25-Wave Optics

TOPIC WAVE OPTICS 25 SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ksoLrqfu"Bizdkj 25.1 Two light waves are given by, E1 = 2 sin (100  t - k x + 30º) and E2 = 3 cos (200  t - k x + 60º) The ratio of intensity of first wave to that of second wave is : nks çdk'k rjax, E1 = 2 sin (100 t - k x + 30º) rFkk E2 = 3 cos (200  t - k x + 60º) ls çnf'kZr gS rks igys dh rhozrk dk nwljs ls vuqikr gksxk & 2 (A) 3 I1 4 (B*) 9 22 1 (C) 9 1 (D) 3 Sol. (B) I  A2  I2 = 32 = 4 / 9 25.2 The wavefront of a light beam is given by the equation x + 2y + 3z = c, (where c is arbitrary constant) then the angle made by the direction of light with the y-axis is : ;fn fdlh çdk'kiq¡tdkrjaxkxz lehdj.k x+2y+3z=c,(tgk¡c,dLoNUnfu;rkadgS)lsfn;ktkrkgSrksçdk'k fdj.ky-v{k lsfdruk dks.kcuk,xh? 1 (A) cos–1 14 2 Ans. (B) cos–1 14 (B*) cos–1 (C) sin–1 (D) sin–1 Sol. x + 2y + 3z = c represents a plane. ry dks O;Dr djrk gS Now angle  is given by : dks.kgS: nˆ . ˆj b  2  cos = | nˆ || ˆj | =   = cos–1    14  where tgk¡ nˆ  aˆi  bˆj  ckˆ = ˆi  2ˆj  3kˆ . 25.3 If the ratio of the intensity of two coherent sources is 4 then the visibility [( is max min )/( max min )] of the fringes ;fnnksdyklEc)L=k s rdhrhozrkdkvuqikr4gSrksfÝUtksadhn`';rk[( -  )/( +  )] gksxh vf/kdre fuEu vf/kdre fuEu (A) 4 (B*) 4/5 (C) 3/5 (D) 9  –  Sol. (B) max min = max min   2   2 1 1 2  1  – 1 – 2  1  (1 2)2 – (1 – 2)2 8 4 = ×     = 2 2 =   1  1  2 2  1    1 –  2 2  1  (1 2)  (1 – 2) 10 5 25.4 As shown in arrangement waves with identical wavelengths and amplitudes and that are initially in phase travel through different media, Ray 1 travels through air and Ray 2 through a transparent medium for equal length L, in four different situations. In each situation the two rays reach a common point on the screen. The number of wavelengths in length L is N2 for Ray 2 and N1 for Ray 1. In the following table, values of N1 and N2 are given for all four situations, The order of the situations according to the intensity of the light at the common point in descending order is : fn[k ;hx;hO;oLFk dsvuqlkjçkjEHkesavkn'kZrjaxnS/;Zovk;keokyhrjaxnhxbZgSaAtksizkjEHkesavyxvyxek/;e esafdlhdykijgSaAfdj.k1gokesapyrhgSrFkkfdj.k2lekuyEckbZL,dsfy,ikjn'khZek/;eesapkjvyx&vyx ifjfLFkfr;ksaesapyrhgSAçR;sdfLFkfresanksfdj.ksinsZij,dmHk;fu"BfcUnqijigq¡prhgSAfdj.k2dsfy,LyEckbZ esarjaxla[;kN2ofdj.k1dsfy,LyEckbZesarjaxla[;kN1gSAuhpsrkfydkesapkjksafLFkfr;ksaesaN1oN2dkekufn;k x;kgSApkjksafLFkfr;ksaesamHk;fu"BfcUnqij]rhozrkdkvuqikr?kVrsØeesgSaA fd j . k 2 fd j . k 1 Situations fLFkfr ; k 1 2 3 4 N1 2.25 1.8 3 3.25 N2 2.75 2.8 2.25 4 (A)  = 4 > 2 > 1 (B)  >  3 = 4 > 2 (C)  > 2 > 3 >  (D*)  > 3 = 4 >  Sol. in cases I, II, III, IV the path differences are respectively fLFkfrI,II,II,IVesaiFk UrjØe'k%   3 2 , , 4 and 4  phase differences are respectively dykUrjØe'k% , 2, /2, 3/2    and  =  cos2     2   the intensity in the four cases are pkjksfLFkfr;ksesarhozrkØe'k% 0,  , 0 , 0 2 2 respetively. 25.5 If the distance between the first maxima and fifth minima of a double slit pattern is 7mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength of the light used is : ;axdsf}fNnziz;ksxesafNnzksadse/;nwjh0.15fe h-gSrFk fNnzksalsinsZdhnwjh50lseh-gSrksblf}fNnziz;ksx esaizFkemfPp"BrFk iapefufEu"Bdse/;nwjh7fe h-gSrksmi;ksxfd;sx;sizdk'kdhrjaxnS/;Zgksxh& (A*) 600 nm (B) 525 nm (C) 467 nm (D) 420 nm (A*)600uSuksehVj (B)525uSuksehVj (C)467uSuksehVj (D)420uSuksehVj Soln : There are three and a half fringes from first maxima to fifth minima as shown. çFke mfPp"Bls ikapos mfPp"Brd 3.5fÝUtsgSA   = 7mm 3.5 = 2mm   = D d = 600 nm. 25.6 In a YDSE: D = 1 m, d = 1 mm and  = 5000 n m. The distance of 100th maxima from the central maxima is: YDSE esaD=1m,d=1mmrFkk=5000nm rks dsUnzh; mfPp"B ls100th mfPp"B dh nwjh gksxh : (A) 1 2 m (B) 3 2 m (C*) m (D) does not exist lEHko ugha gSA Sol. (B) For 100th max. 100th mfPp"B ds fy, d sin  = 100  sin  = 100  5000  109 1  103 5  104 = 103 = 0.5 = 1 2 1  y = D tan  = 1 × tan 30 = 3 25.7 Let S and S be the two slits in Young’s double slit experiment. If central maxima is observed at P and angle S PS = , then the fringe width for the light of wavelength  will be. (Assume  to be a small angle) S1 oS2;axf}fLyViz;ksxesnksfLyVgSA;fndsUnzh;mfPp"BPijçrhrgksrkgSrFk dks.kS1PS2 =gSrksrjaxnS/ ;Zdsfy,fÝUtpkSM+kbZgksxh&¼ekukfddks.k cgqrdegSa½% (A*) / (B)  (C) 2/ (D) /2 Sol. In  S1PO esa :  tan 2 = d/ 2 D As D > > d   is very small. cgqrvYigSA  tan       d  D  1  Fringe width fÝUt pkSM+kbZ = D   Ans. 2 2 2 2D d  d  25.8 Figure shows two coherent sources S1, S2 vibrating in same phase. AB is an irregular wire lying at a far  distance from the sources S1 and S2. Let d = 10–3.  