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Two lamps ol points on the linc passing through them is the illumination ,1,,. 1,1 ,,r' I' (Mudrrt.t l't't'1 of thenl the samc ? from lm CP lamp in bctrvccn llt'' lrr" f Ans. ti) 66.67 cm line joining thc trvo sources (ii) I metre ft11111 ')'r ( l' lamp and 2 nretres tAns- /7.88 lttmetls/rt1 trrr tr' 100 CP and 25 CP are placcd onc metrc ap:r|l /\l \\lr'rr 1 ' from l(X) (il' I'rrrr1' 7.I NEWTON'S CORPUSCULAR propagation THECRY 1 1.1 Write short llotes on : (i) (r,) f'lickcrphotometer [.unrmer-Brodhum photometer. (l'lysort l't"t, Thebrarrchot-cpticsthatdealswiththeproductic'n'emissionand phenomena of interof light, iL nature and the stucty of the optic-s. The basic physical calred ference, criffraction anJ foiarir"tion is of light were formulated in the latter half ;;i;6Lt ;"garding the nature the general belief was of the seventeenth century' Until about this time' These cor,t"iilgf,, consisted of a stream of particles called corpuscles' sun candle' a (an electric lamp' puscles were given out by a lighi''out*" originator The velocities. i".i "ia theyiravelled in straiglt lines rvith large Newton' According lsaac Sir was theory of the emission or .o'p*"olit emits tiny, light and elastic to this theory, a tuminirs body continuously_ particles or corpuscles corpusctes in att directions' These particles called aresosmallthattheycanreadilytravelthroughtheinterstices.oftheparand they possess the property ticles of matter with the verociry of light ofreflectionfromapolishedsurfaceortransmissionthloughatlansparent medium.Whenthesep*,itf*fallontheretinaoftheeye'theyproduce this theory' phenomena like recthe sensation of vision' On the basis of could be accounted for' sattilinear propagation, reflection and refraction high speed from a luminous with isfactorily. Since the pardcles are emitted travel in straight lines according body, they, in the absence of other fu!::' rectilinear propagrtion to Newtoll's seconcl law of motion' This explains of light. REFLECTION OF LIGHT ON CORPUSCULAR 7.2 corpuscle comes within a very approaching the surface S9' Wtren the THEORY Letss'beareflectingsurfaceandlMthepathofalightcorpuscle sntall distanceftomthesurface(indicatedbytherlottedlineAs)it'according of repulsion due to the surface to the theory, begins to exierience a force (Fig.7.l) 244 , A Texrbook ol O1tttt..t Narure of Light 245 parallel component x remains the sarne throughout. Thus at t/, the corpuscle again possesses two comof velocity x and 1. and the resultant direction of the corpuscle ponents lf. its magnitude will be again y at l/ but in th'e opposite direction. The The velocity u of the corpuscre at M can be resorved into two c.rrr ponents x and y paralrel and perpendicurar to the reflecting surfaee. Th. lbrce d repulsion acts perpendicular to the surface .ss' anJ consequenrry the component y decreases up to O and becomes zero at O the point of incidence on the surlace SS'. Beyond O,the perpendicular component of the velocity increases up to As the parallel component of the velocity remains the u sin i = u'sin r same, 'lI I particle moves unaffected by the p.es"n"" surfaces AB and ,S.f, the path of the corpuscle is convex to the reflecting surface.'Beyond the point ry', tht: is along /VR. The velocity of the corpuscle will be u. Between the Ne Fis. 7.1 and the reflected path of the corpuscle and the normal lie plane viz. the plane of the paper. Further, the angres between the incident and the reflectecr paths d the corpuscles with the normals at M and N are equal. Arso, thelncidenr of the surface S,f. x = u sini = usinr, .'.1=r Fig. 7,? . .,r in the samt. srnr _ sln-l u' a velociw of light in the second mediurl velocity of light in the first medium (refractive index ofthe second medium wilh reference to the the first medium) incidence bears a constant ratio to 73 ' THEoRy REFRACTION OF LIGHT ON CORPUSCULAR = ,p, of the angle of incidence, but which is different for different marerial.s ['et u and u' be the velocity of the corpuscre in ihe two media and r tht, angle of refraction in the second medium. Newton assumed that when a light corpuscle comes within a very srnall limiting distance from the refracting surface, it begins to experienct. a force of attraction towards the surface. consequently the component .r ttrc_velocity perpendicular to the surface increases gradualry from AB t, A'Y.'ss is the surface separating rhe rwo media (Fig. 7.2). IM is thc irrcident path of the corpuscre traveliing in the fir.st medium with a verocity u and incident at an angle i. AB to A' E is a narrow region withih which the.corpuscle experiences a force of attraction. NR is the refracted path the corpuscle. Let u sin f and u cos i be the components of the verocity of the coqpuscle at M p,.ailer and pe,rrendicular to the surface. The vc locity parallel to the surface increasei by an amount which is inclependenr Thus, the sine of the angle of the sine of the angle rf refraction. This is the well known Snell's law of refraction. lf i > r, then u' ) a. i.e., the velocity of light in a denser medium Iike water or glass is greater than that in a rarer medium such as air. 'f But the results of Foucault and Michelscrn on the vclticity of light show that the velocity of light in a denser mediurn is less than that in a rarer medium. Newton's corpuscular theory is thus untenable. This is not the only ground on which Newton's theory is invalid. In the 1'ear 1800, Ytrung discovered the phenonrenon of interference of light. He expcrilnentally demonstrated that under certain conditions, light when added to light produces darkne.ss. The phenomena belonging to this class cannot be explained, if following Newton, it is supposed that light consists of nrirterial particles. T[vo corpuscles coming together cannot destroy each other. Another case considercd by Newton was that of simultaneous reflection and refraction. To explain this he assumed that thc particles had fits 246 - A Ta.ttln,'k ol' ( )1ttit..r Nautrc o.l' I'ilitrt )'11 so that some were in a state favourable to reflection and others wcrc ln a condition suitable for transmission. No explanation of interferencc, rlil fraction and polarization was attempted because verv little was kn.w, about these phenomena at the time of Newton. Further, the corpusculirr theory has not given any plausible explanation about the origrn of the lirrt.t. of repulsion or attraction in a direction normal to the surface. 1.5 }IAVE NTOTION 7.4 ORIGIN OF WAVE TIIEORY The test and completeness of any theory consisrs in its ability to ex plain the known experimental facts, with minimum number of hyptthese, From this point of view, the corpuscular theory is above all prejudices antl with its halp rectilinear propagation, reflection and refraction could be ex Bctitre procee.ling to study the various optical characteristics of sinrple hannonic basis of tiuygens rvave the<lry, the lntltion) and the composttton ttr ,nuti,,rn lthc-iinrpl.,st ftlrrtr of wave motions arc disctrssed' The superposition of $vtl ()r nlore sirnple harmonic harmonii wave through a medium ca.n be trrtnsf.op,g"tinn of a simple verseorlongitudinat.lnatransversewilve'theparticlescrfthemediunl propagation ancl in ;r lunsitudinal vibrate perpcnclicular tcl the direction of parallel to the directirln of propavibrate jaflicles of the medium *ou",,to surlace of still water' trilnslerse gation. Wlren a stone is tlropped on the through atlnosphcric"air is in the waves arc llrrxluce<i. t'-p^soiiun of souncl prropagated throu-eh a tnerliunt' fbrnl of longitudinal waves whcn a 'qave is theparticles<rfthen)cdiurnare<lisplacedfrcrmthcirrnearlgrsitirlnstlfrest 'lttese resto;ing forces a'e-due to thc ancl resttlring lirrces c<lme int() piay' gravity ani s'rfac" tension' Due to the ptrirxiic ela-sticity phenomena tln the plained" longitudinal, in which the vibration of the particles is parallel to th* di recfion of propagation of the wave. Assuming that energy is transmitted in the form of waves, f{uygens could sar.i.sfactoriiy explain refleclion, refraction and double refraction nr>ticed in crystals like cluartz or calcite. Howe,ver, the ptrenomenon of p. larization discovered by him c<lukj not be expiaine<J. It **, difficult t, conceive unsymmen'-ical behaviour of longitudinal waves about the axis .r propagatir:n. R.ectilinear propagation of light also coukl not be expiainctl on the basis of wave theory, which otherwise seems to be obviclus ac cording to corpuscular the'ry. The difficulties mentione<l above were clr,cr corne, when F'resnel and Young suggested that light waves arc lransvrirst, and nOt longitudinal as suggested by Huygens. In a transverse wavc, rht. vibrations of the ether particles take place in a dirrction perpendicullr r. the.direction of oropagation. Fresnel could also explain .successlulll, rtr,. rccdlinear propagation of light by combining the effect of all thc sec.ntr.rl waves starting froln the different points of a prim;rry wat,e front. By about the middle of the seventeenth century, vihile the corpusc,rar theory was accepted, the idea that light might be some sort of wave motion had begun to gain ground. rn 1679, christian Huygens proposed the wavc theory of light. According to this, a ruminous body is i roro." of distur bance in a hypothetical medium cailed ether. This mediunr pervaries arr space' The dislurbance from the source is propagated in the fi:rm of wavcs thrcugh space and &e energy is clistributed equally, in all directions. wherr these rvaves carrying eneigy ,:re incident on the eye, the nptic nerves arc excited and the sensation of vision is produced. These vibrations in thc hygnthetical ethe.r medium actording to Huygens a.re similar to those pr<r duced in solids and liquids" They iue of a mechanicar narure. The hyp.thetical ettrer mediun"l is attributed the property of transmitting elasrre waves, which we perceive as light. Huygens assurned these waves to bt, of the medium, motionofthepartictesof-themecliunt,awilvernotjonisl:rcxluctxl'Atany inrtont,thecontourofallthepafficlesofthemediunlconstitutesilwave' [,ctl,beaparticleritlvingonthecircunrf.ertnceofacircleofl,lrriius aitgullrr a with a unitilrnl velrrity u lflg 13)' Lct til be the unifi;rm veiocit), of thc particlc (zr = rl tu )" Ther circie aloirg ',vlrich /r m()vcs is cal led the cirr;le of reference. As the particlt Ij flx)ves rilund the circle continuoirsly rvith urrifttrrn vivelocitl', the iir<lt oi the perpcrrriicular A/' If x ('i:X'i i:r YY' brates aklng thc cliarneter rrtotitx the then unitbrm' the m1;{ion ol f is trl' ,1/ i: pcrirx-liu i.c. il tlkcs lhc sttnt: tirnc to vibrate once betrveen the points l' and )/ At an1,' instant, thc distancc ol" '!/ liont rhr: centre 0 of the circle is calletl the riisFii:. i 1 piacenrent. ti the particie Inoves from X t<r P in rir"nr: r, ihen / POX = /-fu1P0 = 0 = o)I' []rotr: the \MPA' srn {J =' sin or ) OM I = -,,- r)r OM =') = a sin rrl / 0Miscaliedtheclitplacementofthevibrating;rarticl*'"lhcilispllrir: nlentofavibratingpartlcteatanyinstantcanheilelineriilsitsilistiil]cc of t vilrr"iitlr: displacemetrt lrom the rncan pxrsition"6f rest,The nraxirrturn particle is called its amPtirude' DisPlaccment = ) = cslno' ..(tl A kxtbook of Optics Nature of Light 261 *.i-4=a i-t=o ,=FJ, (rr) (a) If a =f,uff , rn" resultant vibration is an oblique elliPse KLMN as shown in Fig. 7.11 ...(v) (, (u\ Thh represents the equation of a straight line BD (Fig. 7.9) i.e., the paticle vibrates simple harnnnically along the line DB. If c = n;sinc coscl = 0; z__:L__-\ u l^-/ =-l Fig. 7.1I ;*fr*ff = o [;.i 1= ' , (,o rf t l,'z ?n on rhe other hand if cr = =-(3,), an oblique ellipse Kl*tN is shown in Fig' is repeated after every time period. (v,) * ", f, ,n" resultant vibration is again 7'll ('r' The cycle of changes This reprcsents oquation of a suaight line AC (Fig. 7.9). 3rc o=;* "r r;:sinc= l; (iv) If t-t--' cosct = 0 ?.12 HUYGENS PRINCIPLE According to Huygens, a source of light sends out waves in all di(r, S recrions, throu;h a trypottretical medium called ether. In Fig. 7.1'2 is a source of light sendXzX ing light energy in the form of waves in all directions. After anY given d*F=', 1t 3r a=i*T a=b of time (t), all the particles of the meinterval dium on the surface This rcpresents the equalion of an ellipse EHGF (Fig. 7.9) with c and D as the semi-rmju and semi-minor axes, will be XI vibrating in phase. Thus, XY is a Por- tion of the sphere of radius ar and centre S. u is and the velocits- of ProPaga' tion of the wave. XY is called the primarY wave- rhen or radius **# = t l+72 = o' This represents the equation of a circle of a (Fig. 7.10). Fig. 7.10 front. A wavefront can be defined as the locus of all the points of the medium which are vibrating in phase and are also displaced at the (0 Fig.7.t2 all the strondary waves from different Therefore' CD forms the corresponding points on Cb at the same time' to the . I It t I In the A s BAC ahd BDC BC is common BD=AC IBAC and = IBDC = 9U . waves from E must turbanc" fr11m F reaches the point C the secondary and radius FC draw centre E as With have travelled a distance EN = FC' sphere' It can be shown a sphere and draw tangents CN and CN' to the congment' are thai the triangles EFC and ENC But and AC = AF = FC = AC = AF+FC ME EN ME+ EN l.la)' By the time the disturbance rethe on B the poina A and flecting surface.12 (i). 7. points on AB reach the Thus.I4 REFLECTION OF A PLANE WAVEFRONT AT A SPHERICAL SURFACE plane llcr APB be a convex refla:ting surface and oPfi the incident R reaches and at Q wavefront (Fig. 3 etc. F'rom geometry' ALB . R and In Fig. is the backward wavefront.I3 REFLECTION OF A PLANB WAVE FRONT AT A PLANE SURFACE I*t XY be. With the point B as centre and radius equal to AC construct a sphere.3 etc.13 wavefront whose centre of curvature is F' SimilarlY. However. But according to Huygens principle. '1. the secondary waves corresponding to the Points lYing on the incident wavefront _+----" lro I . Fig. then any sntall portion of the wavefront can be considered plane [Fig. 7. 7. Y. the secondary wavefiont is confined only to the and not the backward wavefront XrYr. [3t i be the angle of incidence (Fig. Fig.14..is a ctror<t. 12 (i)l the wavefront is spherical. 7. as centres. =RB= PL=PK.f Light 263 the distance of the source is small [Fig.PL) = 2R. Then AKB forms the re- flected sPherical From the point C. After an interval of time . APB is a small arc of a circle of radius PO = the arc."n"r*o plaie'wavefronr and also the angle of incidence is equal X. In Fig. CD tbrms the reflected plane wavefrgnt' In the tirne the disir.2.14 OPR will reach the surface AKB in tbe same time after remirror APB. Wth the points 2. These secondary waves travel through space with the same velocity as the original wave and the envelope of all the sdcondary wavelets after any given interval of time gives rise to the secondary wavefront. IABC=i=IBCD=r.12 (ii)]' Thus rays of light diverging from or converging to a point give rise to a spherical wavefront and a parallel beam of light gives rise to a plane wave front According to Huygens principle. the secondary waves from P must have travelled a distance PK back into the same medium such that OA gens. XY is the primary spherical wavefront and in Fig. the angle of incidence is equal to the angle It can be shown that all Hence. 7. a plane reflecting surface and AMB the incident plane wavefront. The t*o triangles are congruent. Then 3P = BA. 7.( angle of reflection. is the forward wavefront and XrY.12 (ii) XY is the primary plane wavefront. 7. A k"rthook of OPtk's Nature o. PL is called the sagina of Al] = PL(ZR. PF is the focal flection. draw spheres of rwJii at'. F is called the focus of the spherical length of the minor. 7. i=r Fig. When the source is at a large distance. 7.'. X. All the particles on AB will be vibrating in phase.'. The surfaces X. \ In the time the disturbance at A reaches C the secondary waves from the point B must have travelled a distance BD equal to AC.PL- Pl] .13). forward wavefront absence of backward wavefront was given by Huyno explanation to the If of reflection' Tltus. draw tangents CD and CD'. 7.12) ue sources of secondary disturbance.{ and XrY. the secondary waves travel a distance or'. .262 same time. 7. refer to the secondary wavefront. all points on the primary wavefiont (1." poino on CD form the refleited plane wavefronr. This theory can also explain reflection and refraction.1879) He did fundamental work in colour vision and colour photography llr r'.. though the theory does not clearly envisage rvhy... 1'''('irlili.*.i] [t. u is -ve. or AC = KM (aPProximatelY) VKM = LP+YtPQ+QN PIKP+ PQ+QMl = LP+PPQ+8N p [.p=f i = .i ).(] * ) = :_:=or_ r) According to the convention of signs. hr1 h2 h2 h2 h2 or and R.A Texthook oJ' Optics Nature of Lighr 275 ltAC = LP+YPQ+QN But.lllt'-.[_u.I n. Newton assumed that the corpuscles possess fits rr.pendicular to the reflecting or refracting surtace by a corpuscle.ll at (xld :ii. is +ve -[i. gr=& rr=*unaUN=* Substituine these values in equation (i) .* an{"i easy transntission at lhe ".) [i i ) (") y=a.) or If *-: = (p.:= (p_.u-'.i.'. is [+.