Tutorial 6: Solutions1. A stationary radiating system consists of a linear chain of parallel oscillators separated by a distance d. The phase of the oscillators varies linearly along the chain, Find the time dependence of the phase difference ∆φ between neighbouring oscillators such that the principal maximum of the radiation will be scanning the surroundings with a constant angular velocity ω. 1. Figure 1 shows that the radiating system which consists of chain of parallel oscillators separated by a distance d. FIG. 1: Linear chain of parallel oscillators Consecutive sources have a phase difference ∆φ at their position. The path dif- ferences in the radiation from source 1, 3, ......(N − 1) with respect to 0th oscillator, are d sin θ, 2d sin θ, .....(N − 1)d sin θ. 2π {d sin θ, 2d sin θ, .....(N λ − 1)d sin θ}. Let, Consequently, the phase differences are: 2π d sin θ λ = α. So, the resultant phasor at an angle θ, P = E 1 + ei(α+∆φ) + e2i(α+∆φ) + ........... + e(N −1)i(α+∆φ) 1 − eiN (α+∆φ) = E 1 − ei(α+∆φ) 1 ±2.. A fringe pattern appears on a white screen held 1. and d = 0. both numerator and denominator of the intensity expression becomes zero and we have ≡ 00 form and that will give a finite maxi- mum value (using L’Hospital rule). Here. D = 1. for (α + ∆φ)/2 = kπ. 2π d sin(ωt + α) λ ∆φ = 2kπ − 2.). (k = 0.1mm at a distance of 1m is used to illuminate two slits with light of wavelength λ = 550nm.8nm) is incident on a screen containing two narrow horizontal slits separated by 0.8nm. Therefore the first zeros of intensity will occur at (m = 0) y= 2 Dλ 2d (1) . . ym The symbols in the above formula have their usual meanings. So the condition for principal maximum to be observed at an angular position θ is 2π d sin θ λ Now.. (α =arbitrary constant).200mm. The slit separation is d = 1mm. (2m + 1)λ 2d (2m + 1)λD ≈ 2d θm ≈ or. ±1. (b) A small aperture of diameter 0.00m away. if the principal maximum ‘scan’ the surrounding with a constant angular velocity ω. ∆φ = 2kπ − then θ = ωt + α.. λ = 632. So finally we have. What is the fringe spacing and the expected visibility of the fringe pattern? i) We know that for m − th order dark fringe. (i) How far (in radians and in millimetres) above and below the central axis are the first zeros of intensity? (ii) How far (in mm) from the axis is the fifth bright band? (iii) Compare these two results. (a) An expanded beam of red light from a He-Ne laser (λ = 632.0m.200mm...So the intensity I = PP∗ 2 1 − cos N (α + ∆φ) = E 1 − cos(α + ∆φ) = E2 N (α+∆φ) 2 2 (α+∆φ) sin 2 sin2 So. 58 cm (5) iii)The spectrum is equispaced.) a light wave emitted directly by the source S interferes with the wave reflected by the mirror M.(b) Here. As a result an interference pattern is formed on the screen Sc. λ = 550nm. Hence α = 10−4 rad.60mm. Young’s double slit arrangement it becomes (for derivation see the coherence chapter of the book by SB and SPK) V = sin πdα λ πdα λ where α is the angular extent of the source.58 mm (2) θ = 1. Here the diameter of the aperture is 0. The source and the screen are separated by a distance l = 1m. y = 1. 2. 3 . So using all other numerical values we get V = 0. Hence. After the source is moved away from the plane of the mirror by ∆h = 0. At a certain position of the source the fringe width is equal to ∆x = 0. the fringe width decreases by a factor η = 1. So.5 × 10−4 rad ∆θ = Visibility is defined as V = Imax − Imin Imax + Imin For. the position of the fifth bright band is (1.58 × 2 mm away. the first bright band from the axis is 1. Find the wavelength of light. the angular fringe spacing is λ d = 5.95. 3. (a) In a Lloyds mirror experiment (see Figure 2. and d = 1mm.58 × 2) × 5 mm = 1.58 cm away.6 × 10−3 rad (3) ii) The fifth bright band (n = 5) will be at y=n Dλ d (4) So. We know.25mm.1 mm and the distance of the slit position from the aperture position is 1 m.5.y = 1. 