across the circuit. & A. . 4.24 When 100 volt D. The loop contains a 20 V battery with negligible internal resistance. & A. as the rod moves.35 T magnetic field points out of the page.3 A wire is bent into 3 circular segments of radius r = 10 cm as shown in figure .I. In what manner does the emf around the triangle vary with time .C. ab lying in the xy plane.87 t. Current is passed down one coil.1 (i) (ii) (iii) Two straight conducting rails form a right angle where their ends are joined.M.6 Consider the possibility of a new design for an electric train. what is the magnitude of the emf developed in the wire. with B in tesla & t in sec. how much power would be lost for each Ω of resistivity in the rails ? is such a train unrealistic ? A square wire loop with 2 m sides in perpendicular to a uniform magnetic field. There is a uniform magnetic field of induction B normal to the plane of the paper and directed into the paper.5 (i) (ii) Q. if a magnetic field B points in the positive x direction. through another conducting wheel & then back to the source via the other rail. A 0. What is the total emf in the circuit ? What is the direction of the current through the battery ? A rectangular loop of dimensions l & w and resistance R moves with constant velocity V to the right as shown in the figure.2 m/s to the right as shown in figure. into a conducting wheel through the axle. a constant current i flows through R. a distance l apart and each having a resistance λ per unit length are joined at one end by a resistance R. (i) (ii) Q.042 − 0. Sketch the flux. when B increases at the rate of 3 mT/s ? what is the direction of the current in the segment bc. A variable force F is applied to the rod MN such that.0 s. induced emf & external force acting on the loop as a function of the distance. The engine is driven by the force due to the vertical component of the earths magnetic field on a conducting axle. bc lying in the yz plane & ca lying in the zx plane. 8 Page 8 of 16 E. A conducting bar contact with the rails starts at vertex at the time t = 0 & moves symmetrically with a constant velocity of 5. Calculate: The flux through the triangle by the rails & bar at t = 3. It continues to move with same speed through a region containing a uniform magnetic field B directed into the plane of the paper & extending a distance 3 W. If the magnitude of the field varies with time according to B = 0. with half the area of the loop in the field . Each segment is a quadrant of a circle.Q. The emf around the triangle at that time. EXERCISE–II .0 m . Find the velocity of the rod and the applied force F as function of the distance x of the rod from R Q. Q. A perfectly conducting rod MN of mass m is free to slide along the rails without friction. what current is needed to provide a modest 10 − KN force ? Take the vertical component of the earth's field be 10 µT & the length of axle to be 3.4 (i) (ii) (iii) Q.2 Two long parallel rails. The loop is moved away from the wire with side AB always parallel to the wire. with S closed. long time after S is opened.C. Initially. long time later. R2 and R3 respectively. immediately after S is open.12 A variable magnetic field creates a constant emf E in a conductor ABCDA.M. CDA and AMC are R1. Q.8 A square loop of side 'a' & resistance R moves with a uniform velocity v away from a long wire that carries current I as shown in the figure.10 A long straight wire is arranged along the symmetry axis of a toroidal coil of rectangular cross−section. calculate the force required to pull the ring with uniform velocity v. Q. Q.1) ln 2 sec. Q.I. is placed over the rails as shown in the figure. switch S2 is opened. Q. After time t = (0. 9 Page 9 of 16 E. distance between the side AB of the loop & wire is 'a'. & A. Q. A uniform magnetic field B exists in the region of space.2) ln 2 sec. Find the current in the circuit at time t = (0.11 A uniform magnetic field B fills a cylindrical volumes of radius R.14 (i) (ii) (iii) (iv) Find the values of i1 and i2 immediately after the switch S is closed.A rectangular loop with current I has dimension as shown in figure . Find the magnetic flux φ through the infinite region to the right of line PQ. Find the work done when the loop is moved through distance 'a' from the initial position. whose dimensions are given in the figure. A light uniform ring of diameter d which is practically equal to separation between the rails. The resistances of portion ABC. The number of turns on the coil is N. Q.7 . if the current i = im cos ωt flows along the straight wire. Q.9 Two long parallel conducting horizontal rails are connected by a conducting wire at one end. Find the amplitude of the emf induced in this coil. A metal rod CD of length l is placed inside the cylinder along a chord of the circular cross-section as shown in the figure.13 In the circuit shown in the figure the switched S1 and S2 are closed at time t = 0. If resistance of ring is λ per unit length. and permeability of the surrounding medium is unity. What current will be shown by meter M? The magnetic field is concentrated near the axis of the circular conductor. If the magnitude of magnetic field increases in the direction of field at a constant rate dB/dt. find the magnitude and direction of the EMF induced in the rod. Q.19 A box P and a coil Q are connected in series with an ac source of variable frequency. Take R = 4Ω.15 Consider the circuit shown in figure.4 A flows in the circuit. Also find the voltage across P and Q respectively. & A. Find the impedance of P and Q at this frequency.16 Suppose the emf of the battery.C. Q. Q.I. Find the maximum instantaneous current in each capacitor.Q. If the same coil is connected to a 12V. 50 rad/s ac source a current of 2.9 mH and a resistance of 68Ω series. Q. At what frequency the current in the circuit lags the voltage by 45°? 10 Page 10 of 16 E. the current leads the voltage by 60°. Calculate the current and the power dissipated in the LCR circuit. The oscillating source of emf deliver a sinusoidal emf of amplitude emax and frequency ω to the inductor L and two capacitors C1 and C2. Also find the power developed in the circuit if a 2500 µF capacitor is connected in series with the coil.20 A series LCR circuit containing a resistance of 120Ω has angular resonance frequency 4 × 105 rad s–1. L = 6H & find an expression for the battery emf as function of time. When only the inductance is removed. When only the capacitance is removed. . Box P contains a capacitance of 1µF in series with a resistance of 32Ω coil Q has a self-inductance 4. the circuit shown varies with time t so the current is given by i(t) = 3 + 5t. At resonance the voltages across resistance and inductance are 60 V and 40 V respectively. The emf of source at 10 V. Find the values of L and C.18 An LCR series circuit with 100Ω resistance is connected to an ac source of 200 V and angular frequency 300 rad/s. The frequency is adjusted so that the maximum current flows in P and Q. Determine the inductance of the coil. the current lags behind the voltage by 60°.17 A current of 4 A flows in a coil when connected to a 12 V dc source. where i is in amperes & t is in seconds. Q.M. the power given to the two coils is the same. 2] Q. When switch is closed: (A) The initial reading in VL will be greater than in VR.5 m. (ii) Obtain the time dependence of the torque required to maintain the constant angular speed.EXERCISE–III Q. 5] An inductance L. What is the induced emf across the terminals of the switch ? (i) Obtain an expression for the current as a function of time after switch S is closed. it is observed that when the terminal velocity is attained. the induced voltage and the energy stored in the first coil are I1. Page 11 of 16 E. 2] same instant are I2.C. battery B and switch S are connected in series. They are connected at the two ends by resistance R1 & R2 as shown in the figure. Corresponding values for the second coil at the [JEE’94. v2 and W2 respectively. A horizontally metallic bar L of mass 0. the resistance per unit length of the wire is 1 Ω/m. V1 and W1 respectively. the power dissipated in R1 & R2 are 0. The current in the second coild is also increased at the same constant. [JEE '93.6T perpendicular to the plane of the rails. Find the current in segments AE.1 [JEE’93. Then: (A) Q. vertically down the rails under the action of gravity. The entire circuit is placed in a steadily increasing uniform magnetic field directed into the place of the paper & normal to it .3 Two parallel vertical metallic rails AB & CD are separated by 1 m. The rate of change of the magnetic field is 1 T/s. The current in one coil is increased at a constant rate.4 Two different coils have self inductance 8mH and 2mH.2 kg slides without friction. At that time the current. Initially. Neglect the resistance of the ring and the rod. the switch is open.76 W & 1. AEFD is a square of side 1 m & EB = FC = 0. [JEE '95. resistance R. Voltmeters VL and VR are connected across L and R respectively. [ JEE '94. A uniform & constant (a) (b) V2 1 (D) V = 4 1 magnetic induction B is applied perpendicular & into the plane of rotation as shown in figure. An inductor L and an external resistance R are connected through a switch S between the point O & a point C on the ring to form an electrical circuit. BE & EF. The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. (B) The initial reading in VL will be lesser than VR. There is a uniform horizontal magnetic field of 0.I.M. given that the rod OA was along the positive X-axis at t = 0. (C) The initial readings in VL and VR will be the same.2 W respectively.5 I1 1 = I2 4 (B) I1 =4 I2 W2 (C) W = 4 1 A metal rod OA of mass m & length r is kept rotating with a constant angular speed ω in a vertical plane about a horizontal axis at the end O. & A. Q.2 A rectangular frame ABCD made of a uniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. (D) The reading in VL will be decreasing as time increases. 10] 11 . 6] Q. Find the terminal velocity of bar L & value R1 & R2. At a certain instant of time. It is connected to a 10 volt battery. The rod is tied to a massless string which passes over a pulley fixed to the edge of the table. [REE '98.C.6 . A mass m. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic induction B .A solenoid has an inductance of 10 Henry & a resistance of 2 Ω. 3 × 2 = 6 .36 (1 + 2t) × 10−2 A increases at a steady rate in a long straight wire. tied to the other end of the string hangs vertically. 2] Q.10 A current i = 3. the Henry. (i) The SI unit of inductance. The loop are co-planar & their centres coincide. 5] [ JEE '98. 5] Q.M. How long will it take for the magnetic energy to reach 1/4 of its maximum value ? [JEE '96. Find the magnitude & the direction of the induced current in the loop.4 x 10−2 Ω. the acceleration of the mass at the instant when the velocity of the rod is half the terminal velocity.8 Fill in the blank. A small circular loop of radius 10−3 m is in the plane of the wire & is placed at a distance of 1 m from the wire. [JEE '97. 2] A pair of parallel horizontal conducting rails of negligible resistance shorted at one end is fixed on a table.4×2=8] Q.7 Select the correct alternative. (A) (iii) 12 Page 12 of 16 E. & A. A conducting massless rod of resistance R can slide on the rails frictionlessly. The distance between the rails is L. uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. calculate: the terminal velocity achieved by the rod. A metallic block carrying current I is subjected to a uniform magnetic induction B j . At the position MNQ the speed of the ring is v & the potential difference developed across the ring is : (A) zero (C) π RBV & Q is at higher potential Bv π R 2 & M is at higher potential 2 (D) 2 RBV & Q is at higher potential (B) [JEE'96.I. A constant. Select the correct statement(s) from the following (A) the entire rod is at the same electric potential (B) there is an electric field in the rod (C) the electric potential is highest at the centre of the rod & decreases towards its ends (D) the electric potential is lowest at the centre of the rod & increases towards its ends.9 (i) (ii) [JEE '96. [assume the speed of the carrier to be v] Q. The moving charges experience a force F given by ______ which results in the lowering of the potential of the face ______. A constant magnetic field B exists perpendicular to the table. The resistance of the loop is 8. The mutual inductance of the system is proportional to : L 2 L2 (B) (C) (D) L L A metal rod moves at a constant velocity in a direction perpendicular to its length . can be written as : (A) weber/ampere (B) volt − second/ampere (C) joule/(ampere)2 (D) ohm − second (ii) A small square loop of wire of side l is placed inside a large square loop of wire of side L(L >> l).11 Select the correct alternative(s). If the system is released from rest. 3] Q. Q. Loop-A carries a current which increases with time.I. (iv) . the switch is opened.f. & A. What is the potential drop across L as a function of time? After the steady state is reached. B0 and a are positive constants.0 A at approximately the time (A) 500 s (B) 20 s (C) 35 ms (D) 1 ms [ JEE ’99 ] Q.It is found that the maximum value of Q is 200 µC.what is the value of I ? Find the maximum value of I. (Scr)] Q. (a) (b) (c) when Q=100µC. (c) an expression for the speed of the loop. with CM fixed Q. carrying current I.15 A magnetic field B = (B0y / a) k is into the plane of paper in the +z direction.is connected across a charged capacitor of capacitance 5. (d) when I is equal to one half its maximum value. lies in x-y plane with its centre at origin.13 A coil of inductance 8.4 mH and resistance 6Ω is connected to a 12V battery. what is the value of Q Q.0mH.M. starts falling under the influence of gravity. v(t) and its terminal value. The internal resistance of the battery is negligible.m.14 A circular loop of radius R. mass m and resistance R. The current in the coil is 1.and the resulting LC circuit is set oscillating at its natural frequency. In response.17 An inductor of inductance L = 400 mH and resistors of resistances R1 = 2Ω and R2 = 2Ω are connected to a battery of e. (b) the total Lorentz force acting on the loop and indicate its direction.16 Two circular coils can be arranged in any of the three situations shown in the figure.what is the value of dI / dt ? when Q=200 µC . Let Q denote the instantaneous charge on the capacitor.12 Two identical circular loops of metal wire are lying on a table without touching each other.0µF.C. the loop-B [JEE ’99] (A) remains stationary (B) is attracted by the loop-A (C) is repelled by the loop-A (D) rotates about its CM. What is the direction and the magnitude of current through R1 as a function of time? [JEE ’2001] 13 Page 13 of 16 E.An inductor of inductance 2. [JEE ’99] Q. Find (a) the induced current in the loop and indicate its direction. A square loop EFGH of side a. E = 12V as shown in the figure. Note the directions of x and y axes in the figure. in x-y plane. The total magnetic flux through x-y plane is (A) directly proportional to I (B) directly proportional to R (C) directly proportional to R2 (D) zero [JEE ’99] Q. and I the current in the circuit . The switch S is closed at time t = 0. Their mutual inductance will be (A) maximum in situation (a) (B) maximum in situation (b) (C) maximum in situation (c) (D) the same in all situations [JEE ’2001. [JEE 2003] Q. a current IQ2 flows in Q.20 A square loop of side 'a' with a capacitor of capacitance C is located between two current carrying long parallel wires as shown. A constant magnetic field is also present in + Z direction.23 An infinitely long cylindrical conducting rod is kept along + Z direction. (a) (b) Draw a graph between charge on the lower plate of the capacitor v/s time. thickness d (d <<R) and length L.22 In an LR series circuit.21 The variation of induced emf (ε) with time (t) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as (A) (B) (C) (D) [JEE 2004(Scr)] Q. Then current induced will be (A) 0 (B) along +z direction (C) along clockwise as seen from + Z (D) along anticlockwise as seen from + Z [JEE’ 2005 (Scr)] Q. If the resistivity of the material of cylindrical shell is ρ. find the induced current in the shell. If the number of turns were to be quadrupled and the wire radius halved. The switch remains closed for a long time. [JEE 2004] O Vrms = 220 V.Q. P and Q are two coaxial conducting loops separated by some distance. A variable current i = i0sin ωt flows through the coil. 3] (A) halved (B) the same (C) doubled (D) quadrupled Q. [JEE 2005 ] 14 Page 14 of 16 E. R = 11 Ω.19 A short -circuited coil is placed in a time varying magnetic field. It is given that L = 35 mH.C.I. When the switch S is closed. a sinusoidal voltage V = Vo sin ωt is applied. & A.M. the electrical power dissipated would be [JEE 2002(Scr).18 As shown in the figure. 3] . Q. V ω = 50 Hz and π = 22/7. Then the directions of IQ1 adn IQ2 (as seen by E) are: (A) respectively clockwise and anti-clockwise (B) both clockwise (C) both anti-clockwise (D) respectively anti-clockwise and clockwise [JEE 2002(Scr). calculate maximum current in the square loop. Find 2π the amplitude of current in the steady state and obtain the phase difference between the current and the voltage. 24 A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R. Also plot the variation of current for one cycle on the given graph. The value of I in the is given as I = I0sinωt. Electrical power is dissipated due to the current induced in the coil. a clockwise current IP flows in P (as seen by E) and an induced current IQ1 flows in Q. When S is opened. T/4 T/2 3T/4 T t Q. 25 In the given diagram. charge on the capacitor is CV(1–e–1) [JEE 2006] Q. V R S1 C L S2 Q. it can never represent (A) an electrostatic field (B) a magnetostatic field (C) a gravitational field of a mass at rest (D) an induced electric field [JEE 2006] . (A) at t = 0. then for t ≥ 0 t π (A) the charge on the capacitor is Q = Q 0 cos + LC 2 t π (B) the charge on the capacitor is Q = Q 0 cos − LC 2 (C) the charge on the capacitor is Q = − LC (D) the charge on the capacitor is Q = − d 2Q dt 2 1 d 2Q LC dt 2 [JEE 2006] 15 Page 15 of 16 E. energy stored in the circuit is purely in the form of magnetic energy (B) at any time t > 0. charge on the capacitor is CV(1–e–2) (C) the work done by the voltage source will be half of the heat dissipated when the capacitor is fully charged. switch S1 is closed and S2 is kept open. the capacitor was uncharged.M. then (A) after time interval τ. If time constant of this circuit is τ. there is no exchange of energy between the inductor and capacitor (D) at any time t > 0. The capacitor can be connected in series with an inductor ‘L’ by closing switch S2 and opening S1.28 If the total charge stored in the LC circuit is Q0.Comprehension –I The capacitor of capacitance C can be charged (with the help of a resistance R) by a voltage source V. Then.26 Initially. instantaneous current in the circuit may V C L [JEE 2006] Q. S1 is opened and S2 is closed so that the inductor is connected in series with the capacitor. Q. current in the circuit is in the same direction (C) at t > 0. Now. (D) after time interval 2τ.I. by closing switch S1 while keeping switch S2 open. a line of force of a particular force field is shown. & A. Out of the following options.C. charge on the capacitor is CV/2 (B) after time interval 2τ.27 After the capacitor gets fully charged. So as train is deviated then as is move down coil on track repel it and as it move up then coil attract it.32 Match the following Columns Column 1 (A) Dielectric ring uniformly charged (B) Dielectric ring uniformly charged rotating with angular velocity . In this there is a coil on a railway track and magnet on the base of train.I.C. Due to motion of train current induces in the coil of track which levitate it.M.30 What is the disadvantage of the train? (A) Train experience upward force due to Lenz's law. . (B) Friction force create a drag on the train. (C) Retardation (D) By Lenz's law train experience a drag [JEE 2006] Q. & A.29 What is the advantage of the train? (A) Electrostatic force draws the train (C) Electromagnetic force draws the train [JEE 2006] (B) Gravitational force is zero. (D) Dissipative force due to friction are absent Q.Q. Comprehension –IV Magler Train: This train is based on the Lenz law and phenomena of electromagnetic induction. (C) Constant current in ring i0 0 cos ωt in ring ( D ) C u r r e n t i = i (B) Time varying electric field (D) Induced electric field Column 2 (P) Time independent electrostatic field out of system (Q) Magnetic field (R) Induced electric field (S) Magnetic moment 16 [JEE 2006] Page 16 of 16 E. Disadvantage of magler train is that as it slow down the forces decreases and as it moves forward so due to Lenz law coil attract it backward.31 Which force causes the train to elevate up (A) Electrostatic force (C) magnetic force [JEE 2006] Q. 11 1 1 A.8 LE 2 Q.21 π/4 Q.ANSWER KEY Q.6 µ 0ia 2 π 2Rb Q.6 2 2 2 2 3 µ 0 I aV 3a + a n 4 = 2π 2 R 3 1 3 +n 4 Q.22 20 ≅ 2H π2 Q.14 3µV.18 200 rad/sec Q.9 17 µ0 a +b IL ln 2π a 4B2 νd πλ Q.74 V.10 b µ 0 hωi m N ln a 2π Page 17 of 16 E.13 2 e −1 Q.C.22 Tm2.4 (i) 3.4 Q.M.16 kMT2/(R) Q.0 × 1017 W.2 µF 3 π Hz EXERCISE–II Q. + BId Bd B2d 2 Q.12 erk directed along tangent to the circle of radius r.0 µV Q.5 21. 2m e2 Q. 2N Q.3 (i) 2.23 20 V Q. & A.17 L I 2R C πa2 R Q.7 mgR B2l 2 Q. anticlockwise µ02 I2a 2 V Q.15 EL eR 2 1 π Q.3 5.8 V.7 φ = Q.I.3 × 108 A. (iii) linearly Q.1 10 µV Q. EXERCISE–I .24 R = 100W. clockwise Q.10 2R 12 Q.25 C = 9. (iii) totally unrealistic Q.8 4π2 R Q.9 I–1 Q. (ii) 56.20 Q. (ii) 1.2 λVyB0 Q.4 × 10−5 V (ii) from c to b Q.19 q = Q0sin LC t + 2 Q.8 1 A 3 Q.. A 15 10 Q.1 (i) 85. whose centre lies on the axis of cylinder.5 0.2 I(R + 2λx ) 2I 2 mλ(R + 2λx ) . 30 Ω Q. ABDC Q. 17. 97. . 8 × 105 rad/s 32 EXERCISE–III Q.6Ω.9 (i) Vterminal = 2 B Z 2 .Q. (iv) (a)104A/s (b) 0 (c) 2A (d) 100 3 µC Q. 400W 1 µF.14 (i) i1 = i2 = 10/3 A.16 42 + 20t volt Q. (ii) τ = cos ωt + (1 − e−Rt/L) Q.C. R B2a 2t − 0 mgR (c) V = 2 2 1 − e mR B0 a 18 Page 18 of 16 E.17 0.14 D B0av in anticlockwise direction.10 1. (ii) B.7V.15 (a) i = Q.3 pA Q. (ii) Q.13 D Q. R1 = 0.4 ACD [ ] Bωr 2 1−e − Rt / L mgr 1 ωB2 r 4 2 . D. C.11 (i) A.12 C Q.15 C2= ε max C1 1 1 + ωL − ω(C1 + C 2 ) C 2 Q.47 Ω. 7.13 67/32 A Q.5 (a) E = Bωr (b) (i) I = 2R 2 2 4R Q.18 2A.M.1 IEA= 7 3 1 A .8 evB kˆ . IFE = A 22 11 22 Q.6 t = L ln 2 = 3. B. (ii) i1 = 50/11 A .08 H.I.2 A. IBE= A . 