(iii) 7.KL) + (QM + Ml$ AL=CM=fi(aPProximatelY) Here "'(') rr=#.. known for the discovery of the Electromagnetic Thanry ol' Litllrt u.t] - ve _*.Iu i) . o is -ve R.rsy rcflection JAMES CLERK MAXWELL (T&1.hich allorv thcrn e.i: i() i\( wton's . Rectilinear propagation of light is a natural deduction on the basis of corpuscular theory.KP + QM = 6P ..22 NATURE OF LIGTTT (i) corpuscular theory. how and when the force of attraction or repulsion is experienced per. was overconp when Fresnel suggested that th. Show how thc wave theory and the corpuscular thmry of light account for (a) refraction and (b) total intcmal reflection of light. (rrr) t0 Wave theory of light. Rajasthhn 1986) The next ifiportant advance in the nahrre of light was due to the work of Clerk Maxwell. propagation of light is in the form of longitudinal waves.) Huygens PrinciPle. How was the (Raiasthan 1990) issue decided in favour of the wave theory ? Discuss the nanrre of light. polarization of light) cannot be accounted for by the corpuscular theory. with the help Write a short discussion on the nature of light. According to Huygens. one cannot expect the unsymnrctrical behaviour of a beam of light about the axis of propagation. EXERCISES VII What is Huygens principle in regard to the conception of light waves ? Using Huygens conception show that p is equal to the ratio of wave velocities in the two media. bending of lighr round corners or illumination of geometrical shadow cannot be conceive<l according to corprscular theory because a corpuscle travelling at high speed will not be deviated from its straight line path. Explain Huygens principle. on the basis of wave tlr€try it can be shown mathematically. (r) (. But the experimental results of Foucault and Michelson show that the velocity of light in a rarer medium is higher than that in a &nser medium.ther's effect arxl produce darkness. How is'refraction cx(Delhi 1992\ plained on this theorY ? 7. How do you explain the phenomenon of reflertion. lens ftrrmula the What is a wavefront ? How is it produced ? Derive of light' ttreory the of basis vave for a thin lens on the Write a short note on the wave theory of light. the discovery of the phenomenon of interference by Thonns Young in l8m. 2. (i0 Wave theory. Explain how the phenomena of reflection and refraction of light are accounted for on the wave theory and point out the physical signilicance (M)'sore l99l) of refractive index. Newton's corPuscular theorY. thus enabling the understanding of the phenomenon of diffraction. Certain crystals like quartz. (Puniab [Delhi (Ilons. Nanre of Light particle concept. This difficulty. a solid ground to the rvave theory.bending of war*:s round obstacles is possible. Explanation of this has not been possible with the corpuscle concept. that light is regarded as a wave motion at one time and as a particle phenomenon at another tillle l. On the basis of this concept. rectilinear propagation of light can be correlated.hpur 1987) . calcite etc.ration of the ether particles is ransverse ard not longitudinal garre. State Huygens principle for the propagation of light' Using the samc' deduce the formula connecting object and image distances with the constants of a thin lens. ll. Deduce of lenglh focal for the expression an tight. that the velocity of light in a rarcr medium is higtrer than ttrc velocity of light in a denser medium. Derive the refraction formula for a thin lens 6gru 1992) on the basii of wave theory. (Rajasthun l99l) 1985) 19931 9. Finally. 3. refrrction and rectilinear prcpagation of light on the basis of wave thoory ? (Mysore 190 . exhibit the phenomenon of double refraction. in a way. Obtain an expression for refraction of a spherical wave at a spherical surface. Write an essay on thc nature of light. Double refraction can also be explained on the basis of wave theory. This is in rccmdance with the experimental rcollts'on the velocity of light. the expairrcntal results of Foucault and Michelson on the velocity on light in different npdia and the'revolutionary hypothasis of Frxqel in 1816 that fie vib. Applying the principle of secondary wave points. But in the case of longitudinal waves. lnterference could not be explained on the basis of corpuscular theory because two material particles cannol cancel one another's effecl The phenomenon of diffraction vrz. (Agra 1986) 13.the axis of propagation (viz. what is Huygens principle ? obtain the laws o[reflection and refraction on the basis of wave theory of light. (Goral.276 corpuscular A Textbook of Optit-s thery the velocity of light in a denser medium is higher tharr the velocity in a rarer medium.. Similar to sound waves. 5. The unsymrnetrical behaviour of light about. It is very interesting to note. Huygens wave theory could explain satisfactorily phenonrna the of rcflection and refraction. The phenomenon of intcrference can also be understood consideiing that light energy is propagated in the form of waves. At one tine the corpuscular theory held the ground and at another time the wave theory was accepted. TVo wave trains cf equal frequency and arnplitude and differing in phase can annul one ar! . Maxwell's electromagretic theory of light lends support to Huygens wave theuy whereas quantum theory strengthens the 12. The controversy betwen the corpuscular theory and wave ttr theory existed till about the end of the eighteenth century.. the phenonrcnon of polarization can also be understood. of of Huygens wave theory a thin iens in terms of the radii of curvaturc of its two surfaces and the refractive index of the material of which it is made. Write short notes on : (iii) Conchuion.e light waves are transverse and not longitudinal. (Delhi 21.9n twg o interference displacernent the hfiium' irains act simultaneously on any pafiicle in a periment on of {Delhi 20.. INTERFERENCE a.Delhi (Hons.) .) ) i Fig.) I 99 l) t7.) t984j Explain Huygens principle of wave pnrpagation and apply it to prove the laws of reflection of a plane wave at a plane surface.".A 14. Apply Huygens principle to derive the relation Te. Discuss some typicat 18. lt. Thomas Young successfully ference g. 15. Also. What is Huygens principle ? How would you explain the phenomenon of reflection and refrrction of plane waves at plani surfaces on the basis of wave nature of light ? (Sub. State and explain Huygens principle of secondary waves.Sc. 1988) lDelhi .Ii.l of thd particle at any instant is trains.u-'.t i =.) 19g6l ondary wavelets. [Delhi l9M . sta-te the principle of superposition.) B. (Mysore 1990\ Sate and explain Huygens principle of secondary waves. Give the mathematicar theory of interference between two waves of amplitude ar and az with phase dif- case. Apply this principle for explaining the simultaneous reflection and refraciion of a plane light wave from a plane surface of separation of two optical media. r) l for a thin lens. Each wave train behaves as if others are absent. This principle was explained by Huygens as t if they have not interfered at all. Each wave train retains in 1678' . 16. Deduce the laws of reflection with the help fDelhi B. (Rajasthan l9g5) t9. 1992) A.nstrated the wave tireory of light. after the <Iue to the superoosititln of all the wave ffi ".[*.g" its individual characteristics.rtbook of Optic.(Hons. [Rajasian l9g5l of Huygens theory of sec- wave light in 1802'LWt. Sate and explain Huygens principle of secondary waves..rgfdrnoDuCrIoN ol' The phenomenon of interference of light has proved the validity exhis dem. l poins A and B are the two sources which produce waves of equal a[pltude and constantlhase difference. As the intensity (energy) is directly proportional to the square of the amplitude (/ * A) the intensity ar these points is four times the intensity due to one wave. the two sources will act. The points shown by crosses in the diagram will have maximum displacement because. are also bright because the two waves reinforce. the phase difference between-the two waves reaching the point strouu ue an odd number multiple of n or the path difference between the two waves should be an odd number multiple of half wavelength. Young demonstrated the experiment on the interference of light. I-et us consider the waves produced on the surface of water. g. because a large number of difficulties are involved. the amplitude of the displacement is twice the amplitude of either of the waves. In Fig. Therefore. The points shown by circles in the diagram will have minimum displacement because the crest of one wave falls on the trough of the other and the resuttant displacement is zero. The poins such as F are clark because the crest of one falls on the trough of the other and they neutralize the effect of each other. A an<l B are equidistant from . the two sources must be n. He allowed sunlight to fall on a pinhole s and then at some distance away on two pinholas A and B (Frg. two virtual sources formed from a single source can act as coherent sources.fln actuar practice it is not possibre to have two'independent sources which are coherent. Methods have been devised where (r) interference of light takes place betwe€n the waves from the real source and a virtual source (il) interference of light takes place between waves from two sources formed due to a single source.n of two trains within the region of cross over.A 'l'extbook o. * It is not possible to show interference due to two independent sources of liglrt. For minimum intensity at a point. the phase difference = Zrc . at these points the wave.. Since the wavelength of light waves is extremely smajl (of the order of l0-5 cm). In all such cases. and B.)))$ ---CRESTS tiple of 2n or the path difference between the two wavqs is a whole number rnultiple of wavelength. It means that the two sources must emit radiationioFttre same fectly similar in all respects. waves spreacl out on the surfac6-of water which are iircular in shape. Maximum intensity is observed at a point where the phase difference between the two waves reaching the point is a whole number mul- TROUGHS WPH^SE ference Fig. Suppose For a path difference for a path difference x. colour (wavelength). the phase difference is 6 l. either the crest of one will combine with the crest of the other or the trough of one will combine with the trough of the other.W6(. 8. Points.11 \/ /2 YouNG.afii-ilJ {rrygv pofiEi@r ro the points of rnaximum displacement.L spherical waves also spread out tiom y'.s reinforrce with each other. similar to E. The energy is onry t anffi. g. @corEREltT_qgu\gE_L{ Two s'urces are said to be . on the screen interference bands are produced which are altematively clark and bright. These waves are of the same amplitude antl wavelength.6.s EXnERTMENT In the yearJ802. At any instant. The two sources may emit light waves of largely different amplitude and wavelengtly2il the phase difference [etween rhe two may change with time. The points such as E are bright because the crest due to one wave coincides rvith the crest due to the other and therefore they reinforce rvith each other. nearly the _salne amplitude and are always in phase with each other. as if they are per- coherent if they #it tight waves of the same f{equengy. =f. Spherical waves spread out from . But for experimental purposes.l and are close to each other.2).2 If the path difference between the two waves = 2n- DIFFERENCE AND PATH DIFFEREN'X is )r.f Optics Interference 28t The phenomenon of interference of light is due to the superp<.rsiti. where the trough of one falls on the trough of the other. It should be remembered that there is no loss due to interfe*E*. In such a case.urow and must also be close to each other. the particle will be under the action of the displacement due to both the waves. (2n .\ lnterJ'erence 283 . and two niurow pinholes A and B (Fig.or. i. .. the energy cannot be destroye<I.: .rr(l + cos6) + acosoltsin6.) .8"S. = 2t. or the path I =0 Intensity is minimum when the path difference is an odd number multiple of half wa"elength /\ If v " :t and \'. = at sinr6 + a211 + cos:6 + 2cos5) R1 = al sin26 + a2 + a1cos26 + 2alcos6 = 2rt2 + 2ar cos 6 = 2a2 (1 + cos 6) R2 R2 AN_A-LJT. 3n..(iv) Special cases : (i) When the phase difl'erence . For. . + or l"(parhditTercnce)...4 2x3tr4r br which represents the equation of sirnple halgg&-yerate! of amplitude R a2 {1'+ cos 6)2 the law of conservation of energy.(2n + l) n. Here also the energy is not destroyed but only transferred from the points of minimum intensity to the points of maximum intensity....are the disPlacements ---.. = Fig.. 7t..(.) . the Jirinr"r..I"CA..:.tllwok o.T. asino. . .--.2n2 tl 'fi:. it is fount that the intensiiy at bright poins is 4a2 and at dark points it is zero." 2f = phase dift'erence = ..-. 2i. . According to ..._. 8.. Let a be the amplitude of the waves. * The intensity at a point is given by the square of the amptitucie I r = +r'cort i 6 = 0. The phase difference between the trvo waves reaching the point P. . 8. u= x.+-TyI.ffinergy and asin6=Rsin0 .z .\ . .pNToFINTERFERENCE bonsider a monochromatic source of light S emitting wavei of iniiielength l. + + .. -. . .t2. From equation (iv).'. is 6. "nr..[ OPtk'. + osin(ror + 6) ) = asin o)t + asinolrcos6 + acosoltsin6 = asinc. . Dr..(r.and (ii) and adding... ) = RsincorcosO + Rcosorrsin0 l' = Rsin(tor + 0) Squaring (i)..J2' v = asincor lr= asin(trlt + 6) ) =_Yr * !. R2 sin2 0 + fi2cos20 = a2 sin2 6 + '(..3 l. 2 (2n)...[.. Taking a (l + cos 6.. at any instant. .3)' A and B are equi<Jistant frOm S ancl act as two virtual coherent sources..) *-3r I -r -5r -4tt -2tr Fig.._. at bright . nl".. | = 4r. For a path Phase <Jift'erence :. 2r._...r-(zlrf or ttre path dilterence r = 0. = R cos 0 . 8.6 = difference.. . l=4a' Intensity is maximunr when the phase difference is a whole number multiple of 2n or the path difference is a whole number multiple of wavclength... (ii) When the phase difference._.' dhtribution. At C. =Q*P . .. A and B. 8. ' rvhere xdi nL Dl = n=0. A and coherent the from D placed a distance at be a distance d. the path difference is zero and a bright fringe is formed' I* ll Fig.-* bd j] This equation gives the distarrces of the dark fringes from the point C' When.1'' ll Ioenx BRIGHT FRTNGE wrorx nLD Therefore the distance between any two consecutive bright fringes sources. ox * r). The Consider a point P at a distance point P from A and B. n=0. -a6l 37' D -7: . = --i- s. 2..^ ^ xr=-21 n=2. THEORY OF INTERFERENCE FRTNGES U pinh)fes two and S source I and o nu.nt in the absence of the interference phenomenon dUe to the two waves.[. 1.r [@l IHr.. and the average is still 2a2. Here. If the path difference is a whole number multiple of wavelength ir. If the path difference it of half Wavelength.3. It is equal to the uniform intensity ?tf which will be pr"r..*. xr=-Zn=3.o* monochromatic by separated sources coherent two act as B B. 8. xz*x.'. PQ = x-*.(iii\ G) This equation gives the distances of the bright fringes from the point C. the point P is bright. .es is g). (i) Bright fringes. Ttre point C on the screen is equidistant from the path diffeience berween the two wai. (BP)...(D and _ n2n (2n+l\?"D *[#) ..('.. As shown in Fig.(:...) ..(v) - (AP)2 (BP)'.(AP)2 = 2r7 =l*. Therefore.4 the intensity varies from 0 to 4a1..2. Lrt a screen . the formation of interference fringes is in accordance with the law of conservation of energY. equidistant from S.(iv) * *!l-!g*!"t ./ or . the intensity due to the two waves shoul<I be 2a2 but actually it is 4at. l.r=ry ' z)rD n=2. But BP-AP = BP+AP BP = AP = D (approximatel) .r=* n=1. Path diherence Phase difference = BP-AP =# = = # .* = **f. .Interference 285 A Textbook of OPtics points./. wheren = o..3. Tfiii. the point P is dark. tfie point C has waves reach at the XD . Therefore.5 When n=1. # =. 2?\D 7\D d = d (ii) Dark fringes..( o -'eo.. x. multiple maxifrum intensity' x from c. . n?v D . xr= 3)'D ?l 57. .tj ] t. Two coherent sources are 0. d increase in wavelength (D) with increase in the distance D and (c) by bringing the two sources A and B close to each other..ti mm {rom the central fringc.4 rwo the screen hed on Fringes light. Calculate tLw wut'e-len. from equations (v) and (vi).=5x10-5cm L = 50004 . tiifiiii or'aart fringes is known as fringe width. ftr Young's double slit experiment the separation of the slirs is . Moreover. .. D = 287 *.. d = O. (Hons.3. If the overall separation of l0 fringes on a screen 200 crn awa). p1. The fringes are formed on both sides of C.is 2 cm. . find the slit separation.18 mm opart arul the finges are absented on a screen 80 cm away.:. l.1. " = Q..2.r!-9!!l-tiv e- qark fri n The distance uetwLen any'iwo .i"ii. F * D.5A are slits the t0() cm of distance at a from wavelength of light Here ? IDelhi B. alternately bright and dark parallel fringes are fbrmed. Example 8. Glieen light of wavelength 5100 Afrom a narmw slit is incident on. (ii) The widtu of the fringe is directly proportional to the distance of the screen frorn the two sources.:.m" I2 il1lgg. d = ? D = 200cm l0P = 2sm 7.[10 = Therefore. however.2 ?t=ry " '" 0.31 min at a distance of I metre the slits.)l ^ )"D F=Z .011><u2 100 mmapart are Examp te8. [Delhi 19771 = 0'05 cm d= lmm =0.= d- )"D p 51011)x e=+ l0-8x200 0. Calailctte the wavelength of light' Here F=0'3tmm=0'031cm d = 1.05 x 0. the fourth bright fringe is siruated at a.1 100 d = 0. 