Title 28. Ray Optics Easy 2 Ans.pdf ✅

1. (b) Angle of refraction will be different, due to which red and green emerge from different points and will be parallel. 2. (a) Deviation    1   3. (a) 2 sin 2 sin A A  m  + = 2 60 sin 2 60 38 sin + = o o sin 30 sin 49 = 1.5 0.5 0.7547 = = . 4. (d) Using  = i1 + i2 − A  55 = 15 + i2 − 60 o  i2 = 100 5. (b) Sodium light gives emission spectrum having two yellow lines. 6. (c) Colour of the sky is highly scattered light (colour). 7. (a) 8. (c) 9. (c) Man is suffering from hypermetropia. The hole works like a convex lens. 10. (a) 11. (b) In myopia, u = , v = d = distance of far point By , 1 1 1 f v u = − we get f = − d Since f is negative, hence the lens used is concave. 12. (b) 13. (c) Cylindrical lens are used for removing astigmatism. 14. (b) 15. (a) Image formed at retina is real and inverted. 16. (d) Visible region decreases, so the depth of image will not be seen. 17. (a) D f v u P 3 100 3 25 1 100 1 1 1 1 = = − + = − + = = + 18. (c) If eye is kept at a distance d, then , ( ) 0 e f f L D d MP − = MP decreases. 19. (c) For lens u = want’s to see =  v = can see = – 5 m  From f v u 1 1 1 = −  − −  = 1 5 1 1 f  f = −5 m. 20. (a) For improving near point, convex lens is required and for this convex lens u = − 25 cm, v = −75 cmf cm f 2 75 25 1 75 1 1  = − − −  = So power D f P 3 8 75 / 2 100 100 = = = + 21. (b) In short sightedness, the focal length of eye lens decreases, so image is formed before retina. 22. (b) For improving far point, concave lens is required and for this concave lens u = , v = − 30 cm So f cm f 30 1 30 1 1  = −  − − = for near point u cm u 30 1 15 1 30 1 −  = − − = − 23. (c) For myopic eye f = – (defected far point)  f = − 40 cm  P 2.5 D 40 100 = − − = 24. (c) For lens u = want’s to see = −60 cm v = can see = −10 cm f v u 1 1 1  = − ( 60) 1 10 1 1 − − −  = f  f = −12 cm 25. (b) Focal length = – (Detected far point) 26. (c) In this case, for seeing distant objects the far point is 40 cm. Hence the required focal length is f = − d (distance of far point) = – 40 cm Power cm D f P 2.5 40 100 100 = − − = = 27. (b) 28. (a) 29. (a) 30. (a) For viewing far objects, concave lenses are used and for concave lens u = wants to see = −60 cm ; v = can see = −15 cm so from f v u 1 1 1 = −  f = − 20 cm . 31. (d)
32. (a) In short sightedness, the focal length of eye lens decreases and so the power of eye lens increases. 33. (d) Colour blindness is a genetic disease and still cannot be cured. 34. (c) Convexity to lens changes by the pressure applied by ciliary muscles. 35. (b) f = − d = − 100 cm = − 1 m D f P 1 1 1 1 = − −  = = 36. (c) For correcting myopia, concave lens is used and for lens. u = wants to see = −50 cm v = can see = − 25 cm From f v u 1 1 1 = − ( 50) 1 25 1 1 − − −  = f  f = −50 cm So power D f P 2 50 100 100 = − − = = 37. (c) 38. (c) f = − d = − 60 cm D f P 1.66 6 10 60 100 100  = = − = − = − 39. (b) For correcting the near point, required focal length f 50 cm (50 25) 50 25 = −  = So power P 2 D 50 100 = = + For correcting the far point, required focal length f = −(defected far point) = −3m P D 0.33 D 3 1  = − = − 40. (b) Negative power is given, so defect of eye is nearsigntedness Also defected far point cm p f 40 ( 2.5) 1 100 = − = − = − = − 41. (a) In myopia, eye ball may be elongated so, light rays focussed before the retina. 42. (c) 43. (d) 2 1 (defected far point) 1 1 = − − = = f P = −0.5 D 44. (a) Resolving limit of eye is one minute (1'). 45. (d) Because for healthy eye image is always formed at retina. 46. (a) The defect is myopia (nearsightness) As we know for myopic person f = – (defected far point)  Defected far point = – f = – m P 0.5 ( 2) 1 1 = − = − = 50 cm 47. (b) Power of convex lens P 2.5 D 40 100 1 = = Power of concave lens P 4 D 25 100 2 = − = − Now P = P1 + P2 = 2.5 D − 4 D = − 1.5 D 48. (c) 49. (d) 50. (a) As limit of resolution of eye is o       60 1 , the pillars will be seen distinctly if o        60 1  i.e., 60 180 1          x d 60 180    x d  60 180 3.14 11 10 3     d   d  3.2 m 51. (b) 52. (b) 53. (d) 54. (d) f = – (defected far point) = – 20 cm 55. (b) Power of the lens given positive so defect is hypermetropia. 56. (b) Far point of the eye = focal length of the lens cm P 151 0.66 100 100 = = = 57. (c) A bifocal lens consist of both convex or concave lenses with lower part is convex. 58. (a) For lens u = wants to see = – 30 cm and v = can see = – 10 cm  d x