BOA = 0.12°. How many bright spots will be seen on the wire, including points A and B. fp=k uqlkjS1oS2nksdyklEc)L=k srlekudykesadfEirgSAAB,dvleferrkjgSAtksfdS1rFkkS2lscgqrnwj  fLFkrgSA;fnd =10–3. BOA= 0.12° gSrksAoB dksfeykdjrkjijfdruspedhys/kCcsfn[kkbZnsxsA (A) 2 (B*) 3 (C) 4 (D) more than 4 ls vf/kd Sol. Say ‘n’ fringes are present in the region shown by ‘y’ ekuk'y'}kjkfufnZ"V{ks=kesa‘n’fÝatsgS  y = n = n.D d  y  tan (0.06º )  0.06    n D  n = 103    180 180 0.06 =  3 d > 1. Hence; only one maxima above and below point O. So total 3 bright spots will be present (including point ‘O’ i.e. the central maxima). vr%] fcUnqOdsÅijouhps,d&,dmfPp"BgksxkAblizdkjdqy3pedhysfcUnqmifLFkrgksaxs(tgk¡fcUnq‘O’ij dsUnzh;mfPp"Bgksxk) 25.9 The path difference between two interfering waves at a point on the screen is /6. The ratio of intensity at this point and that at the central bright fringe will be : (Assume that intensity due to each slit in same) O;frdj.k djus okyh nks rajxksa ds chp insZ ds fdlh fcUnq ij iFkkUrj  / 6 gSA bl fcUnq ij rFkk dsUnzh; pedhyh fÝUt ij rhozrkvksa dk vuqikr gS ¼;g ekfu;s fd izR;sd fLyV dh rhozrk leku gS½ & (A) 0.853 (B) 8.53 (C*) 0.75 (D) 7.5  Sol. At path difference 6   , phase difference is 3  iFkkUrj 6 ,dykUrj 3  =  + O + 2  cos 3 = 3 O max = 4 So the required ratio is blfy, vko';d vuqikr 3I0 4I0 = 0.75 25.10 In the figure shown in a YDSE, a parallel beam of light is incident on the slits from a medium of refractive index n . The wavelength of light in this medium is  . A transparent slab of thickness ' t' and refractive index 1 1 n3 is put infront of one slit. The medium between the screen and the plane of the slits is n2. Find the phase difference between the light waves reaching point ' O ' (symmetrical, relative to the slits) fp=kesafn[k ,x;s;axdsç;ksxesa,dlekUrjçdk'kiqatviorZukadn1okysek/;elsfLyVijvkifr gksrkgSA blek/;eesaizdk'kdhrjaxnS/;Z1gSA,dikjn'khZifêdkftldheksVkbZ't'vkSjviorZukadn3gSdks,dfLyVdslkeus j[knsrsgSAinsZvkSjfLyVdschpdkek/;en2gSrksfcUnq'O' ¼fLyVksadslkis{klefer½ijigq¡pusokyhçdk'krjaxksa esadykUrjD;kgksxkA 2 2  2 n 1  n 3  2 n 1 (A*) n1  1 2 (n3  n2)t (B) 1 (n3  n2)t (C) 2 1   n 2 1 t (D)   1 (n3  n2) t Ans. n1  1 (n3  n2) t Sol. light wavelength in medium n1 ⇒ Wavelength in vaccum =  is =  = n  0 1 1 The path difference between the lights waves reaching point O = (n3–n2) t = extra path which the light from S1 travelled compared to the path from S2. 2 Corresponding phase difference = 0 2 (path difference) = n11 (n3–n2) t n1ek/;eesaizdk'kdkrjaxnS/;Z= ⇒fuokZresarjaxnS/;Z=0 =n11 O fcUnqijigq¡pusokysizdk'kdse/;iFk Urj=(n3–n2)t=vfrfjDriFkS2dsiFkdhrqyukesaS1lsvkusokykizdk'k r;djrkgS 2 lEcfU/kr dykUrj = 0 2 (iFkkUrj) = n11 (n3–n2) t 25.11 A Young’s double slit experiment is conducted in water ( ) as shown in the figure, and a glass plate of thickness t and refractive index  is placed in the path of S . The magnitude of the phase difference at O is : (Assume that ‘’ is the wavelength of light in air) ;axf}fLyVç;ksxikuh(viorZukad1)esafd;ktkrkgSAteksVkbZdhdk¡pdhifêdk(viorZUkakd2 ]S2 dsiFkesaj[kh tkrhgSrksOijdykUrjdkifjek.kgksxk:(ekuksafd‘’ok;qesaizdk'kdhrjaxnS/;ZgS) (A)  2 Sol. Here path difference will be : iFkUrj x = ( –  ) t   = 2 ( –  ) t 2 1 Hence vr% (C)  2 1 25.12 In the figure shown if a parallel beam of white light is incident on the plane of the slits then the distance of the nearest white spot on the screen from O is: [assume d << D,  << d ] fp=kesalekUrj'osrçdk'krjaxfLyVijvkifr gksrhgSrks;gk¡fcUnqOlsinsZijfudVre'osrfcUnqdhnwjhgksxh: [ ;g ekurs gq, fd d << D,  << d ] (A) 0 (B) d/2 (C) d/3 (D*) d/6 Sol. we know that P will be the central maxima (at which path difference is zero) KkrO;gSfdPdsUnzh;mfPp"Bgksxk(tgk¡iFk Urj'kwU;gS) Now vc OP = d – d  d 2 3 6 25.13 In the figure shown, a parallel beam of light is incident on the plane of the slits of a Young’s double slit experiment. Light incident on the slit, S1 passes through a medium of variable refractive index  = 1 + ax(where ‘x’ is the distance from the plane of slits as shown), upto a distance ‘𝑙’ before falling on S1. Rest of the space is filled with air. If at ‘O’ a minima is formed, then the minimum value of the positive constant a (in terms of 𝑙 and wavelength ‘’ in air) is : fp=kesafn[k ;s;axdsf}fNnziz;ksxmidj.kdsfNnzksadsryijlekUrjizdk'kiqatvkifr