11y tyo-9 9r." ber yv. F * 2'. F B=L3 or ?t=ry . nLD 80 7' = ? '*=*rT ()r xl l.. the width of the fringe increases (a) with +.-r.distunce of l0. The breadth of a bright to half the fringe width and is equal or a dark tringe is. = ff-ff =.*.1. 0= 0.lcm D = l00cm F = 0'2cm .ffi.llmm = l..vith a certain moru> chromatic soltrce of light. a double slit. (rrd) the width of the fringe is inversely proportional to the distance between * Thus. o! OPtics lnterference Herc.Sc. (r) the 'nidth-oPtntt fringe is directly proportional to the wavelength of light.50 mm Here = 5100x l0-Ecm. It is found that '.*id.19 cm I)=lm =100cm the two sources.qth oJ'light.)io'Y:":.286 A Textbook gres.I : 11 :.t. monochromatic b1.:".(ui) cm. Therefore.18 mm = 0'018 cm x = lO. formed illttminated is the What r\m apart' 0. from Example 8.. n=4. i . i.0ttx0. it is clear that the width of the brighi fringe is equal to the width of the dark fringe.9 mm and the fringe spacing is 0.0l8 ^ t"=7iD=--4x80= 60?5 A =6075x10-E cnr **.08cm.9 mm = 0. All the fringes are equal in width and are independent of the order of the fringe.05i cm "Example 8. the tlistanci between the centers of the two slirs is d and the source double the x distance a placed at from entitting tight of wavetength l'. Here '=[ffi] D = l.0144 m Hence. Here IIASI b---.44 cm The fringe width is very large when d is very small' As d increases' of Sr' These the tirst minimum of S. for v'hat the \/ fitnges first disappear'? lDelhi (Hons) 1992\ A tight iource emits light of two wavelengths I. = 4300 A = 4.bright on the screen js 0.6.D) ("i'.4) n"i y =+lr'-r'] =f [.b'uo LLI" I = 60004 = 6xl0-7 m D=lm 0=0. fhe source is ured in a double slit inExample 8. = 5tO0 A'.fl ./ ) u=t. these minima occur at an angle = 1. If now the source ilit is gradually opened up.025 mm. Find the distance between the two co' For source A. The distance between two consecutive.A kxtbook cf OPtics Intederence ( t"*\ d"=lztl ' 289 Example 85.5 m and the distance betv.6 A L. produce an interference pattern on a screen kept at a dis' tonce of m from them. 8.#&l[rr.= The fringe pattern fitst disap'pe-s when the central maximum of one pattern overlaps on the first minimum of the second pattern' The first minimum occurs at a distance ) (nL.lxl0-7m n=3 Here U =-[*) when x >> L '. = 4300 A ancl L. = nLrD *inlurD d -(r''| ^ "'\u *r-xz-x. by distance and B are two extrenre points of the source S separated Iz=5l0OA =5.025 mm = 25 x 10-6 m I.5 mm. Calculate the separation between the third order bright fringes due to these two wcvelengths. The distance between the sources and the screen js . -2-d Also I I=e =* --2d o *! "t-2d .. the separation between the trvo fringes is l'44 cm' Example 8!7. . = or+ft +- sources.1.5m d = O.5mm=5x10-am .3x10-? m Fig.een the slits is 0.=. moves towards the zeroeth maximum two meet when d = d Here I fringes herent e. Two coherent sources of monochromatic liSht of wavelength 6000 A. \. ro-7-a3x ro-?] = 0. A Youngb double slit experiment is arranged sttch that slit. terference experintent. is width will slit. lxtion should take place from . 2. No ref. Here T I d lr- t *=-7n7. d.. apatt. if its distance from C (2n+ 8J FRESNEL'S MIRROR..* '' ) F=0'2cm it# = 0. The two reflected beams from the mirrors Mrand Mroverlap between E and F (shown as shaded in the diagram) and interference fringes are formed' Sor complete theory read Article 8.2x10-3 m 5 .6) Here. The pencil of light from two .290 A Textbook of OPtics Itrterference 291 d=? ^ 7"D F= a A monochromatic source of light s is used. Light of wavelength 5500 A'fiom Q twrrow slit is incident on a double slit.3 erc. l. is ancl it will be at the centre of a dark fringe. the following conditions must be sat.E. D = Yr+Y. The two minors M. 2.3. 8'7)' where n = 0. Fringe width P =ry where n (b) P =: A point on the screen will be at the centre of a bright fringe."D n=5 D=2fficm=2m i.5x10-7 m x=lcm=10-2m d=? (a) as two virtual coherent sources and interference fringes are obtained on the screen. calculate (a) the . A and B are two coherent sources at a distance d apart.055 cm Theory. )"D o= u x l0-a d = 1. and M" arranged at an angle of nearly 18ff so that their surfaces are nrarty (noi exactly) coplanar (Fig.ancJ M. l. 6xl0-7xl d=l. if its disrance from C B=1. Fig. The screen is at a distance D from the virtual sources. d distance at some B virtual sources A antl . anfl silyered on the front'surfaces.S u l\)A Fresnel produced the interference fringes by using two plane mirrors M.5x10-7x2 l0d = 55 x10-a m d = 0. For the fringes to be formed.5x5.slit separation and width. 8.2mm Example 8. These fringes are of equal width and are alternately dark and bright. .s incident on the two mirrors. etc. should be made from optically tlat glass isfied.7 = 5500 A = 5.The overall separation of 5 fringes an a screen (b) the fringe 200 cm away is I cm. after reflection appears to come from B act and A Therefore.nLD x d. CEl o :d bas" to ba. A and B can be calthe two virtr'ral sources intersection between the points of ai"un"" is Y'' "i'tlr"**t 0 is known''tJ ffi. other fringes on ' ' between A and B is 20' E and of equal width are produced between E and F but beyond MN to diffraction' F fringes of large *iotrr ur'" produced *tricIffi rebe can screen is a stlp to limit the rays.' Jf.nu... I i']1T. l.:f:. The point C is equiTheory. ffi.3 a dark fringe if is distance from C is @)))LD- ffiffi1.ru...".*tt igplrJ:t of light are more' a monochromatic source seen in the field "f than xJth white light' uott"'nJli c . | .onbothsidesthe width of oi C.'. Therefore distance between-A :il1Ty. :i:j:.>1 *_yi4 Fig' 8. I' l'icld of view.A 292 Textbook of OPtics Inrerference fringes 293 The polishing.l sourcPs' Suppose tltt: (l' interlcrettcc l-'O A-g-lt screen is pliwetl at f)t a :*-t^:nhoent tr.se' ^{c.-ithasmaximumintensity. l.. rno_:ro exrend the back of the mirrus. for crossing 20 bright fringes fiom the t"rmir" the fringe *iAtt.onH:.: eyepiece = D arxl Suppose the distance between the source and the A and B = d' The eyepiece the disance between the two virtual sources of the bench) to delength ttre to (perpendicular i."Jl'ffi :: g. 88 (Fig. 8.Therefore. 2."i willbeatthecentreofabrightfringeifitsdistancefromCis etc. . The point will be at rhe centre of F =ry.r: Xx.i# on ..Fresnel'sbiprismcanbeused light' to cletermine the wavelength of a given source of monochromatic of light and A fine vertical slit S is adjusted just close to a source therefractingedgeisalsosetparalleltotheslitSsuchthatbcishorizontal $r Y i.tff L'"'il"T#1il'l'"""J'.ffi t'"* Determinafionofwavelengthoflight. *orot horizontally Srppote.i.. 2. where n = 0.ely bright and dark fringes are produced' The bright fringe or dark fringe' g :"l:H*.ffi "r'about 179" u where n = 0.*#lfi.ff N\ Y i.ith = + Moreover' any point on the screen . the eyepiece has moved through a distance .*# uiprir*71'"oiritt" t""u r.a by an eye-piece or a low power microscope and focus of the eyepiece' in tne fieia of view. To observe the./ -FOCAL pulrug I Fig.8). They are adjusted on an optical bench' -A microneter i.{oon::. the tine of intersection of tne two i'"i't* l'O intersection up to the line of must be parallel to the line sornce (slit)' The distance 0""'*n culated as follows' 't'" and tle'sourct S S". For complete theory refer to Article 8'6' distantfromAandB.h" opticat bench at some distance from the prism to view tne fringas in its focal plane (at its cross wires)' eyepiece ffi ffi source A.7A i eve prce f winrsl cBoss I & I ' '. lg lbwer Portton 1'gffill "0J""'**''"J *. If the point C ls at the principal the fringes are observed in the field of view' white whereas the the cenual fringe C is width When white light is used *"-"ot*ttJu""uuse the fringe '"pJ"tion 4 = Z0y.XL'{I. t: t{ g...::1"T.ffi" *:.fringes' the seen are fringes pfo.3 etc- {. ot. fletermination of the distance between the two virtual sources (d).(. and d is the geometical mean of d. . )" can be Here il= 5l= .. (ld t2 lf d is also known.9). dtun dum Therefore ..=.-r1aq . for blue light the distance between the two apparent sources is different to that with red light.i) .) d can be calculated. . Let it be equal to d. angle it be dr In this case."D Here. Fig..294 A kxtbook of Then the fringe width. $ and D in equation (i). Oprics Inte rfe rence 295 p=* B= d 7. Suppose the lens is in the position t. F-. the two coherent virtual sources are prG duced by refraction and the distance between the two sources depends upon the refractive index. then d = 2(tt."uD n th fringe will be. calculated. and dr.. (FiS.9 FRINGES WITH WTIIT:E LIGHT USING A BIPRISM Now move the lens towards the eyepiece and bring it to some other position Lr. Measure the distance between the images of A and B as seen in the eyepiece. Substituting the value of d.) cry. 8.) (ii) and (iii).'L? (p 2 xu Therefore for blue and red rays. the nl. dt -a -m dun I .l)4.--n (\ -h lr V'.(. will be greater than d...) ages field of view of the eyepiece. which in turn depends upon the wavelength of light.. But the fringe width 7)=mandu=tt. The second method to find d is to measure accurately the refracting u. we make use of the displacement method.. so that again the images of A and B are seen clearly in the :: $o Ao +fl Lr \ V -hr 'd. Therefore the total angle between Aa and Ba is 20 = 2(F.(. the centre of the fringe at C is white while the fringes on both sides of C are coloured because the fringe width (0) depends upon wavelength.D ut. = 71uj\ nlu..l)cr. For this purpose. Measure the distance between the two imin this case also.9 t=--' nluD - whered = (Z1t_l)ay. Therefore. 8.l)slr.* fi44 l t-l7l-ra---n#{ When white light is used. l. If the distance between the prism and the slit S is ). Therefore d can be calculated. In a biprism. the deviation produced 0 = (p.(... From equations .) x=Y In equation (i) 0 and D are known. lrt d.'{dA *..i. The distance of the n th fringe from the centre (with monochromatic light) sF-B. As the angle is small. Moreover. the wavclength of the given monochromatic light can be determined. that the images of the virtual sources A and I are seen in the field of view of the eyepiece. 8. *'= 2t1t. A convex lens is placed between the biprism and the eyepiece in such a position. .(.. the fringes obtained in the case of a biprism using white light are different from the fringes obtained with Fresnel's mirrors. = 20cm.424x lO-2 d=0. '\ roo ^ v= a Here Ir = LD d = 2(lt. cr=lo=#radian i ). Interference fringes are obserted.424x l0-2 )t=Y F=0. Distance between the two coherent d=O=0.0195 x0. 5900x 10-8x o P--LD d I 0=? p= l d = Z(tr.atlNcmdistancefromtheslit.ahtetnce betvveen the source and the biprism is 20 cm" _calgulate (il 6900 A and the.Tcmapart.-xo d or Here For a convex lens Here 1.5-l)x22xl0 Here P = l'5 = 0.0195 cm are obseryed at 100 cm from the slit.ll. cm D=5+75=80cm e=ry l= or 5890x10-6x80 9.L2. 2ql Examples. = 10+ 100 = 110 cm 7u ll = 5900x l0-8 cm. What is the distance between the two coherent sources ? (). The inclined faces of a glass prism (1t = l '5 ) make an angle oj l' *ith the base of the prism.Abiprismisplaced5cmfrontaslitilluminatedby sodium light (L = 5890 ii.fr+yr= 20+80 = l00cm .30 _ 5g50x l0-8 cm . on intruducing a convex lens 30 cm away from the slit. t7 tne.) I 99 or or T=#"* O=0.(Hons. %=l00cm .Sc. u+u = O u' u=30cm l(X) cm E:rample &lO. Find the fringe width obsented at a distance of I m from the biprismlDelhi B. two images of the slil are seefl cr=lo=ftradian ). . tNagpur 1984) I l : . ll i.5 on a screen 80 cm avtay frcm it.30cm _ 0.30cm sources. The -slit is '10 cm from the biprism *d is illwninored by 1ight of L = 5900 A.e.5 l.424x l0-2 cm.g. bands of width 0.=Y. = 5850 A Example 8. P.Calculatethewavelength lRajastlaa l9t)51 of sodium light.05cm !.with a biprism of refracing itgle l' and refracrtve index 1. = 80cm O .fringe width when the wavelength of light used is (Karyun 1986) (ii) sss} A.0195cm D = 100 cm. fne width of the fringes obtained on a screen 75 cm from the biprism is 9. F = 9. = 5890x l0-8 cm d = ?.037 cm E:rample E.l) cr yr ll0x 180x7 2(1. In a biprism experiment with sodiwn light.1)slr 1. ).296 A Tbxtbook of OPtics lnterference ii I .tt+Jz.=10cm. 5 l) 25 = 0.l) d. d = 0. . calculate the angle at the tertex of the biprism. g=ry o=Lf (i) and (ii) . t' Here )'r = 25 cm. .D ---- d 5890 x l0-8 x 80 0.05 .888 o. 15.) biprism.16. D = l2O cm sources formed by wilh its base.I)ar'. u ' = ?\. . .05 cm. ftr an experiment with Fresnelb biprism. = 5x l0-5 cm F = 1. fringes for light of wavelength 5 x l0-5 cm are observed 0.I)r.o7s ^ n = Bd T = -t0 *ffi = 5900 A =.02(1.M) radian 0 = l77o 42' Exanrple 8.075 cm. 9= 9.5900x l0-8 cm or 0 = 0. =ftJq:_1411-75r 2x0..075 cm.888 cm width P =:= ff .2 mm = 0.)'r But or From equations .'.13. The distance between the two virrual sources was found to be 0. illwtinated by sodium light (L = 5g90xl0-B cmj ard frum the the dbtance between the virtual sources is found to be 0.(. Find the wavelength of light of the source if the eyepiece has to be moved through a distance l.ggg cm for 20 fringes to ctoss the field of view.5.42.(.14. = 5890 A x = 5890x 10-8 cm LD ^ * Example 8. A biprism is placed at a distance of 5 cm in front of a narrow slit.298 A kxtbook of Optics (. T-he slit source is z0 cm the biprism material a biprism whose inclined faces mal<c angles of I degree away frorn the biprkm and 1t of = 1.2 mm apart at a tlistance of 175 cm from the prism.In a biprism experiment the eyepiece was pracecl at a distance of 120'cm from the source.50 and it is at a distance of 25 cm from th illuminated slit. The prism is made of glass of refractive index 1.) Itle rfe rence 299 ^ 2(p.05 cm D ='5 +75 = 80cm Width of the fringe 5'= Also z(p-r)cr'v.lt x L0-t cm --F__: D = j-r* lz i' o'.l) cr')' Example 8. Here (Mysire t98I) ?\" - 5890x l0-t cm.02 l. Fringe n= 2O I= 1.01687cm Example 8.01976 cm ). Find the width of the fringes obserued in an eyepiece placed at a distance of 75 cm g=? 9 = 0.50 d = Z(tt . calculate the separation between the coherent Here. +lr) = z(lt .2 = 175 cm cm 2(p-1)c)'. 6900x I0-8x l00x 180x7 P=@ (. or o= 2p lo'. " 5890x l0-8 x l00x l80x 7 lr=@ 0=0. )...02 radian The verrex angle 0 = (7r-2a) radian = (1t-0...* d = 0.) l" = 6900 A xD or 6900x l0-8 cm r.+L) (tr.. 1 D= m.5x10-am . glass or mica : = l.5 ul) x l0 m=1. Cabulate the separation between the coherent sources formed by a biprism whose inclined faces malce angles of 2" with irs base.lA= F=ry.Interference 300 301 A Textbook ol Optics d = Z(tt.=? f I .5 x l|'a m. if ttu eye-piece is to be moved tansversely through a distance of 1.een' I cm from the prism [Delhi (Hons) l99l] d = 2fil_ l)slr lIerc.I0DETERMINATIoNoFTIIETHICKNESSoFATIIIN SHEET OF TRANSPARENT MATERHL a given thin sheet of Eansparent material e'g'.49x ro-3 for 20 {ringes.50.: "90 (U:-!.l) crlr a.50 make eingle rtf 2" with the base' A slit illminated If the distance biprism the I0 cm of distance a at from ptr.7llxl0-7 m = " ---l. = 2o = 2x0.01888 x 7. The distance between the virual sources was found to be 7.17.19.=l0cm . )r=lm .5xnxl0 90 +# = ft radian jr : 90 -)'r=l0cm=0.:')o-Zxn=a.=5711A S.18 mm. The inclined faces of a biprism of refractiue monochrcmatic by ct 1.j"01 = 3.ria iiin. = 035 cm Etample E.1 ^gd *=T Here F= 1.l8xl0-3x3.5 x l0-a m zox 1. ^ )LD lt = -.5 d.50 F = 0. 1977) 0r = 1.5-l)rx20 = -Irn radian L ' 0. fbd the wavelength of light used' b'et$.1+1 = 1.P=* ILD nd ^ td *nD 1. Find the wavelength of light. the slit source being l0 cm away from the biprism (Delhi t974.2 '= r8il- 2 x 0.lE.49x10-3 5.l8o -90" ).2 m form the source.2m Here F= 1. ct= lo )'=20cm _ 2(1.18 mm = 0-18 x l0-3 m d = 2(lt.5 x22x2o 7x 180 = 5900x l0-m = 5900 A = 035 cm Example 8.888cm = 0. n=2. the eye-piece is placed at a distance of 1.888 cm i T I )'r+lr= 0.01888 m Thebiprismexperirrrentcanbeusedtodetcrminethethicknessof d=7.l) c). 2(1.) index Example 8. it two clark fringes observed at a distance of is 0.adian -L .5. /n a biprism experiment.0 D=1.1 m 2 l. (Delhi 1985) .=P'j 0. . occurs. therefore (n {znt"\! 2pr cos 7' = (n + l) tr + I ) can . v ()r = 12n + l):.inlirterrt on the upper surface of the film is partly reflected along AZ and partly refiactetl along A8. -'.. .) (2) If the path ditfercncc x = (2n+1) tiestructive interference takes place and the ] *ner" film rt = 0...= p(AB+ nQ-p\Ctut .een light reflected frctm the top and the bottom surface af a thin film. x = ?lttcosr-. At B part of it is reflected along BC qlM ind finally r.1.established correct the onfiei-biJis of electromagnelib theory that.s \ r o ^:.310 of A Tbxtbook of Optics l. cos r = = 2tcosr (AE + EP) cos r ' thin film of oil on the surface of water and also by the thin film of a soap-bubble.. o.AP. appears dark. a rai. It has been. The violet fringes are produced by V and V' while the reQ/ fringes are produced by R and R'. Instead of a diffraction grating.:.CM .1yr1 = 1t.. The white and dark fringes are seen through the eyepiece or can be produced on the screen.ILECTBD LrGHr .(ii) (l) If the path differE-fiCe i--Twhere n = 0. lt has been observed that interference in the case of thin filnrs takes place due to (l) reflected light and (2) transmitted light. 2. = p(zl8+BO-AN i.. l. ..