9. (iv) i1 = i2 = 0 Q.7 D g 2 Q.3 V = 1 ms−1. (iii) i1 = 0. l dB l2 R2 − 2 dt 4 Q.76V Q.19 77Ω. v = velocity at time t.47 sec R mgR Q.6 π × 10–13 A = 50. & A.28 W Q.20 0.11 . i2 = 30/11 A. (b) Fnett=B02a2V/R.2 mH. (iii) B. i2 = 20/11 A. C1= ε max C 2 C1 1 1 + ωL − C1 C 2 ω(C1 + C 2 ) Q.12 E R1 R1R 2 + R 2 R 3 + R 3R1 Q. R2 = 0. D Q. C.30 D Q. & A.31 C 20 O -10 2 T T/8 T/4 T/2 5T/8 (µ 0 ni 0ω cos ωt )πa 2 (Ld) ρ2πR Q. S.24 I = Q.17 12e–5t. (D) Q.26 B Q.S . Q.22 20 A.29 D Q.18 D Q. R. (b) Page 19 of 16 E.23 A Q. I v = 220 2 sin ωt i = 20 sin (ωt-π/4) Q. (B) P.27 D Q. ∴ Steady state current i = 20sin π100t − 4 4 Q. (C) Q. π 1 .C Q.21 B V.16 A Q. Q.Q.I.M.28 C Q. 6e–10t Q.19 B Q.20 (a) Imax = µ 0a π CI 0ω2ln 2 . S 19 9T/8 t .25 A.32 (A) P. Key Concepts 2. from IIT-JEE 8. Answer Key 7. from AIEEE 1 . 10 Yrs. 34 Yrs. Que. Exercise III 5. Que. Exercise IV 6. Exercise II 4.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P7. Exercise I 3. Optics Index: 1. coordinate of focus .1. f v u f = x. 7. (b) The direction of the incident rays is considered as positive x-axis. (a) (b) (c) 5. (i) (ii) 4. MIRROR FORMULA : = + . (c) Vertically up is positive y-axis. v = x-coordinate of image Note : Valid only for paraxial rays. CHARACTERISTICS OF REFLECTION BY A PLANE MIRROR : The size of the image is the same as that of the object. (ii) The angle of incidence (the angle between normal and the incident ray) and the angle of reflection (the angle between the reflected ray and the normal) are equal ∠i = ∠r 2. Concave Convex PARAXIAL RAYS : Rays which forms very small angle with axis are called paraxial rays. if the mirror be rotated through an angle θ the reflected ray turns through an angle 2θ. IMAGE : Image is decided by reflected or refracted rays only. actually converge (real image). (i) LAWS OF REFLECTION : The incident ray (AB). Note : According to above convention radius of curvature and focus of concave mirror is negative and of convex mirror is positive. 1 1 1 8. u = x-coordinate of object . the reflected ray (BC) and normal (NN') to the surface (SC') of reflection at the point of incidence (B) lie in the same plane. 2 Page 2 of 20 GEOMETRICAL OPTICS KEY CONCEPTS . The point image for a mirror is that point Towards which the rays reflected from the mirror. For a fixed incident light ray. (a) (b) OBJECT : Real : Point from which rays actually diverge. OR From which the reflected rays appear to diverge (virtual image) . SIGN CONVENTION : We follow cartesian co-ordinate system convention according to which (a) The pole of the mirror is the origin . For a real object the image is virtual and for a virtual object the image is real. This plane is called the plane of incidence (also plane of reflection). Virtual : Point towards which rays appear to converge 3. SPHERICAL MIRRORS : 6. They are called conjugate positions or foci. (i) REFRACTION THROUGH A PARALLEL SLAB : Emerged ray is parallel to the incident ray. the normal (NN') to the refracting surface (II') at the point of incidence (B) and the refracted ray (BC) all lie in the same plane called the plane of incidence or plane of refraction . NEWTON'S FORMULA : Applicable to a pair of real object and real image position only . 2. Sin i = Constant : Sin r for any two given media and for light of a given wave length. This is known as SNELL'S Law .TRANSVERSE MAGNIFICATION : m = h2 =−v h1 u h2 = y co-ordinate of images h1 = y co-ordinate of the object (both perpendicular to the principle axis of mirror) 10. OPTICAL POWER : Optical power of a mirror (in Diopters) = – 1 . REFRACTION -PLANE SURFACE 1. f f = focal length (in meters) with sign . 3 Page 3 of 20 GEOMETRICAL OPTICS 9. (ii) Lateral shift x = t sin(i − r) cos r t = thickness of slab Note : Emerged ray will not be parallel to the incident ray if the medium on both the sides are different . . X. if medium is same on both sides. XY = f 2 11. Sin i n v λ = 1n2 = 2 = 1 = 1 Sin r λ2 n1 v2 Note : Frequency of light does not change during refraction . DEVIATION OF A RAY DUE TO REFRACTION : 3. (i) (ii) LAWS OF REFRACTION (AT ANY REFRACTING SURFACE) : The incident ray (AB).Y are the distance along the principal axis of the real object and real image respectively from the principal focus . r. 5. R. R.) CONDITIONS OF T. the ray passes symetrically about the prism. (i) (ii) CRITICAL ANGLE & TOTAL INTERNAL REFLECTION ( T. of glass w. 2. of glass . I. 4 .I. where n = absolute R. When δ = δm then i = i′ & r = r′ . 4. Note : When the prism is dipped in a medium then n = R. I. REFRACTION THROUGH PRISM : 1. medium .I.t. n Critical angle C = sin-1 r ni 6. (h′ < h) at near normal incidence h′ = µ2 h µ1 Note : h and h' are always measured from surface. There is one and only one angle of incidence for which the angle of deviation is minimum. & then sin n= [ A + δm 2 sin [ ] A 2 ] . 3.(r + r′) r + r′ = A Variation of δ versus i (shown in diagram) .APPARENT DEPTH OF SUBMERGED OBJECT : Page 4 of 20 GEOMETRICAL OPTICS 4. δ = (i + i′) . Ray going from denser to rarer medium Angle of incidence should be greater than the critical angle (i > c) . nv + nR Net mean deviation = 2 n ′v + n ′R − 1 A – − 1 A′ . u & R are to be kept with sign as v = PI u = –PO R = PC (Note radius is with sign) µ1 v m= µ u 2 LENS FORMULA : 1 1 1 − = v u f (b) 1 = (µ – 1) f (c) m= 1 1 − R1 R 2 v u 5 Page 5 of 20 GEOMETRICAL OPTICS 5.nr′) A′ .(a) (b) 2. REFRACTION AT SPERICAL SURFACE 1.nr) A = (n′v . (i) COMBINATION OF TWO PRISMS : ACHROMATIC COMBINATION : It is used for deviation without dispersion . Condition for this (nv . For a thin prism ( A ≤10o) . δ′ are the mean deviation. 9. (ii) DIRECT VISION COMBINATION : It is used for producing disperion without deviation condition n v + nR for this 2 n ′v + n ′R − 1 A = − 1 A′ . of material for violet.7. I. ω= angular dispersion deviation of mean ray (yellow) For small angled prism ( A ≤10o ) ω= n + nR n −n δv − δR = v R . (a) µ 2 µ1 µ 2 − µ1 − = v u R v. Dispersive power (ω) of the medium of the material of prism . 8. δ = ( n – 1 ) A DISPERSION OF LIGHT : The angular spilitting of a ray of white light into a number of components when it is refracted in a medium other than air is called Dispersion of Light.n= v δy n −1 2 nv. 6. . ω′ are dispersive powers for the two prisms & δ . Angle of Dispersion : Angle between the rays of the extreme colours in the refracted (dispersed) light is called Angle of Dispersion . θ = δv – δr . 2 or ωδ + ω′δ′ = 0 where ω.nr) A = (nv′ .n′r) A′ . nR & n are R. red & yellow colours respectively . 2 Net angle of dispersion = (nv . Q. The ends of the rod are perpendicular 4 to the central axis of the rod. Q. A ball is dropped from the balloon at a height 15 m from the mirror when the balloon has velocity 20 m/s. Find the minimum and maximum height of a man (eye height) required to see the image of the source if he is standing at a point A on ground shown in figure. close to the mirror. What is maximum distance from the center of the mirror and the bulb so that the required area is illuminated? Q.8 m above the ground. A glass plate of thickness 6 cm and index 2. with its lower edge 50 cm above the ground.Q. The mirror is rotated in the 9 direction as shown in the figure by an arrow at frequency rev/sec. Find the speed of the image of the ball formed by concave mirror after 4 seconds? [Take : g=10 m/s2] Q.5 A concave mirror of focal length 20 cm is cut into two parts from the middle and the two parts are moved perpendicularly by a distance 1 cm from the previous principal axis AB. Find the length of the image? Q. visible to him reflected from the mirror.6 A balloon is rising up along the axis of a concave mirror of radius of curvature 20 m. find the length of the floor between him & the mirror.0 is placed between the object and mirror. If his eyes are at a height 1. A man stands infront of the mirror at a distance 2 m away from the mirror. just touches the rod. Find the maximum value of angle θ for which internal reflection occurs inside the rod? liquid of refractive index 6 Page 6 of 20 GEOMETRICAL OPTICS EXERCISE # I .7 A thin rod of length d/3 is placed along the principal axis of a concave mirror of focal length = d such that its image. A circular area of radius R = 1 m on the floor is to be illuminated after reflection of light from the mirror. When the angle of incidence becomes 37° find the speed of the spot (a point) on the wall? Q. π The light reflected by the mirror is received on the wall W at a distance 10 m from the axis of rotation.4 A light ray I is incident on a plane mirror M. There is a light source S at a point on the ground. A bulb is to be placed on the axis of the mirror.3 A plane mirror of circular shape with radius r = 20 cm is fixed to the ceiling.2 In figure shown AB is a plane mirror of length 40cm placed at a height 40cm from ground.9 A long solid cylindrical glass rod of refractive index 3/2 is immersed in a 3 3 . a light enters one end of the rod at the central axis as shown in the figure.8 A point object is placed 33 cm from a convex mirror of curvature radius = 40 cm. The height of the room is 3m. Find the distance of final image from the object? Q. is hung parallel to a vertical wall of a room.1 A plane mirror 50 cm long . which is real and elongated. Find the distance between the images formed by the two parts? Q. 2 cm on the lower face. Q. edge of the bottom surface of the tank is just visible.14 A cubical tank (of edge l) and position of an observer are shown in the figure.11 A ray of light enters a diamond (n = 2) from air and is being internally reflected near the bottom as shown in the figure.16 A ray is incident on a glass sphere as shown. The ray emerges from the sphere parallel to line AB. perpendicular to the mirror's optical axis. and has a refractive index 2 3 . then find the distance of the object from the mirror? . Find maximum value of angle θ possible? Q. Q. Q. If the beam finally gets focussed at a point situated at a distance = 2 × (radius of sphere) from the centre of the sphere.Q. The opposite surface of the sphere is partially silvered. An insect is at the centre C of its bottom surface.10 A slab of glass of thickness 6 cm and index 1. Q. horizontal beam of light is incident upon a quarter cylinder of radius R = 5 cm. Find the refractive index of the sphere. Find the refractive index of the sphere.17 A narrow parallel beam of light is incident on a transparent sphere of refractive index 'n'. A patch on the table for a distance 'x' from the cylinder is unilluminated.5 is place somewhere in between a concave mirror and a point object. To what height a transparent liquid of refractive index µ = 5 must be poured in the tank so that the insect will 2 become visible? Q. find the value of 'x'? 7 Page 7 of 20 GEOMETRICAL OPTICS Q. If the net deviation of the ray transmitted at the partially silvered surface is 1/3rd of the net deviation suffered by the ray reflected at the partially silvered surface (after emerging out of the sphere). When the tank is empty.18 A uniform. The radius of curvature of the mirror is 40 cm. then find n? Q.12 A ray of light falls on a transparent sphere with centre at C as shown in figure. If the reflected final image coincides with the object. The beam width after it goes over to air through this face is _______ if the refractive index of glass is µ.15 Light from a luminous point on the lower face of a 2 cm thick glass slab.13 A beam of parallel rays of width b propagates in glass at an angle θ to its plane face . strikes the upper face and the totally reflected rays outline a circle of radius 3. What is the refractive index of the glass. Find the radii of curvature if the refractive index of the material of the lens is 1. The image is formed at infinity. as shown. One of the refracting surfaces of the prism is polished. For the beam of monochromatic light to retrace its path. find the x and y coordinates of the point where a parallel beam of rays coming from the left finally get focussed? Q. it behaves like a concave mirror of focal length 10 cm. find the value of ‘d’.21 to 24. Irodov.22 In the figure shown.25 A double convex lens has focal length 25.29 An equilateral prism deviates a ray through 23° for two angles of incidence differing by 23°. the focal length of the two thin convex lenses is the same = f.26. Q. If the final image be at infinity.13 to 17.19 A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. A point object is placed at distance ‘d’ on axis of this sphere as shown. Find µ of the prism? Q. If diameter of its aperture is 4 cm.E.0 cm in air. If a glass slab of thickness t and refractive index 1.27 A plano-convex lens. List of recommended questions from I. Q. Find (i) Radius of curvature of curved surface (ii) its focal length in air Q.27.31.5 is inserted between the lens and object. If the object is kept at a distance of 6 cm from the same lens the image formed is virtual. 5. A screen is placed on the other side of the lens. at a distance 20 cm from it. Q.20 An object is kept at a distance of 16 cm from a thin lens and the image formed is real. behaves like a concave mirror of focal length 30 cm. When it is silvered on the convex side.34 to 37 8 Page 8 of 20 GEOMETRICAL OPTICS Q.24 A glass hemisphere of refractive index 4/3 and of radius 4 cm is placed on a plane mirror. on its optical axis. 5.5. 5.5) has a maximum thickness of 1 mm. They are separated by a horizontal distance 3f and their optical axes are displaced by a vertical separation 'd' (d << f). Find the refractive index of the material of the lens. The radius of one of the surfaces is double of the other. Q. The size of real image becomes ‘q’ times that of object when the lens is moved nearer to the object by a distance ‘a’ find focal length of the lens ? Q. find the angle of incidence on the refracting surface. 5.30 A equilateral prism provides the least deflection angle 46° in air. Find the refracting index of an unknown liquid in which same prism gives least deflection angle of 30°.Q. Find the thickness t ? . then find the focal length of the lens? Q.23 A point source of light is kept at a distance of 15 cm from a converging lens. 5.26 A plano convex lens (µ=1. Then find the area of the illuminated part of the screen? Q. Taking the origin of coordinates O at the centre of the first lens. 5. when silvered on the plane side. The focal length of the lens is 10 cm and its diameter is 3 cm. If the size of the image formed are equal. perpendicular to the axis of lens.28 A prism of refractive index 2 has a refracting angle of 30°.21 A thin convex lens forms a real image of a certain object ‘p’ times its size. at a depth of 1 m in the medium. The mirror-liquid system forms one real image and another real image is formed by mirror alone. If the 17ft mark and the surveyor’s eye are both 6ft above the water level.4 An object is kept on the principal axis of a convex mirror of focal length 10 cm at a distance of 10 cm from the pole. coincides. prove that tanθ = a−b .9 A concave mirror has the form of a hemisphere with a radius of R = 60 cm. The disc is just visible to an eye looking over the edge. Q. d + d2 + r2 Q. If air between the glass pieces is replaced by water (µ = 4/3). y = 2x2. assuming that the refractive index of the water is 4/3. Show that d = 2a 2 (µ + 1) A ray of light travelling in air is incident at grazing angle (incidence angle = 90°) on a medium whose refractive index depends on the depth of the medium. with rear portion silvered and the combination of glass pieces is placed at a distance a = 60 cm from a screen.5 A surveyor on one bank of canal observed the image of the 4 inch and 17 ft marks on a vertical staff. the refractive index of the medium and angle of incidence φ. return to P. Find actual speed of jeep rate at which magnification is changing. A small object is placed normal to the optical axis of the combination such that its two times magnified image is formed on the screen. views his own face in a convex mirror of radius r of curvature 'r'.7 (i) (ii) Q. which is sighted by thief in his rear view mirror which is a convex mirror of focal length 10 m. (a) (b) Q. a the distance of P from the centre of the circle and b the distance of the I f centre from the point where the ray in its course crosses the diameter through P. Q.1 An observer whose least distance of distinct vision is 'd'. He observes that the image of jeep is moving towards him with a velocity of 1 cm/s.8 Two thin similar watch glass pieces are joined together. If the magnification of the mirror for the jeep at that time is 1/10. θ be the angle of incidence. A ray enters from P and after two reflections by the circle. A thin layer of an unknown transparent liquid is poured into the mirror.Q. Now. disc is just visible to the eye in the same position. the whole of the (µ 2 − 1) .3 A luminous point P is inside a circle. estimate the width of the canal. Q. front to front.2 A thief is running away in a car with velocity of 20 m/s. with the source in a certain position. A police jeep is following him. calculate the distance through which the object must be displaced so that a sharp image is again formed on the screen. Prove that magnification produced can not exceed . Find. One of them coincides with the source and the other is at a distance of l = 30 cm from source. The object starts moving at a velocity 20 mm/sec towards the mirror at angle 30° with the principal axis. The trajectory of the light in the medium is a parabola.6 A circular disc of diameter d lies horizontally inside a metallic hemispherical bowl radius a. 9 Page 9 of 20 GEOMETRICAL OPTICS EXERCISE # II . Assume that police jeep is on axis of the mirror. The bowl is now filled with a liquid of refractive index µ. which is partially immersed in the water and held against the bank directly opposite to him. What will be the speed of its image and direction with the principal axis at that instant? Q. a+b Q. Find the possible value(s) refractive index µ of the liquid. on the common axis of two lenses. The speed object = 4 cm/s.11 In the figure shown L is a converging lens of focal length 10cm and M is a concave mirror of radius of curvature 20cm. at the moment when u = 30 cm. Q.14 The rectangular box shown is the place of lens. (ii) An object having transverse length 5 cm in placed on the axis of equivalent mirror (in part 1).r. show that θ must be at least tan–1 2 or 25. the moving object. Q. answer the following questions: (i) If X is 5 cm then what is the focal length of the lens? (ii) If the point O is 1 cm above the axis then what is the position of the image? Consider the optical center of the lens to be the origin. The lens are made of difference 3 5 µ1 = and µ2 = as shown in figure.Q. Find. Find the transverse magnification produced by equivalent mirror. at a distance 15 cm from the equivalent mirror along principal axis. The distance between CD & AB is 1 cm. AB is principal axis of the converging lens of focal length F. 4 2 (i) if plane surface of the plano-convex lens is silvered.16 An isosceles triangular glass prism stands with its base in water as shown.t. The focal length of each lens is 15 cm and the lens L2 is to the right of lens A. .10 In the figure shown. The angles that its two equal sides make with the base are θ each. The two lens halves are placed symmetrically w. By looking at the ray diagram. 17 10 Page 10 of 20 GEOMETRICAL OPTICS Q. An incident ray of light parallel to the water surface internally reflects at the glass-water interface and subsequently re-emerges into the air. m a t e r i a l h a v i n g r e f r a c t i v e i n d e x Q.12 A thin plano-convex lens fits exactly into a plano concave lens with their plane surface parallel to each other as shown in the figure. The radius of curvature of the curved surface R = 30 cm. then calculate the equivalent focal length of this system and also calculate the nature of this equivalent mirror.9°. AB and CD are optical axes of the lens and mirror respectively. Q. Taking the refractive indices of glass and water to be 3/2 and 4/3 respectively. A point object O is placed in front of the lens at a distance 15cm. A point object is placed at a distance of 20 cm on the left of lens L1. where a convex mirror of radius of curvature 5 cm should be placed so that the final image coincides with the object? Q.13 In the figure shown ‘O’ is point object. Find the distance of the final image formed by this system from the optical centre of the lens. find the relative speed of approach/separation of the two final images formed after the light rays pass through the lens. Find the distance of the final image from the lens.15 Two identical convex lenses L1 and L2 are placed at a distance of 20 cm from each other on the common principal axis. (ii) A screen is placed normal to the emerging beam at a distance of 2m from the prism combination. Find the refractive index of the glass and the angle of the prism. 