gksrkgSAS1fNnzij vkifr izdk'kifjorhZviorZukad]=1+axokysek/;elsgksdj¼tgk¡fp=k uqlkjxfNnzksadsrylsnwjhgS½nwjh 𝑙 r; djds fNnz S1rd igq¡prk gSA vU; lHkh txg gok gSA vxj fcUnq ‘O’ij ,d fufEu"B curk gS rks /kukRed fu;rkada (nwjh𝑙rFkkrjaxnS/;Z‘’ds:iesa)dkU;wureekugksxk&  (A) 𝑙  (B*) 𝑙2 𝑙2 (C)  (D) None of these Sol. When light passes through a medium of refractive index , the optical path it travels is (t). Therefore, before reaching O light through S1 travels (l + b) distance while that through S2 travels a distance (l+ b) i.e. : path difference = (l + b) – (l + b) = ( – 1)l. For a small element 'dx' path difference x = [(1 + ax) – 1] dx = ax dx For the whole length ; tcizdk'kviorZukaddsek/;elsxqtjrkgS]rksizdk'kiFk(t)gksxkA vr%Oijgq¡pusdsigysS1dkizdk'k(l+b)nwjhpysxkoS2dkizdk'k(l+b)nwjhpysxkA i.e. : iFkkUrj = (l+ b) –(l+ b) = ( – 1)l. NksVs vo;o 'dx' ds fy, iFkkUrj x =[(1+ax)– 1] dx = ax dx dqyyEckbZdsfy,; x = 𝑙 ax dx  O a𝑙2 2 For a minima to be at 'O'. 'O'ijfufEu"Bdsfy, x = (2n + 1)  2 a𝑙2  i.e. : 2 = (2n + 1) 2 . For minimum 'a'; n = 0 'a' ds U;wure eku ds fy, n= 0  a𝑙2     2 2 a = 𝑙2 Ans. 25.14 Interference fringes were produced using white light in a double slit arrangement. When a mica sheet of uniform thickness of refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves through some a distance. This distance is equal to the width of 30 interference bands if light of wavelength 4800 Å is used. The thickness (in 1m) of mica is: 'osrizdk'kdkmi;ksxdjdsf}fNnzO;oLFk esaO;frdj.kfÝUtsacukbZtkrhgSAtc,dlekueksVkbZdhvHkzdifêdk viorZukad1.6 ¼gokdslkis{k½dks,dfLyVlsxqtjusokysçdk'kdslkeusj[kfn;ktkrkgSrksdsUnzh;fÝUtdqNnwjh ijfoLFk firgkstkrhgSA;gnwjh30fÝUtpkSM+kbZdscjkcjgksrhgSA;fnçdk'kdkrjaxnS/;Z 4800ÅgSrksvHkzddh eksVkbZ (1m esa) gksxh : Old] (A) 90 (B) 12 (C) 14 (D*) 24 t D Sol. (D)Shift of fringe pattern fÝUt çfr:i dk foLFkkiu =(–1) d  30 D (4800 1010 ) d D = (0.6) t d 30 × 4800 × 10–10 = 0.6 t 30  4800  1010 1.44  105 t = 0.6 = 0.6 = 24 × 10–6 25.15 Two coherent light sources each of wavelength  are separated by a distance 3. The total number of minima formed on line AB which runs from -  to +  is: nksdyklEc)çdk'kL=k srftudhrjaxnS/;ZgSrFk os3nwjhijgSaAjs[k ABtks-ls+rdgSijcu sokysfufEu"B dhdqyla[;kgSA: (A) 2 (B) 4 (C*) 6 (D) 8 Sol. Lets take any general point S on the line AB. js[k ABijdksbZfcUnqSfy;ktk;s Clearly: for any position of S on line AB; we have for  PQS: PQ + QS > PS {in any triangle sum of 2 sides is more then the third side}  PS – QS < 3. As PS – QS represents the path difference at any point on AB  it can never be more than 3. Now minimas occur at.  , 3 , 5 only. 2 2 2 so 3 minimas below R (mid point of AB) and 3 also above R. Li"Vr%js[kk ABijSdh fdlhHkhfLFkfrdsfy, PQSesa: PQ + QS > PS {fdlhHkhf=kHkqtesa2Hkqtkvksadk;ksxrhljhHkqtklsvf/kdgksrkgS}  PS – QS < 3. pwafd PS – QS,js[kk AB ij fdlh fcUnq ij iFkkUrj n'kkZrk gS  ;g dHkh Hkh 3 ls vf/kd ugha gksxkaA dsoy  , 3 , 5 dsfy,fufEu"BizkIrgksxsaA 2 2 2 vr%R(ABdke/;fcUnq)dsuhps3fufEu"BoÅijHkhrhufufEu"BizkIrgksxsaA 25.16 M1 and M2 are two plane mirrors which are kept parallel to each other as shown. There is a point 'O' on perpendicular screen just infront of 'S'. What should be the wavelength of light coming from monchromatic source 'S'. So that a maxima is formed at 'O' due to interference of reflected light from both the mirrors. [Consider only 1st reflection]. M1rFk M2leryniZ.kgStksfdfp=k uqlkj,dnwljsdslekUrjj[ksgSASdsBhdlkeusyEcor~insZij,dfcUnq OgSA,do.khZL=ksrSlsvkjgsizdk'kdhrajxnS/;ZD;kgksuhpkfg,rkfdnksuksaniZ.ksalsijkofrZrizdk'kdsO;frdj.k ds dkj.k Oij mfPp"B cus A (dsoy izFke ijkorZu ysa)[D>>d,d>>] 3d2 (A) D (B*) 3d2 2 D d2 (C) D 2d2 (D) D Sol. The 2sourcesare. nksL=kksrgSA As O is a maxima, Hence OP = . O ij mfPp"B gS vr% OP =   d  .D ; get   3d2 2 (3d) 2D gy fp=k uqlkjS1oS2 nkslzksrfy,tkrsgSrksS1oS2 dsdkj.kdsfUnz;mfPp"BPfcUnqij 'O'ls d/2nwjhijgksxkA 'O' dsmfPp"Bgksusdsfy,iFkkUrj: iFkkUrj= 3d.d 2D =n   = 3d2 2nD   = 3d2 2d , 3d2 4d 25.17 A ray of light is incident on a thin film. As shown in figure M,N are two reflected rays and P,Q are two transmitted rays. Rays N and Q undergo a phase change of  . Correct order of the refracting indices is: ,dçdk'kfdj.kiryhfQYeijvkifr gksrhgSAfp=k uqlkjMrFk NnksijkofrZrfdj.krFk PoQnksikjxferfdj.k gS rFkk N o Q esa  dk dyk ifjorZu gS rks viorZukad dk lgh Øe gksxk & M N  P Q (A) n2 > n3 > n1 (B*) n3 > n2 > n1 (C) n3 > n1 > n2 (D) None of these, the specified changes can not occur buesalsdksbZugha]vko';difjorZuughagksldrsgSA