Title 28. Ray Optics Easy Ans.pdf ✅

1. (a) d = 2Rh N = d 2  = 2Rh  = 2 × 3.14 × 6400 × 0.1 × 1000 = 2 × 3.14 × 6.4 × 105 = 39.5 × 105 2. (a) Output is available only when both input are available. 3. (d) o  = (360 − 2) = (360 − 2  60) = 240 4. (b) When converging beam incident on plane mirror, real image is formed as shown 5. (b) Incident ray and finally reflected ray are parallel to each other means o  = 180 From  = 360 − 2 180 = 360 − 2 o  = 90 6. (c, d) By keeping the incident ray is fixed, if plane mirror rotates through an angle  reflected ray rotates through an angle 2. 7. (c) Suppose at any instant, plane mirror lies at a distance x from object. Image will be formed behind the mirror at the same distance x. When the mirror shifts towards the object by distance ‘y’ the image shifts = x + y − (x − y) = 2y So speed of image = 2  speed of mirror 8. (c) Number of images 1 5 60 360 1 360  =       = −      = −  9. (d) Fo using distance of image = 4.5 m + 3 m = 7.5 m. 10. (b) Several images will be formed but second image will be brightest 11. (b) According to the following ray diagram length of mirror (10 170 ) 90 cm 2 1 = + = 12. (c) The walls will act as two mirrors inclined to each other at 90 and so will form  =      −1 90 360 4 – 1 i.e. 3 images of the person. Now these images with person will act as objects for the ceiling mirror and so ceiling mirror will form 4 images further. Therefore total number of images formed = 3 + 3 +1 = 7 Note : He can see. 6 images of himself. 13. (b) 60 tan 45 h  =  h = 60 m Plane mirror Real image O I Virtual object P O Q    x x O I O I2 I1 x y 2 1 x (x – y) (x – y) (x + y) O I 4.5 m 3m 3m 90% 100% 90% 10% 10% 10% 10% 80% 9% First image Second brightest image Third image Incident light 10 cm H E F 180 cm 1m 180/2 cm 45o Tower h
14. (d)  = 180  − 60 = 120  15. (a)  i = r = 0 16. (c) When light is reflected from denser medium, a phase difference of  always occurs. 17. (c) Ray after reflection from three mutually perpendicular mirrors becomes anti-parallel. 18. (b) In two images man will see himself using left hand. 19. (c) In plane mirror, size of the image is independent of the angle of incidence. 20. (b) Size of image formed by a plane mirror is same as that of the object. Hence its magnification will be 1. 21. (c) 22. (a) Subtract the given time from . min . 11 : 60 hr 23. (c) Relative velocity of image w.r.t. object = 6 −(−6) = 12 m /sec 24. (b) 25. (c) See following ray diagram The distance focussed for eye = 30 +10 = 40 cm 26. (b) Distance between object and image = 0.5 + 0.5 = 1m 27. (b) Relative velocity of image w.r.t. man = 15 − (−15) = 30 m / s 28. (b) 29. (c) 1 4 72 360 1 360  =        = −      n = − n  30. (c)       = −1 360  n   =        = −1 90 360 3   31. (c) 32. (c) 1 7 45 360 n = − = 33. (b) Diminished, erect image is formed by convex mirror. 34. (a) When a mirror is rotated by an angle , the reflected ray deviate from its original path by angle 2 . 35. (b) 2 R f = , and R =  for plane mirror. 36. (c) Let required angle be  From geometry of figure In  ABC;  = 180° – (60° + 40°) = 80°   = 90° – 80° = 10° In  ABD; A = 60°, B = ( + 2) 60° 30° 30° Incident ray Reflected ray Surface 30° 6m/sec 6m/sec O I Object Image 10cm 30 cm 10cm Object Image 0.5m 0.5m Man Image 15m/s 15m/s I Real image Virtual object O 60° 50° 40°   (90°–)  A B C D
= (80 + 2  10) = 100° and D = (90° – )  A + B + D =180°  60° + 100° + (90° – ) = 180°   = 70° 37. (a) n u v u v n m = + = −  = − 1 By using mirror formula u n f u 1 1 1 + − =  u = −(n − 1)f 38. (c) 39. (d) 40. (c) I cm I f u f O I 0.55 10 ( 100 ) 10 ( ) 5  = − − − − = +  − = 41. (a) For real image m = – 2, so by using f u f m − = u cm u 75 50 50 2  = − − − −  − = 42. (b) By using f u f O I − = I cm I 1.78 ( 40) 2 25 (25 / 2) (7.5)  =  − −      = +  43. (c) 44. (b) ; f u f O I − = where u = f + x x f O I  = − 45. (a) Image formed by convex mirror is virtual for real object placed anywhere. 