f v u 1 1 1  = − ( 30) 1 10 1 − − − =  f = −15 cm 59. (a) Focal length = – (far point) 60. (c) For lens u = wants to see = – 12 cm v = can see = – 3 m f v u P 1 1 1  = = − P D 4 1 ( 12) 1 3 1 = − − − −  = 61. (d) 2 2 1 2 2 2 1 1 I D t = I D t Here D is constant and 2 r L I = So 2 2 2 2 1 2 1 1 t r L t r L  =  t 20 sec (4) 120 10 (2) 60  2  = 2   62. (a) f = − 40 cm and P 2.5 D 40 100 = − − = 63. (a) Focal length of the lens f cm 3 100 = By lens formula f v u 1 1 1 = − v cm m v 100 1 25 1 1 100 / 3 1  = − = − − = − +  64. (d) This is the defect of hypermetropia. 65. (a) For large objects, large image is formed on retina. 66. (d) v = −15 cm, u = −300 cm, From lens formula f v u 1 1 1 = − 300 19 300 1 15 1 1 − = − − −  = f f 15.8 cm 19 300 = − −  = and power cm f P 100 = 300 −100 19 = = – 6.33 D. 67. (d) Time of exposure 2 (Aperture) 1  68. (a) Light gathering power  Area of lens aperture or d 2 69. (b) Time of exposure 2  (f.number) 4 2.8 5.6 2 1 2  =       = t t 50 1 200 1 t2 = 4 t1 = 4  = sec = 0.02 sec. 70. (d) 71. (a) 72. (c) By using o e o e f f L f f D m ( − − ). =   L cm L 15 1 5 ( 1 5) 25 45  =  − −   =   . 73. (b) For a compound microscope o e f f m 1  74. (b) For a compound microscope objective eye piece f  f 75. (b) In microscope final image formed is enlarged which in turn increases the visual angle. 76. (b) 77. (d) Magnification of a compound microscope is given by o e o u D u v m = −   m = mo me | | . 78. (c) Magnifying power of a microscope f m 1  Since violet red violet red f  f ; m  m 79. (a) o e L = v + f  14 = vo + 5  vo = 9 cm Magnifying power of microscope for relaxed eye o e o f D u v m = . or 5 25 . 9 25 uo = or uo 1.8 cm 5 9 = = 80. (b) o e o f D u v m = −  From o o o f v u 1 1 1 = − ( 1.25) 1 1 ( 1.2) 1 − = − +  o v  vo = 30 cm 200 3 25 1.25 30  m  =  = 81. (b) For objective lens o o o f v u 1 1 1 = − ( 4.5) 1 1 ( 4) 1 − = − +  o v  vo = 36 cm 32 8 24 1 4.5 36 1  =      = +           = + o e o D f D u v m 82. (a) For a microscope o e o u D u v  m  =  and o ue L = v + For a given microscope, with increase in L, ue will increase and hence magnifying power (m) will decrease. 83. (b) In compound microscope objective forms real image while eye piece forms virtual image.
84. (b) f D m = 1 + Smaller the focal length, higher the magnifying power. 85. (a) In electron microscope, electron beam (  1Å) is used so it’s R.P. is approx. 5000 times more than that of ordinary microscope (  5000 Å). 86. (c) If nothing is said then it is considered that final image is formed at infinite and o e e o e f f LD f f L f f D m 0 ~ ( − − ). =   2.5 . 0.5 20 25 400 f cm f e e  =    = 87. (d) 11 2.5 25 max = 1 + = 1 + = f D m . 88. (a) 89. (b) DP f D m = 1 + = 1 + (m increases with P) 90. (b) 91. (b) Like Gallilean telescope. 92. (a) o e f f m 1    93. (d) A microscope consists of lens of small focal lengths. A telescope consists of objective lens of large focal length. 94. (c) m = mo me = 25  6 = 150 95. (a) When final image is formed at infinity, length of the tube o e = v + f 15 = vo + 3  vo = 12 cm For objective lens o o o f v u 1 1 1 = − o u 1 ( 12) 1 ( 2) 1 − + = +   uo = −2.4 cm 96. (d) R.P. of microscope  2 sin = 97. (c) m = mo me          =  + e o f D m m 1          =  + e f 25 100 10 1 f e cm 9 25  = 98. (c) A simple microscope is just a convex lens with object lying between optical centre and focus of the lens. 99. (d) In general, the simple microscope is used with image at D, hence 6 5 25 = 1 + = 1 + = f D m 100. (d) 101.(b) Resolving power of microscope  1  102.(a) Cross wire arrangement is used to make measurements. 103.(d) f D f D u f u f L v u e e o o o o o e + + − = + = ( ) L 11 cm (6.25 25) 6.25 25 (2 1.5) 2 1.5 = +  + −   = 104. (d) o e f f LD m~ 500 0.5 1 10 25 =    m = . 105.(c) Intermediate image means the image formed by objective, which is real, inverted and enlarged. 106. (d) o e f f m 1  107. (b) R.P. Blue Red Blue Red ; so ( . .) ( . .) 1     R P  R P  108. (a) f cm m f f D m 5 0.05 25 = 1 +  6 = 1 +  = = 109.(a) Resolving limit 4800 0.1 6000 2 2 1 2 1   =  = x x x x     x 2 = 0.08 mm 110. (b) m = mo me  100 = 5 me  me = 20 111. (d) P f m   1 112. (d) R.P. 4 5 ( . .) 1 ( . .) 1 2 2 1   = =    R P R P 113.(b) Resolving limit (minimum separation)   A B B A P P P P  =   3000 2000 114. (d) Similar to Q.No. 34