Sol. Ray N undergoes reflection at surface II with phase change of  fdj.kNlrgI ls  dykUrj dslkFkifjofrZrgksrh gSA  n > n2 Ray Q undergoes a phase-change of  at II, but there is no phase change when it is reflected from surface . fdj.kQHkhlrg IIijls dykifjofrZr djrhgSA ijUrqtc ;glrg ls ijkofrZrgksrh gSrks dykes ifjorZu ughagksrkgSA  n < n2 25.18 From a medium of index of refraction n , monochromatic light of wavelength  is incident normally on a thin film of uniform thickness L (where L > 0.1 ) and index of refraction n . The light transmitted by the film travels into a medium with refractive index n3. The value of minimum film thickness when maximum light is transmitted if (n1 < n2 < n3) is : n1viorZukadokysek/;els,d,do.khZçdk'krjaxftudkrjaxnS/;ZgS,dL(tgk¡L>0.1) eksVkbZdhiryhfQYe viorZukadn2ijyEcor~vkifr gksrhgSAikjxferçdk'kn3lsviorZukadokysek/;elsfudyrkgSrksfQYedhU;wure eksVkbZ D;k gksxh] ;fn vf/kdre çdk'k ikjxfer gksrk gSA ;fn (n1 < n2 < n3) : n1 (A) 2n2 (B*) n1 4n2  (C) 4n2  (D) 2n2 Sol. Eq. of path diff. for maxima in tansmission (or weak reflection); ikjxeu(;kvkaf'kdijkorZu)esmfPp"Bdsfy,iFkkUrj opt = 2n2L = vacuum , 2 3vacuum ...... 2  n2   3  n1 2   L  , n 2 , ...... 2  L = 4n .  1  2 (notice that  = wavelength in medium is related to  vacuum = n  ) (/;ku nsa  = ek/;e esa rjaxnS/;Z tks  ls lEcfU/kr gS  fuokZr = n  ) 25.19 A broad source of light (l = 680 nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.034 mm in diameter at the other end. The total number of bright fringes that appear over the 120 mm distance is : ,dfoLr`rizdk'kL=k sr(l=680nm) lsvfHkyEcroizdk'knks120mmyEchdkapdhIysVksaijfxjrkgS;sIysVs ,d fljsij Li'kZdjrhgSarFkk nwljsij 0.034mmO;kldsrkj }kjk vyxdj nhxbZgSarks120mm nwjhrd dqy pedhyh fÝUtks dh la[;k gksxh 0.034 mm (A) 50 (B*) 100 (C) 200 (D) 400 Sol. Constructive interference happens when 2t = (m – 1/2). The minimum value of m is m = 1; the maximum 2t  1 2  0.034 103  1 value is the integer portion of  2 = mmax = 100 680 109 2 = 100.5 tc2t=(m–1/2)rkslEiks"khO;frdj.kgksrkgSamdkU;wureeku1 2t  1 = gSaArFk iw.k ZadHk xdkvf/kdreeku  2 2  0.034 103  1 680 109 2 = 100.5 SECTION - II : MULTIPLE CORRECT ANSWER TYPE [k.M-II: cgqy lgh fodYi izdkj 25.20 If the first minima in a Young’s slit experiment occurs directly in front of one of the slits, (distance between slit & screen D = 12 cm and distance between slits d = 5 cm) then the wavelength of the radiation used can be : ;fn;axdsç;ksxesaçFkefufEu"B,dfLyVdsBhdlkeusgSA,(fLyVoLØhudschpdhnwjhD =12cm rFk fLyVksa ds chp dh nwjh d = 5 cm) rks mi;ksx fd;s x;s fofdj.k dk rjaxnS/;Z gksxk & 2 4 (A*) 2 cm (B) 4 cm (C*) 3 cm (D) 3 cm Sol. (A), (C). Path difference iFkkUrj = – D = 1 cm Also ;   D  = (2n – 1)   = 2(1)   2 2n  1 For n = 1,2,3,............. ds fy,  = 2cm, 2 2 cm, cm,.............. 3 5 25.21 If one of the slits of a standard Young's double slit experiment is covered by a thin parallel sided glass slab so that it transmits only one half the light intensity of the other, then: ;fn,dekud;axf}&fLyViz;ksxesa fdlh,dfLyVdksdk¡pdh,diryhlekUrjIysV}kjk<+dfn;ktk;s rkfd ;gnwljhfLyVdhrqyukesa vk/khrhozrkdkizdk'kikjxferdjsrks (A*) The fringe pattern will get shifted towards the covered slit (B) The fringe pattern will get shifted away from the covered slit (C*) The bright fringes will become less bright and the dark ones will become more bright (D*) The fringe width will remain unchanged (A*) fÝzUt izfr:i <+dh gq;h fLyV dh vksj foLFkkfir gksxk A (B)fÝUtizfr:i<+dhgq;hfLyVlsnwjfoLFk firgksxkA (C*) nhIr fÝUtsa de nhIr o vnhIr fÝUtsa vf/kd nhIr gksxhA (D*)fÝUtpkSM+kbZvifjofrZrjgsxhA Sol. max = (    I2 )2   2 2 < 4I min =   > 0  D  = d becuase  D.d. are unchanged so  also remain unchanged. pw afdD.d. lHkhvifjofrZrjgrsgS]vr% HkhvifjofrZrjgsxkA 25.22 A parallel beam of light (  Å is incident at an angle  = 30° with the normal to the slit plane in a young’s double slit experiment. Assume that the intensity due to each slit at any point on the screen is 0. Point O is equidistant from S1 & S2.The distance between slits is 1mm. ;fnlekUrjçdk'kn`<+(Å=30°dks.kvfHkyEcls;axç;ksxesafLyVijvkifr gksrkgSA;fnçdk'kdhrhozrk çR;sdfLyVij0gksrFk fcUnqO,S1vkSjS2lslekunwjhijgSAfLyVdschpdhnwjh1mmgksrks (A*) the intensity at O is 40 Oijrhozrk40gS (B) the intensity at O is zero Oijrhozrk'kwU;gS (C*) the intensity at a point on the screen 4m from O is 40 insZijOls4mnwjhij,dfcUnqijrhozrk40 gSA (D) the intensity at a point on the screen 4m from O is zero insZijOls4mnwjhij,dfcUnqijrhozrk'kwU;gSA