(i) This equation (i). = AP ( pM -. Everybne is familiar *ittr@ uy a 4' "or. = 2 pr cosr L .The difference in path tween the two rays Ar and Qp -to BC... a prism of small angle can also be used.r If -. Also prffiil" CB tJmeet AE producirl-at P. the correct path differenc--in-Ihis case.9J6I"XIPSIERENCE DUE To RI.tle 4arcncc lil V and R. 2 . ----/ d. l. constructiye interference takes place and the film appears bright.(ii. . In the A.. Young was able "t'ir explain the phenomenon on the basis of interference betw./ Suppose.4 etc'.opAcany--.3.'-'| = 15 oTli6tence is i and the angle of refraction is r. and RR' = dz 1'he optical path ditltrc:nce . AE=EP=t) 1 = st. = . when light is reflected from glg-lfr@eq-of ar. . the fringe viidth p will be the same an<I interference = . Interference occurs between the beams from Y R and those f from l/ R'."etc.rini = cM . 1t.1 Here. VV' = d.]Se. I rl/cosr-.T5 INTBRFERENCE IN TTTIN FILMS Newton and Hooke observed and developed the interference phenomenon due to multiple reflections from the surface of thin transparent materials. -.APM.(ir') also be taken . in the case of reflected light does not represent path difference but g!b1!9--appe{ent. Hooke observed such colours in thin films of mica and similar Stin tr:msparent plates. Here rr is an integer only.1. 2Uf 1 I I - ai-ong 6-ecan be calculatedSraw jI/ ntrrmal to *T anrJ AM normal emerges out vsl urvrrE CQ. Newton was able to show the interference rings yhen a convex lens was placecl on a plane glass-plate.r = p (AB+BC-CM) = P"(PC-Cl4) [ '-rin7--m fringes due to different colours will overlap and white achromatic fringes are produced in the field of view. Th6?nge "U --E -n t a. Here"ZAPC = r (FigI-&lS).. Fig"8' 2prcos. 0r as .usuJ 'Therefcrre.PM = 2prcos r .Plt'l 8. (THIN FILMS) Consider a tran$parent film of thickness r and refractive index tl.ren se r change -med i um { ai r-@ / se n equivalent to a path difference . "i*n-incident completely and perfectly vanishei never is transmitted. Therefore..and. BP=2t =2tcosr y = 1t.i iJtpte reflection. 2.(v) According to the principle of reversibility'' " ' etc' tt'= l-f fect anrl their ampJitudes the amount difr"'"nt' The amplitudes will depend on found that been has It films' 0re uilJiJ trrough of light reflected ^"a light is reflected .uv'i'iiiou. At B it is Partly ieflected along BC and PartlY refracted along BR. 8.be to denser medium and t'the transmistransmission coefficient io'''o medium' sion coefficient from denser to rarer ItshouldberememberedthattheinterferencepatternwillnotbeperAT and CQ will not be' the sarne because ttt" int"n*iti"' of the rays e=ffi=o' Thus.3.. Fig. of of the reflected system will be of zero intensity' LIGHr 8.17.4. geometric the terms inside the brackets form a emerges along DQ.theresultantamplitudeof2. change phase a with raY af{ I bY r. the intensity of the minima will shown in Fig' 8'16' The amconsider reflected rays l.. n = 1.4. Thereforc.17. r= . The ray I is reflecie<t at the 4 etc' are all in phase but out of phase 3' 2. no Phase change occurs.[i +E. 0.1 r is less than l.17 INTERFERENCE DUE ro (THIN FILMS) of thickness r ifrd refractive index Consider a thin transparent film u. "f. In the LBPM.' ['et r. (.'as the reflection coef{icient' t the plitude of the inciden.r = P(PD) -P(MD) = 1t(PD-MD) = PPM In Fig.3 etc.. A=atrrlr:r] =[#l Also a = p(BC+ cD)-BN p=#=#orBN-1lMD ZBPC=randCP=BC=CD BC+CD = PD . A ray SA after refraction goes along AB.PM = 2ltrcosr . Draw BM normal to CD and DN normal to BR. rays ftre r. at{+atf{+atf{+"' A=atlrtl+l+l+.*" . 8.t . The raY BC aftet reflection at C finally *" rRANU*ra "' aE atrt" atft" The amplitudes of the reflected rays are: undergoes It medium' d"nt"t u Lf surface etc.3.... the intensity CQ alone' But in the and AT rays the dark fringes will not be'observed for be zero' .A 2 pt cos where Textbook of OPtics Interference nlt."etc'isequalinmagnitude phase with it. Therefore the mininta of the amplirude of .) Fig. 8 16 .. The oPtical Path difference between DQ alrr'J BR is given by.967o for normal incidence. Here at B and C reflection takes Place at the rarer medium (medium-air interface). cosr = PM P# * PM = BP'cosr But. is given by The resultant amplitude of 2' 3' 4 etc' A= As series. 2. tt l8). (l) In the reflccted system. .3.. rhe path difference 5 = 12rr + l) )1 2Prcos'=*lD where rt = 0.'l Tcrrlxxtk rt L o.2. Fig. 8.5) such tlnt the angle of refraction into the plnte is 6tr. 16 srn'.etc. S. Fcrr I I = 0.. = 85'ZlEo 14. the fringes are more distinct.2g.8521 Io {.Snetc. Therefore. This is clear from the following solved example' ' Example 8. Thus the fringes are more sharp in reflected light.3. different colours are seen because r and . rJ is the reflection coefficient and is the maximum intensity.2l%o. the path ditlercncc -r = 2ptcos where .04 'I t__ ' (l -0.It means the visibility of the fringes is rnuch higher in the reflected system than in the transmitted system.314 (i) For bright tiinges..04)'2 I.-4x0.1 ) will be 14.04 .2. Calcuhte the smsllest thickness of the glass plate @unjab 1973) which witl appear dark by ri/tection. the colour will be . For values of 6 = r. the light which comes from any point from it will not include the colour whose wavelength satisfies the equation 2 p t cos r = nlt. In the case Of oil on watbr.79?o 1 = 100-85. in the reflected system. when the angle <lf incidencc is nclrrly '15". Reflectance of 47o I '. 1.f Optk's lntederence ll5 r = nL rt = 0. A parallel beam of tisht (7" = 5890 xl0-8 cm) u incident on a thin glass plate (1t = 1. -" -l*.xirrla (ii) For dark liinges. appear coloured and the colour depend upon the thickness are constant.I8 INTENSITIES OF MAXIMA AND MINIMA IN THB INTBRFERENCE PATTERN OF REFLECTED AND TRANSMITTED BEAMS IN THIN FILI\TS The intensity of the transmitted beam is given by (vide theory of Fabry-Perot In terferometer) l" . . .'14-rin''? (l-r)' t Here 6 is the phase difference. uniform.e .21 = and the angle of inclination.3n.18 / (2) In the transmittcd system.. 1. In the case of transmitted light. the intensity of the maxima will be lNVo and intensity of the minima will be 85. the will film will 0. vary.79o/o of thc incident intensity and the intensity of the minima will be zero (Fig. ctc (. If r and . the intensity of the interference m..-{l [1i. the interference tiinges obtained are less distinct because the difference in amplitude betwcen BR and DQ is very large. = Taking I 8.T9 COLOURS OF THTN FILMS When white light is incident on a thin film. Howcver.-.=l I. Here. I PnQ. meets the surface at solrrc different angle and is reflected along 5 and 6. = Here..20)' the ray I after reflection from the upper and the lower surface of the film emerges as which arc parallel to the line of intersection of the trvo surfaces. and 4 reach the eye whercas the ray 2 .Q. 8. a broad source of light is required... Also ray 2 from sorrE other point of the source after reflection from the upper and the lower surfaces of the film emerges as 5 and 6 which also reach the eye. The interfering rays do no enter the eye parallel to each other but they appear to diverge from a point near the film.. Thp thickness Pn = PnQ. = 1 . 8.) o Qn The next bright fringe (n + r)ccuf of P. As the angle of incidence is small.-.. in the case of such a source of light. When the air filnr is viewed with reflected monochromatic light. The effect is best observed rvhen the angle of incidence is small.n 1 air p and t = 2P O and cosr = I = (2n+D. 8.Q. 8.r.. These fringes can the visibility of the filrn.19).(i. In the case of interference in thin films. The used is narrow. of the air' film at cosr = I Suppose the n th bright fringe occurs at P" (FiC. Therefore.rQn.?l). rays of light are incident at different angles and the reflccted parallel beams reach the eye or the mitiroscope objective. (. With a broad source of light.22). Due to this reason. Subtracting 2 P. Therefore. inclined 821 FRINGBS PRODUCED BY A WEDGE SHAPED T}IIN mirrors and Lloyd's single mirror . for (h+l).22 3 and 4 which reach the eye. 2 I I Pn*r Q.20 I) will (n+ l1 + f ] | 1 Qn*r Qn*m t}-x----{ Fie 8. Similarly we can take other rays incident at diff'erent angles on the film surface which do not reach the eye.20 Textbook of Optics lnterference 321 NECESSITY OF A BROAD SOURCE Interference fringes obtained in the case of Fresnel's biprism.8.. a system of equidistant intert-erence fringes are observed by two coherent sources. The ray I produces interference fringes because 3 Fig.P.) the Thus the next bright fringe will occur at the point where thickness of the air film increase.. The thickness of the air film increases_Jrom O to A (trig.. 5 and 6 do not reach the eye.*. If an extended source of light is used (Fig.2l. 8. 8.(. Applying the relation for a bright fringe. the portion A of the film is visible and not the rest. Each such ray of light lias its origin at a differerrt point on the source. = {2n+3)) 1 . OA Fig.A 8.) (i) tiorn (ii) P". the narrow source limiB FILM Consider trvo plane surfaces OA and OB incline<t at an angle 0 and enclosing a wedge shaped air film. were produced source be obtained on the screen or can be viewed S with an eyepiece.tr = 12 {n.19. to observe interference phenomenon in thin filrns. the rays incident at different angles on the film are accommdated by the eye and the field of view is large. Consider source a thin film and a narrow of light at S (Fig. 2prcos. such that Fig. by 1 Suppose the (n + m) th bright . A glass wedge of angle 0.2 =r ^ m7t . Find the angle of the wedge. Here ). 8. of equal thickness the surface OB is plane.. n = 10. Monochromatic lighi of L = 6. 0 = 0. forming fringes that are 2 mm [Delhi (Hons. the surface OB is face OB is polished and the process is repeated... Light of wavelength 6000 L fath normally on 4 thin film of rcfractive index 1.e.01 radian is illuminated b1' monochromatic light of wavelenth 600A L falling normally on it. 2" = 6000A = 6000x 10-8 cm A. The fringes are of equal thickness because each fringe is the locus of the poinS at which the thickness of the film has a constant value (Fig.(r!') If the distance ^ = o Q. Two glass plates encbse a wedge shaped air film. The tiinge width P.zowedge shaped ^ AR uoA^ l.4x0. it means that the surface OB is plane. (Rajasthan 1989) x = 15 cm. When the fringes observed are of equal width.000 A.The fringes are ob- ^i"^1 s=. 8. Example 836. calculate the fringe width.. the angle of inclinatign between OA and OB can be known.4.38.322 A Textbook qf OPri<-s lnterference 323 Then..09 cm Example 8..005 15 6000x 10-6x 2xo.Qn*^ PnQn 2L .4.24 Or' "20 ntTt .2 served in the field of view and if they are Fig.. plane' surThe not If not.e:- Pn*^Qn*^- x2x z m).(vi) Therefore. p=* cfll n.(v) Fig.(vii) S22 TESTING TtM PLANENESS OF SURFACES If the two surfaces OA and OB are perfectly plane. P--L^=* .$' P=tuu Here tr = 6000A = 6000x l0-E cm lL=1. The standard methd is to take an optically plane surface OA and the surface to be tested OB. 9=2mm=0.1rn^ a broad source falls normally on the 0 = 1.23. 8.. = 6000x l0-8 cm 2t = nlt But o=1 x t = 0x 20x = nX film. o-- 0.37. At what distance from the edge of the wedge.07x10-a radian Example 8. Here." .-P ^ Qn =. ^Qnr.0lradian.B = 0. the air-film gradually varies in thickness from O to A.. r is the distance corresponding to nr fringes.) 19861 apart.05 mm dianeter at a distance of 15 cm from tle edge. will the l0 th fringe be (Punjab 1984) observed by reflected light.t-" 2x I. not of equal thickness it means the surfaces are not plane.005 cm q{fl}sry:i I l .23). there will be nr bright fringes fringe is at P r+D that P n andPt+n such P hetu'een Fringe width.2cm 6000 x l0-8 +. touching at one edge and are separated by a wire of 0. TRINEES OF EOUAL This is an important application of the pheare fringes If the nomenon of interference.oo5 15 = 0. Find the angle in seconds. t = r0 1. This is due to the interference bctwcen the light reflected lrom the ltrwer surfac'e of the lens and the upper surface of the glass plate G. Q ti.25 Example 8.2 Number of fringes Per cm #. Fringe width.0625 t = 2. A horizontal beant of light lalls on the glass plate B at 45".8 x l0-7 x 2xl lStX) = 4.45.=5.14 x=12cm A = 2. bright and dark circular fringes are irroduce.5 p=** = 5.: f8t1l8ll 60x60x lE0 Jd' . The thickness of'the air film is very srnall at the point of contact andiradually increases from the centre @ 0 = 2O seconds ofan arc 2Oxn = ------------:------:]:. = 4800 A = 4. oj the wedge shaped air film between the plates. 1992) 18 bands per cm. Two pieces of plane glass are placed together with a piece of paper between the t*. Thc I'ringcs are uoncsn{ric circles. tvith mon6chromatic light (blue) of wavelength 4800 A there are (Delhi. the tringes are colouretl. l98n lcns of long focal length is placed on a plane glirss plate. ^ Y is a Sourcc rrt rnontrhromatic light at the focus of the lens L.82x10-7 mfatk normally on a glass wedge with the wedge angle of 20 seconds of an arc. = 89 secondsofanarc x l0-5 radian' Example 8.32x I0-a radian (./ I \!/ d/ _( -4 vt Jf 1$ a7 J v / l. B=frcm= 1 1800 * p=* o=+ Here l. Thc 1'ringcs produced rvith lnonochromatic .4i15 x 10-a cm.25). 8.0625x10-s radian.8x l0-7 m outrvtrrls.33xl2 H=- e^ " RINGS Whcn a plano-convex ^sPNEwrONS's 4. A beam of monochromatic light of wavelength 5..o at one edge.32x l0-ox 180x60x60 3. Interfcrence takes place and dark and hiight circular fringes are produced.U.5 = 2x l0-3 m = 0. The reflected beirm from the air filnr is viervecl rvith a microscope. (FiU. .2lr* 2xl. With rtronochronlatic light.5.82x10-7m iP=1. uniform in thickness and with the point ol c()ntact as thc ccntre. a thin film of air is enclosed betrveen the lower surface of thc lens and the upper surface of the plate.82x10-?x60x60x180 2x20xnx cm 1. if on viewing the film normal$.uds the air filr:r cncklsed by the lcns I and the plane glass plate G. 'fhe glass plate B rcflects a part of thc incident light to*'.33 ^ 7'm 6xlO-sxll " . = 5 Percm Fig.l in thc air filni. Ir = 1.326 A Textbook of Optics lntet{ertrct 327 I = 6x l0-5 cm . When vierved with white light. If the refractive in/ex of ghss is 1. find the number of dark interference fringes per cm of the wedge length' (IAS.light are circutar.) = 2. ..27. OE=PQ=t 2R*t = 2R (approximately) n=l I = 2R. = fi}.1 .R When the bright ring is .-(v) I R I r+rq Fig.t.(Zn-D. or where h=n)" n=0. Similarly the radius of the brighi ring is proportional to (i) 2n -.0 {R. .(vi) .0 For air.... the centre is dark.. For the dark rings. {f Rdsult.3.R .(. II I I \ I .fi..(vii) n = 0. .. '2 (2n-l)1...2. (i) Newton's rings by @ggteglg!!.1. 8. The radius of ttie dark ring is proportional to (..26. from the point of contact O. or For bright rings t=fi Substituting the value of t in equations (i0 and (rri).etc... & 2 etc. While counting the order of the dark rings l. ..(. Here. I .. \F-. for the bright rings IT{ T.O 6a (.R ..(. the central ring is not counted..v) R .t D I =2ffi Fig.i0 This corresponds to the centre of the Newton's rings.3ZE A Textbook ol Optics lnterference 329 Theory..={*+^ For dark rings. 2prcos0 = nl. 2.2..(viii) 2t . for the first dark --HE-*r. Suppose the ra<lius of curva.(. and ring. etc. J.26 f = n)'.[iTR n=0 D=2fiLR=o . Alternately dark and bright rings are produced (Fig. 0 is small. cos0 (.. Therefore. interference is due to reflected light.ure of the lens is R and the air film is of thickness / at a distance of OQ = r.. 2prcos0 ='(Zrr* where t. Therefore. ..27).. 8. . (. therefore = fi. In Fig. P= If D is the diarneter of the dark For the cenral dark ring D=2r=2. 8. n = 1. 8.3. . I I . EPxHE = OEx(LR-OE) But and Therefore ring. the radius of the dark ring is zero'hnd the radius of Here. . Dt = 2"|frn For bright rings... l.r-Dn = 8{LR -6.(x) just opposite rcffie"ilng pattern due to reflected light.D.(xi) In equation (ix). '. n and ftrr the n th dark ring. Newtrntls rllgs are t"icwed normallt'l.$ls ott and irt utntatl u'ith rtrt opticalflat.. f = nl'R For dark rings. . the central ring is bright (Fi-e. *n"r" r D..1're' flectiott.28)..46."R -2 \f.r = 5.fi Therefore. = 5460x l0-8 . = 2. j (number of the rin. Ibr irright rings r=0.r lnterference 331 = 2. What i.50 re. sub..(r..Ii6in = 8il"R R the radius of curvature of the lower surface of the lens. Do=2i9l.R ton's rings due to transmitted light.. 8. in the case of New- .R =6.h.^ * 2.l) lR . rt=1. z = 0.q) the radii of the first.29) i. = fi. . 2t = n\" and for dark rings 2t * (2n.s the diameter of the 5 th bright ring "! Thc dirrncter of the l th hright ring is given by llere t). = {.A For the second dark ring. R.R The difference in diameters between the 16 th and the 9 th rings.(rii) Flcre L".rtituting n = 1. the interference frinIes are produced such that for bJEh!_rings. transmitte<l light of tovelength 5460 L. i P=l' and cos0 = I "Ii) . In the case of light (Fig.(xiii\ j f _t_.f =40A cm.50 2prcos0 = and for dark rings a1...r \ for air l. For bright rings. 16 rh and 9 th rings._ D [R .3.\. (2n-l)lR r=z* ff = 2(2n. . (2t.'14?.(xtv) D.50* =-fi /. third etc.. light.etc. 400 =(p-l) = (1.= .. of . Textbook rf Ontic.. Therefore. (2n. second. and using.2. cm p= 1.(Xv) Do. -fr.r) . . --where When. Example 8..2. the radius for the bright and dark rings can be calculated. 