Title 29. Wave optics Easy Ans.pdf ✅

1. (a) Corpuscular theory explains refraction of light. 2. (c) According to Corpuscular theory different colour of light are due to different size of Corpuscules. 3. (b) 4. (c) According to Plank’s hypothesis, black bodies emits radiations in the form of photons. 5. (d) The coherent source cannot be obtained from two different light sources. 6. (c) Huygen's wave theory fails to explain the particle nature of light (i.e. photoelectric effect) 7. (d) Interference is shown by transverse as well as mechanical waves. 8. (c) I ( I I ) ( I 4I) 9I 2 2 max = 1 + 2 = + = I = I − I = I − I = I 2 2 min 1 2 ( ) ( 4 ) 9. (c) 10. (b) The idea of secondary wavelets is given by Huygen. 11. (c) Monochromatic wave means of single wavelength not the single colour. 12. (d) Sound wave and light waves both shows interference. 13. (c) 1 4 1 1 9 1 1 9 1 1 2 2 1 2 1 min max =               − + =               − + = I I I I I I 14. (a) A wave can transmit energy from one place to another. 15. (d) 5 1 25 1 ; 25 1 2 1 2 2 2 1 2 1 =  =  = a a a a I I 16. (a) For interference phase difference must be constant. 17. (c) Interference is explained by wave nature of light. 18. (b) Coherent time c L = = Velocity of light Coherence length 19. (c) 5 3 2 1 = a a 1 16 (3 5) (3 5) ( ) ( ) 2 2 2 1 2 2 1 2 min max = − + = − +  = a a a a I I 20. (c) For constructive interference path difference is even multiple of 2  . 21. (c) Two coherent source must have a constant phase difference otherwise they can not produce interference. 22. (a) Phenomenon of interference of light takes place. 23. (c) Transverse waves can be polarised. 24. (a) 1 9 1 1 4 1 1 4 1 1 2 2 2 1 2 1 min max =               − + =               − + = I I I I I I 25. (a) Reflection phenomenon is shown by both particle and wave nature of light. 26. (b) When two sources are obtained from a single source, the wavefront is divided into two parts. These two wavefronts acts as if they emanated from two sources having a fixed phase relationship. 27. (b) m c 6 8 3 10 100 3 10 =   = =   28. (c) 9 49 1 4 25 1 4 25 1 1 2 2 2 1 2 1 min max =               − + =               − + = I I I I I I 29. (a) Wavefront is the locus of all the particles which vibrates in the same phase. 30. (b) Direction of wave is perpendicular to the wavefront. 31. (d) Origin of spectra is not explained by Huygen's theory. 32. (d) The refractive index of air is slightly more than 1. When chamber is evacuated, refractive index decreases and hence the wavelength increases and fringe width also increases. 33. (a) 2 I  a  1 2 1 4 1 / 2 2 1  =      = a a
34. (a) The essential condition for sustained interference is constancy of phase difference. 35. (c) 1 49 1 3 4 1 3 4 1 1 2 2 2 1 2 1 min max =             − + =             − + = a a a a I I 36. (b) Energy is conserved in the interference of light. 37. (c) 2 I  a 38. (b) 39. (c) 2 I  a 2 2 1 2 1          = a a I I 16 9 4 3 2  =      = 40. (d) 1 100 2 1 = I I Now 2 3 81 121 100 1 100 1 1 1 2 2 2 1 2 1 min max =          − + =               − + = I I I I I I 41. (b)  =  / 3, a1 = 4, a2 = 3 So, 2 1 . 2 cos 6 2 2 2 A = a1 + a + a a   A  42. (b) sin , 1 y = a t and       = = + 2 2 cos sin  y b t b  t So phase difference  =  / 2 43. (c) For 2 phase difference → Path difference is   For  phase difference → Path difference is     2 44. (d) Resultant intensity IR = I 1 + I 2 + 2 I 1 I 2 cos For maximum , R I o  = 0  ( ) 2 1 2 2 1 2 1 2 I I I I I I I R = + + = + 45. (c) Newton first law of motion states that every particle travels in a straight line with a constant velocity unless disturbed by an external force. So the corpuscles travels in straight lines. 46. (d) Diffraction shows the wave nature of light and photoelectric effect shows particle nature of light. 47. (b) At point A, resultant intensity 5 ; 1 2 I I I I A = + = and at point B I I I I I I I B = 1 + 2 + 2 1 2 cos = 5 + 4 I I B = 9 so I I 4I. B − A = 48. (c) 49. (a) Photoelectric effect varifies particle nature of light. Reflection and refraction varifies both particle nature and wave nature of light. 50. (d)       = = = + 2 1 sin , 2 cos sin  y a  t y a  t a  t 51. (c) 1 25 1 1 2 2 1 2 1 min max =             − + = a a a a I I 52. (d) Laser beams are perfectly parallel. So that they are very narrow and can travel a long distance without spreading. This is the feature of laser while they are monochromatic and coherent these are characteristics only. 53. (c) 4 9 1 1 2 1 2 2 1 2 1 min max  =               − + = I I I I I I I I 54. (b) 15 10 8 10 3000 10 3 10 =   = = −   c cycles/sec 55. (a) . 16 25 4 5 1 9 1 1 9 1 1 1 2 2 2 2 1 2 1 min max  =      =             − + =             − + = a a a a I I 56. (c) 57. (d) For destructive interference path difference is odd multiple of . 2  Spherical wave front Point source S