Sol. As d << D,  path difference = d sin (at 0) = 1mm × sin 30º = 0.5 mm if it is a maxima  10–3 × 0.5 = (5000 × 10–10 )m × (n) n must be integer. get n = 1000. Hence O is a maxima of intensity 4I0 pwafd d<k;kA 25.25 A point source of light is placed at origin, in air. The equation of wave front of the wave at time t, emitted by source at t = 0, is (take refractive index of air as 1) ,dçdk'kdkfcUnqlzksrewyfcUnqijgokesafLFkrgSAfdlh{k.kt, ijrjaxdsrjaxkxzdklehdj.kD;kgksxktks t=0,ijlzksr}kjkmRlftZrgSA(ok;qdkviorZukad1ys) (A) x + y + z = ct (B) x2 + y2 + z2 = t2 (C) xy + yz + zx = c2 t2 (D*) x2+ y2 + z2 = c2 t2 Sol. Wave fronts are spherical in shape of radius ct. rjaxkxzctf=kT;kdsxksyh;lajpukdsgSA Hence (D). 25.26 Spherical wave fronts shown in figure, strike a plane mirror. Reflected wave fronts will be as shown in xksyh;rjaxkxzfp=k uqlkjleryniZ.klsVdjkrsgSrksijkofrZrrjaxkxzçnf'kZrgksxk& (A) (B) (C*) (D) Sol. The wave fronts are always perpendicular to the light rays. rjaxkxzges'k çdk'krjaxksadsyEcor~gksrsgSA Hence, (C). 25.27 Wavefronts incident on an interface between the media are shown in the figure. The refracted wavefronts will be as shown in nksek/;eksadhlrgijvkifr rjaxkxzxksyh;niZ.kijfp=k uqlkjvkifr gSrksijkofrZrrjaxkxzgksxk& (A) (B*) (C) (D) Sol. Using snell's law ; LuSydsfu;els sin(450 ) 1  sinr =  r = 300 sinr 1 2 Hence, (B) is correct. Note : The shown lines are wavefronts and not rays. vr%,(B)lghagSA uksV:fn[kbZxbZjs[k,sarjaxkxzgSfdj.kughaA 25.28 Plane wavefronts are incident on a spherical mirror as shown. The reflected wavefronts will be in the figure ,dleryrjaxkxzxksyh;niZ.kijfp=k uqlkjvkifr gSrksijkofrZrrjaxkxzgksxk& (A*) (B) (C) (D) Sol. After reflection by mirror the parallel rays concentrate at the focus. niZ.klsijkorZudsi'pkr~lekUrjfdj.kQksdlijdsfUnzrgksrhgSA Hence the plane wave front becomes spherical concentrated at the focus. vFk Zr~leryrjaxkxzQksdlijxksyh;rjaxkxzesaifjofrZrgkstk,xkA Hence, (A). 25.29 Certain plane wavefronts are shown in figure. The refractive index of medium is dqNfuf'prleryrjaxkxzfp=kesaçnf'kZrgSrksek/;edkviorZukadgksxk& (A*) 2 (B) 4 (C) 1.5 (D) Cannot be determined Kkr ughfd;k tkldrk gS 2 Sol. In  ABC ; sin (i) = d 1 In  xyz ; sin (r) = d sin i  sin r = 2 =  . Comprehension # 2 vuqPNsn#2 CENTRAL BRIGHT FRINGE 1 2 3 4 5 The figure shows the interference patttern obtained in a double-slit experiment using light of wavelength 600 nm. 1, 2, 3, 4 and 5 are marked on five fringes . 600nmrjaxnS/;Zdsçdk'kdslkFkfd;sx;sf}&fNnzç;ksxlsçkIrO;frdj.kçØefp=kesan'k Z;kx;kgSA 1, 2,3,4rFk 5vadik¡pfÝUtksaijn'k Z;sx;sgSA 25.30 The third order bright fringe is rhljs Øe dh pedhyh fÝat gS & (A) 2 (B) 3 (C) 4 (D*) 5 Sol. Order of the fringe can be counted on either side of the central maximum. For example fringe no. 3 is first order bright fringe. fÝaUtksa dkØe] dsUnzh; mfPp"B dsfdlh Hkh rjQ ekik tkldrk gSAmnkgj.k ds fy;s fÝUt uEcj3, igysØe dh pedhyh fÝUt gSA 25.31 Which fringe results from a phase difference of 4 between the light waves incidenting from two slits? nksuksafNnzksalsvkifr çdk'krjaxksadschp4dsdykUrjlsmRiUufÝUtdkSulhgS? (A) 2 (B) 3 (C*) 4 (D) 5 Sol. (C) Since, 2nd fringe represent central bright fringe hence, 4th fringe results from a phase difference of 4 between the light waves incidenting from two slits . Sol. (C)D;ksfd2nd fÝaUtdsfUnz;mfPp"BdkscrkrhgSAvr%4th fÝaUtnksfLyVlsvkusokyhizdk'kdhrjaxksadschp 4dykUrjds ifj.kkeLo:icusxhA 25.32 Let  X and  X represent path differences between waves interfering at 1 and 3 respectively then (| X | – | X |) is equal to 1rFkk3ijO;frdj.k djrhgqbZ rjaxksads chpiFkkUrj Øe'k%XA rFkkXC gksrks (|XC| –|XA|) cjkcj gS& (A) 0 (B*) 300mn (C) 600 nm (D) 900nm.  Sol.  X =  ;  X = 2  X –  X  = 2 = 300 nm . SECTION - V : MATRIX - MATCH TYPE [k.M-V:eSfVªDl&lqesyizdkj 25.33 A double slit interference pattern is produced on a screen, as shown in the figure, using monochromatic light of wavelength 500 nm. Point P is the location of the central bright fringe, that is produced when light waves arrive in phase without any path difference. A choice of three strips A, B and C of transparent materials with different thicknesses and refractive indices is available, as shown in the table. These are placed over one or both of the slits, singularly or in conjunction, causing the interference pattern to be shifted across the screen from the original pattern. In the column-I, how the strips have been placed, is mentioned whereas in the column-II, order of