11 Page 11 of 20 GEOMETRICAL OPTICS Q. Find the distance between red and violet spot on the screen.18 The refractive indices of the crown glass for violet and red lights are 1.77 and 1. A beam of white light is incident at a small angle on this prism. A prism of angle 6° is made of crown glass.73 respectively. and emerging from it in a direction making an angle 6°30' with the reversed direction of the incident beam. The refracted beam is found to have undergone a deviation of 1°15' from the original direction.51 and 1.17 A parallel beam of light falls normally on the first face of a prism of small angle.5° anticlockwise. (i) Determine the angle of the flint glass prism. The other thin flint glass prism is combined with the crown glass prism such that the net mean deviation is 1. the reflected beam striking at the first face again.Q. . Which is the topmost colour on screen. At the second face it is partly transmitted and partly reflected.49 respectively and those of the flint glass are 1. The refractive index of air is 1.5 A small fish. Indicate the path of the ray subsequently . Find the image of the fish seen by the observer . Obtain a relation between the slope of the trajectory of the ray at a point B (x . The lens is kept at 0. is viewed through a simple converging lens of focal length 3 m .6(i)An eye specialist prescribes spectacles having a combination of convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm .352 . The magnification of the image formed by one of the half lenses is 2. The power of this lens combination in diopters is : (A) + 1.4 m below the surface of a lake. 0. where k = 1.3 A thin plano−convex.2m above the water surface such that the fish lies on the optical axis of the lens. The refractive index of the water is 4/3. Calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated . Find the focal length of the lens and separation between the two halves. The medium has a variable index of refraction n(y) given by : n (y) = [ky3/2 + 1]1/2. where the ray the ray intersects the upper surface of the slab-air boundary .2 (i) (ii) A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1. Lens of focal length F is split into two halves.0 . Determine the coordinates (x1 . one of the halves is shifted along the optical axis.1 (i) (ii) (iii) (iv) Q.5 (C) + 6. Draw the ray diagram for image formation.Q. y) in the medium and the incident angle at that point. O). Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle.67 (D) − 6. The assembly is in air. The separation between object and image planes is 1. The point of incidence is the origin A (O. Obtain an equation for the trajectory y (x) of the ray in the medium.5 (B) − 1. [JEE ’96] Q. y1) of the point P. [REE ’96] Q. [JEE ’96] Q.0 m−3/2 .4 Which of the following form(s) a virtual & erect image for all positions of the real object ? (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror [JEE ’96] Q.0 (see figure).8 m. Assuming n = 1.67 [JEE '97] 12 Page 12 of 20 GEOMETRICAL OPTICS EXERCISE # III . [JEE ’95] A right angle prism (45° − 90° − 45°) of refractive index n has a plate of refractive index n1 (n1 < n) cemented to its diagonal face. a ray is incident on AB (see the figure) . the image will be: (A) real. The ray undergoes total internal reflection . the wavelength of (i) (ii) Q.8 m . The angles of the prisms are as shown . Find the position (relative to the lens) of the image of the object formed by the system. find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum . as shown in figure . The line PQ cuts the surface at a point O and PO = OQ . also located at C . n1 & n2 depend on λ.8 A prism of refractive index n1 & another prism of refractive index n2 are stuck together without a gap as shown in the figure.6 (iii) A spherical surface of radius of curvature R separates air (refractive index 1.5 (D) 1. Find the maximum value of ratio 13 Page 13 of 20 GEOMETRICAL OPTICS (ii) . & located at a point between C & ∞ (C) virtual. Let O be the pole of the mirror & C its centre of curvature .5) and of square cross-section is bent into the shape shown in figure. A small object is placed outside the tank in front of the lens at a distance of 0. [REE '98] semicircle. Referring to the diagram.45 + λ2 λ2 where λ is in nm . with its axis directed vertically upwards.3 (B) 1. select the possible value(s) of n from the following : (A) 1. a mirror is placed inside the tank on the tank wall perpendicular to the lens axis. A parallel beam of light falls perpendicularly on the plane flat surface A.0) from glass (refractive index 1. If n is the refractive index of the medium with respect to air. (ii) A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence of 45º . If the mirror is now filled with water. [JEE ' 97] Q. & located at a point between C & O (D) real. A point object P placed in air is found to have a real image Q in the glass .7 (i) Select the correct alternative(s) : [JEE '98] A concave mirror is placed on a horizontal table. Calculate the wavelength λ0 for which rays incident at any angle on the interface BC pass through without bending at that interface .8 × 104 1.4 (C) 1. For light of wavelength λ0.20 + & n2 = 1. [JEE '98] A rod made of glass (µ = 1. The centre of curvature is in the glass . & will remain at C (B) real. A point object is placed at C .80 × 104 light according to n1 = 1. & located at a point between C & O .9 10. d is the width of a side & R is the radius of inner d so that all light entering the glass R through surface A emerge from the glass through surface B. The separation between the lens and the mirror is 0.5 R (A) 5 R (B) 3 R Q. On the opposite side of the lens.A thin equiconvex lens of glass of refractive index µ=3/2 & of focal length 0.5) . The distance PO is equal to : (C) 2 R (D) 1.9 m from the lens along its axis .3 m in air is sealed into an opening at one end of a tank filled with water (µ = 4/3). It has a real image. 4.75. A ray of light is incident at the surface AB of the slab as shown. find the location of its image while the whole system remains inside the liquid.5 R (D) divergent lens of focal length 3. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. (C) divergent lens of focal length 3.6 respectively.10 A concave lens of glass. [JEE '2000 (Scr)] (a) A diverging beam of light from a point source S having divergence angle α. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown.14 Select the correct alternative. falls symmetrically on a glass slab as shown.13 Two symmetric double-convex lenses L1 and L2 with their radii of curvature 0. [JEE ’99] Q. If the thickness of the glass slab is t and the refractive index n. such that the ray comes out only from the other surface CD is given by −1 n 2 −1 n1 (A) sin n cos sin n 1 2 1 (B) sin −1 n1 cos sin −1 n 2 n (C) sin −1 1 n2 n (D) sin −1 2 n1 A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall. The greatest distance over which he can see the image of the light source in the mirror is (A) d/2 (B) d (C) 2d (D) 3d 14 Page 14 of 20 GEOMETRICAL OPTICS Q.5 is placed on a table.5.3m from L1. has both surfaces of same radius of curvature R. The lenses with a separation of 0. is immersed in water of refractive index n2(n1> n2). [REE ’99] Q. refractive index 1.11 The x-y plane is the boundary between two transparent media. Find the unit vector in the direction of refracted ray in medium -2. The maximum value of the angle of incidence αmax. On immersion in a medium of refractive index 1. Find the focal lengths of lens L1 and L2. of refractive index n1. Medium-1 with z > 0 has refractive index 2 and medium – 2 with z < 0 has a refractive index 3 .5R (B) convergent lens of focal length 3. The angles of incidence of the two extreme rays are equal.0 R . [JEE '99] Q.2m each are made from glasses with refractive index 1.12 A quarter cylinder of radius R and refractive index 1. then the divergence angle of the emergent beam is (A) zero (B) α (C) sin−1(1/n) (D) 2sin−1(1/n) (b) (c) A rectangular glass slab ABCD.2 and 1.Q. A point object P is kept at a distance of mR from it. A ray of light in medium –1 given by the vector A = 6 3 ˆi + 8 3 ˆj − 10 kˆ is incident on the plane of separation.0 R. it will behave as a [JEE ’99] (A) convergent lens of focal length 3.345 m are submerged in a transparent liquid medium with a refractive index of 1. An object is placed at a distance of 1. (B) air and immersed in L1. The distance between the lens and mirror is 30 cm. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light.51 and 1. On repeating with another liquid. An isosceles prism of angle 6° is made of crown glass. [JEE 2001] Q. If A′ B′ is the image after refraction from the lens and reflection from the mirror.16 A thin equi biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the figure.18 An observer can see through a pin-hole the top end of a thin rod of height h.19 Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams. he can see the lower end of the rod. [JEE 2001] Q. Also locate positions of A′ and B ′ with respect to the [JEE 2000] optic axis RS. The lens will diverge a parallel beam of light if it is filled with (A) air and placed in air. (C) L1 and immersed in L2. Q. Calculate the net dispersion of the combined system.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens .73 respectively. the object and the image again coincide at a distance 25cm from the lens. (D) L2 and immersed in L1.17 The refractive indices of the crown glass for blue and red lights are 1. An upright object AB of height 1. Calculate the refractive index of the liquid. Then the refractive index of the liquid is (A) 5/2 (B) 5 / 2 (C) 3 / 2 (D) 3/2 [JEE 2002 (Scr)] Q. find the distance A′ B′ from the pole of the mirror and obtain its magnification. the object coincides with its own image. The beaker height is 3h and its radius h.15 A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0. [JEE 2002 (Scr)] (A) (B) (C) (D) 15 Page 15 of 20 GEOMETRICAL OPTICS (d) . A beam of white light is incident at a small angle on this prism. When the beaker is filled with a liquid up to a height 2h.77 and 1. Q. It is found that when a point object is placed 15cm above the lens on its principal axis. placed as shown in the figure.A hollow double concave lens is made of very thin transparent material. Determine the angle of the flint glass prism. It can be filled with air or either of two liquids L1 or L2 having refractive indices n1 and n2 respectively (n2 > n1 > 1).49 respectively and those of the flint glass are 1.6 cm as shown. The space between the lens and the mirror is then filled with water of refractive index 4/3. 25 (D) None Q.25 A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1. red (B) violet. Q. Both its surfaces have radii of curvature R.20 Two plane mirrors A and B are aligned parallel to each other.Q. Calculate its focal length for µ1 < µ2 < µ3.28 The ratio of powers of a thin convex and thin concave lens is 16 Page 16 of 20 GEOMETRICAL OPTICS Q.23 White light is incident on the interface of glass and air as shown in the figure.22 A meniscus lens is made of a material of refractive index µ2. For minimum deviation which of the following is true ? [JEE 2004 (Scr)] (A) PQ is horizontal (B) QR is horizontal (C) RS is horizontal (D) Either PQ or RS is horizontal.26 Figure shows an irregular block of material of refractive index 2 .5. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of 0. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is [JEE 2002 (Scr)] (A) 28 (B) 30 (C) 32 (D) 34 . – 50 (B) 75. A light ray is incident at an angle of 30° at a point just inside one end of A. blue (C) all colours (D) all coloure except green [JEE 2004 (Scr)] Q.27 An object is approaching a thin convex lens of focal length 0. [JEE 2004] Q. as shown in the figure.3 m with a speed of 0.21 A convex lens of focal length 30 cm forms an image of height 2 cm for an object situated at infinity. – 15 (D) – 75. If a convcave lens of focal length 20 cm is placed coaxially at a distance of 26 cm in front of convex lens then size image would be [JEE 2003 (Scr)] (A) 2. It has two different media of refractive indices µ1 and µ3 respectively. After refraction it is incident on a spherical surface CD of radius of curvature 0. Find the distance OE upto two places of decimal. [JEE 2004] 3 and equivalent focal length of their 2 combination is 30 cm. orange.24 A ray of light is incident on an equilateral glass prism placed on a horizontal table.01 m/s. If green light is just totally internally reflected then the emerging ray in air contains (A) yellow.4 m and enters a medium of refractive index 1. The distance of the virtual image from the surface of the sphere is [JEE 2004 (Scr)] (A) 2 cm (B) 4 cm (C) 6 cm (D) 12 cm Q. when light is incident on it as shown. [JEE 2003] Q. indigo. 50 Q.4 m from the lens.514 to meet PQ at E. 50 (C) 10.5 cm (B) 5. A ray of light strikes the face AB as shown in the figure. The plane of incidence coincides with the plane of the figure. on its two sides (see figure). Then their focal lengths respectively are [JEE' 2005 (Scr)] (A) 75.0 (C) 1. 50 ± 0. [JEE 2005] Q. (in cm) : (A) 10 (B) 15 (C) 20 (D) 25 [JEE' 2005 (Scr)] .29 Figure shows object O.10 cm (C) 5. (a) the angle of incidence.32 A point object is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The image will form at (A) 60 cm left of AB (B) 30 cm left of AB (C) 12 cm left of AB (D) 60 cm right of AB [JEE 2006] Q.05 cm (D) 5. so that the emergent ray from the first prism has minimum deviation.00 ± 0. A light ray strikes the first prism at face AB. respectively. (D) If lower half of the lens is covered with black paper area of image will become half.33 Graph of position of image vs position of point object from a convex lens is shown. Then which of the following statement is correct? (A) Area of image πr2 directly proportional to f (B) Area of image πr2 directly proportional to f2 (C) Intensity of image increases if f is increased.31 Two identical prisms of refractive index 3 are kept as shown in the figure. Then [JEE 2006] Column 1 Column 2 (A) Intensity of light received by lens (P) Radius of aperture (R) (B) Angular magnification (Q) Dispersion of lens (C) Length of telescope (R) focal length f0.30 What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces? [JEE 2005] Q. focal length of the lens is (A) 0.10 cm [JEE 2006] Q.Q. Final image I is formed after two refractions and one reflection is also shown in figure. [JEE 2006] Q. (b) through what angle the prism DCE should be rotated about C so that the final emergent ray also has minimum deviation. if the plane surface is silvered. Then. Find the focal length of mirror. fe (D) Sharpness of image (S) spherical aberration 17 Page 17 of 20 GEOMETRICAL OPTICS Q.50 ± 0.05 cm (B) 0.34 Parallel rays of light from Sun falls on a biconvex lens of focal length f and the circular image of radius r is formed on the focal plane of the lens.35 A simple telescope used to view distant objects has eyepiece and objective lens of focal lengths fe and f0. Find.00 ± 0. 2 m . 10.8 Q.9 1 sin −1 3 Q.17 µ = Q.2 (i) sin−1 Q.18 5 cm Q.6 Q.EXERCISE # I Q.5 On the object itself Q. separation = 0. 75/2 cm Q.17 4/3 Q.9 1.4 tan–1 2 7 with the principal axis.3 f = 0.r=450 ) Q.2 µ = 3.15 5.23 m 2 cm Q.10 8/5 cm/s Q.12 sin–1 2 41 4 15 cm Q.18 (i) 2° .5 1.1 (a) tan θ = dy dx 4 = cot i x (b) y = k (c) 4.2 Q.11 Q. 1 4 1 2 2 n − n1 − n1 2 Q.15 Q.4 m Q.6 m 2 (d) It will become parallel to x-axis (ii) r1 = sin−1 (n sin 45º) = 72.20 11 cm Q.10 42 cm Q.13 l= (3f − 2d )fd 4fd − 2d 2 − f 2 Q.13 160cm.19 Q.25 75/4 cm.16 3 3 Q.1 Q. n = 1.24 3 cm Q.5 16 feet Q.7 (i) D.9 Q.2 Q.75) = 48. A = 2° 8 Q.9 cm.9 cm 13 .5 or ( 5 −1) Q.12 + 60.28 45° Q. + 4/5 4π mm 9 EXERCISE # III Q.14 h = l Q.22 (5f. (b) 1 × 10–3 /sec Q.7 Q.8 15 cm towards the combination Q.4 1000 m/s 42 cm 3 −1 Q.10. (ii) C.30 5 2 EXERCISE # II (a) 21 m/s.26 (i) 0.11 Q. C Q.59º 3 ˆ 2 2ˆ 1 ˆ i+ j− k(angleof incidence=600 .5 8 Q.(ii) 0.21 43 5 Q.0.6 (i) B.3 d/2 75 cm Q.(ii) Q.27 1. 320cm 80 m/s Q. (ii) 90 cm from the lens towards right Q.29 Q.23 (π/4) cm2 b(1 − µ 2 cos 2 θ)1 / 2 sin θ Q. sin–1(1/3) 6 26 cm Q.12 m = 4/3 Q.4 m. (iii) A Q.7 apq (q − p ) Q.10 A Page 18 of 20 GEOMETRICAL OPTICS ANSWER KEY .94º Q. 2d) Q.5 (ii) i = sin−1 (0. D.11 r = 5 5 2 2 18 1 r = 2 R max Q. cm/sec 3 4 Q.14 10cm.4 B.8 (i) λ 0 = 600 nm. 14 (a) B (b) A (c) D (d) D Q.5 cm below RS.06 m correct upto two places of decimal.29 C Q. (b) 60° (anticlockwise) Q.21 A Q.17 40 and –0.19 C Q.1 Q.23 A Q.Q.6 Q.16 1. Magnification is 1.20 B Q.32 C Q.31 (a) i = 60°.25 C Q. V= 560 cm to the right of L2 .09 m/s . (D) P. 0.34 B Q. (C) R. Magnitude of the rate of change of lateral magnification is 0.24 B 1. S 19 Page 19 of 20 GEOMETRICAL OPTICS Q.4 = 6.33 C Q.15 A′ B′ at 15 cm to the right of mirror .35 (A) P .040 Q.26 Q. (B) R.5 Q.27 0. B′ is 0. Q.28 C Q. Q. f2 = 70cm.3 cm above RS and A′ is 1.22 f =v= Q.514 × 0.18 B µ 3R µ 3 − µ1 Q.13 f1 = -70cm.30 60° Q.3 s–1. Answer Key 7. Optical Instruments Index: 1. from AIEEE 1 . from IIT-JEE 8. Key Concepts 2.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P8. 34 Yrs. Que. Exercise IV 6. Que. 10 Yrs. Exercise I 3. Exercise III 5. Exercise II 4. the distance between the objective and eyepiece in cm is (A) 110 (B) 55 (C) 50 (D) 45 Q. Eyepiece is movable and image is always needed at 24 cm from the eye.Q.5 Which of the following statement(s) about a simple telescope (astronomical) is/are true (A) the objective lens forms a real image. The distance between the foci (which are between the lenses) of objective and eyepiece is 18 cm. The total magnification of the microscope is(Consider normal adjustment and take D = 25 cm) (A) 562.8 The magnifying power of a telescope in normal adjustment can be increased (A) by increasing focal lengths of both lenses equally (B) by fitting eyepiece of high power (C) by fitting eyepiece of low power (D) by increasing the distance of object 2 Page 2 of 8 OPTICAL INSTRUMENTS EXERCISE–I .7 An astronomical telescope has an eyepiece of focal-length 5 cm. Find the minimum and maximum magnification which can be produced by the microscope.5 Q.5 (B) 625 (C) 265 (D) 62.1 A distant object is viewed with a relaxed eye with the help of a small Galilean telescope having an objective of focal length 15 cm and an eye piece of focal length 3 cm (A) The distance between the objective and the eyepiece lens is 12 cm.8cm between lenses. Q.3 A distant object is viewed with a relaxed eye with the help of a small Galilean telescope having an objective of focal length 12 cm and an eyepiece of focal length -3 cm.0cm and 6cm respectively. (C) the maximum magnification is 30 and corresponds to the separation 9. Q.8cm between lenses. Focal length of the objective and the eyepiece are 1. (B) The distance between objective and eyepiece lens is 15 cm. If the angular magnification in normal adjustment is 10.8cm between lenses. (D) the maximum magnification is 30 and corresponds to the separation 11. (A) The distance between objective and eyepiece lens is 9 cm. (B) The angular magnification of object is 5 (C) Image of the object is erect (D) The distance between objective and eye piece lens is 18 cm Q. (C) The image of the object is inverted (D) The angular magnification of the object is + 4.8 cm to 11.2 A microscope consists of an objective with a focal length 2 mm and an eye piece with a focal length 40 mm. D = 24cm. The distance between the two lenses in normal adjustment will be (A) 150 cm (B) 100 cm (C) 98 cm (D) 200 cm Q.6 The separation between the objective and the eye piece of a compound microscope can be adjusted between 9.8 cm.4 A Galileo telescope has an objective of focal length 100 cm & magnifying power 50 . (B) The eyepiece acts as a magnifying glass (C) the focal length of the objective lens is short (D) the final image is inverted Q. (B) the minimum magnification is 20 and corresponds to the separation 11.8cm between lenses. (A) the minimum magnification is 20 and corresponds to the separation 9. (B) The tube lengths of the two telescopes differ by 2f.14 In an astronomical telescope in normal adjustment. real image of S will be formed behind the eyepiece.5 m Q. They have the same magnification. if the tube length is increased slightly from its position of normal adjustment (A) a virtual image of S will be formed at a finite distance (B) no image will be formed (C) a small.5 m (B) convex lens with f = 2. The magnifying power of the telescope is (A) L l (B) L +1 l (C) L –1 l (D) L +1 L −1 Q. real image of S will be formed behind the eyepiece. close to it (D) a large. The lens he is using is (A) concave lens with f = 0. (A) The tube lengths of the two telescope differ by f. real image of S will be formed behind the eyepiece.15 An astronomical telescope and a Galilean telescope use identical objective lenses. In the position of maximum angular magnification. (D) the Galilean telescope has longer tube length. close to it. far away from it.13 In the previous question. Q. The length of this image is l.2 m (D) convex lens with f = 0.16 A single converging lens used as a simple microscope. (B) No image will be formed (C) A small.11 An astronomical telescope has an angular magnification of magnitude 5 for distant objects.A person with a defective sight is using a lens having a power of +2D. (D) A large. The eyepiece forms a real image of this line. Q. (B) the image formed by the objective is real. far away from it. 3 Page 3 of 8 OPTICAL INSTRUMENTS Q. real image of S will be formed behind the eyepiece. Q. The separation between the objective and the eyepiece is 36 cm.12 An astronomical telescope in normal adjustment receives light from a distant source S. (C) The Galilean telescope has shorter tube length. The focal length fo of the objective and fe of the eyepiece are (B) 50 cm and 10 cm respectively (A) 45 cm and – 9 cm respectively (C) 7. (A) the object is placed at the focus of the lens (B) the object is placed between the lens and its focus (C) the image is formed at infinity (D) the object and the image subtend the same angle at the eye. (D) none of the above Q.2 cm and 5 cm respectively (D) 30 cm and 6 cm respectively Q. a straight black line of length L is drawn on the objective lens.10 In a compound microscope (A) the object is held slightly beyond the focal point of the objective. when both are in normal adjustment.9 . (C) the image formed by the eye piece is virtual. The tube length is now decreased slightly (A) A virtual image of S will be formed at a finite distance. The eyepiece of the astronomical telescope has a focal length f. The Final image is formed at infinity.0 m (C) concave lens with f = 0. If the object subtends an angle of 2° at the objective.21 In a reflecting astronomical telescope. erect and magnified (C) real.17 When an astronomical telescope is in normal adjustment.5 (C) 1 (D) can't see 4 Page 4 of 8 OPTICAL INSTRUMENTS Q. is focused on a distant object in such a way that parallel rays emerge from the eye lens. the magnification produced by it M.19 When length of a microscope tube increases. he puts a simple microscope of magnifying power 5X is normal adjustment before his eyes.20 In a compound microscope. the magnification produced will be . its magnifying power (A) decreases (B) increases (C) does not change (D) may increases or decreases Q. erect and reduced Q. then (A) The final image will be erect (B) The larger image will be obtained (C) The telescope will gather more light (D) Spherical aberration will be absent Q. if the objective (a spherical mirror) is replaced by a parabolic mirror of the same focal length and aperture.24 A man is looking at a small object placed at near point. the angular magnification achieved is : (A) 5 (B) 2. erect and magnified (B) real. Without altering the position of his eye or the object. If this is now turned around with the eyepiece facing a distant object and the eye placed close to the objective. inverted and magnified (D) virtual. consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm. the angular width of the image is (A) 10° (B) 24° (C) 50° (D) 1/6° Q. Q.