1m. From the above equation we find that lλ ∆x = η d + 2∆h (7) Since d increases to d + 2∆h when the source is moved away from the mirror plane by ∆h Thus. Behind the biprism there is a plane parallel plate.0cm and b = 130cm.λ = 550nm. r = 0. Hence number of possible maxima 2bα +1≈9 ∆x 4 (9) . Therefore ∆x = (b + r)λ 2α (8) putting b = 1. (iii) The maximum width δmax of the source slit at which the fringe pattern on the screen can still be observed sufficiently sharp. d is the separation between the source S and its image. The wavelength of light is λ = 0. (ii) The shift of the interference pattern on the screen when the source slit S is displaced by δl = 1.70m falls normally on the base of a biprism made of glass (n = 1.3m. 2∆h η−1 2∆x∆h λ = (η − 1)l d = and Finally λ = 0.520) with refracting angle θ = 5degrees.55mm. Find: (i) The fringe width on the screen and the number of possible maxima. (c) A plane wave of light with wavelength λ = 0.(b) Consider the experiment with Fresnels mirrors. the angle between the mirrors α = 12.6µm (b) (i)Here S ′ S ′′ = d = 2rα and l = b + r.1m.0mm along the arc of radius r about the centre O.500). α = 12′ we get ∆x ≈ 1. Find the fringe width on a screen placed behind this system. for Lloyd’s mirror arrangement. with the space between them filled up with benzene (n = 1. the distances r and b are r = 10. (a) We know the general formula for fringe spacing ∆x = lλ d (6) Here. The full patern will not be sharp at all if the pattern due to narrow slits are 12 ∆x apart because then the maxima due to one will fall on the minima due to the other. Each of these prisms are oppsitely placed back to back with each halves of the Biprism. The fringe pattern due to wide slit is the superposition of the pattern due to these two narrow slits. n = 1. θ = 5o = π/36 rad.(ii) When the slit moves by δl along the arc of radius r the incident ray on the mirror rotates by δl . n′ = 1. Here we have to assume that the space between the biprism and the glass plate. filled with benzene. Then the two prisms being oppositely placed.2mm 5 . where a is the distance of the position of the source (S) behind the Biprism and δ is the angle of deviation for each refracting halves of the Biprism. So δx ≈ 0. constitutes two complementary prisms. Then above equation becomes ∆x = Given that λ = 0.500. the net deviation produced by them is δ = (n − 1)θ − (n′ − 1)θ = (n − n′ )θ Hence d = 2aθ(n − n′ ) So ∆x = (12) (a + b)λ 2aθ(n − n′ ) (13) λ 2(n − n′ )θ (14) for plane incident wave a → ∞.520. r Therefore the shift of fringe will be of magnitude b δl = 13 mm r (10) (iii) If the width of the slit is δ then we can imagine the slit to consist of two narrow slit of separation δ. we know that d = 2aδ. Thus we can say that the condition for the fringe pattern be observed sufficiently sharp is δ ∆x b ≤ r 2 So r λ = 42µm δmax = (1 + ) b 4α (11) c) For a Fresnels’s Biprism arrangement.70 µm. Hence.33 at which light with wavelength 0. For the smallest integer solutions. (b) A thin film of alcohol (n = 1. So.65µm (b) Both of the reflections for interference occur from rarer to denser medium–one reflection from air(n0 = 1. Equatting above two equations we get an equation 5k ′ = 4(2k + 1). Given that λ1 = 640 nm. condition for maximum and minimum are 2n1 d cos θt = pλ1 (for maximum) 1 = (q + )λ2 2 (for minimum) For normal incidence cos θt = 1. Hence we equating the condition we get 5p = 2(2q + 1). For the smallest integer solutions p = 2. For air medium n0 = 1 and given that θi = 30o .36) and the other one from alcohol (n1 = 1.51). What is the thickness of the film? (a) For interference of reflected wave from a thin film of thickness d and refractive index n.47 µm. in both of the reflected waves. λ2 = 512 nm. Then the minimum thickness is d = 0.4.51). k ′ are integers.36) lies on a flat glass plate (n = 1. the reflected light is a minimum for λ = 512nm and a maximum for λ = 640nm. Given that n = 1.0) to alcohol(n1 = 1.64µm experiences maximum reflection while light with wavelength 0. whose wavelength can be changed. for maximum reflection in illumination r 2d n2 − 1 1 = (k + )0. θt is the angle of transmission.40µm is not reflected at all.64µm 4 2 and for minimum illumination in reflection r 1 2d n2 − = (k ′ )0.36) to glass (n2 = 1. is incident normally. If θi is the angle of incidence. k ′ = 4. The angle of incidence is equal to 30degrees. k = 2. (a) Find the minimum thickness of a film with refractive index 1. 