the fringe at point P on the screen that will be produced due to the placement of the strip(s), is shown. Correctly match both the column. ,df}&fNnz(doubleslit)O;frdj.kizk:idksinsZijfp=k uqlkjmRiUufd;ktkrkgSAblesa,do.khZ(monochromatic) izdk'kdksiz;ksxesafy;ktkrkgSftldhrjaxnS/;Z500nm gSAfcUnqPdsUnzh;nhIrfÝUtdhfLFkfrdkscrkrkgSA ;gnhIrfÝUtrccurhgStcizdk'krjaxsdykesafcukiFk Urjdsogk¡igq¡prhgSA;gk¡ijrhurjgdhikjn'kZd ifêdk, A, B rFkk CmiyC/k gS ;s ifêdk, vyx&vyx eksVkbZ dh gS rFkk vyx&vyx inkFkksZ ls feydj cuh gSA mudheksVkb;koviorZukadlkj.khesansj[k gSAbudks,d;knksuksafNnzksijblrjgj[k tkldrkgSfdfdlh fNnzijvdsys;k,dlkFkbuifêdkvksadksj[k tkldrkgSAftldsdkj.kO;frdj.kizfr:iviuhizkjfEHkdizk:i dhfLFkfrlsfoLFk firgkstk,ALrEHk-Iesaifêdk,fdlrjgj[khtkrhgS;gfn;kgqvkgSrFk LrEHk-IIesainsZij fcUnqPijfÝUtdkØefn;kx;kgStksfdifêdkvksadksMkyusdsdkj.kmRiUugqvkgSALrEHk sadkslqesfyrdfj,A Slit I Slit II P Screen Column - I Column - II (A) Only strip B is placed over slit-I (p) First Bright (B) Strip A is placed over slit-I and strip C is (q) Fourth Dark placed over slit-II (C) Strip A is placed over the slit-I and strip B and (r) Fifth Dark strip C are placed over the slit-II in conjunction. (D) Strip A and strip C are placed over slit-I (in (s) Central Bright conjuction) and strip B is placed over slit-II. LrEHk-I LrEHk-II (A) dsoy Biêh dks fNnz-I ijj[kk tkrk gSA (p)izFkenhIr (B) iêhAdksfNnz-Iij j[kktkrk gS o iêh (q)prqFkZvnhIr CdksfNnz-II ijj[kk tkrkgSA (C) iêhAdksfNnz-Iij j[kktkrk gSrFkk iêhB (r)iape¼ik¡poh½vnhIr o Cdks la;qDr :i ls fNnz -II ij j[kk tkrk gSA (D) iêhAoCdks la;qDr :i ls fNnz-I ij j[kk tkrk (s)dsUnzh;nhIr gSrFkk fNnz-II ijiêh B dksj[kk tkrkgSA Ans. (A) r, (B) r, (C) s, (D) p Sol. By using ( – 1)t = n , we can find value of n, that is order of the fringe produced at P, if that particular strip has been placed over any of the slit. If two strips are used in conjuction (over each other), path difference due to each is added to get net path difference created. If two strips are used over different slits, their path differences are subtracted to get net path difference. 4B Sol. (µ–1)t=ndk iz;ksxdjus ij]ge ndk eku Kkr djldrs gSa] ;g fcUnqPijfÝat dkØe 3B gksxk;fnogfof'k"Viêhogk¡ijfdlhfNnzijj[knhtk,A;fnnksifê;kala;qDr:ilsfNnz 2B ijj[knhtkrhgSrksdqyiFk UrjtqM+tkrkgSA;fnnksifê;kavyxvyxfNnzksaijj[knhtkrh 1B gSrksmudkiFk UrjdkvUrjfd;ktkrkgSrkfdifj.k ehiFk UrjizkIrgks CB vc n1 = (1  1) t1  = 5 1B n2 = 4.5 2B n3 = 0.5 3B (a) dsfy,fÝUtdkØe4.5vFkkZr~5ohadkyhfÝat (b) dsfy,ifj.kkehØe5–0.5=4.5vFkkZr~5ohadkyhfÝat (c) ds fy, ifj.kkeh Øe 5 – (0.5 + 4.5) = 0 vFkkZr~ ;g fcUnq P ij fQj ls dsUnzh; nhIr fÝUt gSA (d)dsfy, ifj.kkehØe (5+0.5)–4.5=1vFkkZr~izFke nhIrfÝUtgSA 25.34 A monochromatic parallel beam of light of wavelength  is incident normally on the plane containing slits S1 and S2. The slits are of unequal width such that intensity only due to one slit on screen is four times that only due to the other slit. The screen is placed perpendicular to x-axis as shown. The distance between slits is d and that between screen and slit is D. Match the statements in column- with results in column-. ,do.khZ izdk'k dh lekUrj fdj.k ftldh rjaxnS/;Z  gS bldks fNnz S1 o S2 ds ry ds vfHkyEcor~ vkifrr djk;k tkrk gSA fNnzksa dh pkSM+kbZ cjkcj ugha gS ij bl rjg gS fd dsoy ,d fNnz ds dkj.k insZ ij rhozrk dsoy nwljs fNnz ds dkj.k insZ ij rhozrk dh pkj xquh gSA insZ dks x-v{k ds yEcor~ fp=kkuqlkj j[kk tkrk gSA fNnzkas ds e/; nwjh d gS rFkk insZ o fNnz ds e/; nwjh D gSA LrEHk- esa fn;s x;s oDrO;ksa dks LrEHk- esa fy[ks x;s ifj.kkeks ls lqesfyr dfj;sA Column-I Column-II D (A) The distance between two points on screen (p) 3d having equal intensities, such that intensity 1 at those points is 9 th of maximum intensity. D (B) The distance between two points on screen (q) d having equal intensities, such that intensity 3 at those points is 9 th of maximum intensity. 2D (C) The distance between two points on screen (r) d having equal intensities, such that intensity 5 at those points is 9 th of maximum intensity. 