46. (b) Given ( ) 1 u = f + x and ( ) 2 v = f + x The focal length ( ) ( ) ( )( ) 1 2 1 2 f x f x f x f x u v uv f + + + + + = + = On solving, we get 1 2 2 f = x x or 1 2 f = x x 47. (d) The image formed by a convex mirror is always virtual. 48. (b) Object should be placed on focus of concave mirror. 49. (b) u cm f u u f m 90 ( 30) ( 30) 4 1 ( )  = − + − +  =       + − = 50. (b) Size is . 5 1 It can’t be plane and concave mirror, because both conditions are not satisfied in plane or concave mirror. Convex mirror can meet all the requirements. 51. (c) Plane mirror and convex mirror always forms erect images. Image formed by concave mirror may be erect or inverted depending on position of object. 52. (d) Virtual image is seen on the photograph. 53. (b) u v m = − also f v u 1 1 1 = +  = +1 v u f u  f u v u − = 1 − f u f u v − = −  so f u f m − = . 54. (b) To make the light diverging as much as possible. 55. (a) Let distance = u. Now = 16 u v and v = u +120 u u cm u u 16 15 120 8 120 =  =  = +  . 56. (a) Virtual image formed is larger in size in case of concave mirror. 57. (a) Real, inverted and same in size because object is at the centre of curvature of the mirror. 58. (b) Image is virtual so m = + 3. and cm R f 18 2 = = So from f u u f m − − −  = − = ( 18) ( 18) 3  u = − 12 cm. 59. (d) 20 , 2 cm R f = = m = 2 For real image; m = −2, By using f u f m − = , u cm u 30 20 20 2  = − − − − − = For virtual image; m = +2 So, u cm u 10 20 20 2  = − − − − + = 60. (d) Convex mirror always forms, virtual, erect and smaller image. 61. (b) When object is placed. Between focus and pole, image formed is erect, virtual and enlarged. 62. (b, c) Convex mirror and concave lens form virtual image for all positions of object. 63. (c) Here focal length = f and u = − f On putting these values in f u v 1 1 1 = + 2 1 1 1 f v f f v  = − +  = Point image
64. (b) Erect and enlarged image can produced by concave mirror. f cm f f f u f O I 6 1 ( 4) 3  = − − − = + +  − =  R = 2f = −12 cm 65. (a) 66. (b) f u f m − = ( 20) 3 − −  − = f f  f = −15 cm 67. (d) When object is kept at centre of curvature. It’s real image is also formed at centre of curvature. 68. (c) u = −20 cm, f = +10 cm also f v u 1 1 1 = + ; 3 20 ( 20) 1 1 10 1 v cm v  = − = + +  virtual image. 69. (a) Mirror formula . 3 20 ( 10) 1 20 1 1 1 1 1 f cm f v u f  = − + − = +  = If object moves towards the mirror by 0.1cm then. u = (10 − 0.1) = 9.9 cm. Hence again from mirror formula v cm v 20.4 9.9 1 1 20 / 3 1   = − +  = − i.e. image shifts away from the mirror by 0.4 cm. 70. (d) Image formed by convex mirror is always. Erect diminished and virtual. 71. (d) R cm R f 40 2 =  = 72. (b) f = −15 cm, m = +2 (Positive because image is virtual) v u u v m = −  = −2 . By using mirror formula u cm u u 7.5 1 ( 2 ) 1 15 1 +  = − − = − 73. (d) u = −30 cm, f = +30 cm, by using mirror formula ( 30) 1 1 30 1 1 1 1 − = + + = +  f v u v v = 15 cm, behind the mirror 74. (d) R = −30 cm  f = −15 cm O = +2.5 cm, u = −10 cm By mirror formula 30 . ( 10) 1 1 15 1 v cm v  = − = + − Also ( 10) 30 ( 2.5) − = − + = −  I u v O I  I = +7.5 cm. 75. (d) 76. (a) f u f O I − = 6 f ( 4 f) I f − − − − = +   I = − 2 cm. 77. (d) Convergence (or power) is independent of medium for mirror. 78. (d) 2 1 20 20 20 2 = +  = − = I f u f O I  I = 1mm 79. (a) m =  3 and f = – 6 cm Now f u u f m − − −   = − = 6 6 3 For real image − −u − − = 6 6 3  u = − 8 cm For virtual image − −u − = 6 6 3  u = − 4 cm 80. (a) Focal length of the mirror remains unchanged. 81. (d) 82. (a) blue  red 83. (b) r v    ,  1 84. (a) Å air medium 4000 1.5 6000 = = =    85. (d) Velocity and wavelength change but frequency remains same. 86. (a) 1.5 4 10 5 10 3 10 14 7 8 =     = = = −   c v c 87. (c) To see the container half-filled from top, water should be filled up to height x so that bottom of the container should appear to be raised upto height (21–x). As shown in figure apparent depth h' = (21 − x) Real depth h = x O I 30cm 15cm (21 – x ) x (21 – x ) 21 cm