(A) 1 M (B) 1 M +1 (C) 1 M −1 (D) M −1 M +1 Q.18 In a simple microscope.22 A simple telescope. if the final image is located at infinity then its magnifying power is (A) 25/F (B) 25/D (C) F/25 (D) (1 + 25/F) Q. the intermediate image is (A) virtual.23 A man wearing glasses of focal length + 1 m cannot clearly see beyond 1 m : (A) if he is farsighted (B) if he is nearsighted (C) if his vision is normal (D) in each of these cases. 1 mm from the objective. The last image is formed at a distance of 25 cm from the eyepiece. Also find the separation of the lens. The focal length objective is ______ & that of ocular is ______ .Q.2 A telescope has an objective of focal length one meter and adjustable eyepiece. and are kept at a distance of 20 cm. Q. Q. Q.02 m. Eyepiece of the same telescope consists of two plano convex lenses each of focal length f separated by 2/3 f as shown in the figure.95 cm and 5 cm respectively. The final image is formed at 25 cm from the eye. find out the distance by which the eyepiece of the microscope must be moved to refocus the image.7 The focal lengths of the objective and the eyepiece of a compound microscope are 2.8 In a compound microscope the objective and the eyepiece have focal lengths of 0. from the objective lens. Find the value of l for which final image will be formed at infinity with its angular magnification 100/3.03 m from its objective which consists of several convex lenses in contact and has focal length 0.piece Fe = 5 cm. Q. What should be the separation of the two lenses when the virtual image of a distant object is formed at a distance of 24 cm from the eyepiece? What is the magnifying power of telescope under this condition? Q. What is the minimum separation between two points d of the objects which can now be distinguished.10 If the focal length of the objective and eyepiece of a microscope are 2 cm and 5 cm respectively and the distance between them is 20 cm. (Adjustment at ∞). of the eye.3 An eye can distinguish between two points of an object if they are separated by more than 0. Q.The object is now seen by a compound microscope having 20 D objective and 10 D eyepiece separated by a distance of 20 cm. Also find f. The final image formed by the eyepiece is at infinity.0 cm. Find the magnification of the microscope for a normal eye.0 cm and 3.0 cm respectively. Q.5 A Galilean telescope of angular magnification 10 has the length of 45 cm when adjusted to infinity. what is the distance of the object from the objective when the image seen by the eye is 25 cm from eyepiece? Also find the magnifying power. if the final image is 25 cm from the eye.6 A compound microscope is used to enlarge an object kept at a distance 0.1 The focal length of the objective of a microscope is Fo = 3 mm. Q. Calculate the position of object and the total magnification.4 The objective of an astronomical telescope consists of two thin lenses in contact. An object is at a distance of a = 3.9 A Galilean telescope consists of an objective of focal length 12 cm and eyepiece of focal length 4 cm. 5 Page 5 of 8 OPTICAL INSTRUMENTS EXERCISE–II . of focal lengths +20 cm and – 25 cm respectively. Find the distance of object and image produced by the objective. If a lens of focal length 0.22 mm when the object is placed at 25 cm from the eye . Objective eyepiece → → l 2f/3→ → Q. How much motion must be given to the eye piece to focus an object lying between 5m and infinity. The distance between the objective and the eyepiece is 15.1 m is removed from the objective. are separated by a distance of 1. the telescope is focused to see an object 10 km from the objective. Find the angular magnification achieved due to the converging lens. (b) Find the linear magnification (c) Find the angular magnification. (ii) the angular magnification produced.18 An object is seen through a simple microscope of focal length 12 cm. Q.0 cm and an eyepiece of focal . Q.0 cm (a) Find the position of the image.2 m. Where should a small object be placed so that the eye is least strained to see the image? Find the angular magnification produced by the microscope. The eyepiece and the objective are to be interchanged such that angular magnification of the instrument remains same in normal adjustment. (a) What is the distance of his retina from the eye-lens ? (b) What is his near point ? Q. (i) the separation between the objective and eyepiece.13 A 10 D lens is sued as a magnifier.0 cm.16 The image of the moon if focused by a coverging lens of focal length 50 cm on a plane screen. Q.11 .0 m away from the objective.0 cm.12 The eyepiece and objective of a microscope. of focal lengths 0. (b) If the telescope is to focus an object 2. The telescope is focussed for distinct vision at near point on an object 200 cm away from the objective. the image again forming at infinity? Q.2 cm (a) At what distance from the objective should an object be placed to focus it properly so that the final image is formed at the least distance of clear vision (25 cm) ? (b) Calculate the angular magnification in this case.4 m respectively. what should be the tube length and angular magnification. Q. The distance of distinct vision is 25 cm.15 A Galilean telescope is constructed by an objective of focal length 50 cm and an eyepiece of focal length 5. Q.6 cm from a magnifier of focal l length 4.19 A small object is placed at a distance of 3. length 5. Q.0 cm separated by 12.14 The separation L between the objective (f = 0.17 A young boy can adjust the power of his eye-lens between 50 D and 60 D. The final image is formed at infinity.. The image is seen by an unaided eye from a distance of 25 cm. His far point is infinity. Where should the object be placed to obtain maximum angular magnification for a normal eye (near point = 25 cm)? Q. (a) Find the tube length and magnifying power when it is used to see an object at large distance in normal adjustment.20 A compound microscope consists of an objective of focal length 1.21 An astronomical telescope has an objective of focal length 200 cm and an eyepiece of focal length 4. What is the new separation between the lenses? Q. Find the angular magnification produced if the image if formed at the near point of the eye which is 25 cm away from it. Find the length of the tube and the angular magnification produced by the telescope. 6 Page 6 of 8 OPTICAL INSTRUMENTS Q.5 cm) and the eyepiece (f = 5 cm) of a compound microscope is 7 cm.A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm.3 m and 0. Calculate. 5 cm and the tube length is 6. What power of lens should he now use ? Q.5 cm. Q. The objective has a focal length of 0. what is the range of the power of the eye-lens.28 A professor reads a greeting card received on his 50th birthday with + 2. she is lost from her team. Q. Ten years later.5 D glasses keeping the card 25 cm away. If the retina is 2. he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. The telescope is focused for normal vision of distant objects when the tube length is 1.29 The near point and the far point of a child are at 10 cm.23 A simple microscope is rated 5 X for a normal relaxed eye. Find the focal length of the objective an d the magnifying power of the telescope.27 The eyepiece of an astronomical telescope has a focal length of 10 cm. While on a mountaineering trip.22 the near and for points of a person are at 40 cm and 250 cm respectively.26 A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. Find the focal length of the eyepiece.25 A compound microscope has a magnifying power of 100 when the image is formed at infinity. what maximum distance is clearly visible ? . She tries to make an astronomical telescope from her reading glasses to look for her teammates.30 A lady cannot see objects closer than 40 cm from the left eye and closer than 100 cm from the right eye. Q. What should be the separation between the lenses so that the microscope projects an inverted real image of the object in a screen 30 cm behind the eyepiece ? Q. Find the power of the lens he /she should use while reading at 25 cm.24 Find the maximum magnifying power of a compound microscope having a 25 diopter lens as the objective.Q.0 m. a 5 diopter lens as the eyepiece and the separation 30 cm between the two lenses. and 100 cm. Q. the least distance for clear vision is 25 cm. What will be its magnifying power for a relaxed farsighted eye whose near point is 40 cm ? Q.0 cm behind the eye-lens. an object is placed at a distance of 0.5 cm from the objective. With this lens on the eye. (a) Which glass should she use as the eyepiece ? (b) What magnification can she get with relaxed eye ? 7 Page 7 of 8 OPTICAL INSTRUMENTS Q. EXERCISE–I Q.1 A, B, C Q.2 A Q.3 A, D Q.4 C Q.5 A, B, D Q.6 A, D Q.7 B Q.8 B Q.9 D Q.10 A, B, C Q.11 D Q.12 A Q.13 D Q.14 A Q.15 B, C Q.16 B, D Q.17 A Q.18 A Q.19 D Q.20 C Q.21 D Q.22 B Q.23 D Q.24 D EXERCISE–II Q.1 -180,13.46 cm Q.2 25 cm Q.3 0.04 mm Q.4 101 cm , f = 4 cm Q.5 50, – 5 cm Q.6 9 cm Q.7 12 cm Q.8 – 95/94 cm, – 94 cm Q.9 L = 7.2 cm, M = 2.5 Q.11 (i) 70.80, (ii) 2 Q.12 1.6 m Q.10 – 190 cm , – 41.5 83 Q.15 (a) 10, (b) 185 5 cm, 3 3 Q.13 7.1 Q.14 – 15 Q.16 – 2 Q.17 (a) 2 cm, (b) 10 cm Q.19 7.0 Q.20 (a) – Q.22 -53 cm Q.23 8 X Q.24 Q.25 2 cm Q.26 5 cm Q.27 90 cm, 9 Q.28 + 4.5 D Q.29 + 60 D to + 51 D Q.30 right lens, 2 241 cm , (b) 42.2 211 8 Q.18 3.08 Q.21 −50 67 8 Page 8 of 8 OPTICAL INSTRUMENTS ANSWER KEY STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P9. Wave Optics Index: 1. Key Concepts 2. Exercise I 3. Exercise II 4. Exercise III 5. Exercise IV 6. Answer Key 7. 34 Yrs. Que. from IIT-JEE 8. 10 Yrs. Que. from AIEEE 1 1. (i) (ii) (iii) If two coherent waves with intensity I1 and I2 are superimposed with a phase difference of φ, the resulting wave intensity is I = I1 + I2 + 2 I1I 2 cos φ For maxima, optical path difference = nλ [optical path = µ (geometrical path)] 1 1 For minima, optical path difference = (n – )λ or (n + )λ 2 2 2π (optical path difference) Phase difference φ = λ 2. The phase difference between two waves at a point will depend upon (i) (ii) (iii) (iv) the difference in path lengths of two waves from their respective sources.( geometrical path difference) the refractive index of the medium (media) phase difference at source (if any). In case, the waves suffer reflection, the reflected wave differs in phase by π with respect to the incident wave if the incidence occurs in rarer medium. There would be no phase difference if incidence occures in denser medium. 3. Young's Double Slit Experiment (i) If d << D ∆x = S2P – S1P = d sin θ If λ << d then sin θ ≈ θ ≈ tan θ as when P is close to D so θ is small. dy ∆x = D dy For maxima = nλ D Dλ 2 Dλ or y = 0, ± ,± ,± d d dy For minima = [n + (1/2)]λ D Dλ 3Dλ or y = ± ,± , ±, so on 2d 2d λD Fringe width, β = d (ii) (iii) (iv) 4. Displacement of fringe Pattern When a film of thickness 't' and refractive index 'µ' is introduced in the path of one of the source's of light, then fringe shift occurs as the optical path difference changes. Optical path difference at P. ∆x = S2P – [S1P + µt – t] = S2P – S1P (µ – 1) t = y. (d/D) – (µ – 1)t ⇒ The fringe shift is given by ∆y = D(µ − 1) t d 2 Page 2 of 12 WAVE OPTICS KEY CONCEPT where φ = λ 2 Interference at thin film optical path difference = 2µt cos r = 2µt (in case of near normal incidence) For interference in reflected light Condition of minima 2µt cos r = nλ 1 Condition of maxima 2µt cos r = n + λ 2 3 Page 3 of 12 WAVE OPTICS 5. the resultant intensity at a point where they have a phase difference of φ is I = 4I0 cos2 6. .Intensity Variation on Screen If I0 is the intensity of light beam coming from each slit. (i) (ii) 2π(d sin θ) φ . Q. Q. The rays AB and A′ B′ undergo interference.2 cm apart and are illuminated by yellow light (λ = 600 nm).11 The distance between two slits in a YDSE apparatus is 3mm. the intensity at the position where the central maximum occurred previously remains unchanged. find the minimum thickness of the glass plate. If the micasheet is removed from earlier position and placed in front of S2 find the number of fringes crossing O. where central bright fringe was formed before the introduction of the glass sheets.2 In Young’s double slit experiment.8 Light of wavelength 520 nm passing through a double slit. Find the minimum distance between two points on the screen having 75% intensity of the maximum intensity.5 is introduced. The distance of the screen from the slits is 1m. find the minimum value of d/λ for which there is destructive interference with almost zero resultant intensity at points on the x-axis having x > > d. 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. Q. Q. d = 1mm.1 In a Young's double slit experiment for interference of light.12 The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of 4th bright fringe after a glass sheet of refractive index 1. Find the wavelength of the light used. Assuming that both the glass plates have same thickness and wavelength of light used is 4800 Å. Q. introduced in front of S1.3 In the ideal double slit experiment. find the separation d between the slits. It is found that the 9th bright fringe is at a distance of 7.6 A ray of light of intensity I is incident on a parallel glass-slab at a point A as shown in figure. It undergoes partial reflection and refraction.4 One slit of a double slit experiment is covered by a thin glass plate of refractive index 1. Q. when a glass plate (refractive index 1.Q. The slits produce same intensity on the screen. Q. is now occupied by the 5th bright fringe. 4 Page 4 of 12 WAVE OPTICS EXERCISE – I .4 and the other by a thin glass plate of refractive index 1.7.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ). What would be the fringe width on a screen placed 1 m from the plane of slits if the whole system is immersed in water of index 4/3? Q. Q. the slits are 0. find their thickness. Microwaves of wavelength 1 mm are incident on the plane of the slits normally. Find the thickness of glass sheet. At each reflection 20% of incident energy is reflected. 0 and d.10 In a YDSE apparatus. Q.5 mm apart and the interference is observed on a screen at a distance of 100 cm from the slit. If the wavelength of the light is changed to 400 nm. find the number of fringes observed in the same segment.9 In Young's double slit experiment the slits are 0.7 The figure shows the Young's double slit experiment with a mica sheet of thickness t and refractive index µ.5 Three identical monochromatic points sources of light emit light of wavelength λ coherently and in phase with each other. Find the distance of the first maxima on the screen from the central maxima.5 mm from the second dark fringe from the centre of the fringe pattern on same side. The point on the screen. λ = 600nm and D = 1m. Q. produces interference pattern of relative intensity versus deflection angle θ as shown in the figure. They are placed on the x-axis at the points x = – d. Find the ratio Imax/Imin. If the screen is moved by 5 × 10−2 m towards the slits. 1 5 A m b r o a d e e t a t b r i g h t Q . List of recommended questions from I. The interference pattern produced by the slit and its image is viewed on a screen distant 1m from the slit. The wavelength of light is 600nm.19 A monochromatic light of λ = 5000 Å is incident on two slits separated by a distance of 5 × 10−4 m . 1 and S2 separated by distance 2λ emit light of wavelength λ in phase as shown in figure.18 In a two − slit experiment with monochromatic light. Find the intensity at the centre of the screen.Find the angular positions θ on the wire for which intensity reduces to half of its maximum value. i n d F 2 0 m t h e m l o n g n u m t h a t b e r o f d i s t a n c e .17 In a biprism experiment with sodium light. Find the thickness of the plates. before the plates were introduced is now occupied by the third bright fringe. Q.13 A lens (µ =1.2(a) A thin glass plate of thickness t and refractive index µ is inserted between screen & one of the slits in a Young's experiment.7) of the same thickness. If the distance between the slits is 10−3 m.70.Q .7 cm apart at 100 cm distance from the slit. assume l >> d. 1 6 T w o s o u r c e o n e o f e n d f r i n g e s c o h e r e n t l i g h t a n d f o r m s o a r e e d o f o v e r u r c e s S w a v e l e n g t h s e p a r a t e d t h e 1 b y 2 0 m a m 6 8 w 0 n m i r e 0 i l l u m . between the screen and the slits. D >> d). the change in fringe width is 3 × 10−5 . A circular wire of radius 100λ is placed in such a way that S1S2 lies in its plane and the midpoint of S1S2 is at the centre of wire. Assume y >> d & d >> λ. 5. between the slits & the source S. 5 Page 5 of 12 WAVE OPTICS Q. 0 4 8 m i n a t e s m i n n o r m d i a m a l l y e t e r a t t w o g l a s s p l a t e s 1 t h e o t h e r e n d . 5. Calculate the wavelength of sodium light. Irodov.74 to 77.80 EXERCISE – II Q.72. In this case find the minimum distance between the points on the screen where the intensity is half the maximum intensity on the screen. Find the minimum thickness of the film which will minimize the intensity of the reflected light. 5. What will be the phase difference at P if a liquid of refraction index µ is filled .5 × 10−6 m & refractive index µ = 1.65. Q.4) and the other by another glass plate (µ2 = 1. O Q.79. bands of width 0. 5.2 in order to reduce the reflection from its surface at λ = 4800 Å.1 If the slits of the double slit were moved symmetrically apart with relative velocity v.0195 cm are observed at 100 cm from slit. two images of the slit are seen 0. if the intensity there is I0 in the absence of the plate . (wavelength of light in air is λ due to the source.14 A long narrow horizontal slit lies 1mm above a plane mirror.E.69. . 5.71(a) & (b).5 is placed between one of the slits & the screen. the wavelength of light used is 4000 Å. calculate the rate at which fringes pass a point at a distance x from the centre of the fringe system formed on a screen y distance away from the double slits if wavelength of light is λ.67. Q. The interference pattern is seen on a screen placed at a distance of 1 m from the slits . Q.5) is coated with a thin film of refractive index 1. fringes are obtained on a screen placed at some distance from the slits . Find the distance of first maximum above the mirror. The point of central maxima on the screen. 5. If the intensity at the centre of the screen is I. what was the intensity at the same point prior to the introduction of the sheet. 5. calculate the wavelength of the light used. On introducing a convex lens 30 cm away from the slit between biprism and screen. 5. 5.3 (i) (ii) n e s l i t o f a Y o u n g 's e x p e r i m e n t i s c o v e r e d b y a g l a s s p l a t e ( A source S is kept directly behind the slit S1 in a double-slit apparatus. (b) µ1 = 1. Q. A thin glass plate of thickness 1. Also find the lateral shift of the central maximum. The central maxima is above O. (vi) intensity at 4mm from 0 upwards. Q. The other end of the spring is fixed to ground. The interference is observed on the screen.2 mm. It absorbs 50% light energy and transmits the remaining.I. µw = 4/3) 6 Page 6 of 12 WAVE OPTICS Q. 3/2 is placed infront of s1. In the final position.8 A central portion with a width of d = 0. Find the x & y co-ordinates of the nth maxima on the plate as a function of time.4 µm. A monochromatic light beam of wavelength λ = 5000 Å is incident normally on the diaphragm. the fringe width decreased to n = 2/3 of the original value. the distance between the slits & the screen is 100 cm . Q. a thin glass plate of refractive index 1. Assume that spring is light & plate always remains horizontal. The distance between the slits is d (>> λ). Q. Thickness of A is 20.25 mm. (Refractive index of water. For a certain distance between the slits.5 mm is cut out of a convergent lens having a focal length of 20 cm. calculate ratio of intensity at C to maximum intensity of interference pattern obtained on the screen.6 In a YDSE experiment. where C is foot of perpendicular bisector of S1S2.5.5 is kept in front of one of the slits & the shift of central maximum is observed to be 20 fringe width. Point M is the mid point of the line S1S2 & this point is considered as the origin.9 A screen is at a distance D = 80 cm from a diaphragm having two narrow slits S1 and S2 which are d = 2 mm apart. Water is filled in space between diaphragm and screen. A thin transparent film of thickness 4µm and R. . Light coming through A & B have intensities I & 4I respectively on the screen. Both halves are tightly fitted against each other and a point source of monochromatic light (A = 2500Å) is placed in front of the lens at a distance 10 cm. The widths of s1 and s2 are w and 2w respectively. The interference pattern is observed on a horizontal plate (acting as screen) of mass M. A & B are two thin transparent films each of refractive index 1.40. When the distance between the slits is increased by ∆d = 1. Assuming intensity of beam to be uniform and slits of equal width. (a) What is the maximum thickness of B to do so.7 A plane wave of mono chromatic light of wavelength 6000Å is incident on the plane of two slits s1 and s 2 at angle of incidence α = (1. The plate is left from rest from its initial position .8/ π)0.5 In a YDSE a parallel beam of light of wavelength 6000 Å is incident on slits at angle of incidence 30º. which is attached toone end of a vertical spring of spring constant K.5 µm and S2 by another sheet of thickness t2 = 1. Intensity at point O which is symmetric relative to the slits is 3 I . If the maximum intensity on the screen is I then find (i) intensity at 0 (ii) Minimum intensity (iii) fringe width (iv) Distance of nearest maxima from 0 (v) Distance of central maxima from 0. Assuming thickness of B to be that found in part (a) answer the following parts. Find the maximum possible number of interference bands that can be observed on the screen. The slits are in horizontal plane.Q. maximum intensity & minimum intensity on screen. Both sheets are made of same material having refractive index µ = 1. (c) Distance of nearest minima from O. (d) Intensity at 5 cm on either side of O. Point O is equidistant from s1 and s2.25 µm as shown in figure.4 Two slits S1 & S 2 on the x − axis & symmetric with respect to y-axis are illuminated by a parallel monochromatic light beam of wavelength λ . Slit S1 is covered by a transparent sheet of thickness t1 = 2. an interference pattern is observed on the screen with the fringe width 0. Find the thickness of the plate & wavelength of the incident light. (b) Find fringe width. Q. At t = 0 the plate is at a distance D(>>d) below the plane of slits & the spring is in its natural length. Q. are 4I & I respectively.13 Two coherent monochromatic sources A and B emit light of wavelength λ. Each glass plaate reflects 25 % of the light incident on it & transmits the remaining . find the value of b. Separation between the slits is d = 2 mm. The screen at O is normal to SO . Q. If the intensities of the upper & the lower beams immediately after transmission from the face AC. A thin lens of circular shape and focal length 0. find the distance OA. The distance between A and B is d = 4λ. The two halves are placed symmetrically about the central axis SO with a gap of 0. What minimum thickness is required if light of wavelength 600 nm in air reflected from the two sides of the film is to interfere constructively? It is found to be difficult in manufacture and install coatings as thin as calculated in part (a) What is the next greatest thickness for which there will also be constructive interference? Q. A convergent lens is used to bring these transmitted beams into focus. (ii) If the gap between L1 & L2 is reduced from its original value of 0.10 In a Young's double slit experiment a parallel light beam containing wavelength λ1 = 4000Å and λ2 = 5600Å is incident on a diaphragm having two narrow slits. (ii) distance between two consecutive black lines. (i) If a light detector is moved along a line CD parallel to AB. The refractive index of the prism as a function of wavelength is given by the b relation.15 Two parallel beams of light P & Q (separation d) each containing radiations of wavelengths 4000 Å & 5000 Å (which are mutually coherent in each wavelength separately) are incident normally on a prism as shown in figure.11 (a) (b) A plastic film with index of refraction 1.10 m is cut into two identical halves L1 and L2 by a plane passing through a diameter .5 mm. decrease or remain the same ? 