6 . q = 2. the condition for maximum and minimum of the fringe pattern is 1 2nd cos θt = (k + )λ (for maximum) 2 = k ′ λ (for minimum) where.33.40µm 4 (15) (16) where k . Hence minimum thickness will be d = 0. Hence. additional π phase difference will be introduced. then from Snell’s law p cos θt = 1 − n20 sin2 θi /n2 . When monochromatic light. 5nm 4n1 (17) ii) and the thickness for observing minimum reflection is d= λ = 125nm 2n1 (18) (b)For maximum intensity. Now. from Eq.5. i)Hence the coating thickness of this film to observe maximum reflection is d= λ = 62.5. the conditions for maximum and minimum reflection are those given in the solution of 4.(a).60m falls on a thin file of refractive index n = 1. (19). we have −2nd sin θt δθt = ∆kλ (21) For neighboring maxima ∆k = −1. 5(a) A glass plate(n2 = 1.5) is coated with a polymeric film (n = 2). (b) Diffuse monochromatic light with λ = 0. Here both sides of coating film are rarer medium. (20). (a) A glass plate of (n = 1. (20). p λ n2 − sin2 θi d= sin 2θi δθi 7 (22) (23) . From Eq. δθt = p cos θi n2 − sin2 θi δθi So.0 degrees. Hence. Determine the film thickness if the angular separation of neighboring maxima observed in reflected light at angles close to 45 degrees to the normal is equal to δθ = 2. Calculate the coating thickness so as to observe (i) maximum and (ii) minimum reflection using a light of wavelength λ = 500nm.(21) and (22) we get the finally. n0 = 1. 1 2nd cos θt = (k + )λ 2 (19) n0 sin θi = n sin θt (20) From Snell’s law For air to film surface reflection.5) is coated with a polymeric film of n1 = 2 and the wavelength is λ = 500nm. using Eqs. 00022radian = . The experimentalist observes 12 fringes per cm. n = 1.013degree 2n∆x (25) 7. for normal incidence we know. and λ = 546nm If 11 th bright ring(k = 10) of λ1 coincide with the 10 th ring (k = 9) of λ2 . What is the refractive index of the liquid? (7)(a) The formula for a bright ring in the fringe pattern of Newton’s ring arrangement is 1 λR rk2 = (k + ) 2 n (26) For air film.83mm. radius of curvature R = 1m.Given that. then 21λ1 = 19λ2 8 (27) .60 cm. Here. (6)For wedge shaped film d = αx.Under the influence of gravity. So. θi = 45o . hence ∆x = 0. One of the wavelengths is 546 nm. A coherent and monochromatic light of wavelength λ = 488nm falls near-normally on this wedge. ∆x = λ 2nα (24) where α = wedge angle of soap film. If the 11th bright ring of the 546 nm fringe system coincides with the 10th ring of the other.6 µm. the diameter of the eighth dark ring decreases from 2.34. λ = 0. Hence α= λ α = . Determine the wedge angle of the soap film. The experimentalist observes 12 f ringe/cm. λ = 488nm.5.74 µm. n = 1. δθi = 2o = π/90 rad. 6. (i) What is the second wavelength? (ii) What is the radius at which overlap takes place and the thickness of the air-film there? (b) When a Newtons ring apparatus is immersed in a liquid.shaped film is formed when a metal is vertically dipped inside a soapy solution (refractive index = 1.92 cm to 2.34). d = 22. n = 1. i)Here it is given that. a wedge. where x is the distance of the point of incidence of light from the apex. *(a) A Newtons ring apparatus (comprising of a spherical surface of radius 1 m) is illuminated by light with two wavelength components. 