2prcos $ = (2n-l) Here./1"R = 2{trR Similarly the difference in the diameters between the fourth and first rings.") 'I=(t-..e. the fringe width decreases with the order of the fringe and the fringes got closer with increase in their order. . 8. (il) Newton's rings by transmitted.. For bright rings.lln-R I I (t l\ L'. bright rings can be obtained directly. A thin equiconvex lens of focal length I metres arul rel'ie<:ti"-e inrlcx 1.. is the radius of the ring and D' = 2{n}'R Thke the case of Taking the value .l) ). & 2 . .R = 2 fi. the diameter of the 5 th ring was 0.(.(. = n)'R Dn 2l 1ar- -. 8. m=10. can be calculated.t) xs+oox to-rx400 D = 0. the diameter -:o Boy's nrethod. Suppose the diameters ard the 15 th ring are detemined.590 cm..ht.590 cm anrl tlwt of the 5 th ring was 0. L is a plano-convex lens of large focal length. = lA. = 0. is reflected by the glass plate B incline<l at an angle of 45o to the horizontal.49. l./zxtzxs.332 A Textbook o[ Optics ln!erJerctrce 333 R=400cm p = . Circular bright and dark rings are seen with the centre dark.as found to be 0..36 cm f of the n th ring = p.* of the 5 th ring of light used is Here 5890 A'. Suppose.) Example 8.(iv) and m The radius of curvatrre of the lower surface of the lens is determined with the help of a spheronrter but rnore accurately it is determined by (D^. lDelhi B'sc (Hon^s) 19861 systerr.s)2 _ = l5-5 =_ 10. if the wwelength Hence. m (4r ^ = -. pl9g':-(0.48.. Then..590 cm. . Hence the wavelength of a given monochrorrntic source of light can be determined. (2n =''* rt Here R=300cm. A plano-convex lens of radius 300 cm is placed on ttn optically' flat gkt. i1y l) l. (D.R D = 0. tf the rarliw of the plano-convex lens is IAO cm.*. R = ? .rs plate and is illuminated by monot'hntmatic lig. l. = 5890x 10-E cm. .627 cm zdu DETERMINATToN oF TIm wAVELENGTTI oF SODIUM LIGHT USING NEWTONS'S RINGS The arrangement used is shown ih Fig..(D)' qr.. D. = 0. A parallel beam of light from the lens L.336 cm D...t is a source of sodiunr light..: 1. = 4alJR F. of the z + m th dark ring. qr. With the help of a travelling microscope..(. The tliamcter oJ the 8 th dark ring in the trarcmitted system is 0. = 5880 A ^= * 4rrrR .*^)' = 4(n+m) lI ..=? n ^= = 8. ad--. l. measure the diameter of the n th dark ring.72 cm.4 x=-7d--=4xloxr.336 cm and the diameter of the 15 th ring = 0...336 cn. calculate the wot.. Here Ds = 0.)r_(D.). Ds = 0. = 5760 A 2 . r = 0.330'z )uuux 1_ cm lu ' r n = {RRnv ro-s = -Jif O* tOO.72 cm' Culculate the *'av'elength of light For rhe transmirte<J used. 4mR - (D_). .590 cm..) (i) ftom (rt) (D^r^)'-(4. it be D1+n R=l00cm.2x(0-161'? p = tz x s: rlJtlo cm But.) Example 8.25. Newton's rings are viewed through B by the travelling microscope M focussed on the air film. :5 (Dl2 4 = nlrR or I-et -4_: Mrfosure the diameter Jni* - . (D_.^)'. Find thi radius of curvature of the pLrno-convex lens.elen|th oJ light used.----'-_. 1_-4+m"n' ^or Subtracting -).4 -4xloxR (D.336 cm. In a Newton's rings experiment the diameter of the 15 th ring v. rr= or = 5760x l0-8 l.47. In a Navton's rings experiment. Example 8.Jm^ (D. For dark rings passe. the diameter of the third dark throug.. I'-or dr. The diarneter of the n th and the (n + nl) th dark rings are determined rvith the help of a travelling microscope (Fig..6mm -.334 A Textbook of Optics lnlerfcrencc 335 ^ ^ = +* (0.lu. These are kept in a metal container C.f. The diameters of the dark rings are determined for various orders.R : 4 = 4r7'n . The light is falling at such an angle that it film at an angle of zero degree to the normal..r o.9 cm If I)..Butr=l:o'= 4 n)'R p Example 8.. r=fi ll .)' = V= ry 4m7" R . known. along the where 1'-axis and rn along the x-axis..'.3x5890x10-t R = 144.. The diameters of the n th ring ancl the Qt + m) th ring are determined.ed by the liquid.)r4D4) If l' .*^)' = 4 (rt+n) l.(0 . or lf *?1U . varying from the n th ring to the (rr + nr) th ring' first with air as graph is plotterl bctween 1--:a' r i I I I l&:'--= AB A.30 = l.r rings e. are known p can be calculated..-JQJ-Q)1-.(D'.. The air filnr between the lorver surface of the lens and the upprcr surface of the plate is repla.+ m 2pt=n7.) @.But.3l ...f curvature of the lens surface in contact with the glass plate when with a light of wavelength 5890x /0. 8. then divide (rrr) BY (i) (D. A 0r23456789 . 8.2 mm.=5890x10-8cm n=-.tperiment.8.. = *I mm R= I nlt Fig....R The liquid is poured in the container C without disturbing the arrangement.*. (D.t lRajasthan.0.590)r - (0.h the air ring is 3..(iii) is not m.)'= and D'n*^ is the diameter of INDEX oF A LIeuID usING *ftRnFRtcrIvE RINGS NEWTON'S The experiment is performed when there is an air film betwecn the pirrno-convex lens and the optically plane glass plate. find the radiu. In a Newton'.30). n=3i1.' is the diameter of the n th ring (D'^t^)' = (D'.B' the me<Jium and then with the liquid.16 cm the (n+rn) th ring then. l9t)71 l=nluR: Here . Fig.')'= # .82 n=e.336): t0rsast)*-to--tr cm or or M= 2R nlu = 99. or 2 p r cos0 = nL for dark rings Graphical method.. For the liquid.50.(iv) D1 n+B -D1 n = 4m).8 cm. D'n* n and D' .(. R. . _ (2x5.179 Here D.u*ri rurjbces.=? ' n+fr' 1 /1=-_-. U)ethi lett3\ D. = 0.53..40 cm Example 8. (5.90x lo-3 m m R=100cm=lm. 1. rt D? or R = l.u Example 852. z =0.54. (2n.6cm n = 10.-.(3.. 15cm ? Dividing (rr) by (r) .4x l0x i = 5. gives the refractive index of the liquid. R= 100 medium Dl= 4r7'n p= .590 cm = 5.2. 1984) Here _=(i*?J=. calculate the wavelength of light used.. L = 6000A = 6x10-5 crn l.36x l0-r): ^=-='-.. = 5880 A Example 8.nd the plate.=]rn.=-5895x10-8 cm. = [i'l 1. The wavelength of light used is 60U) ir. 0. In a Newtor|s rings experiment the diameter of the 15 th ring wa. calculate the refractive index of the liquid. The ratio of the slopes of the two lines (air and liquid).. 1n a Newtonb rings experiment the diameter of the 12 th ring clnnges from 1.tox6xTo-= 2{X) cm 4 x (0. (. If the radius of the plano-convex lens is 100 cm.15 -5 o. In a Newton's rings arrangement.!r&B "lL A. Here..1)x58/5x l0-8x 100 P-E tt = 1. calculate the refrective index oftLe liquid. fbr the A with a liquid between ine pbne ond .1 +n lu .R tL - ZF-- For liouid mcdium Dz = tl For air 4rr ).336 cm = 3.55. the diameter of the l0 th ring is found to be 0.3. . 1. If of the 5 th bright ing k 3 mm and theradius of 'curvature (Gorakhpur 1986) rings are formed b1.-- (D \2-D2 a 4mR 4xl0xR *D2 D: .(rl) cn.27 = cm. d a drop o. .f the fitce of the lens in contact with the plate.es from 1..36x l0-3 D. etc.. Newton 's length 5895 Y=-Ay Example 8.D. (Delhi .) Here n=5.=0.3x4. (Delhi For liquid medium 1990) R=? ^ .(i) .27 cmwhert a liquid is inrroduced befiueen the lens and the plate.9 x l0-r)2 .3-16 cm. Example 8.40 cm n 1. reflected light af wave- n th bright ring... l.880x l0-7 m Here tr=j.6 cm.l'w'dter (p = +l:) be placed in benreen the len. (Nagpur 1985) AB 'the diameter of the curveil surface is 100 cm. D. Calculate the refr"active indcx of rhe liquid.51..6)2 Dl = fu7'3 For air met]ium A = lnlvR P . = 0..336 A Textbook ol OPtics Interference 31'7 rz .r and the Jtlute.I) l.. /n a Newton's rings e4teriment tlu diatneter ttf thc l0 th ring chang.35 cmwhen a tiquid is introduced Ltet*'een the lens a.590 cm and that of rhe 5 th ring was 0.50 cm to r.s found to be 0. = 1. Obtain thc raditrs of cunature o.(ii) tt . R . 2 t= 8. 8.. 8. -=[4jl = l'3-5 cm D.) ^ (2.etc. = @* P=l = I =---.A Tetthot* Dividing (i..9 x l0-5 cm = 5. tt OT t+Fig. Suppose the radius of the n th dark ring = r. = n?tR = 5.{ t.L^.32. 2prcos R = 1.tt+ r] i.) consider two curved surfaces of radii of curvature R. *. l99l) zprcos0 = nI.rlr I t[&-\)=l:-i-)=*.= l'50 crn D. = 5.)= "n n = 0.32).. The dark ancl bright rings are formed and can be viewed with a travelling microscope.so r \ .95 l0x5.-&)= " .'. 1. ) = [ tz(. For the .. 2.etc.56. (Delhi. r..) by (r) oJ Optics lnterference From geometry.26 2. Newton s rings are observed h reflected light oJ l. l. 2t = n)r.5 x l0-3)' l0 x 5. A thin air film is enclosed between the trvo surfaces (Fig.J = ^:i But PQ=t ry-4.'- '("0 . in contact at the point O. {hs thickness of ttre air film at P. For n th bright ring..3.9x l0*7 cos0 - x10-6 m o=@y NEWTON'S RINGS FORMED BY TWO CURVED SURFACES or where (2n+t)L f.3. I-+l For reflected light. I 7'. n)t t=_T . the value of z= l0-l =9 .. [ -: I n)u (i) Here. r.059 m (ir) Thickness of the air film = | For bright rings.(.l I ._ I )_ *r[E o )= n = 0.(. PQ = PT.l _ tzr. = (D ) Pr=L and u'*' Qf=q pO / [. P = l'235 Example 8. fordarkrings..9 x tb-5 cm. and R..2.. The diameter of the l0 th dark ring is 0'5 cm' Find the radius of curuature of the lens and the thickness of the air film.". is l0 th bright ring.QT L_.9 x l0-7 m or where ?s = n=10 tr== '(*. Here' for air .9 x l0-' ' ^[ r . rurk. the geo. one would expect 6. itrdicating thereby that explained easily be can ii silaigfrt [nes.[ . size compaied to the wavelength castbyanextendedSource.t "p"rtir".1 INTRODUCTION r_t.AB isrQe illuminated i. XI'is the screen placed in the path of light'. . the shadows ex- light ro. Rectilinear propagation of light been observed that on the basis of Newton. l92l refractive index of the material of the biprism. This bending of geomerical wi*rin the light of edges of an obstacle or the encr@chment is placed MN obstacle an opaque If irrfoo* is called diffraction. foundtheedgesofanopaqueobstacleorilluminationofthegeornerical produces Each progressive wave. l92l (r) 96.l)0 where d is the distanoe between the source and the biprism. if un opuqu" obstacle is placed in the light travels shar[ shadow is cast on ih". ft tt ffiil.-if seconoary wavefronts are drawn.gy of light a beam of bending sound *"r"i. in g. sources in Fresvirtual coherent two the Show that the distance between nel's biprism arlangement is 2d(a . (r) What are coherence length and coherence time ? Why is it imposiibte to observe interference between light waves emitted by independent sources ? lDelhi (Hons\. r. *"ni"uf-rtrrOonir"ion AB the light sdurces shadow of.l illuminaion be should there o).th€geo.ilr-" h. 0 is the angle of the biprism and n is the lOsmania. Discuss how the wavelength of monochromatic radiation can be determined in the' laboratory with the help of this interferometer.i .uigf. S is a source of monochromatic light and MN is a.point sources and obstacle is in "r. gmnrerical the in illumination the But also.s.il*'ililio shadow... losmania.s corpusiular theory. according to Huygens wave theory wavefront. secondary the forms which of envelope secondary .*ro*hment of light in the geometrical shadow. 1921 DIFFRACTION't 9. l92l Discuss briefly the various methods for obtaining coherent sources of lOsmania. the (o).meni"atshadow.s are not sharp. Similarly.i$erigeMNzstheprimarywavefront. portionofthescreenandaboveAandbelowEistheregioNE. lDelhi (Hons).. Give a complete description of Michelson's interferometer.i of shadow a If of light. in 6.an obstacle is not commonly observed because of very large are used the obstacles secondly not. l92l light in the laboratorY. in propagated is metical shadow also..C-g.lightfromevery (bright point on the surface of the bulb forms its own diffraction pattern . one would p*"".accordingto rruvg"*' .nd the i'orm"O by small obstacle. 97.screen. similarly.r 99.sayafrostedelectricbulb.small In Fig.A Textbook of OPtics What do you mcan by the tcrm coe{I'icient of finesse ? (c) Prove that the fringcs obtaincd with Fabry Perot interferometer are sharper than those obtained with Michelson interferometer. If iight "n". Th. But it has (a small circular opening U"u." p"tt of light tFig. #frffilnon experience that the path of light entering adarkroomthroughaholeinthewindowilluminatedbysunlightisa path of light' .o*t ortiori. of light passes through a small of the georegion into'the a narrow stii) it spreaas to sorne extant of waves' the form ".. (a) Distinguish between spatial and temponl coherence. can be identified.I and I + cos 180 = I .: a wavefront can be divided into a large number of strips or zones calle<l Fresnal's zones of smalt area and the resultant effect-[f any point will depend on the combined effea of a[ the r"*na.xlr is the screen and so is perpendicular to xy.ungecl in the form of a cross) that one sees while viewing a disiant source oi tigt t through a fine piece of croth are ail examples of diffraction effects. they should give a-wave travelling forward as well as backward. (3) the effect at P will also depend on the obliquity of the point with reference to the zone under consi&ration. from the. To obtain the resultant effect at a point p on the screen. S is a point souroe of monochromatic light and MN is a small lpertnre.fi. The luminous bader that surrounds the profile of a mountain just beforc the sun rises behind it. Similarly.-aG to the part of the wavefront at c. the effect is maxirnum at O because 0 = Oandcos0 = l.nunuting will depend on the.-{isqance of the point from the . combined in a sriking manner Huygens wavelets with the principle of interference and could satisfactorily explain the bending of light round obstacles and also the rectilinear propagation Fis. particurai zone €) ttre effect at a point due to any ".9.g. The term diffraction is referred to such proutems in which one consiclers the resurtant effect prcxruced by a rimitei portion of x and dark diffraction bands) and these overlap such that no single partern Difiraction 405 effect will be maximum at O and decrea. \rc FRESNEL'S ASSUMPTIONS C0 becatise 0 = 90" and cos 90 = 0. Considering an elementary --t--L_ --/o-*. the light streaks that one sees while looking at a srrong source of light with half shul eyes and the coloured rp""to lrr.y *uu".oni . that if the secondary waves spread out in all directions liom each F)int on the primary wavefront. It is maximum in a direction radially outwards from C an it decreases in the opposite direction.I = 0. Augustin Jean Fresner in l8ls. the resultant effect at an external point due to a wavefront will depend on the factors discussed below :In Fig.3 RECTILINEAR PROPAGATION OF LIGHT ABCD is a plane waVefront perpendicular to the plane of the b+ b+ b+ perper { + lIg Mz M.various zones .q4 4 Te.1 Diffrrction phenonrcna are part of our comrmn experience. as the amplitude at the rear of the wave is zero there will evidently be no back wave.es with increasing obiiquity.r I 2 P t o ti trh .. in a direction tangential to the primary wavefront at C (along CQ) the resultant effect is one half of that along r- 1- l: Y Fg. Accmding to Fresnel. the 9. MCN is the incitlent spherical wavefront due to rhe point source . the resultant effect is zero since 0 = 180" and cos 180 = . 9.s.rrbook of Optics a wavefront.. 9. (a) (b) --\o-t olo}- wavefront at . C. e. Fresnel assumed til. The eft'ect at a point due to the obliquity factor is proportional to (l +cos0) where Z PCO = 0. In this direction CS.2. This property of the secondary waves eliminates one of the difficulties experienced with the simpler form of Huygens principle viz.2 of light. (vi) incoming light is €!4e-L{_peg{e. O is the centre of the aperture and r is the radius of the aperture. f..et the distance of the source from the aperture be a (. O is the pole of AB be a small aperture (say a pin hole) and S is a point source .... the principal focal length L .is the incident sphericzrl rvavefront and with reference to the point p.l \df . A zone plate is made by arranging the radii of the circles which define the zones such that they are the same as the radii of newtonb rings formed between a plane surface and the surface having radius of curvature 200 cm Find the principal focat length of the zone plate.3m and a.74 x 10-{ Diffraction phenornena can conveniently be divided into two groups viz. [Delhi (Hons) 19921 9... In the Fresnel class of diffraction.) For Newton' rings.. rr = From equation (i) 2...5xlo-7 -!-*l0.= o.(. Here \-n=tr . s..(. the source and the io"en on which the pattern is observed a.(.4t8 A Tbxtbook of Optics Dtl[raction Ir=i fr=* lll 7)uf r2 ..6.(.e atjgfinite iistgnlg from rhe aperrure or rhe obsracle causing diffraction..) For a zone plate. Substituting the value of u and l" in equation (iii) .3 t.r = 0. a = -03m of the .t5 m tt 5x lo-7 Example 9.. the source or the screen or both are at Jinite distances from the aperture or obstacle causing diffraction. Fraunhofer diffraction pattem can be easily observed in practice..ir)