7 Page 7 of 12 WAVE OPTICS Q. where λ is in Å & b is a λ positive constant.5 mm . what is the number of maxima observed ? Q.Q. The value of b is such that the condition for total reflection at the face AC is just satisfied for one wavelength & is not satisfied for the other.30 m . find the resultant intensity at the focus. while that from L1 & L2 to O is 1. calculate : (i) distance of first black line from central bright fringe. Another identical glass plate is kept close to the first one & parallel to it .14 Two identical monochromatic light sources A and B intensity10–15W/m2 produce wavelength of light 4000 3 Å. (i) If the third intensity maximum occurs at the point A on the screen. Find the ratio of the minimum & the maximum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate.16 In the figure shown S is a monochromatic point source emitting light of wavelength = 500 nm . The distance along the axis from S to L1 and L2 is 0.12 A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in figure. If distance between diaphragm and screen is D = 40 cm. what is the maximum number of minima observed ? (ii) If the detector is moved along a line BE perpendicular to AB and passing through B.20 + 2 .15 m.Calculate resultant intensity at focal point F of the lens. .80 is put on the surface of a car window to increase the reflectivity and thereby to keep the interior of the car cooler. The glass has a variable refractive index n = 1 + x where x ( in mm) is distance of plate from left to right.60. µ (λ) = 1. A glass of thickness 3mm is placed in the path of the ray as shown in fig. The window glass has index of refraction 1. Q. will the distance OA increase. It is found that the point P on the screen where the central maximum (n = 0) fell before the glass plates were inserted now has 3/4 the original intensity .5 m . It has slit separation of 1 mm & distance between the plane of the slits & screen is 1. The slits are separated by a distance d = 150. [JEE '97(I)] Q. find the y − coordinates of all the interference minima on the screen . the central fringe shifts to a position initially occupied by the 6th bright fringe due to red light . Calculate the maximum distance between the slits so that the fringes are clearly visible. each of frequency 106 Hz.0 m from it. [JEE '98] Q. Find the thickness of the plate . the upper slit is covered by a thin glass plate of refractive index 1. (a) Calculate the fringe width . as shown in the figure. The sources are synchronized to have zero phase difference.5 at this wavelength is put in the path of one of the interfering beams.0 mm. The intensity of microwaves is measured on screen placed parallel to the plane of the slits at a distance of 1.53 . the slits S1 & S2 are illuminated with coherent microwave sources. The intensity I(θ) is measured as a function of θ. Page 8 of 12 WAVE OPTICS EXERCISE – III Q.1 In an interference arrangement similar to Young's double. The screen is placed 1 m from the slits . find the y − coordinates of the first minima on either side of the central maximum. (b) One of the slits of the apparatus is covered by a thin glass sheet of refractive index 1. The student's eye can distinguish two neighbouring fringes if they subtend an angle more than 1 minute of arc. If I0 is the maximum intensity then I(θ) for 0 ≤ θ ≤ 90º is given by : [JEE '95] I0 I0 (A) I(θ) = for θ = 30º (B) I(θ) = for θ = 90º 2 4 (C) I(θ) = I0 for θ = 0º (D) I(θ) is constant for all values of θ . lies below the point P while the 6th minimum lies above P. Calculate the thickness of the glass plate. When a thin glass plate of refractive index 1.6 A coherent parallel beam of microwaves of wavelength λ = 0. the source is red light of wavelength 7 × 10−7 m . (b) If the incident beam makes an angle of 30º with the x − axis (as in the dotted arrow shown in the figure). Find the refractive index of glass for the green light . are observed by a student sitting close to the slits.0 m .4 In Young's experiment. Find the wavelength in the visible region that will have maximum intensity on the screen at 10−3 m from the central maxima.7 In a Young's double slit arrangement. Fringes formed on the screen. Also estimate the change in fringe width due to the change in wavelength . [REE '98] 8 . a source of wavelength 6000 Å is used. When the source is now changed to green light of wavelength 5 × 10−7 m. The slits are illuminated by a parallel beam of light whose wavelength in air is 6300 Å . Q. [REE '96] Q. Find the smallest thickness of the sheet to bring the adjacent minima on the axis.7 . where θ is defined as shown .33 . A light of wavelengths in the range 2000 − 8000 Å is allowed to fall on the slits . The separation between the slits is 1. Using this information calculate the position of 3rd bright & 5th dark fringe from the centre of the screen.5 mm falls on a Young's double slit apparatus. Interference pattern is observed using light of wavelength 5400 Å . It is further observed that what used to be the 5th maximum earlier.4 while the lower slit is covered by another glass plate having the same thickness as the first one but having refractive index 1. [JEE '97 (II)] Q. (a) If the incident beam falls normally on the double slit apparatus.5 In a Young's experiment.3 A double − slit apparatus is immersed in a liquid of refractive index 1. (Absorption of light by glass plate may be neglected) .33 m .slit experiment.Q. Also find the wavelengths that will have maximum intensity at that point of screen in the infra − red as well in the ultra-violet region.2 In YDSE the separation between slits is 2 × 10−3 m where as the distance of screen from the plane of slits is 2. the central bright fringe shifts by 10−3 m to the position previously occupied by the 5th bright fringe . find the wavelengths of the light that form maxima exactly at point O .16 Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. Then.14 A glass plate of refractive index 1.8. (C) the intensity of the maxima decreases and that of the minima increases. Light of wavelength λ travelling in air is incident normally on the layer . y can represent (A) electric field (B) magnetic field (C) displacement Q. A transparent sheet of thickness t = 1. The observed interference fringes from this combination shall be [JEE '99] (A) straight (B) circular (C) equally spaced (D) having fringe spacing which increases as we go outwards. Q.5 ×10–7m is placed over one of the slits.) 2001] (A) 2I (B) 4I (C) 5I (D) 7I 9 Page 9 of 12 WAVE OPTICS Q.45 mm separation. [JEE '99] Q. [JEE '2000] Q. The phase difference between the beams is π/2 at point A and π at point B. one slit is made twice as wide as the other. [REE '2000] Q. (A) the wave intensity remains constant for a plane wave (B) the wave intensity decreases as the inverse of the distance from the source for a spherical wave (C) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave (D) total intensity of the spherical wave over the spherical surface centered at the source remains same at all times. Find the light intensity at point O relative to the maximum fringe intensity. Now. instead of taking slits of equal widths. Obtain the positions of maxima on a circle of large radius lying in the xy-plane and with centre at the origin. The slice is placed on a flat glass plate as shown. in the interference pattern [JEE '(Scr.A young’s double slit experiment is performed using light of wavelength λ = 5000Å. Write the condition for their constructive interference. [All wavelengths in this problem are for the given medium of refractive index 4/3. Where does the central maximum of the interference pattern now appear? [REE ’99] Q.10 In a wave motion y = a sin(kx-ωt). (B) the intensity of the maxima increases and the minima has zero intensity. The refractive index of the material of this sheet is µ = 1.15 Two coherent light sources A and B with separation 2 λ are placed on the x − axis symmetrically about the origin.9 As a wave propagates. They emit light of wavelength λ.12 A thin slice is cut out of a glass cylinder along a plane parallel to its axis. Ignore dispersion] Q. A light of 600 nm wavelength is falling on the slits having 0.5. which emerges in phase from two slits a distance d = 3 ×10–7m apart.5 is coated with a thin layer of thickness t and refractive index 1.11 (a) (b) (c) [JEE '99] (D) pressure The Young's double slit experiment is done in a medium of refractive index 4/3. Q.13 In a double slit experiment.4 µm and refractive index 1. obtain the least value of t for which the rays interfere constructively.) 2000] (A) the intensities of both the maxima and the minima increase.8 .5 m from the slits as shown [JEE'99] Find the location of the central maximum (bright fringe with zero path difference) on the y-axis. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. If λ = 648 nm. if 600 nm light is replaced by white light of range 400 to 700 nm. The lower slit S2 is covered by a thin glass sheet of thickness 10. (D) the intensity of the maxima decreases and the minima has zero intensity. The interference pattern is observed on a screen placed 1. Then the difference between the resultant intensities at A and B is [JEE (Scr.17. 18 A vessel ABCD of 10 cm width has two small slits S1 and S2 sealed with identical glass plates of equal thickness. [JEE'2002] (a) What is the shape of the interference fringes on the screen? (b) Calculate the ratio of the minimum to the maximum intensities in the interference fringes formed near the point P (shown in the figure).6 mm (B) 14 mm (D) 28 mm Q. number of fringes observed in the same segment of the screen is given by [JEE (Scr. 40 cm below P and 2 m from the vessel. A monochromatic light source is kept at S. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP.Q. A thin film (µf = 2. If the wavelength of light is changed to 400 nm.) 2001] (A) 12 (B) 18 (C) 24 (D) 30 . The central bright fringe is found to be at Q. The minimum distance between two successive regions of complete darkness is [JEE' 2004 (Scr)] (A) 4 mm (B) 5. 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. [JEE (Scr. [JEE'2001] Q.23 In a Young's double slit experiment. a liquid is poured into the vessel and filled up to OQ.17 In a young double slit experiment.19 A point source S emitting light of wavelength 600 nm is placed at a very small height h above the flat reflecting surface AB (see figure). the corresponding two rays. the middle point of S1 and S2. (c) If the intensities at point P corresponds to a maximum.8 mm. The distance between the slits is 0.22 In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. POQ is the line perpendicular to the plane AB and passing through O.21 A prism (µP = 3 ) has an angle of prism A = 30°. CP represents a wavefront and AO and BP. two wavelengths of 500 nm and 700 nm were used. Light of wavelength 550 nm is incident on the face AB at 60° angle of incidence. Calculate the refractive index of the liquid. to illuminate the slits as shown in the figure below. Symbols have their usual meanings. The distance between the slits is 0.2) is coated on face AC as shown in the figure. Q. Now.1 mm and the distance between the plane of the slits and the screen is 1 m. Find [JEE' 2003] (i) the angle of its emergence from the face AC and (ii) the minimum thickness (in nm) of the film for which the emerging light is of maximum possible intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. Calculate the position of the central bright fringe on the other wall CD with respect to the line OQ. calculate the minimum distance through which the reflecting surface AB should be shifted so that the intensity at P again becomes maximum.) 2003] 3λ λ (B) cosθ = 4 d (A) cosθ = 2 d λ 4λ (C) secθ – cosθ = (D) secθ – cosθ = d d Q. The intensity of the reflected light is 36% of the incident intensity. [JEE 2004] 10 Page 10 of 12 WAVE OPTICS Q. Q. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d = 103.20 In the adjacent diagram. 18 6000 Å 11 Q.9 5000Å Q.16 ± cos 8 8 Q.25 In Young’s double slit experiment an electron beam is used to form a fringe pattern instead of light.5 1/3 Q.35 cm Q.11 2(µ − 1) t λ 35.5 mm Page 11 of 12 WAVE OPTICS Q.6 81 : 1 Q. 1.2 mm Q. 2.10 0.7 Q.15 141 −1 2 n + 1 −1 2 n + 1 .225 mm Q. 3 Q.3 2λ Q. If speed of the electrons is increased then the fringe width will : [JEE' 2005 (Scr)] (A) increase (B) decrease (C) remains same (D) no fringe pattern will be formed ANSWER KEY EXERCISE – I Q.2 18 Q. 3 & π ± cos n = 0.15 mm Q.becomes I is: 4 λ (A) sin–1 d [JEE' 2005 (Scr)] λ (B) sin–1 3d λ (C) sin–1 2d λ (D) sin–1 4d Q.17 λ = 5850 Å Q.4 8 µm Q.8 µm Q.12 4.98 × 10–2 mm Q. 2.14 0.8 1.1 0.19 0 . n = 0. 1.24 In Young's double slit experiment maximum intensity is I than the angular position where the intensity .13 (25/24)×10−7 m Q. 1. 5 nλ D ' . D = distance between screen & slits Q.Y−coordinate =− D' . (v) 8mm down.3 µm 1 .16 (i) 1 mm (ii) increase EXERCISE – III Q. Imax = 9I.575 µm Q. D min = = . 0) R R 3 − .19 (a) circular.10 (i) 280 µm. tminimum = = 90 nm 7. 2 2 .5 9.2mm downwards .6 (a) ± 400 µm (decrease) 7 1 3 .13 (i) 8.5 × 10 m Q.± 15 7 (b) + 8000 Å .085 D . (ii) 4 Q.23 3. (b) 2. Where D’ = D + Mg/K (1– cosωt) d (a) tB = 120 µm (b) β = 6mm.11 (a) y = − 13/3 mm.12 A Q. 2 2 .6 . 0). mm 3..2 (a) I0 = I sec 2 ..33 × 10 m. D Q.33 nm Q.18 (i) y = 2 cm. I (at 5 cm below 0) = 3I Q. (iv) 0. B. 125 nm Q. mm .5 mm λ 3λ λ . Imin = I (c) β/6 = 1mm (d) I (at 5cm above 0) = 9I.1 π(µ − 1) t Q.21 0. 1.3 2 2 β λD 1 µ πd µ 1 πd ∆ φ = + ∆ φ = + .9 3/4 Q. 2 ... 15 3 7 Q. (iii) 0.14 4 × 10–15 W/m2 Q.6 λ = 600 nm.15 (0.63 mm.2 7.0 × 105 Å2 . 433. 8000 Å . 9 I Q.1 A. 2000 Å 3 Q. (b) 16.− 2 2 Q.2 7. C.11 (a) 8.12 1 : 49 Q. C Q. 4 π Q.3 0.14 t = R R 3 Q.2 4000 Å.13 A Q. (ii) µ = 1.2 R R 3 − . (R. ( – R.16 B B Q.75Imax (c) 650 nm.20 B Q.− 2 . (b) intensity at O = 0.coordinate = Q. (ii) I/9. (c) 3000Å Q.10 A.0016 Q. 6 2.7 π π 6. Q.22 D Q. .17 Q.8 5 Q.x v λy Q. (i) (ii) 2 2d l D λ l D λ Q. (b) 4 µm λ Q.48 .. Q. .4 7 µm . (ii) 560 µm –8 –7 Q.15 8. – R).24 B Q. (vi) I Q. . C Q. t = 24 µm Q.7 (i) I/3.6 mm. 1.25 B 12 Page 12 of 12 WAVE OPTICS EXERCISE – II .9 A.8 y = 0. R) Q.4 X . R R 3 . (0. Key Concepts 2. 34 Yrs. Que. Exercise IV 6. Exercise III 5. Que.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P10 Modern Physics Index: 1. Exercise I 3. Exercise II 4. from AIEEE 1 . 10 Yrs. from IIT-JEE 8. Answer Key 7. 1 mv 2 + φ 2 φ = Work function = energy needed by the electron in freeing itself from the atoms of the metal . particle accelerated by a p. ν0 is different for different metals . 5.V. h = plank 's constant = 6. (a) (b) (c) (d) CATHODE RAYS : Generated in a discharge tube in which a high vaccum is maintained . of electron + work function . Electrons are emitted if the incident light has frequency ν ≥ ν0 (threshold frequency) emission of electrons is independent of intensity . EINSTEINS PHOTO ELECTRIC EQUATION : Photon energy = K. ( 10 to 15 K. each photon having a frequency ν and energy = E = h ν . 2 Page 2 of 20 MORDERN PHYSICS KEY CONCEPTS . The wave length corresponding to ν0 is called threshold wave length λ0 .) 1 2 P2 = eV . mv = 2 2m Can be deflected by Electric & magnetic fields . Number of electrons emitted per second depends on the intensity of the incident light .d. when metals are exposed to light (of a certain minimum frequency) is called photo electric effect. E.W. (i) (ii) (iii) (iv) (v) The phenomenon of the emission of electrons . PLANK'S QUANTUM THEORY : A beam of EMW is a stream of discrete packets of energy called PHOTONS . in vacuum C = 3 × 108 m/s = ν λ 3.M. V is 2.1. WAVE NATURE OF MATTER : Beams of electrons and other forms of matter exhibit wave properties including interference and diffraction with a de Broglie wave length given by λ = h p (wave length of a praticle) . φ = h ν0 STOPPING POTENTIAL OR CUT OFF POTENTIAL : The minimum value of the retarding potential to prevent electron emission is : eVcut off = (KE)max hν= (vi) Note : The number of photons incident on a surface per unit time is called photon flux. K. of C. They are electrons accelerated by high p.63 × 10-34 Js . ELECTROMAGNETIC SPECTRUM : Ordered arrangement of the big family of electro magnetic waves (EMW) either in ascending order of frequencies or of wave lengths Speed of E.d. PHOTO ELECTRIC EFFECT : 4.R.E. Results : Can be explained only on the basis of the quantum theory (concept of photon) . n = 1 . This concentration is called the atomic nucleus .En1 = Energy emitted when an electron jumps from n2th orbit to n1th orbit (n2 > n1) . En α 2 . which does not happen in an atom. mvr = n . 13..6 ev Binding Energy (BE)n = .ATOMIC MODELS : THOMSON MODEL : (PLUM PUDDING MODEL) Most of the mass and all the positive charge of an atom is uniformly distributed over the full size (i) of atom (10-10 m) .e. (v) For hydrogen like atom/spicies of atomic number Z : n2 Z2 . (c) BOHR ATOMIC MODEL : Bohr adopted Rutherford model of the atom & added some arbitrary conditions... n 2 ν = frequency of spectral line emitted .particle scattered by thin foils of matter . (ii) A stable orbit is that in which the angular momentum of the electron about nucleus is an integral (n) multiple of (iii) (iv) h h . Enz = (– 13. The energy emitted or absorbed is a light photon of frequency ν and of energy . n 1 ∆E = hν . i. 3 . .e. An accelerating charge radiates the nucleus spiralling inward and finally fall into the nucleus . of waves in unit length (1m)] = R 1 2 − 1 2 λ n2 n 1 .6) 2 ev Z n Rz = RZ2 – Rydberg's constant for element of atomic no.(n ≠ 0).529 Aº) Note : If motion of the nucleus is also considered . (iii) Failed to explain the large angle scattering α . (b) RUTHERFORD MODEL : ( Nuclear Model) (i) The most of the mass and all the positive charge is concentrated within a size of 10-14 m inside the atom . (ii) The electron revolves around the nucleus under electric interaction between them in circular orbits.6 ev 1 i.. (iii) En Energy of the electron in the nth orbit = 2 n n Note : Total energy of the electron in an atom is negative .529 Aº) n2 .e.. n2 (iv) En2 . Z .. i. [ no. Where R = Rydberg's constant for hydrogen = 1. 1 = ν = wave no. (ii) Electrons are studded in this uniform distribution . rn α n2. These conditions are known as his postulates : mv 2 k ze 2 = 2 r r (i) The electron in a stable orbit does not radiate energy . rnz = Bohr radius Z n2 = (0.6 ev) 1 2 − 1 2 . This could not be explained by this model . E = hν ν . 2π rn = radius of nth circular orbit = (0. 2 . − 13. (a) . (ii) ∆E = (13. indicating that it is bound . then m is replaced by µ . 3 Page 3 of 20 MORDERN PHYSICS 6.En = . 2π 2π The electron can absorb or radiate energy only if the electron jumps from a lower to a higher orbit or falls from a higher to a lower orbit . FOR HYDROGEN ATOM : (Z = atomic number = 1) (i) Ln = angular momentum in the nth orbit = n h . (1Aº = 10-10 m) .097 × 107 m-1 . µ n 2 me SPECTRAL SERIES : Lyman Series : (Landing orbit n = 1) . n 2 = n1 + 1 is the α line = n1 + 2 is the β line = n1 + 3 is the γ line . IONIZATION ENERGY : The energy required to remove an electron from an atom . 2 n 2 4 n2 > 4 1 1 − 2 n 22 3 Bracket Series : (Landing orbit n = 4) In the mid infrared region ν = R (v) Pfund Series : (Landing orbit n = 5) In far infrared region ν = R 12 − 1 2 5 In all these series n2 8. n2 > 3 1 1 − 2 .. Are produced when a metal anode is bombarded by very high energy electrons . 