Title 29. Wave Optics Hard Ans.pdf ✅

1. (a) By using 1 9 1 1 4 1 1 4 1 1 2 2 2 1 2 1 min max =               − + =               − + = I I I I I I . 2. (d) By using n11 = n2 2  92  5898 = n2  5461  n2 = 99 3. (b) By using I = I1 + I2 + 2 I1 I2 cos At point A : Resultant intensity I I I I I I A 5 2 = + 4 + 2  4 cos =  At point B : Resultant intensity I I I I I I B = + 4 + 2  4 cos = . Hence the difference I I I A B = − = 4 4. (b) By using 2 1 2 cos 2 2 2 A = a1 + a + a a  37 6 3 (4) (3) 2 4 3 cos 2 2 = + +   =   A . 5. (d) For destructive interference, path difference the waves meeting at P (i.e. ) S1P − S 2P must be odd multiple of /2. Hence option (d) is correct.] 6. (d) Path difference     =  2      11 22 2  = i.e. constructive interference obtained at the same point So, resultant intensity I I I I I I R ( ) ( 9 4 ) 25 2 2 = 1 + 2 = + = . 7. (b) By using 1 7 1 5 12 1 9 12 1 81 144 1 81 144 1 1 min max min max 2 1 =             − + =               − + =               − + = I I I I a a 8. (c) Time difference T.D.   =  2 T    =  2 2 3T T   = 3; This is the condition of constructive interference. So resultant intensity ( ) ( ) 2 . 2 2 1 2 I I I x y x y xy R = + = + = + + 9. (b) For dark fringe path difference ; 2 (2 1)   = n − here n = 3 and  = 6000  10–10 m So 15 10 1.5 . 2 6 10 (2 3 1) 7 7 =  m = microns   =  −  − − 10. (b) Distance of n th minima from central maxima is given as d n D x 2 (2 − 1) = So here x 1.25 mm 2 10 (2 3 1) 500 10 1 3 9 =   −    = − − 11. (b) Distance between n th bright and m th dark fringe (n > m) is given as d D x n m n m          = − +      = − + 2 1 2 1  x 1.63 mm 1 10 6.5 10 1 2 1 5 3 3 7 =           = − + − − . 12. (b) By using 2 4 cos 2  I I R = {where I = Intensity of each wave} At central position  = 00 , hence initially I0 = 4I. If one slit is closed, no interference takes place so intensity at the same location will be I only i.e. intensity become s th 4 1 or 4 0 I . 13. (a) By using d   =  2 1 2 1     =  2 5890 (0.20 10% of 0.20) 0.20  = + o o  2 5890 0.22 0.20  =   2 = 6479 So increase in wavelength = 6479 – 5890 = 589 Å. 14. (c)    air medium =  medium 0.4mm 1.5 0.6  = = . 15. (b)
SBy using phase difference ( ) 2 =     For path difference , phase difference 1 = 2 and for path difference /4, phase difference 2 = /2. Also by using 2 4 cos 2 0  I = I  cos ( / 2) cos ( / 2) 2 2 1 2 2 1   = I I  1 / 2 1 2 / 2 cos cos (2 / 2) 2 2 2 =       =   I k  . 2 2 k I = 16. (a) By using shift t p x = ( − 1)     (1.5 1) 2 10 2 5000 10 6 10 −   =   = − − x Since the sheet is placed in the path of the first wave, so shift will be 2 fringes upward. 17. (a) Shift due to the first plate x ( 1)t 1 = 1 −   (Upward) and shift due to the second x ( 1)t 2 =  2 −   (Downward) Hence net shift = x2 – x1 ( )t  2 1   = −  5 p = (1.8 − 1.5)t    t 8 10 m 8 micron 0.3 5 4800 10 0.3 5 6 10 =  =   = = − −  . 18. (a) By using shift 4 cos ( / 2) 2 I = I0   4 cos ( / 2) 2 I0 = I0   2 1 cos( / 2) = or 2 3   =  3 2  = Also path difference     = =  D 2 xd  3 2 2 6000 10 10      =       − D d x  2 10 2 . 3 x =  m = mm − 19. (a) Initial phase difference 4 0   = ; Phase difference due to path difference ( ) 2 ' =     where  = d sin   4 (sin 30 ) 4 2 ( sin ) 2 '         = =  = o d Hence total phase difference 4 ' 0   =  +  = . By using 0 2 0 2 0 2 2 / 2 I 4 I cos ( / 2) 4 I cos  = I      = =   . 20. (a) According to given problem angular fringe width 180  60 =     d ] 180 60 [As 1' rad  =  i.e.  6 10 180 60 7     − d i.e. d m 3 2.06 10 −    d max = 2.06 mm 21. (a) In case of interference of two wave I = I1 + I2 + 2 I1 I2 cos (i) In case of coherent interference  does not vary with time and so I will be maximum when cos = max = 1 i.e. 2 max 1 2 1 2 1 2 (I ) I I 2 I I ( I I ) co = + + = + So for n identical waves each of intensity I0 0 2 2 0 2 max 0 0 (I ) ( I I ......) (n I ) n I co = + + = = (ii)In case of incoherent interference at a given point,  varies randomly with time, so (cos)av = 0 and hence 1 2 (I ) I I R Inco = + So in case of n identical waves 0 0 0 (I ) I I ....... nI R Inco = + + = 22. (b) Amax = 2A + A = 3A and Amin = 2A − A = A . Also 1 3 9 2 2 min max min max  =      =         = A A A A I I 23. (c) By using       = − c v '  1  5802 Å 3 10 4.5 10 ' 5890 1 8 6 =            = − . 24. (b) S1 S2 d Screen D C 1 2
By using c v  =   (3737-3700)= 8 3 10 3700   v  v 3 10 m / s 6 =  . 25. (a) By using ; c v =    where 100 0.4 =    and c = 3  108 m/s  8 100 3 10 0.4  = v  v = 1.2  106 m/s Since wavelength is increasing i.e. it is moving away. 26. (d) Luminous flux = 4 L = 4  3.14  42 = 528 Lumen Power of lamp Luminous efficiency Luminous flux = 264 W 2 528 = = 27. (b) 2 cos r L I  = cos 2 I r L   = 40 Candela cos 60 5 10 10 2 4 4 2 =     = − 28. (d) 2 r L I = r dr I dI 2  = − (  L = constant)  100 2 100    = − r dr I dI = −2  1 = −2% 29. (c) For equal fogging 2 2 1 1 I  t = I  t 1 2 1 1 2 2 2 2 t r L t r L   =  10 1 20 4 16   t 2 =  50 . 2  t = Sec 30. (d) The illuminance at B 2 1 L IB = .........(i) and illuminance at point C 5 1 ( 5) ( 5) cos 2 2 = =  L L IC  5 5 0 L  I = ......... (ii) From equation (i) and (ii) 5 5 0 I I B = 31. (b) 2 1 r I  So, 2 2 (Length of object on slide) (Length of image on screen) Illuminanc e on screen Illuminanc e on slide = 10 : 1 35 3.5 4 2 =         = mm m 32. (a) The illuminance at A is 3 / 2 1 2 (13) 3 13 3 13 cos ( 13 ) L L L IA =   =  = The illuminance at B is 3 / 2 2 2 (17) 3 17 3 17 cos ( 17 ) L L L IB =   =  =  3 / 2 13 17       = B A I I 1 2 17m 13m 2m A B 2 2 3m  5 m A Lamp 1 m B  2 m 60o 2 m Screen Normal