[email protected]) '.Oe. = 0.lO = a) and the distance of the screen from the aperture be b (Op = b)..6m . The m .(v) zone Negative sign shows that the point source is to the left plate and its distance is 0. Fresnel class of diffraction phenomena are treated frst in this chapter. -l Multiplying equation (iir) by 3 and equating with (rv).3 ' 0.=2m FRESNEL = .w$r_3 lers and the diffracted beam is focussed on the screen with an6*Giteni.*'a (i) and (ii) r=T=* But R=200cm=2m u-. radius of the n th ring. o.y'' al. SP is perpendicular to the screen.(.. (i) Fresnel diffraction phenomena and (id) Fraunhofer diffraction phenomena. From " = fi'?.. the phenomenon observed on the screen is called Fresnel diffraction patterfl In the Fraunhofer class of diffraction phenomena..8 DIFFRACTION AT A CMCULAR APERTURE L-et of monochromatic light.t. p.=7 -l ll)r ll3i. = 0-06 -.) ll1 _-_=_ ur'f. 0"06 -.. = QJlz1eit. [.o-bservation of Fresnel diffraction phcnomena Qo lo! requqe _any tgry&. In this case. In such a case..(iv) n. 131l -. the effect at a specific point on the screen due to the elpg5gd_lgcfCggly4yglrg_!! is considered and no modification is made by lenses and mirrors. . = ''liTF .3 m. rtr*i"ti""r ii"it*"it Fraunhofer diffraction phenomena is simpler. xll is a screen perpendicular to the plahe of the paper and P is a point on the screen.*#i*rdMi'd{d*! f. .33 plane of the paper. " tne directions of asin0 = (2rr+ I)l . Tbnhook 448 A I )ilfruct ion 449 .9.t and this again gives rhe position . ft ur. and the path difference between the secondary waves from these p_oints i. In this .ABL il. A plane wavefront is incident on the srit AB and each point on this wavefront is a source of secondary disturbance. The secondary waves frompoints equidistant from o and situated in the upper and lower halves oA and oB of the waterfront traver-rhe same disiance in reaching una the . . slnun *tro*"" (2n + l\ )" = .r r i"i"l the secon<iary nraxima can be obtained. The secondary *ur". and L. mum intensity. L. . is the collimating ter.or"... Now.of a central bright miximum at P followed by serondary maxima and minima on both the sides. Thus. diff. horvcver. The whole wavefiont can be considered to be of two halves OA and OB and if the path difference betwen the secondary waves from A and B is L then the path difl'erence between the se irdary waves from AL = asin0 where a is the width of the slit and AL A and O will be .34.the parh diffcrence between the secondary '"vaves originating from .'n*_ elling in this direction reach the point F on the screen. in the A. The intensity distribution on the screen is given in Fig. the diffraction pattern due to a single slit consists. terference takes place and the point P will be of minimum intensity. The intensity distribution due to Fryet's criffraction at a srraighr edge is given in ril. of mini- sp is perpendicular to the . The screeri is perpendicurar to is a stit of width a. Ae and BV corne to focus at where e gives the dira:tion of the n th minimum. If this path difference is equal to l.il*y . All r#r".. the path difi"erence is ocld rnultiples <. be prane and the diffracted right is co[ected on the screen with the a lens."n = l. the source of tight shourd eirher be srn0 = AB AL AL =. In Fig. the incident wavefront Then. 2. 9. the wavelength of light used. inclined ar an angte 0 to the direttion op. -L . This is so.. The secondary waves reinforce one another and P will be a point of maximum intensity. ourelling in the direction parallel to Op viz. destructive in- lower half... and Ag ing through the srit {r_! Ag is incident on il* rens L.v/ dT. Here n is an integer.er in path by Fig. Thus.. ].2 1 Similarly tbr every point in the upper half OA. lyst help FRAUNHOFER DIFFRACTION AT A SINGLE SLIT To obtain a Fraunhot'er diffraction pattem. from 'f the slit or a collimating lens irust be used. asin0 = n l" nX P and a bright centrar image is observed.ol Optics in the region of the geometricar shadow.t is a narrow srit perpendicurar to the prane of the paper and illuminated by monochromatic tigtrt. The rine are achromatic lenses. screen ] In generai stnU = t.2 or where bllveen the secondary waves originating'from tf. Ttre point p wil be of maximum or minimum intensity depending on the pif." poirs of the waveiront. and the nnJ'rrru"t"o beam is observed on the screen Miy'. f. Dtaw OC and ff p"rrtnd"ular to "o*rponaing AR. disrance or ii"-iigtt purrthe is . consider the secondary waves travelling in the direction AR. because the secondary waves from the ctrrresponding points of the ).path difference is zero.. then F will be a point of minimum intensity.1 and B.3 etc. If the direction of the secondary waves is such that AL = 2 L then also the point where they meet the screen rvill be of minimum intensity. If. rhere is a corresponding point in the lower half oa. the incideni sphe}icar wavefront. 9'33' . xl on page +zo. lfr'l i.r0... of the tentrctl Example 9.citivr.Butwithanarrowslit.stinctdiffractiontnaximaandtninirrtir on both the sides of P.!tl. hlrli rttquler width of the central maxlnlulil' i-rP'.n ts = T iurr/rlri. Find tht' hatf trngttlttr width a slit of v.iJ2 crt x = 5mm = 0.24... is very n*ar the slit or thc s(]Ic{rll 15 l.1. /lr 'l s...xrsition of the central bright ntaxinlutn and the points ancl B on the screen for which the path difl'erence between the points A (ii)' if the width a of the slit is large' sin0isbrrrallandhence0issrnall'Themaximaandminimaare\rery closet()thecentralmaxinrumatP.Thisresultsadi.s:ttnit.i siri n a J :_'- I t fi {t x is the disunce of the secondall minirnunr frorn thc point i' Thus.\r l-'. the wave- D=2m=200cm 0.\.itl i' t|.:tt frcrt fh':'/('flr 'rr it:t. the width of the central maximum is more.hen the slit i. rrii. From equation t{).irt iit': po[ttrf iJ the r{ir .J . The diffraction light' With consists of alternate bright and dark bands with rnonochromatic white light. i-he secrs 1. I l2x lere rir1 0 =- I LI wherc tl is.r illumirttttcrl hi nr.trvelrn8th 60tt0 A.ti. =. il':e iens Lr.tr . t) Zr = 1 11 Lt.n.n..2 mm = 0. the width of the central 1. = 5000 A ."eislarge.'''!e"g.9.l'. With a narpattem row slit.qhr' the nutxinunr.a\ l -= (.fi)\..il lDclnt {SuLt) iv77l o. apefture.3'1 a = llx i0-r'cm.J m rlr({i1 F.=5x10-5cm l.r) srno =...02 x 0.5cm The width of the central maximum is proportional to length of light.va1. L Ll n='i asin0 . on<1ary maxirna ar* of much lcss intensity. thel ()r ll)xamplc ^!r 1) = .ii. j fitrtl "..idtlt 0f patkril hriglrt nw.tider.2r'm undthr:fr.!'thetrtrtr.i ^ A ---2Cn *uiirnu* ir rnor" than with violet light (shorter wavelength).)' = oO{)ili = 6x l0'1ctri' (''\ l()-5 _ L r.rrtittilr.tl' In the case of Fraunhofer diffraction al s narrow rect...aisslrralland t.l()" }. the centrai maximum is white and the rest of the diffraction with red light (longer wavelength).! !i.t. etc.ttrtt-hrttntttlic lilil* oJ lrancls are coloured.\ .t ---a l" ri) Here a = 0.5rnnr . lr. corresponcl tt: the pttsitirins of secondary mintr:ta' iiil rapitil" i"iiis The intensil'.rinrLtnt itr the Frctunhrfer diffractit.t.l)ilf rnct iort 45t 450 A Texrbook of OPtk's p corresponds to the... t..rt.ingrrlar r:lsin0 = nI rvhere / is the tocal lenglh of thi: lerls sin [.r = x J I- But. {iom the point P (}Lltwiirrds.rciuttlttlt:t 'itifrit.\!it {l :i''t'.tt j cm w. the wiiith of the central rnaximurn = 2x' where ^ 9t =)" D . {rr'lti lf" the tens 1. .456 A . ..dslnU .u-Ttor porygon method. as a function of cr or sin 0 (along 2a .rngth of the arc OM = Ka whcrre /( is a constant and a is the width of.i. . the ampriru<les are sma. the resultant ampritude .The value of !ll3 fn. Chord OM = 2 OL = 2r sin cr "'(') The length of the arc oM is proportional to the width of the srit .et a monochromatic parallel beam of light be incident on the slit AB of width a. (.ri i'np (along the Y-axis).(iv) intensity at any point on the screen is proportional to )". The secondary waves fiavelling in the same direction as I ar rhe 6rnr is given by #.y srna. Fig. 9. and the phase difference increases by infinetesima..tr* circurar *.35)" 0. 9'27)' As the waverrontlavT is divided into a rarge number of strips.[ " ) The . amounts from strip to strip' Thus.) equation (i) of 2r in CT Chord oM = {9...[ .tuxbook of Optics Diflraction 457 of the circular arc... 9. O.(iv) ZMNP=2a /.(-I. OM = A where A is the amplitucie of the resultant 4 = 16.sin2cr l=A:=A'ri=1.irJulo. Here.the resultant ampritude due to alr the individuar smalr IsinoY .36 L. Thus. strips can be ohtaine<r by the. Thus'thevalueoftrdependsontheangleofdiffraction0.!!9eU radius Ka fnauxgpFng D^BrB*SrloN Ar A sINcLE. $tq and the intensity 'f vibration at a point on rhe scrcen Fig.. Therefcrre a phase difference of 2a is given by centrc 2s=+-asino where e and .11!!!9 CT llLE a = I . L OKL sincx = 4 : Ot = rsind a. is the ^K at an angre 0 from the point g upwards.. cli (t ig.t ) is givcn O.. the path difrbrence changes antr hence the phase difference arso increases. 935).oP gives the direction of the initiar vector and NM rhe direcrion of the finar vector due to the secondary waves from A. . .. the vibration porygon coincides *i*.st:IT (CALCULUS MBTHOD) 2r=&' (x Substiruting this value ...'" l-.34 represents the intensity distribution. rin o But.. 9. Sr 4 = l. (.the slit Also.. Let cr be the phase difference between the secondary from the points g and A irf the slit (Fig.. different values of 0 gives the intensily at thc point a' under consideration.OKM ht the __ (I sin 0 is the path difference between the socondaly waves from 2a r (Fig. a graph ir It is .(v) where r is the rariius of th" . -= I *. 9. \': = .l p'=il +--*#+1 L r r) In the case of distance from the slit. . .-. the wavefront rvirh c[&diiiEi G.. r asin0'\ --r------ .** [. are (1. T'he seconclary rvavLrs travelling at an angie 0 come to focus ar p. .fgt_glgprdinates. the screen is at very large r >> .t. therefore I .li ' i . -..il r \ nasin0 /" 2\/ U r'ii' -.Al^iI .'ilur.-71 f .-.j i ' r. The cenrre of the slit O is the Consider a small elemenr r/..:: i : )i { -- t. r :sinO\l +Lj^:': lld.' S.* ]sin rn l/ fi.' .458 A Textfunk o.r.! r !.t 'ir -. f o""1 P'=tL .fi) t..t' ". r . is p.: rvavelitint.jj d'11 i rr sin 0 'lj tlr ..i:r. Substituting this value of p in equation (li) Fig.-'-' t"t lsln t i /" I I p.l).(iu) a Fraunhofer diffiaction.. . consider the screen to be at a distance r frorn the slit..n .J6).-?] 1 ao p r But.=-. ) 3 . ona lr is negligible.i..l' Otiict l)ilfractiott 4-59 the incident light come to focus at the point p. ".n i . : tA ilt . * t/r Ar/:sin2ni' ui \rri zir -n I I i... iiisplaceinent at I' rlue til the rvlrol. The clistance of the elemenr from the point p .ilr t'i j r 'l ..t sin0 r = (v) P = r-rsin0.. rn) [Fig... ! I .. = K lsinl 'Ihe resuirant Pl .J 'i' .i . {..i Slnl-l1'' (l i ./fislnu t i i ------L \ -cos2nl rl 2)..t .: ]a.- . 9... I ) t': trJ s.: z "l(' 'o L ^ ^ )) -2 ]" . I .r.-..:. I rr:'-. p=.irrg ihe valur: oi { in cqultiun (iii} r. 9.Icos}nl^l. 1. I 2zz^ -..37]...37 The displacement at the point p' *1 is given clu* to the elemenr de at anv instant : by. *l ' r / !.r of.ilffir'oririnates of the point p.[ . (Fig. i.- . 0 = 0.. the secondary maxima are of decreasing irrtensity given aklve' equation the rer:tiqns of these ntaxima ere obtainer.-...3rl "'(vii) [.Dilfracrnn 461 A Textbook of Optics ./ --) 5:tl | =i . seconclar. and hence Ttre value .f /\l *= and 5n z W when o+0 is equal to I.v maxima are givenYy the equation p{S*ondary . 2' 2'..(vi) given by is tf ' *= 3I.l trom Substituring this vatue .I Ltl 0 -I = *-lt)) (b) For the scrondarY maximum.. Hence.J I . 3 etc. o'.. The directions of (2n + Io l) l" (v) (page 4-57) =aJ and the diThus. o=3n 2 I^[ttt' 2 *( Y (. u' / I / -1 \ ' t . 3 .t ttre intensity at p. u i = 1o ) . 2. For the point / O. I l+l='[*l={* L. "4g Ia') ) and (sinc)' I=lu . in equation (vii). Cr = 0. = K.'.' ) =/. V4 C"n**t Maximum.-+0 p on rhe screen (Fig.[Y j'.) ."d. l :'"" | -"1 . rhe intensity r=r(!1!gl G '[ ) at P = /" I ... (I= ^equarion (2n TE u + ll7. j:LT]-zsru' =[ r l' -." I' = Ka2f (a) For the lirst secondary maxinrum I ) an. the P' ^(.l---t Here 1. 9..tI= . Ibr cr 0 = stn (I -__l I.. Nlaxima. = l. wtrich is maximum." The amplitude at values of ct are Substituting tt = L.27)... - l3a2 and is the value of the intensity at p.(sinrg ) (. The intensitY at P' is given bY i' 2 - 2a (<)r = ( r'iii) t' = llry-" "l. J:r i :r tti i':: l"' r:.t' li'r:. .A Textbook of Optics Di.l.: 'l? 1Y * ".cntr' []ig...r '1i.{.'aiurs tf .'a ''l'.1'1.i''l { 3 .ir.lL:lurl lii llii) i:i':l riir':r .= i ii t.c-qualion r'rsin0 tllirsriturrnil ihr-: =:ri .it !.. given l.ii-t lr1 litc. 'j.r.:ilt. t".t1.i'i iii r'rilli!'! i' . Q -ti1 g*-(Uff1_t1)?gl .ril ii.. o't r-lR"l'L'H. tt ". mum (l'rig.&l j. ur (2 sin 0 cos cr) - (sin?cx) 2u = 0 (1i. 463 n x t *r & the tancl = ct that (r Ii' Sraphs are plottcd for 'v = fi ancl') = tiln it rviil be found The tuo minirnr' s.11 f) rn l"'iitlltiiltl ii'l' ')lil 1llir. l' jir' r: [.i.:s t .rllsl "':'.'.-) ..- # u":hlH(). -i tllt llt '-'tliriiltir'i illi' r. .r-'i-llndery iitiltir.'l-::-Y!'1"' :U!Y lrlg Ct' For I' to be maxirnun'l 'ldr *----+ E i " T x I 'dt' _ n da-' l *I ?c fl l.'i .iii i l.rr/il ^ il' lr'l:I .i r . lhe .':i:n tltl:.fJi'at'tutn .Ir}' niirtinll tlir) secondarl' Nlinirna. .i !r11 ralur . littr pi:.{' '---si!. '). .ct.-18)' 'Ihe dirc{tions oi t}rc 5e\:i)rl{li. i t:.i 'lr' (i ' ) ' I ln l:ig tire '..) i.lla arc nct exactly midway betri'ecn are ptlsitionsrllthescctindarytnaxinraareslightlyt0wiudsthccenit.a '..altnlluiP :.i ..'lri.nilar)'maxi.hi t ll til il] t .v thu. l.2x 'l - !).[ ti. 0 = nl" .) wi.22.r = o. All the secondary waves travelling in a direction parallel to OP come to f<rcus at P. ail the seconirary waves emanating from points equidistant from o travel the same clistancl before reaching P and hence they arr reinforce one another.. to ttre position of the centrar maximum. cp is perpendicurar t.. the radius of thc central bright ring rJe.".. sinO f I { .orr"rp*rn.16. The width of each slit is n and the width of the opaque portion is b.('v) In Fig. be of minimum intensity this parh difference is equal to inlegyal multiples of )v i.r 0. Therefore.rthook of Optics come to foius at p. . The irt.ru- I)ifiiiu tiott 465 Fronr e<luations (iii) and (iv) -{ co f ()f fx ..4& A Tb.x = fiom P 't = l00x5tJ93xl0-n ttffit=-- = o'5893cm Further.ndary waves ernanaring from the points g and A (extremities of a <liamerer) i. the.. .th 58t)-l it.(.. :l .+.11. and lying on a circle of radius x wi. the diffraction paftern due to a circular aperture consists of a cental bright disc cailed the Airy's disc.(.r.cteases.nd p is a point on the screen..t = .. Now consider the secondary wavqs traveiling in a direction incrined at an angre 0 with the direction cP. dsin The point P..nrrty'nr tn.(ri) l)' d of minirnum intensity.s AD... be of maximum intensity if the path difference is i.IT sin0 = 0 = dsin 0 = ).point p.. ? and bright concentric rings cailed the Airy's rings.sc in the diarnetcr ol'the aperture..d ."r""n. From rhe b ABD. t.rt se(ptrdzrt mitttmum. elling in the direction of the aperture . .) AIso. I{ere. The screen is perpendicular to the plane of the paper. p . 9.s illu' minatt'd bl ntrttrochromati(' li1ht o..f = lfi)cnr.(vi ) of the Airy's disc. Accorilin. ficre .t is the radius .rtreen i: (tt o di. But actually. equal to odd muttiptes t ! 'fhe tiiscussion o[ the intensity tlistribution of the bright and tlrrk rinls is sinrilar to lhc onc given tbr a rcctangular slit./ rnrn. ln [-raunhrl'er difJrrt<tittn pdttent due ttt ct . rrn tf. the distance PP.l' rvaveleng. 'fn" widrlt of rlte slit i. ptane wavefront is incident on the circurar aperture. lrt a plane rvavefront be incident on the surface of XY.: first dark ring is slightly more than that given by equation to Airl'.e.26 FRAUNHOFBR DIFFRACTION AT DOUBLE SI.22 fL ..t. L is a collecting lens and MN is a screen perpendicular to the plane of the paper. AB and CD are two rectangular slits parallel to one another and perpendicular to the plane of the paper. tlu. dsin P-. surrounded by arternate dark dark rings is zero and that of the bright rings decrJase. is a point o = Q!ll__U?' . for the first secondary minimum. Rrr a rectangular slit. Therefo.xample 9. The path clifference between the sec.( r') radius o1' the (u). be x.. With incrca.01 cm. C-rrftllate the separatiotr betwectr the centrul nuutirnunt atul ( Mtsorc) the fir. =e= d . F. if is at a large distance from the the coilecting rens is very near the srit or when the screen lens.40.r. d= 0. The secontiary'*ur". gracruarty out*ora. be of minimum lntenslty.rtance oJ llM ctn frcm the . I = 58934 = -51193 x l0-E cm. the /D = dsin0 if Arguing as in Articre 9. then all the points at the same dislance from P as p. P corresponds to the position of the cenral bright nuximum.tirtglc slit..e. P is a'point on the screen such lhat OP is perpendicular to the screen. 9. wi. the screen. 'lhus. I rnnr = 0. it rs gtvcn try rvherc 1.slit and the slit t. . Arl these secondary waves meet at p. When B is ut tight ungle. allows the wave to pass through it when it is paraller to .I.RIZATION OF' TRANSVERSE WAVE.l Fig. will stop the other components. becarse the slit S. A wave emerges along cD and it is due to transverse vibrations paraller to the srit s. are at right angles to each other the rope will not show any vibration beyond Sr.ARIZATION"" . No change is observed in the light coming out of I [Fif:. and "1". tf S.A Tcxtbook of Optics 563 If the end A is moved in a circular manner. Keep the crystal A fixed and rotat€ the crystal B. The crystal A will act as the slil . the waves will pzu. . 10. rope is anached ro a fixe<l point at s [Fig.? (i)].gsr u'hen the crystel B is rotated. The phenr:menon *f polzuization has herperi to esrabiish beyond doubt that light waves are waves are set up by moving the rope forward and backward along the string. C)n rotating the crystal A. it will show only linear vibration. POLAntzrr. trrnrr"rre) 'rr'vheaher the vibrations are irnuar. and S.sity of light coming oul rrl'it gr. The light is slightly coloured due to the natural colour of the crv-stal. r. Beyond S. 30 P${. whether the liEiri wilues are longitudinal o. Hord the enci'r. and 5. r0.l L.ulually increases and is ma.