1 1 ν=R 2 − n 22 1 Ultraviolet region (ii) 1 1 − 2 n 22 2 .6 ev . V = accelerating potential νm = maximum frequency of X .Where µ = reduced mass of electron .. etc ..13. Balmer Series : (Landing orbit n = 2) Visible region ν = R (iii) Page 4 of 20 MORDERN PHYSICS 7.( . n2 > 2 ...RAYS : Short wavelength (0. (i) 2 In this case En = (–13. n2 > 5 where n1 = Landing orbit EXCITATION POTENTIAL OF ATOM : Excitation potential for quantum jump from n1 → n2 = E n 2 −E n1 electronch arg e .nucleus system = mM/(m+M) . (i) (ii) (iii) (iv) X .radiation 4 −E n electronicch arg e .. The energy required to ionize hydrogen atom is = 0 .. . They cause photoelectric emission .1 Aº to 1 Aº) electromagnetic radiation . IONIZATION POTENTIAL : Potential difference through which an electron is moved to gain ionization energy = 11. 10.. 9.6) = 13. Are not affected by electric and magnetic field .6 ev) Z .. Characteristics equation eV = hνm e = electron charge .. n2 > 1 Paschan Series : (Landing orbit n = 3) In the near infrared region ν = R (iv) . Binding energy = . 5 Page 5 of 20 MORDERN PHYSICS (v) (vi) . γ RADIATION : α − particle : (a) Helium nucleus (2He4) (c) Velocity 106 − 107 m/s (ii) β − particle : (a) Have much less energy .(vii) (viii) Intensity of X . Number of nuclei disintegrating per second is given . of nuclei present at time t (ii) N = No e− λ t . b = 7. (disintegration /s /gm is called specific activity) .b)2 [ MOSELEY'S LAW ] .rays (iv) Series limit of series means minimum wave length of that series. RADIOACTIVITY : The phenomenon of self emission of radiation is called radioactivity and the substances which emit these radiations are called radioactive substances . dt dt Where N = No. A z + 1Y + ν (antinuetrino) STASTISTICAL LAW : The disintegration is a random phenomenon . 1 1 1 =R(z−b)2 2 − 2 λ n1 n 2 NUCLEAR DIMENSIONS : R = Ro A1/3 Where Ro = empirical constant = 1. .[ Total Mechanical Energy ] c 137 n (ii) Vel.rays depends on number of electrons hitting the target . (i) dN dN αN → =−λN = activity . Whcih atom disintegrates first is purely a matter of chance .d. (d) low penetration . (c) higher velocities than α particles (iii) γ − radiation : Electromagnetic waves of very high energy . A = Mass number of the atom 13. . 14. It can be natural or artificial (induced) . λ = decay constant No = number of nuclei present in the beginning . where v (in volts) is the p.1 × 10−15 m . applied to the tube 12400 Aº . Cut off wavelength or minimum wavelength. λ min ≅ V Continuous spectrum due to retardation of electrons . υ = a (z . (B) A− 4 z − 2Y (i) α − emission : zXA → + 2α4 + Energy (ii) β − emission : zXA → β + (iii) γ − emission : emission does not affect either the charge number or the mass number . Characteristic Spectrum due to transition of electron from higher to lower ν α (z . c = speed of light . β . 15. of electron in nth orbit for hydrogen atom ≅ (iii) For x .4 for L series b = 1 for K series Where b is Shielding factor (different for different series) .b)2 . Note : (i) 12. (i) α . (b) energy varies from 4 Mev to 9 Mev . (A) LAWS OF RADIOACTIVE DISINTEGRATION : DISPLACEMENT LAW : In all radioactive transformation either an α or β particle (never both or more than one of each simultaneously) is emitted by the nucleus of the atom. (b) more penetration . 19. greater is the stability of the nucleus . fuse together . (i) (ii) (iii) NUCLEAR FISSION : Heavy nuclei of A . above 200 . A Greater the B. the B. c = speed of light .E. These reactions take place at ultra high temperature ( ≅ 107 to 109) Energy released exceeds the energy liberated in the fission of heavy nuclei . per nucleon increases and hence the excess energy is released . B.693 λ . The total B. z decreases by 1 . Note : (i) (ii) In emission of β− .u. 1 λ (iv) MEAN (v) CURIE : The unit of activity of any radioactive substance in which the number of disintegration per second is 3.7 ×1010 . Q value of reaction = energy released in a reaction .E. 1 amu = 931 Mev MASS DEFECT AND BINDING ENERGY OF A NUCLEUS : The nucleus is less massive than its constituents . In emission of β+ . z increases by 1 .m. Σlifetimeof allatoms total number of atoms No . of the fission fragments . Tav = ATOMIC MASS UNIT ( a. eg. LIFE OF AN ATOM = .6603 × 10−27 kg 12 MASS AND ENERGY : The mass m of a particle is equivalent to an energy given by E = mc2 . 6 Page 6 of 20 MORDERN PHYSICS (iii) .E. ∆ M = mass defect = [ Zmp + (A − Z) mn] − MzA . B. The difference of masses is called mass defect . . OR U) : 1 amu = 17. (Positron) (iv) The energy released in fusion is specified by specifying Q value . The man point of the fission energy is leberated in the form of the K.E. i.E.Half life of the population T1/2 = 0. 411P→14 He+ 0+1e . 18. increases and excess energy is released . at the end of n half−life periods the number of nuclei left N = 16. = (∆ M)C2 . (i) (ii) (iii) 92 1 235 1 236 141 92 U + o n → 92 U→ 56 Ba + 36 Kr +3 o n + energy NUCLEAR FUSION ( Thermo nuclear reaction) : Light nuclei of A below 20 . per nucleon = ( ∆ M) C 2 . eg . break up onto two or more fragments of comparable masses. 2n 1 × (mass of carbon − 12 atom) = 1. 20. Total energy required to be given to the nucleus to tear apart the individual nucleons co mp o s in g the nucleus . away from each other and beyond the range of interaction forces is called the Binding Energy of a nucleus .E.e. ...5 Å making 1016 revolution per second...14 The positron is a fundamental particle with the same mass as that of the electron and with a charge equal to that of an electron but of opposite sign.. Find the longest wavelngth photon that will be emitted λ (in terms of the Rydberg constant R. light of wavelength 220 nm..... monochromatic light of wavelength 2640 A has an intensity of 200W/m2.1 A parallel beam of uniform.. In the same setup.. The magnetic moment associated with the orbital motion of the electron is _______.. they may annihilate each other. [Take : mass of electron = (0...6 × 10–34 Js and 1 eV = 1......... If only 1 of each 5 × 109 incident photons is absorbed and causes an electron to be ejected from the surface. If the same source is placed 0.0 V with respect to the surrounding..85eV makes a transition to a state of excitation energy 10...13 In a hydrogen atom.4 An isolated metal body is illuminated with monochromatic light and is observed to become charged to a steady positive potential 1... Q. Find the wavelength of the radiation emitted..2×10–12 MeV.) Q. Q. then find the stopping potential (b) the saturation current Q. the ejected photoelectrons have maximum kinetic energy Ta eV and de Broglie wavelength λa.(Take plank's constant..... Apply the Bohr atom model and consider a possible transitions of this hypothetical particle to the first excited level.2 m from a photoelectric cell...7 A hydrogen atom in a state having a binding energy 0...6 volt and 18.Q. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.2eV... Q..6 Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage V0 to prevent a collector... The wavelength of the photon emitted in this process if the electron is assumed to have had no kinetic energy when it combines with nucleus is . Find the numerical value of voltage V0....... Q.8 A hydrogen atom is in 5th excited state....eV..nm.6 × 10–19 J) Q... the electron moves in an orbit of radius 0.3 (a) When a monochromatic point source of light is at a distance of 0.... If the De Broglie wavelength of these photoelectrons is λb = 2λa. the total photocurrent in the circuit is ________. The frequency of the incident light is ______________.. When a positron and an electron collide.. The work function of the metal is 3. Q.. h = 6...0 mA.11 Three energy levels of an atom are shown in the figure.5/C2)MeV and hC = 1.. then find The work function of a (b) The work function of b is (c) Ta and Tb (a) Q.7eV is Tb = (Ta – 1.6 m away from the photoelectric cell. The wavelength corresponding to three possible transition are λ1. the cut off voltage and the saturation current are respectively 0. The wave length of emitted photon is ...... Q....m where h is the Plank's constant and C is the velocity of light in air] 7 Page 7 of 20 MORDERN PHYSICS EXERCISE # I ..9 The ratio of series limit wavlength of Balmer series to wavelength of first line of paschen series is ....nm. m/s and the energy of the photon is . ejects electrons which require twice the voltage V0 to stop them in reaching a collector. When the electron jumps to ground state the velocity of recoiling hydrogen atom is .. λ2 and λ3...0 eV..12 Imagine an atom made up of a proton and a hypothetical particle of double the mass of an electron but having the same charge as the electron.. Q.5 663 mW of light from a 540 nm source is incident on the surface of a metal. The value of λ3 in terms of λ1 and λ2 is given by ______. Q..2 When photons of energy 4.10 An electron joins a helium nucleus to form a He+ ion.5) eV.. Q...25eV strike the surface of a metal A. The number of photons in 1mm3 of this radiation are . The energy corresponding to their mass appears in two photons of equal energy..... 4 11H → 42 He with 26 MeV of energy released.particles emitted by the sample. Q. 8 Page 8 of 20 MORDERN PHYSICS Q.11 × 105 β . potential energy and angular momentum in the first Bohr orbit (v) the radius of the first Bohr orbit.2eV to excite the electron from the 2nd Bohr orbit to 3rd Bohr orbit.active sample is present near the counter.particles at t = 108 s. The energy to excite the electron from the second Bohr orbit to the third Bohr orbit is 47. Find the time required to produce 80% of the equilibrium quantity of this unstable nuclide.0 MeV respectively. Q. The counter registers 1 × 105 β . Q.15 A small 10W source of ultraviolet light of wavelength 99 nm is held at a distance 0. Find the ratio of the two radii of corresponding orbits. Q. . Alternatively.5nm.2 eV.particles at t = 36 s and 1. It requires 47. Find (i) the value of Z.05 nm.26 Suppose that the Sun consists entirely of hydrogen atom and releases the energy by the nuclear reaction.1 m from a metal surface. (ii) energy required to excite the electron from the third to the fourth orbit (iii) the wavelength of radiation required to remove the electron from the first orbit to infinity (iv) the kinetic energy.20 Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third of the corresponding Balmer series is equal to 108.1 MeV and 7. a sample is placed in a reactor.16 The surface of cesium is illuminated with monochromatic light of various wavelengths and the stopping potentials for the wavelengths are measured.19 A hydrogen like atom (atomic number Z) is in higher excited state of quantum number n. Find age of rock Q. Find the values of n and Z. Q. This nuclide β— decays with half life τ. Take the mass of the Sun as 1. (ii) A sample of rock taken from the moon contains both potassium and argon in the ratio 1/7.9 × 1026 W.95eV and 5. If the total output power of the Sun is assumed to remain constant at 3. An unstable nuclide is produced at a constant rate R in the sample by neutron absorption.24 An isotopes of Potassium 19 Ar which is stable.22 The binding energies per nucleon for deuteron (1H2) and helium (2He4) are 1. Q. A β . (ii) the number ofphotoelectrons emitted per unit area per second if the efficiency of liberation ofphotoelectrons is 1%.23 A radioactive decay counter is switched on at t = 0. The radius of an atom of the metal is approximately 0. The energy released when two deuterons fuse to form a helium nucleus (2He4) is ________. Estimate the value of work function of the cesium and Planck’s constant.Q. Q.7 × 1030 kg. The atomic number of this nucleus is ______________. The counter registers the number of β .7eV respectively.4 × 109 year and decays to Argon 18 (i) Write down the nuclear reaction representing this decay.17 A hydrogen like atom has its single electron orbiting around its stationary nucleus. Find (i) the average number of photons striking an atom per second.4eV and 8. Find T½ of this sample 40 40 Q.18 A single electron orbits a stationary nucleus of charge Ze where Z is a constant and e is the electronic charge. The results of this experiment is plotted as shown in the figure. Q.25 At t = 0. K has a half life of 1. the atom from the same excited state can make transition to the second excited state by successively emitting two photons of energies 2.15eV respectively.21 Which level of the doubly ionized lithium has the same energy as the ground state energy of the hydrogen atom. find the time it will take to burn all the hydrogen. This excited atom can make a transition to the first excited state by successively emitting two photons of energy 22. .. & the stopping potential for the photoelectrons ejected by the longer wavelength.52. 6. 6.C13 + 1H1 → 7N14 6 N14 + 1H1 → 8O15 → 7N15 + +1e0 7 N15 + 1H1 → 6C12 + 2He4 Find how many tons of hydrogen must be converted every second into helium . 6.277 6.. 6.38.28. (assume no transmission) Q.0 cm2 of a clean metallic surface of work function 2.49..247.00814 amu.28 An electron of mass "m" and charge "e" initially at rest gets accelerated by a constant electric field E.249. The beam falls normally on an area 1.22. 6.4 A beam of light has three wavelengths 4144Å. 6.266.134. 6... 6. 5.42. 6.094 × 107 m–1) Q.3 Electrons in hydrogen like atoms (Z = 3) make transitions from the fifth to the fourth orbit & from the fourth to the third orbit.137. Find the radius of the largest circular path followed by the emitted photoelectrons. 5.263. 6.289 EXERCISE # II Q. 6..233. Q. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron.33. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.133.40. 6. 5.141. If a constant uniform magnetic field of strength 10–4 tesla is applied parallel to the metal surface.270. 6. 7 Q.136. Assume that hydrogen forms 35% of the sun's mass .264.. The light falls normally on the plate. Calculate the work function of the metal. 5.35.43. Find the number of photons striking the metal plate per square meter per sec. 5.262.95 volts. 6.37.32.138.264.5× 1011m. (Rydberg constant = 1.8 × 10-7 m and power 1. 9 Page 9 of 20 MORDERN PHYSICS Q. 6. Irodov.273. 6.50. 6. 6.2 A small plate of a metal (work function = 1.265. 1 amu = 931 MeV. atomic masses mH=1. List of recommended questions from I.17 eV) is placed at a distance of 2m from a monochromatic light source of wave length 4. 6..8. 6. 5.00388 amu.41. incident on a cylinder (height h and base radius R) placed on a smooth surface as shown in figure if: surface of cylinder is perfectly reflecting surface of cylinder is having reflection coefficient 0. 5. Calculate the number of photoelectrons liberated in two seconds.m–2 equally distributed amongst the three wavelengths.39.31.51. distance between the sun and the earth= 1. mHe=4. Calculate in how many years this hydrogen will be used up if the radiation of the sun is constant .E. 6... 4972Å & 6216 Å with a total intensity of 3. The resulting radiations are incident normally on a metal plate & eject photo electrons.53.. 5. 6.260. 6. me = 5. 6.. 6. mass of the sun=2 × 1030 kg.49 × 10-4 amu.21.30.27. 6. The rate of change of DeBroglie wavelength of this electron at time t is . 6. 5. 6. The solar constant is 8 J / cm2 min.135. 6.27 Assuming that the source of the energy of solar radiation is the energy of the formation of helium from hydrogen according to the following cyclic reaction : C12 + 1H1 → 7N13 → 6C13 + +1e0 6 .6×10–3 W. 6.1 (a) (b) Find the force exerted by a light beam of intensity I.. 5. 6. 6.. 5.214.3 eV.0 watt.249.. Some have energy more and some have less than 2. 10 Page 10 of 20 MORDERN PHYSICS Q.6 m is inserted in the aperture as shown. (ii) saturation current in second case. Some of the emitted photons have energy 2. Find the De Broglie wavelength of each particle in the frame of their centre of mass. A photoemissive detector D of surface area 0. calculate the values of the stopping potential in the two cases (without & with the lens in the aperture). If a concave lens L of focal length 0.9. λ1 & λ2 . Find : the number of photons striking an atom per second. the number of photoelectrons emitted per second if the efficiency of liberation of photoelectrons is 1%. The atoms of the gas make transition to a higher energy level by the absorbing monochromatic light of photon energy 2. calculate threshold wavelength for the cell.12 A gas of identical hydrogen like atoms has some atoms in the lowest (ground) energy level A & some atoms in a particular upper (excited) energy level B & there are no atoms in any other energy level.5 cm2 is placed at the centre of the screen.13 A monochromatic light source of frequency ν illuminates a metallic surface and ejects photoelectrons.7eV.10 A stationary He+ ion emitted a photon corresponding to the first line its Lyman series. possessing De Broglie wavelengths.05 nm. Find the de Broglie wavelengths of both particles in the frame of their centre of mass.1 m from a metal surface.7 eV. Find the work function of the metal and the frequency ν.6 (i) (ii) (iii) Q. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases. Q. Find the principal quantum number of the initially excited level B. A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. it is found that maximum velocity of photoelectrons increases n = 2 times. Find the ionisation energy for the gas atoms.8 A neutron with kinetic energy 25 eV strikes a stationary deuteron. The radius of an atom of the metal is approximaterly 0. Subsequently.7 eV. an aperture A of diameter 0.8 × 10–3 A/w. The efficiency of the detector for the photoelectron generation per incident photon is 0. A monochromatic point source S radiating wavelength 6000 Å with power 2 watt. If the work-function of the photoemissive surface is 1 eV. Q.1 m & a large screen SC are placed as shown in figure .9 Two identical nonrelativistic particles move at right angles to each other. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. find the new values of photon flux density & photocurrent Assume a uniform average transmission of 80% for the lens . Find the velocity of the photoelectron. The corresponding spectral sensitivity of photocell is J = 4. the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength of 1215 Å .11 (i) (ii) (iii) Q. the atoms emit radiation of only six different photon energies. When another monochromatic radiation of wavelength λ2 = 1650Å and power P = 5 × 10–3 W is incident.(i) Q.5 . The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. Find the maximum and the minimum energies of the emitted photons. Calculate the photon flux density at the centre of the screen and the photocurrent in the detector . Q. When the whole experiment is repeated with an incident radiation of frequency (5/6)ν .7 (i) (ii) Monochromatic radiation of wavelength λ1 = 3000Å falls on a photocell operating in saturating mode. Q. A small 10 W source of ultraviolet light of wavelength 99 nm is held at a distance 0. with what kinetic energy will the electron be ejected ? Q. Assuming that the entire photon energy is taken up by the electron. Find the amount of radon (Rn222) originally introduced into the source.5 mg of the substance.15 A classical model for the hydrogen atom consists of a single electron of mass me in circular motion of radius r around the nucleus (proton). What is the mass of 37 Li ? Assume that particles are free to move after the collision.Q.5 × 109 yrs & that of U235 = 7. The atom is bombarded with high energy electrons. [T1/2 of Rn = 3. Q. The atomic weight of the substance is 230. Since the electron is accelerated. Find the value of Z.01167. Calculate an expression for the radius r (t) as a function of time. Calculate the age of the earth.18 The kinetic energy of an α − particle which flies out of the nucleus of a Ra226 atom in radioactive disintegration is 4.21 When thermal neutrons (negligible kinetic energy) are used to induce the reaction . This energy may appear as X-ray or may all be used to eject an M-level electron from the atom. Q. Assume that at t = 0. if it produces 1. The yield of this reaction is (1/ 4000) i.19 A small bottle contains powdered beryllium Be & gaseous radon which is used as a source of α−particles.E.e. 1. (ii) energy released when L-level electron moves to fill the vacancy in the K-level.6 days.13 × 108 yrs) Q.2 × 106 neutrons per second after 7. the kinetic energy of the electron in the first Bohr orbit and the wavelength of the electro magnetic radiation required to eject the electron from the first Bohr orbit to infinity. 2 e2 1 −15 · = r ≈ 3 × 10 m Take : e 2 3 4 π ε 0 m e C Q. (iv) K.16 Simplified picture of electron energy levels in a certain atom is shown in the figure.83 MeV. α − particles are emitted with an energy of 1. The nuclear charge Ze. the atom continuously radiates (i) (ii) (iii) 4 electromagnetic waves.17 U238 and U235 occur in nature in an atomic ratio 140 : 1.8 days] Q. Assuming that at the time of earth’s formation the two isotopes were present in equal amounts.4 β are emitted each second by 2.. 11 Page 11 of 20 MORDERN PHYSICS Q. (Half life of u238 = 4. only one α−particle out of 4000 induces the reaction. Given the masses of boron neutron & He4 as 10.00894 & 4.20 An experiment is done to determine the half − life of radioactive substance that emits one β−particle for each decay process. Neutrons are produced when α−particles of the radon react with beryllium. . Hence or otherwise find the time t0 when the atom collapses in a classical model of the hydrogen atom.14 An energy of 68. Q.78 MeV.00386 u respectively. the radius is r0 = 10–10 m. The total power P radiated by the atom is given by P = P0 r where e6 (C = velocity of light) P0 = 96 π3ε 0 3C 3m e 2 Find the total energy of the atom. Find the total energy evolved during the escape of the α − particle. of the electron emitted from the M-level. (iii) wavelength of the X-ray emitted. 10 5B + 10 n → 37 Li + 42 He . Find the half life of the substance. The impact of one of these electron has caused the complete removal of K-level is filled by an electron from the L-level with a certain amount of energy being released during the transition.0 eV is required to excite a hydrogen like atom from its second Bohr orbit to the third. Find : (i) the minimum potential difference through which electron may be accelerated from rest to cause the ejectrion of K-level electron from the atom. Measurement show that an average of 8. 