92 cm to 2. Calculate the index of refraction of the gas at its final density. a total of 186 dark fringes are counted to move past a reference line.60 cm Since we know the diameter of the m th dark ring is (Dm )2 = 4mλR (30) when film is air if the intermediate medium is liquid of refractive index of n.(a) An interference filter is designed for normal incidence of 488 nm light.5nm ii) Therefore the radius at which overlap takes place is r = 2. When a gas is allowed to slowly fill the container. If an observer uses a spectrometer and looks at the reflection of white light from the surface of the filter at 30 degrees with respect to surface normal.26 8. what wavelength should the observer see? *(b) One of beams of an interferometer passes through a small glass container containing a cavity 1. Then λ = λ0 r 1− Hence observed wavelength will be λ = 453. The refractive index of the spacer is n = 1.87 µm 2R (29) b) The diameter of the eighth dark ring decreases from 2. The light used has a wavelength of 610 nm. 8 a)interference filter is designed for normal incidence of λ0 = 488 nm light. then diameter of the m th dark ring will be (Dm )2 = 4mλR n (31) From the above two equations we get the refractive index of liquid is n = 1. The refractive index of the spacer is 1.4mm (28) thickness of air film at that point will be t= r2 = 2.30 cm deep.2 nm 9 sin2 α n2 (32) .35. assuming that the interferometer is in vacuum.35 If the observer wants to look the reflection at α = 30 degrees with respect to surface normal.Hence λ2 = 603. 30cm.2945mm and the mean wavelength of the two source is λav = 548. the order of sixth dark ring from the center will be (m0 − 6) = 79994.0088 9. 2d = (λav )2 λ1 λ2 ≈ λ1 − λ2 λ1 − λ2 10 (36) . calculate the difference between the two wavelengths. m0 = 2d ≈ 80000 λ (35) ii) The equation for a dark ring is 2d cos θm = mλ m = m0 corresponds to the central disc. Since. the distant traveled by the mirror is d = 0. (b) In an experiment with Michelson interferometer. we get the value of index of refraction of the gas is n = 1.3 nm. the distance travelled by the mirror for two successive positions of maximum distinctness is 0. One arm of the device is 2 cm longer than the other.3nm.(a) Looking into a Michelson interferometer one sees a dark central disc surrounded by concentric bright and dark rings. As the central fringe is dark. and the wavelength of the light is 500 nm. If the mean wavelength for the two components of the source is 548. Determine (i) the order of the central disc and (ii) the order of the 6th dark ring from the center. i) Hence. we can write 2d = m0 λ (34) where m0 is the order of the central disc. 9 a) Here one arm of the device is d = 2cm longer than the other and the operating wavelength is λ = 500nm. Hence.8 b) Here it is given that l = 1.2945 mm. We know (n − 1)l = mλ (33) putting all the given values. b) In this experiment. λ = 610nm and order of fringes move are m = 186. we know for concordance. c) The chromatic resolving power is defined as √ λ = 1. The uncoated surfaces of the slab have a reflectance (r2 ) of 0.hence λ1 − λ2 ≈ 0.5cm and the reflactance of uncoated surface is r2 = 0.5 cm. 10) the refractive index of the slab is n = 4.5nm (37) 10. (b) the ratio (Imax /Imin ). determine: (a) the highest order fringe in the interference pattern. (c) the chromatic resolving power.5 and thickness is d = 2.90.515m0 F ∆λ (40) where coefficient of finesse (F) can be calculated from F = 4R = 360 (1 − R)2 (41) Hence resolving power comes out as λ = 2632332 ∆λ 11 (42) . a) the highest order fringe in the interference pattern is m0 = b) 2d = 91575 λ (1 + R)2 Imax = 361 = Imin (1 − R)2 (38) (39) Where R is reflactance and R = r2 .5) and a thickness of 2. If the slab is used in the vicinity of wavelength λ = 546nm Then.90.A Fabry-Perot etalon is designed from a single slab of transparent material having a high refractive index (n = 4. If the slab is used in the vicinity of wavelength 546 nm.