(Tects rlo not tell us about rhe ryie of wnl/e n]otion i.s. Rntate both the crystals together so that their axes an: aluai's parallel..1(D)]. 10. the inrcn.l move the rope up and down perpendicular to AB.he slit rrrrtrc'd thrr.IGHT If longitudinal SOUT]CE ii) a f's:rn of. It iJ observed that the sht q does not allow the wave to pass through it *h"n it is at right angles to the stir S. These Experiments on interference .c puallel to the slit S...fyg_rngilon.rallcl . l{).qr. [l-we consider the propagation of light . [FiS. irrespective <lf their prlsitiort.e.S .ximunt again when the two crlslais are yl.:iltj dirfraction have rhown rhat hght is transverse wavesa \=- POLA. (l) (2) -_l ... s.rugh /o rt.t nr) light emerges out of B [Fig.. The ligirt tr-ans8 trecomes dimmer and dimmer. the rope will show cir'' cular motion up to the slit S. 10..2 A similar phenomenon through a tourmaline crystal. *rrcuiar or torsional. no remarkable change is noticed' Now place the crystal B parallel to A.Itre Let a rope AB be passed through two parallel slits S.rs a krngitudinal wave motion then no extinction of light slxruld s.2 (ri)]- Il'the crystal B is further rotated.s through S. 10. .i and TO-2 / Fig.l(a)]. '[his rxperiment shows conclusively that light Ls not propagated as longitudinal or compressional waves.2). l0. has been observed in light when it passes Let light fiom a source S'fall on a tourmaline crystal I which is cut parallel to its axis (Fig. iti"#r-occur plane of polarization and the bration. 10.y'0s pLaNE oF poLARIzATIoN when ordinary right is passe! through a tourmarine crystar. the vibrations of the particres are represented parailer (arrow heads) and perpendicular to thl plane of tf. Moreover. PI. @-iIJnu.e papur'(. (10.ANE OF VIBRANON ffi0. t0.ight is polarized and vibrarions ari confined to"onry one directioripeip"nai"rr* to the direction of propagation of light. vibrate onry It is crear that after.i.prace a tourmarine crystat uno . or-8" waves crystar A is said to be porarized because it has acquired "ri the property of one sidednoss with regard to the direction of the rays. ffi@ (itr) lrYiPI-ANEOF Fig. the vibrations may be linear. (Fie.5) (rfl the vibrations arc represenred only perpenP dicular to the plane of the paper.frLARTZATION BY REFLECTToN v" Polarization of right by reflection fmm the surface of gtass was discovercd by Malus in l g0g. He found that porarized right is obtui*d *h"n ordinary light is reflecred by a prane shiet of grassl consider tte rigtrt incident atong the parh AB oi the gtass . r0. has acquired the property of one sidedness. in one <tirection. 10. *utl. reflected arong 10.. Therefore any vibration can be resolved into two component vibrafions at right angles to each other..o. porarization .t Fig. 10.6). arso perpendicurar to the direction . ogcq{rtre it pi"* of at righr angtes to the in which vibrations oicur is known -or as plane'of vibration.3 is trre ptane urprane fO: . 10. This is.4)..5 tY.'." slowly. Simirarty polarized light by reflection can be produced iro. In the parh of BC.. a 57'5" for a glass surface and is known as the-potarizing "qr.5U pobrization This experiment prov_es that light waves are transverse waves.u. The prane is that plane in which re_:tU_rgr_lns. and yy.tot l. \uT. From our idea oi wave modon. This angle of incidencJ l. circular or elliptical vi_ circular Ondinary light from a source has very large number of wavelengths. (10. otherwise light coming out of B could-never be extinguished by simply ro_ tating the crystal g.fac. Fig. As light waves are transverse waves the vibrations can be resolved into two planes rx. s. Ir win be observed ttrat tigit is compretery extinguished onry at one panicular angle of incidence. tight is brations consist of two linear vibrations at right angles to each other and having or elliptical.passrng through the crystal the light A Textbook of Optics 565 ar right angres to each other and of prqpagation of tighi (Fig.A. it . The plane EFGH in Fig. prane porarized light and In Fig.f angre. I 2.3 Por-ARtzAIoN In Fig. rtrrrerorJ right corning.3) (r0 the vibrations are shown onry parailer to the prane of qrp"r In Fig.5(r).uri*. v a phase difference of .6 gc. the . When the mirror M. the intensity of the reflected beam along CD decreases and becomes zero for 90" rotatirln of Mr. IO5 BIOTS POLARISCOPE It consists of two glass plates M. When the upper plate M.. tt = From Brewster's law sin i .f *re mediu.\W BREIYSTER's LAw In l8ll.i\ 'i"' ..5' by the second glass plate M.r.5o. minimum al nO' and again maximum at 360o.8 the He found that ordinary light is completely polarized in the plane of incidence rvhen it gets leflected from a transparent medium at a paflicular angle known as the anffiTpolarization. This light is again reflected at 57. ' He was able to prove that the tangent of the angle of polarization is numerically equal to rhe refracrivel.on the first glass surface at B and is reflected along 8C (Fig. 10'7)' The glass plates are painted black on their back surfaces so as to avoid any reflection and this also helps in absorbing refracted light.. is known as the polarizer and Mrtas ttre analyser. 10"9). It is reflected along BC and refracted From Snell's law Fig. keeps the angle of incidence constant and it does not change with the rotation of Mr.. A beam of unpolarized light AB is incident at an angle of about 57. the rotation of the plate M. placed parallel to the first. Thus IA \. about BC. the light reflected by glass is plane polarized and can be detected by a tourmaline crystal.8). Light due to the components parallel to the glass surface is reflected whereas light due to the components perpendicular to the glass surface is transmitted. alaag BD (Fig. the displacements are confined to a certain direction at right angles to the ray and we get polarized light by reflection. Brewster perftrrmed a nunrber of experiments to study polarization of light by reflection at the surfaces of different media.(. Remember. c Thus.(. Fig r0. is rotated about BC. /.. = aorr.o. = stnr = ) cos[r-rJ (r . The polarized light has been analysed by using another mirror by Biot. and Mz(Fig.lE reflected and the refracted rays are perpendicular to eabh other. Comparing (i) and (rr) The above experiment proves that when light is incident at an angle '.) .56 Polarization A Texrbook of Optics 567 The production of polarized light by glass is explained as follows. is incident at an angle equal to the po.. unpolarized light larizing angle on the glass surflace.. The glass plxe M. The vibrations of the incident light can be resolverl into components parallel to the glass surface and perpendicular to the glass surface. 10.dex . the reflected light consists of waves in which cos.Jvl. 10.5o on a glass surface. is rotated further it is found that the intensity of CD becomes ma:rimum at 180'.7 we find that light travelling along 8C is plane polarized. sini P = tan. Suppose. Thus. After about hundred transmissions. In the case of a gas laser frlled with mirrms outside the windows.#*. only a small fraction df the incident light is reflected. if we use a pile of plates and the beam of ordinary'light is incident_at the polarizing angle on the pile of plates.:i^:. whed this beam of light is rcflected by the second platc. about 8vo of light is lost by reflection on its two surfaces and about 92vo intensity is transmitted. Hence it . =.-t. ul ordinary glass. 10.91. Brewster's windows are used rangement it will in gas lO. . times..ll'. sz. IO. It means the transmitt€d beam has To overcorne this difficulty. f. for monoctromatic lilnt. the value of i is given by However. not widery different to. i. the porarizing. It is clear that the right vibrating in the plane of inciden-ce is not reflected along BC tFig. the p-orarizing angre wil be different ror right . Tfrerefore. After about 100 reflections at the Brewster window. such a type of window is used in lasers and it is called a Brewster window. But it is richer in vibrations in thi plane of incidence. z2 As i + lEn or i+r = 2 r = Z CBD isalso equal to !.52.568 Polarization A Textbook of Optics S@ . The refracted ray will have both the v-ibrations (r) in the plane of incidence and (i0 at right angres to the plane of incidence. polarization wilt be comprete'onriio. again sone of the vibrations perpendicular to the POLARIZATION BY REFRACTION It is found that at a single glass surface or any similar transparent ruedium. rigtt of a particular waverength at a tinre i.lnt gtasser. vibrating parallel to the plane of inciderce is transmitted whereas for the perpendicular vibrations only 857o is ransmitted ar. the reflected and When an ordinary beam of light is incidcnt normally on a glass window. whereas vibrations at right angles to the prane of incidence can contribute for the resurtant intensity.r. The net effect of this type of aris that half thc arnornt of light intensity has been discarded and the other half is completely retained.E lasers. rhe refracted rays iue at right angles to each other. the final beam will be plane polarized.e. For a refractive index of L7 the porarising anite iJauout 59'5'i.r oifferent wavelengths. light travels ttuough the window about a hundred practically no intensity. For glass (lr = 1. [n this way the intensity of the final beam is abour 3 x l0-abecause (0.:. ttre window is tilted so that the light beam is incident at Brewster's angle. sonre of thc vibrations perpendicular to the plane of incidence are reflected by the first plate and the resr are ransmitted through it. we get plane oglarizea lighr along BC. the transmitted beam will have 50vo of the intensity of the incident beam and be complerely plane polarized. is partially plane-polarized. 10.e. From Brewster's raw. As the refractive index of a substance varies with the wavilengtr of the incident right. tn the ieflectea Lam the vibrations arong 8c cannot be observed.*org angre for Fig.. lA}Vo of the light .92)tm = 3 x l0-a.5) at the polarizing angle.10 The light component vibrating at right angles to the plaire of incidence is reflected.angre is. Jir".7 BREWSTER WINDOW one of the important apprications of Brewster's law and Brewster's angle is in the design of a grass window that enabres r00% transmiss-ion of right.d 15% is reflected. Therefore. Therefme. it is clear that for crown grass of refractive rndex I. e.2 E.fl=0. *. fr* la-. and 4 * cos'0 it Note. .. \ ") Example trS.is transmiued through the analyser. But where E . When When $ = A i. a 'Resolve #. In the Biot's polariscope it was found that the intbnsity of the twice reflected beam is maximum when the two planes are parallel and zeno'when the two planes are at right irngles to each'other.Intensity of the exraordinary ray The sarp is also true for the twice ransrnitted bcam from'the polarizer ' /' = A2coSO . Thus. The process continues and when the beam has traversed about 15 or 20 plates.. = (a c*s $)2 = a2 cos2 0.tt| E is the intr. the tnnsmitted light is completely free from the vibrations at dght angles to and analyser.uizer and the analyser e the (Fig. l0. 10.5o to the axis of the tube.nsitv r:f lncident polarized light' ff.570 Polarkorion A Textbook of Ap:iLs 57t plane of incidence are reflected by it and the rest are transmitted. A beam of rnonochromatic light is allowed to fall on the pile of plates at the polarizing angle. the two planes are parallel E.F into tw(] c0mp()nents.=ti[ *'. Frg.ll Only the a cos 0 conrponcnt. Ecaa^ S.'.g MAT. The proof rrf the ialv is trased on the fact that any polarized vibration may be resolved into two recratgular components : (i) parallel to the plane of transmission sf th* anall. compare the buansities of enraordinary and ordinary light. Lrct OP = a tre rhe anplitude of the vibrations transrrutted or ref"lected by the polarizer and 0 is the angle bctrveen o planes of the pol.'r of fula$us states that the@ atyser vaIl€s as the souarre of the cotight transmitted through t the plane of incidence and is. it means vibrations are2erpendicular to the plane of incidencg. in Fig. the vibrations being in the plane of incidence as shown 'E-or and thr plane of the polariztr. k ah-gle ih " the case of the Biot's polariscope this is beru'een tl1s ir?i] rcflccting planes. polarized by reflection at one plane surface is allowed to fall on the second plane surface at the polarizing angle the intensity of the twice reflected beam varies with the angle between the planes of the two surfoces. because cos 0 = I g= (r0 ff light is polarized in the plane of incidence. (i) If light is polarized perpendicular to the plane of incidence.12). . (i) a cos S aiorrg OA ard o Fie. means vibrations are in the plane of incidence.l {f the plane of vibratbn of the ilrcident begn malui an angle of 3U witk the optic *ris.12 (ir) a sin 0 alorrg 0d. = If.ll. having vibrations only in the plane of incidence. \l0. we get plane-polarizd light by refraction with the help of a pile of plates.US LAW When a beam of light.rer (ii) at right angles to it. The pile of plates consists of E. 10. Ttre transmitted light is polarized perpendicuiar to the plane of incidence and can be examined by a similar pile of plates which works as an analyser. IntensiB' of the lrarrsmitted light through the analyser number of glass plates (microscope cover slip) and are supported in a tube of suitable size and are inclined at an angle of 32. l0. / X the twoplanes areat rightangles toeach other tl. t = ' #1.14 (r). 10. 10. e9ggsgives@tic eg&r g{E_ge diErtirn-Tf1t e-6-pric axis. then tal are equal.-eq6Ftnqt ratsiejl gives lyo refracred rays.gt!9-e!-gp!E-4rg\Thereforc.*ith angles equal and 78'5) Eg. of the faces are obtuse [Fis. The ordinary ray has a rcfractive index p^ = u oritrJ-ttril iT i stn rt and the extraorthe ordinary dinary ray has a refractive index t. Thercfone the velocity of tight for the ordinary ray lnsHe tUe crys.Inf@is-iito axis.." t"tp of the following experiment : Mark an ink dot on a piece of paper. it is refracted along two paths inside the crystal. calcite or Iceland spar is crystallised calcium carbonate (ca ce) and was found in targ" f. It crystaflises in many forms and. Fig. because r. This 1t. [Fig.6rndrs d'and H ar. Only in a special case. -bounded by six pgralISIOC*ry.-. the line joining the two & > p..jnf $..13 .. 10. It is found that one image remains stationary and the second image rotates with the rotation of the cryqtal. its refractive index varies with the angle of incidence and it is not fixed.. ln the case of calcite blunt corners A and H coincides with the crystallographic axis of the crystal and it gives the dirction of the optic axis [Fig. tuJf DoUBLE REFRACTTON ///// Erasmus Barthorinus discovered. that when a ray of right by a. the extramdinfry ray travels faster as comparcd to the ordinary ray. in Iceland as very large transparent crystals. is less than r. Due to this reason calcite is also as Iceland spar. -optic axrs rs not a line bft it is a_clinretidn.r5af.14 (ii)1.]<nown u"-r"au""a by cleavage or breakage into a -rho:nbohedrgg.1"1. rf the angles 9.rilr6-= . ' e Fr the two blunt corners. slnr2 It is found that ray obeys the laws of refraction and is refractive index is constant. In fact anlline paranelm-IfiiS + eggat angles-wirh eaih known as the biint Eorners oIiHE cist4.u-iiti". In the case of the exraordinary ray. compared to the velocity of light for the extraordinary ray. place a calc-ite'crystal over this dot on the paper. line drdwn throueh A --'-"le . o[ the rhombohedron all 10.la (rr)1. the phenomenon of double refraction is absent when light is al-lowed to enter the crystal along the optic phenomenon of double refraction can be shown *itn tr..A Polariution Textbook of Optics 573 it will Intensity of the ordinary ray /o = 42 sin2 o not split into two raysf-rnus.1@1" slowly as .': : i -. in r6f9.13-a.6ptic Axis.13 (D)]. 10. The stationary im4ge is known as the ordinary image while the second one is known as the extraordinary image. In calcite. .o*r phenomenon is cailed double refraction.. Now rotate the crysal "*9*" I_ = a:ssfg Here 0= 3(P d . when the three edges of the crys_ @oining tJ. 10.14 t. A.makinp When a ray of Iight AB is incidenr on rhe catcite crystal rnaking an angle of incidence = i. Moreover" the velaity of the exhaordinary ray is differbnt in tal will be less different directions because its refractive index varies with the angle of incidence. These two rays emerge out along DO and CE which arc parallel tFig. l0.shown in Fig. Tivo images will be observed. At two opposite corncrs A and I/. Place your eye vertically above the crystal.= cos2 0 'q=. If a ray of light is incident along thc optic axis r in a direction parallel O the optic axis. (t) along 8C making an angle of refraction = 12 and (ii) along BD making an angle of refraction = r. . In general. the ordinary and the extraofdinruy rays travel along the same direction with the same velocity. In this case.. the two rays are plane polarised. only the extraordinary ray is transmised through the prisrn. = 1'55 Fis. pRrsM It is an optical device used for producing and analysing plane po_ luized light. in 182g. their vibrarions being at right angles to eash other. the section ACGE of the crystal is shown.TX PITINCIPAL SECTION OF THE CRYSTAL A plane which enntains the optic irxis and is perpendicular to the the ordinary anrl the extraordinary rays for .calciF: Refractive index for the ordinary " (! J . ground in such a way that the angle AEG-[Ccomes = 68' instead of 7l'.16. In a particular case.cut alons the olane AKGL.658 ttt. ray is defined as the principal plane of the ordinary ray. the ray of light is not split up into ordinary and extraordinary components. 