00785 u 2 1 1 1 .(i) Two deuterium ( D) nuclei fuse to form a tritium ( T )nucleus with a proton as product. P = product nucleus. Calculate the energy of these neutrons and energy and angle of recoil of the associated Beryllium atom.014102 u ( P )= 1. O = outgoing particle. The energy release in the combined reaction per deuterium & What % of the mass of the initial deuterium is released in the form of energy. Q. The half life of C14 is 5730 yrs.23 A wooden piece of great antiquity weighs 50 gm and shows C14 activity of 320 disintegrations per minute. n) α. The reaction 2 1 3 1 may be represented as D (D. 1 0 Q. 12 Page 12 of 20 MORDERN PHYSICS Q.016049 u . Estimate the length of the time which has elapsed since this wood was part of living tree.01784 amu . . Find the velocity of body after time t. assuming that living plants show a C14 activity of 12 disintegrations per minute per gm.00776 amu. Find the increase in temperature of m mass of water in time t. the Q-value is expressed as mP m – K 1 + I Q = KP 1 + I MO MO Where. Assume that there is no loss of energy through water surface. neutrons are observed to emerge at right angle to the direction of incident beam.002603 u 4 2 . Find : The energy release in each stage . The reaction is represented as T(D .24 Show that in a nuclear reaction where the outgoing particle is scattered at an angle of 90° with the direction of the bombarding particle.26 A body of mass m0 is placed on a smooth horizontal surface. Given : ( D) = 2. ( T) = 3.008665 u 3 1 ( He)= 4.00893 amu . 20% of this energy is utilised in increasing the temperature of water.22 In a fusion reactor the reaction occurs in two stages : .25 When Lithium is bombarded by 10 MeV deutrons. T = target nucleus.27 A radionuclide with disintegration constant λ is produced in a reactor at a constant rate α nuclei per sec. Assuming that the mass is ejected backward with a relative velocity v. I = incoming particle. p) T. m (3Li7) = 7. Q. Q. The mass of the body is decreasing exponentially with disintegration constant λ. Specific heat of water is S. Given that : m (0n1) = 1. Q. (ii) (a) (b) (c) ( ) A tritium nucleus fuses with another deuterium nucleus to form a helium 42 He nucleus with neutron as another product. and m (4Be8) = 8. Initially the body was at rest. During each decay energy E0 is released. m (1H2) = 2.01472 amu . ( n )= 1. 2 (C) 51. in a hydrogen like atom .75 eV [JEE’94] Q.2 × 10-3 W emits mono energetic photons of energy 5. The efficiency of photo electrons emission is one for every 106 incident photons. ionization energy of H atom =13. Assume that the sphere is isolated and initially neutral. [JEE'96] 13 Page 13 of 20 MORDERN PHYSICS EXERCISE # III . Alternatively .4 In a photo electric effect set-up. Find the torque experienced by the orbiting electron. It is scattered at an angle of 90º with respect of its original direction.6 An electron.rays generated is ________ .1.8 A potential difference of 20 KV is applied across an x-ray tube. The maximum kinetic energy of photo electrons liberated from another metal B by photons of energy 4. Why ? Evaluate the time t. Find the ratio of the wavelength of incident light to the De .00 eV (D) TB = 2. Find the allowed values of the energy of the neutron & that of the atom after collision.8 m from the centre of a stationary metallic sphere of work function 3.0 × 10-3m . then : (A) the work function of A is 2.20 eV & 17.Broglie wave length of the electron. Determine the values of n & Z. The minimum wave length of X . a point source of light of power 3. is in an excited state .Broglie wave length of the fastest photo electrons emitted.5 An energy of 24. It is observed that the photo electron emission stops at a certain time t after the light source is switched on. It has a total energy of − 3. Calculate: (i) The kinetic energy & (ii) The De .20 eV (C) TA = 2. [JEE 96] Q.70 eV is TB = (TA .95 eV respectively.6 eV) [JEE '93] Q. Calculate the number of photo electrons emitted per second. If the de-Broglie wave length of these photo electrons is γB = 2γA. If the atom gets de-excited subsequently by emitting radiation .0 eV & of radius 8. such that the plane normal to the electron orbit make an angle of 30º with the magnetic induction.0 eV.6 eV is required to remove one of the electrons from a neutral helium atom. Obtain an expression for the orbital magnetic dipole moment of the electron. The energy (In eV) required to remove both the electrons form a neutral helium atom is : (A) 38.25 eV & 5. The source is located at a distance of 0.50) eV.2 A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. (Ionisation energy of hydrogen atom = 13.00 eV respectively.8 (D) 79.1 (i) (ii) A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest . the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.Broglie wave length γA. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10. [JEE’95] (a) (b) (c) (d) Q.7 An electron in the ground state of hydrogen atoms is revolving in anti-clockwise direction in a circular orbit of radius R.6 eV) [JEE’94] Q. and that photo electrons are instantly swept away after emission.Q. [JEE'96] (i) (ii) Q. The atom is placed in a uniform magnetic induction. the ejected photo electrons have maximum kinetic energy TAeV and de.4 eV.3 Select the correct alternative(s) : When photons of energy 4.225 eV (B) the work function of B is 4. find the frequencies of the emitted radiation.2 (B) 49.25 eV strike the surface of a metal A.0 [JEE’95] Q. (Given : Mass of he atom = 4×(mass of neutron). Find the energy of the electrons in electron volts and the least value of d for which the standing wave of the type described above can form. The longest wavelength of light that can cause photoelectron emission from this substance is approximately : (A) 540 nm (B) 400 nm (C) 310 nm (D) 220 nm (ii) The electron in a hydrogen atom makes a transition n1 → n2. [JEE' 97] Q. Find the number of photoelectrons emitted per sec and their minimum and maximum energies in eV. [JEE’ 2000 (Scr)] Q.14 × 10-15 eV−s ] [JEE ’99] Q. These photoelectrons pass through a region containing α-particles. potential and total energies decrease. n2 = 3 [JEE ’98] Q.0 (D) √m2/√m1 [JEE ’99] Q. that are likely to be emitted during and after the combination.8 eV is emitted. Ground state energy of hydrogen atom is − 13.4 . lying in the 2 to 4eV range.12 Photoelectrons are emitted when 400 nm radiation is incident on a surface of work function 1.0 eV . is (A) m1/m2 (B) m2/m1 (C) 1.10(i) The work function of a substance is 4. h = 4. λ1/ λ2. Find the energies in eV of the photons. Z and the ground state energy (in eV) for this atom.6 eV. (C) Its kinetic and total energies decrease and its potential energy increases. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. [JEE' 2000] (b) When a beam of 10.like atom of atomic number Z is in an excited state of quantum number 2 n.8 (D) 122.51 (B) 13. (B) Its kinetic energy decreases. calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation.11 A particle of mass M at rest decays into two particles of masses m1 and m2. It can emit a maximum energy photon of 204 eV. (D) Its kinetic. having non-zero velocities. [JEE' 2000] 14 Page 14 of 20 MORDERN PHYSICS Q. 0. The longest wavelength photon that will be emitted has wavelength λ (given in terms of the Rydberg constant R for the hydrogen atom) equal to (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R [JEE’ 2000 (Scr)] (b) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Assume the Bohr model to be valid . a photon of energy 40.6 eV photon of intensity 2 W/m2 falls on a platinum surface of area 1 × 104 m2 and work function 5.9eV. A maximum energy electron combines with an α-particle to form a He+ ion. potential energy increases and its total energy remains the same. n2 = 2 (B) n1 = 8. He+ ions thus formed are in their fourth excited state.6 (C) 40.9(i) As per Bohr model. emitting a single photon in this process. The possible values of n1 & n2 are : (A) n1 = 4.5 Å but not for any intermediate value of d. where n1 & n2 are the principal quantum numbers of the two states .53% of the incident photons eject photoelectrons. n2 = 1 (D) n1 = 6. [Take . It is found that one such standing wave is formed if the distance 'd' between the atoms of the array is 2 Å.(ii) Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. The ratio of the de-Broglie wavelengths of the particles. the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is (A) 1. If it makes a transition to quantum state n. Find n.6 ev. n2 = 2 (C) n1 = 8. The time period of the electron in the initial state is eight times that in the final state . A similar standing wave is again formed if 'd' is increased to 2. Also.13(a) Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron.14(a) A hydrogen . Which of the following statements is true? (A) Its kinetic energy increases and its potential and total energies decrease. the deuteron supply of the star is exhausted in a time of the order of : [JEE ’93] (A) 106 sec (B) 108 sec (C) 1012 sec (D) 1016 sec Q. (Neglect the time taken by photoelectron to reach plate B) [JEE' 2002] Q. originating from all possible transition between a group of levels. then (A) lH > lLi and |EH| > |ELi| (B) lH = lLi and |EH| < |ELi| (C) lH = lLi and |EH| > |ELi| (D) lH < lLi and |EH| < |ELi| [JEE 2002 (Scr)] Q.18 Two metallic plates A and B each of area 5 × 10–4 m2. Using Moseley's law. A sample of the blood of volume 1 cm3 taken after 5 hours shows an activity of 296 disintegrations per minute. photons of energy 5 eV falls on the cathode having work function 3 eV. The radius r of the nth Bohr's orbit depends upon principal quantum number n as : (A) r ∝ n (B) r ∝ 1/n2 (C) r ∝ n2 (D) r ∝ 1/n [JEE' 2003 (Scr)] Q. [JEE' 2002] Q. Determine (a) the number of photoelectrons emitted up to t = 10 sec. (b) the magnitude of the electric field between the plates A and B at t = 10 s and (c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s when it reaches plate B. and EH and ELi their respective energies. If the average power radiated by the star is 1016 W.24(i) Fast neutrons can easily be slowed down by : (A) the use of lead shielding (B) passing them through water (C) elastic collisions with heavy nuclei (D) applying a strong electric field 15 Page 15 of 20 MORDERN PHYSICS Q.2 mA.20 Frequency of a photon emitted due to transition of electron of a certain elemrnt from L to K shell is found to be 4. Q. v0 and r0 are constant and r is the radius of the orbit. given that the [JEE' 2003] Rydberg's constant R = 1. are placed at a separation of 1 cm.85 eV and – 0. These levels have energies between – 0. If lH and lLi are their respective electronic angular momenta. (1 Curie = 3.Q. Determine the total volume of blood in the body of the person.)] (A) 2 × 1016 (B) 5 × 1016 (C) 1 × 1017 (D) 4 × 1015 . Plate B carries a positive charge of 33. the processes 1H2 + 1H2 → 1H3 + p & 1H2 +1H3 → 2He4 + n. It produces energy via. Also assume that all the emitted photoelectrons are collected by plate B and the work function of plate A remains constant at the value 2 eV.22 A star initially has 1040 deutrons. Then the number of electrons striking the target per second is [JEE' 2002 (Scr. Assume that the radioactive solution mixes uniformly in the blood of the person. Assume that one photoelectron is emitted for every 106 incident photons. starts falling on plate A at t = 0 so that 1016 photons fall on it per square meter per second.0 microcurie is injected into the blood of a person.19 The attractive potential for an atom is given by v = v0 ln (r / r0 ) .16 A Hydrogen atom and Li++ ion are both in the second excited state. with photons of energy 5 eV each.7 × 1010 disintegrations per second ) [JEE’94] Q. (a) If the saturation current is iA = 4µA for intensity 10–5 W/m2.21 In a photoelctric experiment set up.7 × 10–12 C. (b) Also draw a graph for intensity of incident radiation of 2 × 10–5 W/m2 ? [JEE' 2003] Q. then plot a graph between anode potential and current. (b) Calculate the smallest wavelength emitted in these transitions.ray tube is 5 kV and the current through it is 3.1 × 107 m–1. A monochromatic beam of light.17 A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths.2 × 1018 Hz.23 A small quantity of solution containing 24Na radionuclide (half life 15 hours) of activity 1.15 The potential difference applied to an X . find the atomic number of the element.544 eV (including both these values) (a) Find the atomic number of the atom. 28 Consider the following reaction .25 Which of the following statement(s) is (are) correct ? [JEE'94] (A) The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons. (iii). the former with a probability of 8% and the latter with a probability of 92%. α Q.064100 u & 2 He = 4 . Mass of the helium atom = 4. β (C) β . of the physical quantity in the answer book. Calculate the power output from a sample of 1020 Cm atoms.25 % of the reduced number. In your answer. against the number (i).072220 u . β .75 MeV .97 MeV & that of required to remove a neutron from 17O is : (A) 3. α (D) γ . Q. (B) The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons. 16 Page 16 of 20 MORDERN PHYSICS (ii) . [JEE 96] Mass of the deuterium atom = 2.0024 u This is a nuclear ______ reaction in which the energy Q is released is ______ MeV. (C) In nuclear fusion. After 10 sec the number [JEE 96] of undecayed nuclei remains to 12. Choose the appropriate value of energy from column II for each of the physical quantities in column I and write the corresponding letter A. γ . etc. B.23 17O is 7.002603 u .26 The binding energy per nucleon of 16O is 7. the sequence of column I should be maintained .52 (B) 3. Each fission releases 200 MeV of energy . energy is released by fusion two nuclei of medium mass (approximately 100 amu). β . 94 Pu = 244 . γ . column I lists some physical quantities & the column II gives approx. (ii). The energy in MeV [JEE’95] (D) 7. Column I Column II (i) Energy of thermal neutrons (A) 0. the radiations are : [JEE’94] (A) α . The masses involved in α decay are as follows : 248 244 4 96 Cm = 248 .30 Select the correct alternative(s) . The stopping potential in Volts is : (A) 2 (B) 4 (C) 6 (D) 10 (b) (c) In the following. we can assert that : (B) no nucleus will decay before t = 8 days (A) no nucleus will decay before t = 4 days (C) all nuclei will decay before t = 16 days (D) a given nucleus may decay at any time after t = 0. Its primary decay modes are spontaneous The element Curium 248 has a mean life of 10 Cm 96 fission and α decay. (l u = 931 MeV/c2) [JEE'97] Q. Calculate : (i) mean − life of the nuclei and (ii) The time in which the number of undecayed nuclear will further reduce to 6.5 eV (iii) Binding energy per nucleon (C) 3 eV (iv) Photoelectric threshold of metal (D) 20 eV (E) 10 keV (F) 8 MeV 13 seconds.27 At a given instant there are 25 % undecayed radio − active nuclei in a sample.Consider α − particles . energy is released by fragmentation of a very heavy nucleus. each having an energy of 0.29(a)The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. [JEE '98] 20 10 (i) Let mp be the mass of a proton. energy values associated with some of them. Q.5 % . mn the mass of a neutron. Then : (A) M2 = 2 M1 (B) M2 > 2 M1 (C) M2 < 2 M1 (D) M1 < 10 (mn + mp) (ii) The half − life of 131I is 8 days. γ (B) α . C etc. Increasing order of penetrating powers . 2H1 + 2H1 = 4He2 + Q .5 MeV . M1 the mass of a Ne nucleus & M2 the mass of a 40 20Ca nucleus.0141 u .86 Q.64 (C) 4. β − particles & γ rays . (D) In nuclear fission. Q. Given a sample of 131I at time t = 0.025 eV (ii) Energy of X−rays (B) 0. 37 The volume and mass of a nucleus are related as (A) v ∝ m (B) v ∝ 1/m (C) v ∝ m2 [JEE 2003 (Scr)] (D) v ∝ 1/m2 Q.31 Nuclei of a radioactive element A are being produced at a constant rate α . X. W. what is the mean life of the sample? [JEE 2003] 17 Page 17 of 20 MORDERN PHYSICS Q. In next 2 sec it emits 0.Q. decays into two α−particles and an unknown nucleus. then the kinetic energy of α-particle is : [JEE 2003 (Scr)] (A) 5. Initially both of them have the same number of atoms.32(a) Binding energy per nucleon vs.5 MeV. If initially they have the same number of nuclei. The element has a decay constant λ . [JEE '98] .33 Two radioactive materials X1 and X2 have decay constants 10λ and λ respectively. mass number curve for nuclei is shown in the figure. The time taken for the radioactivity of a sample of 215At to decay to 1/16th of its initial value is [JEE 2002 (Scr)] (A) 400 µs (B) 6. Then (A) X & Y have the same decay rate initially (B) X & Y decay at the same rate always (C) Y will decay at a faster rate than X (D) X will decay at a faster rate than Y [JEE '99] Q. (C) Alpha particles are singly ionized helium atoms (D) Protons and neutrons have exactly the same mass (E) None (e) The half−life period of a radioactive element X is same as the mean−life time of another radioactive element Y.39 A radioactive sample emits n β-particles in 2 sec.38 The nucleus of element X (A = 220) undergoes α-decay.67 × 10−27 kg] (A) 1020 kg/m3 (B) 1017kg/m3 (C) 1014kg/m3 (D) 1011kg/m3 (c) 22Ne nucleus. [mP = 1. If Q-value of the reaction is 5.34 The electron emitted in beta radiation originates from [JEE’2001(Scr)] (A) inner orbits of atoms (B) free electrons existing in nuclei (C) decay of a neutron in a nucleus (D) photon escaping from the nucleus Q.7 MeV (D) None Q.36 Which of the following processes represents a gamma . there are N0 nuclei of the element. (a) (b) If α=2N0λ.35 The half . then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time (A) 1/(10λ) (B) 1/(11λ) (C) 11/(10λ) (D) 1/(9λ) [JEE ' 2000 (Scr)] Q.75 n β-particles. after absorbing energy.life of 215At is 100 µs. Y and Z are four nuclei indicated on the curve.8 MeV (C) 2.3 µs (C) 40 µs (D) 300 µs Q. calculate the number of nuclei of A after one halflife of A & also the limiting value of N as t→∞. The process that would release energy is (A) Y → 2Z (B) W → X + Z (C) W → 2Y (D) X → Y + Z (b) Order of magnitude of density of Uranium nucleus is. The unknown nucleus is (A) nitrogen (B) carbon (C) boron (D) oxygen (d) Which of the following is a correct statement? (A) Beta rays are same as cathode rays (B) Gamma rays are high energy neutrons. Calculate the number N of nuclei of A at time t . At time t = 0.decay? [JEE 2002 (Scr)] A A A 1 A– 3 (B) XZ + n0 → XZ –2 + c (A) XZ + γ → XZ – 1 + a + b (C) AXZ → AXZ + f (D) AXZ + e–1 → AXZ – 1 + g Q.4 MeV (B) 10. 4 eV and one electron of 1. (A) one photon of 10.41 A photon of 10.4 eV [JEE' 2005 (Scr)] Q.2 eV and an electron of energy 1. Find (a) atomic number of the nucleus (b) the frequency of Kα line of the X-ray produced.4 eV (B) 2 photons of energy 10.45 Given a sample of Radium-226 having half-life of 4 days. Find the probability. (A) 1 (B) 1/2 (C) 3/4 (D) 1/4 [JEE 2006] Q.834 amu and mHe = 4.42 Helium nuclie combines to form an oxygen nucleus.47 In hydrogen-like atom (z = 11). (A) Ratio of work functions φ1 : φ2 : φ3 = 1 : 2 : 4 (B) Ratio of work functions φ1 : φ2 : φ3 = 4 : 2 : 1 (C) tan θ is directly proportional to hc/e. where h is Planck’s constant and c is the speed of light (D) The violet colour light can eject photoelectrons from metals 2 and 3.Then what can be detected by a suitable detector.40 The wavelength of Kα X-ray of an element having atomic number z = 11 is λ .24 MeV (D) 4 MeV (A) 10.44 Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. a nucleus disintegrates within 2 half lives. nth line of Lyman series has wavelength λ equal to the de-Broglie’s wavelength of electron in the level from which it originated.2 eV (C) 2 photons of energy 3. [JEE 2005] Q. (R = 1. After few microseconds one more photon of energy 15 eV collides with the same hydrogen atom. The ratio of radii of nucleus to that of helium nucleus is (14)1/3. What is the value of n? [JEE 2006] Q. The binding energy per nucleon of oxygen nucleus is if m0 = 15.0026 amu (B) 0 MeV (C) 5. φ2 and φ3 in an experiment of photoelectric effect is plotted as shown in the figure.2 eV energy collides with a hydrogen atom in ground state inelastically. The wavelength of Kα X-ray of another element of atomic number z' is 4λ.Q.46 The graph between 1/λ and stopping potential (V) of three metals having work functions φ1.4 eV (D) 1 photon of 3. Which of the following statement(s) is/are correct? [Here λ is the wavelength of the incident ray]. Then z' is (B) 44 (C) 6 (D) 4 [JEE' 2005 (Scr)] (A) 11 . find λ1/λ2. when 0 ≤ x ≤ 1 and x > 1 respectively. [JEE 2006] Q.48 Match the following Columns Column 1 (A) Nuclear fusion (B) Nuclear fission (C) β–decay (D) Exothermic nuclear reaction [JEE 2006] Column 2 (P) Converts some matter into energy (Q) Generally occurs for nuclei with low atomic number (R) Generally occurs for nuclei with higher atomic number (S) Essentially proceeds by weak nuclear forces 18 Page 18 of 20 MORDERN PHYSICS Q.43 The potential energy of a particle of mass m is given by E 0 ≤ x ≤1 V( x ) = 0 0 x >1 λ1 and λ2 are the de-Broglie wavelengths of the particle. If the total energy of particle is 2E0.1× 107 m–1 and c = 3 × 108 m/s) [JEE 2005] Q.24 MeV [JEE' 2005 (Scr)] Q. 384 % Q.18 (i) 5. 0.6 (b) 7.3 (a) 0.10 22. (iii) 4 → 1 .8 λdeutron = λneutron = 8.16 (i) 1. n = 7 Q.48 Q.5 Q. (iii) 0.14 × 1018 sec Q.1 × 106 m/s Q.6 volt.5 eV. 16 80π Q.26 8 3 × 1018 sec Q.17 Q.7 × 10–15 J.2eV.06 volt Q.28 Å 1/ 3 Q.24 eV Q.26 v = uλt Q.3 × 10−6 g 10 1. 4 → 3 Q.27 α 0.21 7.0 eV.13 6.875 × 104 V.87 MeV Q.20 Q.12 11.3 1.1 38 I R h 8IhR/3C 15 C Q.24 (i) ln 5 τ Q.9 λ = λ12 +λ 2 2 Q. (ii) 13.25eV.19 3.18 4.0 cm Q.14 2.53 × 10–34 J-s h 1. (ii) r0 1 − (i) – 3 8πε 0 r r 0 .4 eV Q.4 1.21 n = 3.04 × 109 yrs Q. 5 × 1015 Hz Q.27 40 19 K m Q.10 3.956 × 1015 photons/m2s .2 µA Q.5 (i) 4125Å.99 eV.9 885 Q. Angle of recoil = tan–1 (1.096 µÅ (ii) 2. (b) 2.11 (i) 2 . 16. (ii) 5/1600 electrons/sec Q. 25.760 V Q.1 Q.4 A. – 680 eV.33 × 1016 photons/m2 − s .0007 MeV.04 ×10–19J . 0. (ii) 23.5 eV Q.6 15/8 V Q.2 (a) 2. 0.22 23.13 1. (iv) 2.16 Q.23 ( T1 / 2 = 10.2 MeV (c) 0.20 54. 