58. (d) 6 1 1 1 2 1 2 2 2 1 2 1 min max = − +              − + = a a a a a a a a I I 7 : 5 1 2 = a a 59. (b) max 1 2 2 1 2 I = I + I + I I So, Imax = I + 4 I + 2 I.4 I = 9I 60. (b) In interference energy is redistribution. 61. (a) For interference frequency must be same and phase difference must be constant. 62. (c) When a beam of light is used to determine the position of an object, the maximum accuracy is achieved if the light is of shorter wavelength, because Wavelength 1 Accuracy 63. (d) Intensity 2 (Distance) 1  64. (d) Huygen’s theory explains propagation of wavefront. 65. (c) Wave theory of light is given by Huygen. 66. (a) When light reflect from denser surface phase change of  occurs. 67. (a) Photoelectric effect explain the quantum nature of light while interference, diffraction and polarization explain the wave nature of light. 68. (a) Light is electromagnetic in nature it does not require any material medium for its propagation. 69. (c) For viewing interference in oil films or soap bubble, thickness of film is of the order of wavelength of light. 70. (b) 71. (d) For maximum intensity  = 0  I = I 1 + I 2 + 2 I 1 I 2 cos = I + I + 2 I I cos 0 = 4I 72. (a) 73. (c) In interference of light the energy is transferred from the region of destructive interference to the region of constructive interference. The average energy being always equal to the sum of the energies of the interfering waves. Thus the phenomenon of interference is in complete agreement with the law of conservation of energy. 74. (c) m m mm d D 10 1.0 10 5 10 2 3 3 7 = =   = = − − −   . 75. (c) Slit width ratio = 1 : 9 Since slit width ratio is the ratio of intensity and intensity  (amplitude)2 I1 : I2 = 1 : 9 : 1 : 9 1 : 2 1 : 3 2 2 2  a1 a =  a a = 4 1 ( ) , ( ) max 2 min min 1 2 2 max = 1 + 2 = −  = I I I a a I a a 76. (a) d 1    If d becomes thrice, then  become becomes 3 1 times. 77. (c) 2 1 2 1     = or 6000 1.0 5000 2 =  or 1.2 mm 5000 6000  2 = = . 78. (a) 3 10 2 10 6000 10 25 10 − − −     = 150000 10 0.15 10 m 0.015 cm 9 3 =  =  = − − . 79. (c) For brightness, path difference = n = 2 So second is bright. 80. (a) If one of slit is closed then interference fringes are not formed on the screen but a fringe pattern is observed due to diffraction from slit. 81. (d) 82. (d) =  d D  If D becomes twice and d becomes half so  becomes four times. 83. (c) Suppose slit width’s are equal, so they produces waves of equal intensity say I . Resultant intensity at any point  2 I 4 I cos R =  where  is the phase difference between the waves at the point of observation. For maximum intensity I I I o  = 0  max = 4  = ...(i) If one of slit is closed, Resultant intensity at the same point will be I only i.e. O I = I ...(ii) Comparing equation (i) and (ii) we get O I = 4 I
84. (b, d) 9 9 3 1 2 1 2 2 1 2 1 2 min max = − +   =        − + =  a a a a a a a a I I 2. 3 1 3 1 2 1 2 1  = − +  = a a a a Therefore I1 : I2 = 4 : 1 85. (b) 86. (c) Distance of n th bright fringe   n = i e yn  d n D y . . 5460 4360 (Green ) (Blue) 2 1 2 1  =  = x x x x n n    x (Green) > x (Blue). 87. (c) m m cm d D 5 10 0.5 0.1 10 5000 10 1 3 3 10 =  =    = = − − −   . 88. (a) We know that fringe width d D  = L xd d L x =   =  89. (a) In the normal adjustment of young’s, double slit experiment, path difference between the waves at central location is always zero, so maxima is obtained at central position. 90. (a)    = B  d D ; ' 0.3 mm 4 / 3 ' 0.4 =   =   . 91. (b)     1  ,   92. (a)   , v = minimum. 93. (c) 0.4 mm 1.5 air 0.6 medium = = =    . 94. (d) n = 3, 2n = 23 = 6 95. (c) Slit width ratio = 4 : 9; hence I1 : I2 = 4 : 9 3 2 9 4 2 1 2 2 2 1  =  = a a a a 1 25 ( ) ( ) 2 1 2 2 1 2 min max = − +  = a a a a I I 96. (d) d D  =  D  d = 0.2 . 0.012 10 6000 10 (40 10 ) 2 10 2 = cm     = − − − 97. (a) As                         =  = 1 2 2 1 2 1 2 1 d d D D d D       1 4 2 1 2 1 1 2 1 2 1   =                      = D D D D 98. (b) 99. (c) Fringe width     =   d D ( ) As , red  yellow hence fringe width will increase. 100. (d) d D  = 1 1 2 2 2 1 1 2 D d D d      = m 4 2 2.5 10 −   =  . 101.(d) For interference,  of both the waves must be same. 102. (d)   D 103.(a)    ; d = can be increased by increasing , so here  has to be increased by 10% i.e., % Increase 5890 589 Å 100 10 =  = 104. (b) m mm D d 10 0.1 5 10 1 5 10 4 3 7 = =    = = − − −   . 105.(b) If intensity of each wave is I, then initially at central position I 4I. o = when one of the slit is covered then intensity at central position will be I only i.e., 4 o I . 106.(a) Shift      (1.5) 2 10 2 (5000 10 ) ( 1) 6 10   =  = − = − − t i.e., 2 fringes upwards. 107. (b) d D  = 108.(b) Separation th n bright fringe and central maxima is d n D x n  = So, 3.5 . 0.5 10 3 6000 10 1 3 10 x 3 = mm     = − − 109.(d) 11 22 n = n  62  5893 = n2  4358 84.  n2 = 110.(b) Angular fringe width     =   d