10. The two cut surfqc_qs are grounded arid'lo[shed filLeallyjaqand then cepented together by Canada balsjrm wtlose-Lsflaatiyg index lies belweenlhelefractive indices for of the crystal is called the principal section of the crystal.t when a beam of light is transmitted through a calcite crystal. A principal section always cuts the surface of a calcite crystal in a parallqlogram with angles 109' and 7l' optrxrsite faces IO.o"rffi. a plane in the crystal drawn through the optic axis and the extraordinary ray is defined as the principal plane of the extraordinary ray. similarly. It was invented by William Nicol..the ordinary ray is eliminated and. from a portion @ <----. The vibrations of the ordinary ray are perpendicular to the principal section of the crystal while the vibrations of the extraordinary ray are in the plane of the principal secrion of the crystal. The crystal is -.*. The faces A'BCD ancl with are EFG'H on the calorie crystal.cte. (2) When a ray of light is incident perpendicular to the optic axis A calcite crystal whose length i?. it breaks up into two rays : (1) the ordinary ray which has its vibrations peryrendicular to the principal section of the crystal and (2) the extraordinary ray which has OROINABY Fig. The extaordinary ray iis not @ Canad-a fa. The nicol prism is made in such a way that it eliminates one of ttte two rays by total internal reflection. [rt A'BCDEFGH represent such a'crystal havingA' and G' as its blunt corners and A'cG? is one of the principal sections lA'CG' = 70o.. the two planes do not coincide.ra [rn balsam 10. (l) It should be remembered that a ray of light is not split up into ordinary and extraordinary components when it is incident on calcitg.three times its brPadlh is taken. i0. the ray_is_torally . 10.**-rr9m u?fir". as shown in then Fig.parallel to its optic axis.15. when the plane of incidence is a princ'ipal section then the principal section of the crystal and the pnyipal planes of the ordinary and the extaordinary rays coincide.n. It means that the ordinary and the extraordinary rays travel in the same dircction but with different velocities.rs Nrcot. therefore.nt?*[y reflected internally ._j tu = 1.. Spechl Cases. Refractive iral*1ilil-u. we have discussed that"*p".i and ilo -sreut"r is not transmitted. As a crystal has six faces. who ** an in cutting and polishing gerns and crystals. Thus.15. 10.486 V6. for every point there are three princlpal sections.. its vibrations parallel to the principal section. i0. I 574 Pobrizstion A Textbook of Oprics 575 It has been found that both the rays are plane polarized.fuii.15 Refractive index for the extraordinary 'y-r- Pn = E 1. . It is generally found that. lt is clear that Canada ray and it acts fr[en tlie ordinary ray balsam acts as a rarer ntsdium for an passes of to u . The diagonal AC represents the Canada balsam layer in the plane ALGK of Fig. wh"o tt" urF {j@Slgg_fr than the critiCfarngte.12 PRINCIPAL PLANE A plane in the crystal drawn through the optic axis and the ordinar.16 In Fig. and p. .' this position both the ordinary and the extraordinary rays are transmitted through the prism.658. such crystals ***. for which the velocity of light is the. the extraordinary ray will also be totally internally reflected at the canada balsam layer. a ray of unpolarized lilht on passing through the nicol prism in this position becornes plane-polarized. Therefore. it is totally internally reflected and only the exraordinary ray passes through the nicol prism.trfr6.t7 "1*T plane-polarized-light and the prism an ordinary ray and is totally internally reflected E ttre Canada balsam layer and so no light oul of pr. The working of the prism is clear fr. . when they are at right angles to each other. .lrdt HUycENs EXrLANATToN oF DoTJBLE EFRACTION IN UNIAXIAL CRYSf. its refractive index is the same as that of the ordinary ray and it is equal to I. depending upon the direction of propagation of the extraordinary ray p lies between 1.and P. Therefore. The combination of p. *. produces p. A point source of light in a double refracting medium is the origin of two wavefrronts.of light .ili**.I4 NICOL PRISM AS AN ANALYSER . are placed a_djacent to each other optic exb.?*U3 | not.AO.658.s5o I tl t. . If the extraordinary second prism ray P.om lhe foi ffi.486 when the extraordinary ray is travelling at right angles to the directi<ln of the optic axis. the relocity vries with the direction and the wavefronr is an eilipsoid of revotution.576 pobrimtion A Textbook of Optics 5n lf$td and is thgrefg-e lglsl1tted through the prism. Fon tlre ordinary ray. surtirihi within the exuaof. Huygens explained the phenomenon of dotrble refraction wirh the help of his principle of secondary wavelets. Therefore for a particular case p may be more than 1. In (3) The extraordinary ray also has a limit beyord which it is totalry internally reflected by the canada balsam surface.rTade ri.]t sourc:. Fig. is gradually rotated.1. it is not reflected and is transmitted through the prism. . will be more rhan the critical angle. the vdocities of the ordinary and the extraord[rary rays are 6e same along the as shown in Fig.55 and the Fig. is plane polarized When the po larizel extraordinary ray enters the prism in this positions it acts as { sino= r _ t. the intensity of decreases in the accordance with Malus law and when ruft..e. If the exfiaorclinary ray travels along the optic axis.. (2) If the angle of incidence is less than lhe critical angle for the ordinary ray.dina{rvarte surracc.Crli.. the prism p.658 \ POLABIZER Pr ANALYSEB Pz 0=6e. Thereforc. same in all directiom the wavefront is spherical.*ai"rry". 10. It means that light coming out of P. is called a polarismpe. Then. 10. Fig. with respect tocanada barsam =p =i# { ) the two prisms are crossed [i. I When two nicol prisms P. For crystals like quartz..17 (r). tive crystals. and P.iffi#lt is rhe ithro s. detects it. . The sides of the nicol prism are coated with black paint to absorb angle of incidence the ordinary rays that are reflected towards the sides by the canada balsam layer.4g6 and 1. il'-r' P""rbtry ' the wave rirface for fticxraordrnary r-aordlnary ray. then no light comes out of the second prism {.17 (i) shows the position of rwo parallel nicols and only the extraordinary ray passes through both the prisms. For the extraordinary ray. n.Hg* P.AIS F. are called the polarizer and the analyser respectively. which arc known"il as pcitive "JLif cryntals. 10.16 (r). ' Nicol prism can be used for the production and detection of plane-l polarizer light. one of them acts as a polarizer and the other acts as an analyser. The refractive index for the extraordinary ray .S in a catcite crystal [Fig. t0. If the angte of incidence for the ordinary ray is more than the critical angle. The ordinry ordinr'wavc *"re. is the ellipsoidal secondary wavefront for the extraordinary ray. AB is the incident plane wavefront of the rays falling obliquely on the surface MN of the negarive crystal.) and Here. BC. BC AP VVV oO.. Suppose. CP is the tangent meeting the spherical wavefront at p arrd CQ is the tangent meeting the ellipsoidal wavefront at e. Ot is the spherical secondary wavefront fm the ordinary ray and E. and the velocities of light for the ordinary ray along AP and the extraordinary ray along AQ are Vn and V respectively. in air is V. the semimajor axis of the ellipsoid is 4 *6 19 lr' and the semi-minor axis is E. 10. Here. It is maximum and is equal to the velocity of the ordinary ray along the optic axis.(. Note. 10..19 .V Dr AQ=-(=T . The direction of the line joining these two points (Wherc the sphere and the ellipsoid touch each other) is the optic axis. '"*@. According to Huygens' construction. whereas the direction AO of the ordinary ray is perpendicular to the I I . Fig. the velocity of light reaches from B . from Huygens' theory. p" are the refractive indices for the ordinary and the extraordinary rays along AP and AQ respectively..(r) Dr2 lto Thereford AP = BC.19. 10. 10.19.(iii) 10. Pt and *here P. The velocity of the extraordinary ray varies as the radius vector of the ellipsoid. oPTrct AXIS I (a) I I POSITIVE CRYSTAL I I oPTrc4 AXIS T (b) Fig.V --:) vo = . Hence.18 E (l) For the negative uniaxial crystals. the ordinary and the extraordinary rays travel with different velocities along different direction. Therefore.16 OPTIC AXIS IN THE PLANE OF INCIDENCB AND INCLINED TO THE CRYSTAL SURFACE (a) Obtique incftlence.. AQ .4Q.The direction AE of the exraordinary ray is not perpendicular to tlie . The velocity of the extraordinary ray is least in a direction at right angles to the optic axis.. It is least and equal to the velocity of the ordinary ray along the optic axis but it is maximum at right angles to the direction of the optic axis. (2) For the positive uniaxial crystals [L > tro. The crystal is cut so that the optic axis is in the plane of incidence and is in the direction shown in Fig.578 Polariution A Textbook of Opti<:s 579 the extraordinary wave surface lies within the ordinary wave surface [Fie.18 (b)]. by the tinre the incident wave to C. CP and CQ are the ordinary and the extraordinary refracted plane wavefronts fespectively in the crystal. po > FE.. 10. In Fig. rs the principal refractive index for th€ exEaordinary ray h<P"<Po tangent C@ tangent CP. In this case. the ordinary ray travels the distance Ap and the extraordinary ray travels the distance . the walefronts or surfaces in uniaxial crystals are a sphere and an ellipsoid and there are two points where these two wavefronts touch each other.. 40 f.0 I. o further phase difference of Fig. I is introduced between the componenrs when falls normally upon the quartz plate. When an elliptically potarizcd light is passed through a Fresnel's rhomb. A quarter wave plate is used only for light of a particular wavelength. These components after reflection at the point I undergo a phase angle difference or OUARTZ (i0 f.39 Its components (i) parallel to the plane of incidence and (li) perpendicular to the plane of incidence are equal. This phenomenon or the propcrty of rotating the plane of vibration by certain crystds or substances is known as opthal light DE is circularly polarized. When a quartz plate cut wit]t its POLARIZED Nz gle which is more than the critical of incidence at I is I along BC. or a path difference * *. The action of turning the plane of vibration occurs inside the body of the plate and not on its surface.a0 (i)].41. A ray of. one of the polaroids is fixed while the other can be rotated to control the amount of light coming inside. 10. |. Ler the incident light be plane polarized and let the vibrations make an angle of 45" with the plane of incidence. 10. The plane polarized light enters the quartz plate and its plane of vibration is gradually rotated as shown in Fig. is totally internally reflected back along cD. the emergent The quartz plate turns the plane of vibration. o. The apount of rotation through which the plane of vibration iS turned depends ripon the thickness of the quartz plate and the wavelength of light. a fur- ther path difference or I ir inuoduced between rhe component vibrations.40 light emerges out of N. light enters normally at one end of the rhomb and is totally internally reflected at the point Thus. such that light upath differeno of 8c. . Therefore the final emergent ray DE has two cornponents. based upon the fact that a phase difference of the emergent light is plane polarised and its vibrations make an angle of 45' with"the plane of incidence. a further path differen. air int-:rfrce when the angle of incidepce is 54". (.6M and as glass windows in trains and aeroplanes. Fig.38. Fresnel's rhomb wcirks similar to a quarter If the light entering the Fresnel's rhomb is circulady polarized. whereas a Fresnel's rhomb can be used for light of all wavelengths' ' \ry| - oPTICAL %CTMTY When a polarizer and an hnalyser are crosscd. and N.ecl. vibrating at right angles to the ray each other and they have a path differen"" or wave plate. 10. the lFie. The an54o. no light PLANE emerges faces out of the analyser [Fig. parallel to the optic axis is introduced between N. Therefore. pohriwtion # A Textbook of Optics |. 10." of I is intro<Juced between the componcnt i is vibrations (parallel and perpendicular to the plane of incidence)' The total path difference betwecn the component vibrations is introduced berween the component vibrations (prallel and perpendrcular to the plane I ana the emergent of incidence) when light is toally internally reflected back at glass- light is plane polariz. Fresnel's rhomb behaves just similar to a quarter wave plate. 605 aeroptanes. 10. ln The total path difference between the component vibrations is Ttrerefore IO3O FRESNEL'S RHOMB as shown ft Frcsnel constructe8 a rhomb of glass whose angles are 54o and 126" in Fig. of glass. 10. ungui* velocity rwtr of the crystal. the resultant vibrations'are along A'B'. The substances that rotate the plane of vibration to the left (anti-clockwise from the point of view of the observer) are known as laevo-rotatory or left-handed. In the case of quarz. 10. S'mc of the substances rotate the plane Fig.42 right-handed optically active crystal. The resultant vibration will be along A/J. The rotation of the prane of vibration in a sorution depends upon the concentration of the opticalry active substance in the sorution. 10. In the case of a Fig.dTTON A linearly poiarlzed light can be considered as a resultant of OR and OL is along OA'. with the same angular velocity. The resultant of these two vectors It has been found that some quartz crystars are dextro-rotatory whire others are laevdrotabry.ESNEL'S EXPI. Befora entering the crystal. through which thc plane of vi- b'ration is rotaterl. 10.*. have the same velocity. one is the mirror image of the other in their orientation. Considering a right-handed qriartz crystal (FiS. the circularly polarized vibrations. Therefore. circularly fxllarized vibrations rotating in opposite direiticiis.. Thus. This helps in finding the amount of cane sugar present in a sample of sugar solulion. turpentine. Analytical "l'reatment for Calcite. crystals like calcite do not rotate the plane of vibration. on entering the crystal is resolved into two circul. In a crystal like carcite. Right hancred r. the clockwise rotation travels faster while in a lefrhanded optically active crystal the anti- in a clockwise direction. the linearly polarized light. The resultant vector of OR and OL is OA.ANATION OF ROT. = tl COS0)' xz F'or clockwise circular vibrations.43) the clockrvise component travels a greater angle 6 than the anticlockwise component when they emerge out of the crystal.rized iight on entering u. Th" angle. 10.ung" in the prane of vibration of the plane porarised right. sodium chrorate and cinnabar are optically active. the plane of vibration is along AB and after emerging out crf the crystal it is along A'B'.606 pobrization { A Textbook of Optics 607 activity and the substance is known as an opticaily active substance. Fig.s traver with the same angular velociry. Fresner assumed that a prane pnr.i. 1032 FR. I.ive. = asino.tation means that when the observer is rooking towards light travelling towarcts him. it is not opticalry a. sugar sorution. the plane of vibration is rotated nf clockwise rotation travels faster. tlepends upon the thickness or frequency. According to Fresnel.4t Substances like sugar crystars. the plane of vibration has rotated through an angle 6 . It has been found that carcite does not produce'rny . rotating in oppo. site directions. Therefore. In Fig.42. t0.:'y:tul along the optic axis is resorverr inio two circularry polarized vio"rations rotating in opprsite directions with the . when lineraly polarised light enters a crystal of calcite along the optic'axis. llterefore. OL is the circularly polarised vector rotating in the anti-clockwise direction and oR is the circularly polarized vector rotating in the clockwise direction.rly polarized vibrations rotating in opposite directions. the two circularry porarized vibratio. Circularly polarised light is the resultant of two rectangular componenls having a phase difference of ].43 . vibration to the right and they are calletr dextro-rotatory or right handed. .SCS{a vtc-L ' 3 n !f S$* \^$-k ' h sahk .! .I 5 ? o\o'r'r .}ar.t.cilfic^\ o$ G fo\'a''\ r^^c'{o''r I"dI uo\$. lr f-lluotn Cout'iwi n1 to\^-{ t\M a. tfu to'*'t' ir . 'no}.u$ FL Li v"o*rty fu1^d ?sA i.. t'.!ta.ru-ot'\it'''o- + f oLont r'QL^" J-t 'i:l Ii {l SwL S.n- o. eu'dt oc{^Vr. li li .H -*P ttt'*2'L X [" rJ\'^:c']^ G'-.1Il* 'PLa'vtn.. M .t4 ".c-a- 'V' r."t .*'r-\.i o-* hX 1ue. 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