6.2E 0 α t − (1 − e −λ t ) λ ∆T = mS 19 .6 pm 2λ1λ 2 Q.67 × 10–15 J Q. (iii) 10–10 × 100 sec 81 Q. 13.11 λ + λ Q.ANSWER KEY EXERCISE # I Q. Q.0 eV 5.12 18/(5R) 1 2 Q.7 × 10 years Q.6 MeV 40 → 18 Ar + +1e0 + ν (ii) 4.7 487.034) or 46° Q.4 Q. Energy of Beryllium = 5.01366 amu (a) 4 MeV .7 (i) 5/16 photon/sec.8 × 1016.737 Å.8 eV.06 nm Q. 17. (c) 2. (ii) 2.23 5196 yrs Energy of neutron = 19.17 5 Q.0 mA when the potential is steady.2 × 109 years 1.6 eV. 340 eV.8 4.06 × 10–111 m 2π Q.8 nm Q.2 eV λ1λ 2 7 : 36 Q. 3 : 1 2 eV. 0.0213 µA (iii) 1.8 sec) Q. 36.257 × 10–23 Am2 ×10–12 Q.14 489.2 4.25 6.768 MeV . (b) 4.76 × 10–11 A Q.19 z = 3.26 m/s. photo electric emission just stop when hυ = (3 + 1)eV = 4. 4.1 × 1012 Q.15 3C re 2 t 1 e2 .22 Q.15 5 10 20 .6 (i) 1.25 t = ln 2 Q.28 – h/eEt2 EXERCISE # II Q. (ii) KE ≅ 151 eV.312 eV .4 (a) 105 s–1 .3 B.47 n = 24 Q.39 1. (ii) − E. energy = 10. Z = 3 ehB he Q.21 Q.84 eV and 16.546 × 1018 Hz 2 Q.38 A Q. (b) 6.35 A Q.34 C Q.18 5×107. (d) 111 s Q.C Q.12 during combination = 3. (d) E . (b) B .7 (i) (ii) 4πm 8πm Q.36 eV and 0.3 nm Q.45 C Q. (e) C Q. (iv) − C.27 (i) t1/2 = 10 sec. R 20 .6 eV.14 (a) n = 2.43 Q.31 (a) N = Q. (B) P.328 eV . .11 C Q. tmeans = 14.61 Å Q. G. R. dleast = 0. (c) ≅ 33. 4052. 24 Q.23 6 litre Q.5 D Q.40 C Q. 5 eV Q.8 0.5 Å Q. (C) S.23 x 1015 Hz .26 C Q. 6.88 eV (5 → 3) & 2.298 µW Q.20 z = 42 Q.846 x 1015 Hz . 2000N/C.6 x 1015 Hz Q.365 eV. (b) (i) − A. z = 4.1 (i) Allowed values of energy of neutron = 6.95 sec.75n = N0(1 – e–4λ).4 eV.17 3. (b) 286.33 D Q. Allowed values of energy of He atom = 17.24 (i) B.13 (a) C. Q.32 (a) C . 23 eV Q. 11. Q.25 A . (D) P.46 A. Q. (iii) − F.48 (A) P.10 (i) C (ii) A.44 ν = 1. Min.S.15 A Q.36 C 2 4 ln 3 Q.63 eV (4 → 3) Q.9 (i) D.19 A Q.22 C Q. C Q. P.41 A Q.EXERCISE # III Q.29 (a) B.28 Fusion . (ii) A Q. D 1 3 N0 .16 B Q. after combination = 3. (c) B .37 A Q.58 eV. − 217. (b) A Q.E. D Q. (ii) λ = 6. 2 N0 [α (1 − e −λt )+ λ N0 e−λt] (b) 2 λ Q. 0. (ii) 18.2 n = 6.30 (i) C. D (ii) D Q.18 .42 A Q.25×1019 per sec.66 Å Q. 9.6 (i) KE = 3.43 s (ii) 40 seconds Q. Que. Que. 34 Yrs. 10 Yrs. from IIT-JEE 8. from AIEEE 1 . Semiconductors Electronics Index: 1. Exercise III 5. Exercise IV 6. Answer Key 7.STUDY PACKAGE Target: IIT-JEE (Advanced) SUBJECT: PHYSICS TOPIC: XII P11. Exercise I 3. Exercise II 4. Key Concepts 2. ’ Energy bands. interactions occur between neighbouring atoms. with gaps in between. The wave function belonging to lower bands are not so highly localised. The band gap (Eg) and the nature of the filling of the energy levels (according to the Pauli Exclusion principle) are cheifly respeonsible for the electrical properties of solids. repeating mathematical points extending throughout space. therefore. Another reason for the existence of electrons within the conduction band is the thermal excitation of electrons from the valence band. and are responsible for the conduction of electricity by metals. A third type crystalline bond is a metallic bond: one or more of the outermost electrons in each atom become detached from the parent atom and are free to move throughout the crystal. This is very different from the potential feld by an electron within a hydrogen atom. there are important changes in the electronic energy levels and these changes lead to the varied electrical properties of solids. Band structure of solids As isolated atoms are brought together to form a solid. Crystalline solids & their electronic properties A crystalline solid is built around a lattice. this leads to conduction. 2 . Their wave functions a localized to a single atom and they are. Here we will deal with crystalline solids only. The electrons in these bands are responsible for the formation of inter-atomic bonds: the band is referred to as the valence band and the electrons. have progressively decreasing widths and have properties similar to atomic levels. valence electrons. Ge) having band gaps Eg–1eV behave as semiconductors at ordinary temperatures. An electron moving within a crystal lattice is subjected to a periodic potential due to the ionic cores present in the regular arrangement of the lattice. These electrons are known as “free electrons”. insulators and semiconductors The energy gap between the lowest level of the conduction band and the highest level of the valence band is known as the band gap (Eg). Materials (like crystalline Si. the regular. The energy levels (or the single atom: they are distributed in a band like structure.Electronics The study of matter in the solid state and its physical properties has contributed a lot to modern living– particularly. that are lower in energy than the valence band. When there exists a large number of electrons within the conduction band as a result of the filling process. The highest energy band wave functions are highly delocalized: an electron in one of these bands tends to be free enough to move over the entire body of the crystal. This band is known as a conduction band. Metals. where E0 is the band gap and k is Boltzmann’s constant. tightly bound. while those having large band gaps (Eg < 5 eV) behave as insulators at ordinary temperatures. Solids may be crystalline or amorphous –crystalline solids have long-range order in their structure while amorphous solids do not have such order. The probability of electronic at a temperature T(in Kelvin) varies as the factor e–Eg/2kT. Thus materials having very small gaps (E0 < 1 meV) behave as conductors. As this process occurs. they are localised to within a few neighbouring atoms of the lattice. The forces responsible for the regular arrangement of atoms in a lattice are similar to those in molecular bonds – covalent and ionic. The attractive and repulsive forces between atoms find a proper balance when the proper inter-atomic spacing is reached. the science of electronics. This current may be thought of as due to the motion of “hole” within the valence band. the “holes” imagined to possess a positive charge equal in magnitude to that on an electron. this motion leads to an effective uniform drift velocity for electrons within the electric field. Further. When an external electric field is applied to a semiconductor sample. the electrons ‘hop’ from one vacancy (“hole”) to another in the electric field E. a= The electron accelerates for a time τ. electrons from the valence band pick up thermal excitation from atomic motion within the lattice and this leads to a transition to the conduction band.67 eV Sol: At T = 300 K (or 273 + 27) 3 . Conduction occurs also in the valence band. the electrons within the conduction band experience a force proportional to the electric field. For each electron that transits to the conduction band. qe E . In this model (the Drude–Lorenz model). The acceleration of the electron is. m Due to collisions between the electron and and ionic cores within the lattice. before it loses its energy to the lattice in a collision. which is related to the probability of a transition of an electron from the valence band to the conduction band. referred to as a hole. if sufficient energy is transferred to the electron. m being the effective mass of the electron in the conduction band. vd = q eτ E = µeE (by definition) me where µe is the ‘mobility’ of the electron in the conduction band. This accounts for the fact that the “holes” move in an opposite direction to electrons within the valence band. within the valence band. This vacancy. The net current density within the semiconductor is given by: ne qe2τ e nn qe2τ n j = m + m E e n = qe(neµe + nnµn)E where nh and µh are the concentration of holes and hole mobility. a valency is left within the valence band. for two temperatures at 270C and 1270C. respectively. helps in conduction as well.67 eV. F = qeE. Here. ∴ σ = qe(neµe + nhµh) Illustration 1: Germanium has a band gap of 0. the collision time. where qe is the electronic charge. Calculate the value of the quantity e–Eg/kT. the current density j is given by n e q e2τ j = neqevd = E me ≡ σE (by definition).As the temperature is raised in a semiconductor. the average drift velocity of the electron is vd = aτ = qe E E me If there are ne electrons per unit volume. The band gap of Ge = 0. causing an electric current. 135 m2/V-S.135 + 5 x 1020 x 0. Such semiconductors are known as p-type semiconductors. these semiconductors do not have any electorns in the conduction band or holes within the valence band examples are pure crystalline Si. 4 . etc.048 m2/V-S). The concentration of electrons in the conduction band gets correspondingly reduced. the concentration of electrons within the conduction band (ne) equals that of holes within the valence band (nh) Intrinsic semiconductors are usually those which do not have any impurities within them.1 x 300 11600 ≈ 0. such that.026 eV & at T = 400 K (1270C) kBT ≈ 0. An energy level P (or As) lies just below the conduction band of Si.45 x 1012/ m3 and a hole concentration of 5 x 1020 / m3. If number density of atoms in the silicon sample is 5 x 1028 atoms/m3 then find the number density of Indium atoms in silicon per cm3.45 x 1022 x 0. on an average on Indium atom Exercise 1: per 5 x 107 silicon atoms. As) as impurities into the Si – lattice. These impurities are added in very small concentrations so that they do not change the Si-lattice. Ge. This leads to electrons getting transferred from the valence band into this acceptor level. Addition of small quantities of acceptor type impurities (trivalent group III elements like B) leads to an empty ‘acceptor’ level just above the filled valence band.0345 eV e–Eg/kT = 6. the introduction of holes into the valence band. σ = e(ne µ e + nh µ h) = 1. which form bonds) in P0. In Sb. also holds good here. The relation nenh = n12. Extrinsic semiconduction occurs due to the introduction of excess holes or. and thus. A silicon sample is made into a p-type semiconductor by doping.048) = 3.4 x 10–12 at 270C and 3. At absolute zero (T = 0). However. As. electrons into a semiconductor (Si for example).6 x 10-19 (o. this also results in a reduction in the number of holes. Being pentavalent. µ h = 0. there exists an excess electron (in addition to the four. Sol: The conductivity. Ga. Find its conductivity.7 x 10–9 at 1270C kBT = Intrinsic and Extrinsic Semiconductors For intrinsic semiconductors. Illustration 2: A semiconductor has an electron concentration of 0. nenh = n12 This type of Si with excess electrons is known as n-type Si.84 Ω -1-m-1. ( µ e = 0. The excess electron (in this donor level) of P is immediately transferred to the conduction band of Si: this results in an increase in the concentration of conduction electrons – ne. This is done by introducing microscopic quantities of Group V elements (P. This region is known as the depletion region. The forward current becomes significant only after V > 0. the p-n junction has a saturation current IS = 0. Therefore. Where the RHS represents the emf applied to the circuit. The excess elelctrons in n-type Si diffuse into the p-type Si and fill up the holes in the adjacent p-region. Illustration 3 : At a temperature of 300 K. 100 mV and -1 V. Find the current when the voltage across the diode is 1 mV. therefore devoid of electrons and holes therefore has very high resistivity.6 mA Rectifiers (i) Half wave rectifier A half-wave rectifier circuit consists of a diode D and the load resistance RL in series. applied across the diode and Is is the reverse saturation current. V = 100 mV. Thus the p-n junction. The reverse saturation current (Is) also depends on temperature. a p-n juction has a saturation current of 0. qe is the electronic charge (in magnitude).026 eV For V = 1 mV. a p-n junction acts like a diode (or a rectifier). through this dependence is rather weak.6 x 10-3A The current voltage relation for the diode is i = iS (eqV/kBT . When the p end is connected to a negative electrode an the n end to the positive electrode of a circuit the depletion region widens and the resistance increases tremendously due to the withdrawals of charge carries. Ge) the device is known as a p-n junction.6 mA. in practice. Voltage relation for a diode is i = Is e qeV / kT − 1 Where V is the forward p. k. i = 23 µ A. almost.1) kBT (at 300k) ≈ 0. If Vk is the knee-voltage of the diode ( ≈ 0. i = 27. a current flows across the junction easily. ∴i= V0 sin ωt − VR and i > 0 RL 5 . ( ) The current vs. as shown in the adjacent diagram.d. Is is of the order of a few µA to a few mA depending on the diode. does not conduct in the reverse direction. and this is known as the knee voltage. A small region adjoining the junction is.7 V (for si-diodes).7 V for si diode) and I is the current flowing during forward bais: iRL + VR = V0 sintωt.p-n Junction When a p-type semiconductor is joined to an n-type semiconductor (both Si. Solution: At a temperature of 300 K.5 mA and V = -1V. On the application of a forward electric field (p to positive & n to negative) the width of the depletion region is reduced and concequently. the Boltzmann constant and T. the absolute temperature. i = -0. 65 x 10-3J Exercise 2: In the full-wave rectifier circuit. 6 .7 V in the forward biased condition. A voltage waveform as shown in figure is applied to the circuit. D2 are ideal and identical. sin–1 (Vk/V0) < ωt < π – sin–1 (Vk/V0) dueing the 1st half cycle. when sinωt > VR V0 or. An ac–voltage Vs = V0sinωt is applied across the circuit as shown.7 The current (maximum) = = 9. the potential drop across the diode is 0. and the rest of the p.d. What is the maximum current? Calculate the average heat lost in the resistance over a single cycle. Solution: (a) In forward bias.3 x 10-3)2 x 103 x 10-1J = 8.7 V. (ii) Full wave rectifier A full wave rectifier circuit is shown in the adjacent diagram. the diodes D1. Illustration 4: (a) (b) A p-n juction forms part of a rectifier circuit. Plot the current through the resistor as as function of time.The diode is in forward bias. The current i flowing in the circuit. The maximum reverse voltage across a diode is twice the peak forward voltage. It consists of two diodes D1 and D2 connected to a load resistance RL. If the diode is ideal except for a drop of 0.3 mA 1000 (b) The average heat lost in the resistance over a single cycle is i2R ∆ t=(9. is dropped across the resistance R (=1k Ω ) 10 − 0. The emf Vs = 100 sin (100 π t) volt is applied as shown (t is in sec). The current through RL is just a in the case of the half-wave rectifier except that it flows during both the half-cycles. Calculate the voltage across the diode D1 as a function of time. The current through the load resistance is not a smooth dc. in the emitter current IE. pnp or npn. The npn transistor consists of a very thin piece of p-type material sandwiched between two pieces of n-type. which constitute the base current IB.Transistor Transistors are semiconductor devices capable of power amplification. Basically. The base is lightly doped compared with the emitter and the collector. two circuits are formed. These are the active state. the cut off state and the saturation state. Also known as the bipolar function transistor (BJT). The active state is the basic mode of operation. A transistor consists of a thin central layer of one type of semiconductor sandwicthed between two relatively thick pieces of the other type. vix. The pieces at either side are called the emitter and the collector respectively while the central part is known as the base. Almost all the electrons emitted from the emitter are collected by the collector. Since the collector junction is reverse biased. The thickness of base 7 . and is only about 3-5 µm thick.E. depending on the voltage across its junctions. The cut-off and saturation states are typical of transistors operation in the switching mode. The electrons that make up this current are injected from the emitter into the base and diffuse through the base into the collector region. it can be of two types. thereby boosting the collector current.. while the pnp transistor has a central piece of n-type.e. One is the input and the other is the output circuit. (ii) Working of a transistor In amplification we bias B. Base emitter junction is forward biased hence electrons are injected by the emitter into base (n-p-n). junction in forward and C-B junction in reverse. i. But a small fraction of electrons recombine in the base region. (i) Biasing of a transistor A transistor can operate in any one of the three states. the electrons are swept to the collector. It is utilized in most amplifiers and oscillators. in any application using a transistor. State Active Cut off Saturation Junction Emitter Base Base collector FB RB RB RB FB FB Where FB – Forward biased. Operation of an npn transistor: An increase in the forward input voltage VBE (across the emitter-base junction) brings about a fall in the height of the potential barrier at the emitter junction and an increase in the current flowing across that junction. RB – Reversed biased. R i BE where RL is load resistance RBE is input resistance. not by the baising circuit. Generally we use transistor for amplification in common emitter mode and common base mode. The reverse biased CB junction sweeps off electrons as they are injected into the junction. the load is connected between the collector and the emitter through d. we can write. the emitter-base junction is forward biased (FB) and the base collector junction is reverse biased (RB). signals are amplified. In a common-emitter (CE) amplifier. Since IE = IB + IC ∴ 1 α IE IB = +1 IC IC = 1 β +1 α 1−α In an amplifier a. Therefore. the collector – emitter current gain is defined as IB α= IC . Power gain = voltage gain x current gain RL =β R BE ∆I C Thrans conductance is defined as gm = ∆V BE (iii) Transistor as an Amplifier In order to use a transistor as an amplifier. IE = IB + IC Where IE is emitter current. IE The parameters β and α for a transistor are decided by the constriction. the doping profile and other similar manufacturing parameters.region in very small. We get β = β ac = ∆I C ∆I B Vo RL We get voltage gain V = β . supply. current gain is defined as β= IC . By using Kirchhoff’s law. but less than 1.c. IC is collector current. Since the current gain is β . as a result most of the electrons diffusing in to base region cross into the collector base juction. IB is base current.c. β is very large (nearly 100) and. α is very close to 1. The collector-base. 8 . 2)V = 0. if VS= 0.VC = 0 ⇒ VCE = (5. This changes VBE by an amount ∆VBE =Vs. Thus..β acRL ∆ IB The voltage gain AV = vo ∆VCE β R β = = − ac L = − g m RL Where g = ac = transconductance.3V .5... m ri vi ∆VBE ri Illustration 5: In the following circuit the base current IB is 10 µ A and the collector current is 5. ∆ VC = ∆ VCE + RL ∆ IC=0 vo = ∆ VCE = -RL ∆ IC = ..2)V .(1) ⇒ vC < vB loop . then VB + VS = VBE + ∆VBE when The change in VBE can be related to the input resistance ri and the change in IB..5. VS = ∆VBE = ri ∆I B The change in IB causes a change in IC. β ac = ∆I C ∆I B (current gain factor) The change in IC due to a change in IB causes a change in VCE and the voltage drop across the resistor RL because VC is fixed. VC = VCE + ICRL Similarly. Assuming VBE ≈ 0. calculate (a) Base current IB (b) Potential difference between collectors and emitter terminals.3 V ⇒ VC > VE as emitter is grounded..2mA.An a.5 .. Thus. In the given question we are required to just check this loop-1 IBRB .05 . Applying Kirchoff’s voltage law on the output loop..(2) from (1) and (2) VB = 5. Can this transistor circuit be used as an amplifier? In the circuit RB = 5 Ω and RL = 1 K Ω Solution: We know that for a transistor is CE configuration to be used as an amplifier the BE junction must be forward biased & base collector junction must be reverse biased. VB = VBE VS ≠ 0 .2 x 10-3 x 103)]V ⇒ (0. 9 .. input singnal VS is superimposed on the bias VBE.VCB = 0 ⇒ VCB = [(5 x 103 x 10 x 10-6) .5 V ⇒ BC junction is forward biased & hence the given is transistor would not word as an amplifier Exercise 3: In the transistor circuit shown in figure direct current gain of the transistor is 80. VC = 0. The output is taken between the collector and the gournd.2 VCE + ICRL ..ICRL .(5.c.. vo = Avvi = Av(vs + vf) or vo + kvo vo = Av A'v or A'v = Av 1 − kAv By properly adjusting the feed back. (v) Analysis of transistor circuit Input KVL.. The voltage across the Z1 is feedback in the input of oscillator. This results in a change of 0. The voltage gain of the amplifier is Av = vo vi vo and the overall gain is A'v = v s Now. The output frequency of an oscillator is the resonating frequency of L-C network which is given as f o = 1 2π LC From the figure... vi = vs + vf Z1 where vf = kvo = Z + Z vo 1 2 k is feedback constant which represent the fraction of output voltage which is to be feedback to input. it is possible to get kAv = 1 which gives A’v = ∞ . Exercise 4: In a silicon transistor.. VBB = IBRB + VBE . The amplifier is just a transistor used in common emitter mode and the LC network consists of an inductor and a capacitor. Therefore. VBB = IBRB + VBE + IERE usually RE = 0. Find the voltage gain of the amplifier. or we get an output without applying any input.. Z1 and Z2 works as voltage divider. An amplifier and an LC network are the basic part of the circuits. The oscillator generates an ac signal. This can be achieved by feeding back a portion of the output voltage of an amplifier to its input terminal as shown in the figure.(iv) Transistor used in an oscillator circuit The function of an oscillator circuit is to produce an alternating voltage of desired frequency without applying any external input signal.. the base current is changed by 20 µA .02 V is base to emitter voltage and a change of 2 mA in the collector current. (a) Find the input resistance ri and β ac of the transistor.(1) 10 . (b) If this transistor is used as an amplifier with the load resistance 5 kΩ . RC Vo I C RC Therefore voltage gain AV = V = I R = β R i B B B . Vi = IBRB. Find the voltage gain of the amplifier.. assume β = 60 and input resistance Rin = 1000 Ω . current gaing of the transistor output voltage. VCC = ICRC + VCE + IERE for RE = 0.(4) IC = α IE Where α is the a..c. Vo = ICRC.. VCC = ICRC + VCE .(3) Where β is d...For silicon transistor VBE = 0. 11 ...(5) Exercise 5: In the circuit shown. ... IC = β IB ... input voltage..c..7 V Output KVL. current gain of the transistor..(2) current relationship... Documents Similar To Doc 126 B.P.S. 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