3D (D) The distance between two points on screen (s) d having equal intensities, such that intensity 7 at those points is 9 th of maximum intensity. LrEHk-I LrEHk-II D (t) 2D 3d (A) insZijfLFkrnksfcUnqvksaftudhrhozrk,acjkcj (p) 3d bl rjg gS fd bu fcUnqvksa ij vf/kdre rhozrk dh 1 9 xqukrhozrkgSrksbunksuksafcUnqvksadse/;nwjh (B) insZ ij fLFkr nks fcUnqvksa ftudh rhozrk,a cjkcj ( q ) D d bl rjg gS fd bu fcUnqvksa ij vf/kdre rhozrk dh 3 9 xqukrhozrkgSrksbunksuksafcUnqvksadse/;nwjh 2D (C) insZijfLFkrnksfcUnqvksaftudhrhozrk,acjkcj (r) d blrjggSfdbufcUnqvksaijvf/kdrerhozrkdh 5 9 xqukrhozrkgSrksbunksuksafcUnqvksadse/;nwjh 3D (D) insZijfLFkrnksfcUnqvksaftudhrhozrk,acjkcj (s) d blrjggSfdbufcUnqvksaijvf/kdrerhozrkdh 7 9 xqukrhozrkgSrksbunksuksafcUnqvksadse/;nwjh Ans. (A) q,r,s (B) p,q,r,s, t (C) q,r,s (D) p,q,r,s Sol. Initially at a distance x from central maxima on screen is (t) 2D 3d  =  + 4 + 2 2x D cos  , where  = d  = 9 and  =  1 (A) At points where intensity is 9 th of maximum intensity, minima is formed  Distance between such points is , 2, 3, 4, ..... 3 2x 1  (B) At points where intensity is 9 th of maximum intensity, cos  = – 2 or x = 3 .  Distance between such points is  , 2 , ,  +  ,  + 2  , 2, ..... 2x 3 3 3 3  (C) cos  = 0 or x = 4 .  Distance between such points is  2 , ,  +  2 , 2, ..... 2x 1 (D) cos  = 2  or x = 6 .  Distance between such points is  , 2 , ,  +  ,  + 2  , 2, ..... 3 3 3 3 gy% dsUnzh;mfPp"BlsnwjhxijinsZijrhozrkgS& 2x D I = I0 + 4I0 + 2 4 0 cos  , tgk¡  = d Imax = 9I0 and Imin = I0 1 (A) tgka rhozrk vf/kdre rhozrk dh 9 th gS ogk¡ fufEu"B curk gSA mu fcUnqvksa ds e/; nwjh , 2, 3, 4, ..... 3 2x 1  (B) tgk¡ rhozrk vf/kdre rhozrk dh 9 th gS ogka cos  = – 2 or x = 3 . bu fcUnqvksa ds e/; nwjh  , 2 , ,  +  ,  + 2  , 2, ..... 2x 3 3 3 3  (C) cos  = 0 or x = 4 . bu fcUnqvksa ds e/; nwjh  , ,  +  , 2, ..... 2 2 2x 1  (D) cos  = 2 or x = 6 . bl izdkj ds fcUnqvksa ds e/; nwjh  , 2 , ,  +  ,  + 2  , 2, ..... 3 3 3 3 25.35 Two coherent point sources of light having wavelength  are seperated by a distance d. A circle is drawn in space surrounding both the point sources as shown. The plane of circle contains both the point sources. The distance d between both the sources is given in column-I and the total number of corresponding points of maximum intensity and minimum intensity on the periphery of the shown circle are given in column-II. Match each situation of column-I with the results in column-II izdk'kdsnksdyklEc)fcUnqL=ksrksadse/;nwjhdgSrFk izdk'kdhrjaxnS/;ZgSA,do`ÙkdksmunksuksafcUnqL=ksrksa dspkjksvksjfp=kuqlkjvuqjsf[krfd;kx;kgSAo`ÙkdsryesanksuksafcUnqL=ksrgSAL=ksrksadse/;nwjhd LrEHk-Iesa nhxbZgSrFkkmldslaxro`Ùkdhifjf/kijvf/kdrerhozrkoU;wurerhozrkokysfcUnqvksadhdqyla[;kLrEHk- IIesanhxbZgSALrEHk-IdhizR;sdfLFkfrdslaxrLrEHk-IIifj.k efn;sx;sgaSA Column-I Column-II (A) d = 99.4  (p) 398 points of maximum intensity (B) d = 99.6  (q) 400 points of maximum intensity (C) d = 100  (r) 396 points of minimum intensity (D) d = 100   (s) 400 points of minimum intensity (t) 402 points of maximum intensity LrEHk-I LrEHk-II (A) d = 99.4  (p) o`Ùkijvf/kdrerhozrkokysfcUnqvksadhla[;k398gSA (B) d = 99.6  (q) o`Ùkijvf/kdrerhozrkokysfcUnqvksadhla[;k400gSA (C) d = 100  (r) o`ÙkijU;wurerhozrkokysfcUnqvksadhla[;k396gSA (D) d = 100  (s) o`ÙkijU;wurerhozrkokysfcUnqvksadhla[;k400gSA (t) o`Ùkijvf/kdrerhozrkokysfcUnqvksadhla[;k402gSA Ans. (A) p,r (B) p,s (C) q,s (D) s,t Sol. (A) When d = 99.4  398 points of maximum intensity are formed on periphery of circle and 396points of minimum intensity are formed on periphery of circle (B) When d = 99.6  398 points of maximum intensity are formed on periphery of circle and 400 points of minimum intensity are formed on periphery of circle (C) When d = 100  400 points of maximum intensity are formed on periphery of circle and 400 points of minimum intensity are formed on periphery of circle (D) When d = 100.4  402 points of maximum intensity are formed on periphery of circle and 400 points of minimum intensity are formed on periphery of circle (A) tcd=99.4gS]o`Ùkdhifj/khijvf/kdrerhozrkds398fcUnqoo`Ùkdhifj/khijU;wurerhozrkds396 fcUnqcusaxsA (B) tcd=99.6gS]o`Ùkdhifj/khijvf/kdrerhozrkds398fcUnqoo`Ùkdhifj/khijU;wurerhozrkds400 fcUnqcusaxsA (C) tcd=100gS]o`Ùkdhifj/khijvf/kdrerhozrkds400fcUnqoo`Ùkdhifj/khijU;wurerhozrkds400 fcUnqcusaxsA (D) tcd=100.4gS]o`Ùkdhifj/khijvf/kdrerhozrkds402fcUnqoo`Ùkdhifj/khijU;wurerhozrkds400 fcUnqcusaxsA SECTION - VI : INTEGER TYPE [k.M-VI:iw.k Zadizdkj 25.36 An interference pattern is obtained by using a Fresnel’s biprism. If the f ringe width is 4 mm when air is the surrounding medium, then find the fringe width(in mm.) if water is the surrounding medium. Keeping the same source. Assume nglass = 1.5 , nwater = 4/3, nair = 1. ,dO;frdj.kçfr:iÝsuyf}fçTeç;ksxlsçkIrfd;kx;kgSAfÝUtpkSM+kbZ4mmgStcgokpkjksarjQdkek/;e gSrks;fnikuhpkjksarjQdkek/;egSrksL=k srvifjofrZrj[krsgq,fÝUtpkSM+kb(mm.esa)ZfdruhgksxhAnglass =1.5 , nwater = 4/3, nair = 1. Ans. 12 (a b)  (a  b) Sol.   2( 1)   '    where  and  are refractive indices of prism and surrounding medium  '  1   1  1.5  1 0.5 respectively.    = 12 mm. Ans.   ' 4 = 1.5  4 / 3 = 0.5 = 3 3 (a b)  (a  b Sol.   2( 1)   '    tgkarFk Øe'k%fçTerFk pkjksarjQdsek/;edsviorZukadgSA  '  1   1  1.5  1 0.5 Øe'k%     ' 4 = 1.5  4 / 3 = 0.5 = 3 3  = 12 mm. Ans. 25.37 In the figure shown S1 and S2 are two coherent sources emitting light of wavelength ‘’ and having no initial phase difference. S1 and S2 oscillate simple harmonically with amplitude ‘a’ each and frequency ‘f’ each on the line PQ which is perpendicular to the x-axis. The initial position and initial direction of motion of ‘S1’ and ‘S2’ are shown in the figure. S1 and S2 are at their mean position at t = 0 sec. if the y-coordinates of 3rd maxima at time ‘t’.is xD 2a(2  sint) then x is Assume that  << a and a << D fp=k uqlkjS1rFk S2nksdyklEc)çdk'kh;lzksr‘’rjaxnS/;Zdkçdk'kmRiUudjrsgSAçkjEHkesanksauksdschpdksbZ dykUrjughagSAx-v{kdsyEcor~PQjs[k dsvuqfn'kL=k srS1rFk S2'a'vk;kerFk 'f'vko`fÙklsljyvkorZxfrdj jgs gSA ‘S1’ rFkk ‘S2’ dh çkjfEHkd fLFkfr rFkk xfr dh fn'kk fp=kkuqlkj gSA t = 0 sec. ij S1 rFkk S2 ek/; fLFkfr ij gSA 3rd mfPp"B dk y funsZ'kkad ‘t’ le; ij xD 2a(2  sint) Ans. 3 gks rks x gksxk ekuk  << a rFkk a << D [4] 3D Sol. y-coordinate of 3rd order maxima = d 3D r`rh;Øeds mfPp"Bdk y-funsZ'kkad= d where d is the distance between both the sources at any time t tgkafdlhle;t ijnksuksL=k srksdse/;nwjhdgSA d = 4a – 2x = 4a – 2a sin t = 2a (2 – sin t) 3D  y = 2a(2  sint) LONG SUBJECTIVE fucU/k Rediz'u 25.38 In the figure shown ‘S’ is a monochromatic source of light emitting light of wavelength  (in air). Light falls on slit ‘S1’ from ‘S’ and then reach the slits ‘S2’ and ‘S3’ through a medium of refractive index ‘µ1’. Light from slits S2 and S3 reach the screen through medium of refractive index µ3. A thin transparent film of refractive index µ2 and thickness ‘t’ is placed infront of ‘S2’. point ‘P’ is symmetrical w.r.t. ‘S2’ and ‘S3’. Using the values d = 4 1 mm, D = 1 m, µ1 = 4/3, µ2 = 3/2, µ3 = 9/5 and t = 9 × 10–5 m. Find the [4] (i) Distance of central maxima from P.(in mm.) (ii) If the film in front of S2 is removed, then x/27 mm. distance and central maxima shift then x is fp=kesafn[k ;svuqlkj,do.khZ;izdk'kL=k sr‘S’rjaxnS/;Z¼gokesa½dkizdk'kmRlftZrdjrkgSAizdk'k]fLyV‘S1’ij L=k sr‘S’ls]rFk cknesafLyV‘S2’,oa‘S3’ij]vioZrukad‘µ1’okysek/;elsxqtjdjfxjrkgSAizdk'k]fLyVS2rFk S3ls]viorZukadµ3okysek/;elsxqtjdjLØhu¼insZ½rdigq¡prkgSAviorZukadµ2rFk eksVkbZ‘t’okyh,diryh ikjn'khZ fQYe]fLyV ‘S2’ds lEeq[kj[krsgSaAfcUnq ‘P’fLyV ‘S2’rFkk ‘S3’lslefer gSA;fn d=1mm,D=1m,µ1 4 =4/3, µ2 =3/2,µ3 =9/5rFkk t= 9 ×10–5 m gSas] rks fuEu dk eku Kkr dhft;s& (i) PlsdsUnzh;mfPp"BdhnwjhA(inmm.) (ii) ;fnS2dslEeq[kj[khfQYegVknhtk,s]rksdsUnzh;mfPp"Bx/27mm.foLFk firgksrksx gksxkA Ans. (i) 0 (ii) 40 Sol. (i) For the central order bright to be formed at Q dsUnzh;pedhyhfÝUtQijcuus dsfy, (SS1) µ1 + (S1Q) µ3 = (SS2) µ1 + (S2Q – t) µ3 + µ2t or (S1Q – S2Q) µ3 = (SS2 – SS1) µ1 + (µ2 – µ3) t d sin  µ3 = (d sin ) µ1 + (µ2 – µ3) t (1) d2 = D µ1 + (µ2 – µ3) t 12 = 103  3 9  × +   3  5  4 9 × 10–2 = 0   = 0 or the central order bright is formed at P only.  =0;k dsUnzh; fÝUt dsoy P ij cuus ds fy, (ii) In absence of slab from equation (1) ; t = 0 ifêdkdhvuqifLFkfresalehdj.k(1);t=0 d sin  µ3 = (d sin ) µ1 x µ = d µ or x = 2D 3 D 1 2µ1d µ3 = 40 27 mm. Ans. (i) 0 (ii) 40 mmdownwardsuhps dh vksj. 25.39 An equil convex lens of focal length 10 cm (in air) and R.I. 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1 and m2 and placed at the end of the tube. m1 & m2 are placed such that their principal axis AB and CD respectively are separated by 1 mm each from the principal axis of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5  1014 Hz. The light reflected from m and m forms 2 interference pattern on the left end EF of the tube. O is an opaque substance to cover the hole left by m1 & m2. Find : ,dlekumÙkyySalftldhQksdlnwjh10cm (gokesa)vkSjviorZukad3/2dks,duyhdkdsNksVslsfNnzesayxkfn;k tkrkgSAuyhdhyEckbZ1m gSrFk blesa4/3viorZukaddkrjyinkFkZHkjkgSA,dvoryniZ.kftldhoØrkf=kT;k 20cm dksnksHk xksaesadkVdjuyhdkdsvaresam1vkSjm2blrjgyxkfn;ktkrkgSfdeq[;v{kABvkSjCDysUl dh eq[; v{k ls 1 mm dh nwjh ij gSA ,d fLyV tks ysal dks izdk'k nsrh gSA ftldh vko`fÙk 7.5  1014 Hz. gSaA niZ.k m vkSj m ls izdk'k ijkofrZr gksus ds ckn uyhdk ds ck;sa Nksj EF ij O;frdj.k izk:i nsrs gSaA O ,d vikjn'khZ inkFkZ gS ftlds }kjk fNnz dks