Tiêu đề 28. Ray Optics hard Ans.pdf ✅

1. (d) By using f u f O I − = ; where I = ? , O = + 2.5 cm. f → − f , u = – 1.5 f  2.5 f ( 1.5 f) I f − − − − = +  I = −5 cm. (Negative sign indicates that image is inverted.) 2. (c) By using f v u 1 1 1 = +  ( ) 2 1 1 1 f v f v f  = − = + + 3. (b) By using A B A B B A f u f u O O I I f u f O I − −   = − = 10 ( 50 ) 10 4 1 1 1 − − − − −   = uB  uB = −20cm . 4. (a) By using o i A A m = 2 ; where f u f m − = Hence from given values ( ) 3 2 10 25 10 − = − − − − m = and 2 Ao = 9 cm  2 2 9 4 3 2 Ai   = cm      − = 5. (a) No change in focal length, because f depends only upon radius of curvature R. 6. (c) Let the mirror be placed at a distance x from wall By using ( ) 3 ( 3) 9 − − − − = + −  − = x x u v O I  x = −4.5m = − 450 cm. 7. (b) If end A of rod acts an object for mirror then it's image will be A' and if 3 5 3 2 f f u = f − = So by using f v u 1 1 1 = + 3 5 1 1 1 f v − f = + −  v f 2 5  = −  Length of image 2 2 2 5 f = f − f = 8. (d) 9. (c)   1   1 2 2 1     =  4200 3 4 1 2 =  2 = 3150 Å 10. (b) From figure o r = 30  3 sin 30 sin 60 sin sin = = = o o r i  11. (b) From figure it is clear that object appears to be raised by cm (2.5 cm) 4 10 Hence distance between mirror and O' = 5 + 7.5 = 12.5 cm So final image will be formed at 12.5 cm behind the plane mirror. 12. (c) o C C 30 2 1 7000 3500 sin 2 1 1 2 = = = =  =     13. (d) When total internal reflection just takes place from lateral surface then i = C i.e. C = 600 From sin C 1  = 3 2 sin 60 1   = = Hence time taken by light traverse some distance in medium C x t  = ( ) 3.85 . 3 10 1 10 3 2 8 3 t =  sec     = 5 cm 10 cm O cm O' Object image O Initially C Object O Finally Image C A' C F A 2f f / 3 v u = 2f – (f/3) x (x–3)m 3cm 3m
14. (a) From figure it is clear that Total internal reflection takes place at AC, only if  > C  sin  sin C  g   1  sin  9 / 8 1  sin  9 8  sin   15. (a) At face AB, i = 0 so r = 0, i.e., no refraction will take place. So light will be incident on face AC at an angle of incidence of 450 . The face AC will not transmit the light for which , C i   i.e., C sin i  sin or sin 45  (1 / ) o i.e.,   2 (= 1.41) Now as  R   while  G and   , B so red will be transmitted through the face AC while green and blue will be reflected. So the prism will separate red colour from green and blue. 16. (c) Let thickness of slab be t and distance of air bubble from one side is x When viewed from side (1) : x cm x 9 6 1.5 =  = When viewed from side (2) : ( ) t cm t x t 15 4 9 1.5 4 ( ) 1.5  = −  = − = 17. (d) From the figure shown it is clear that For lens : u = 12 cm and v = x = ? By using f v u 1 1 1 = −  12 1 1 16 1 + = − + x  x = 48 cm 18. (a) By using 1 2 1 1 1 F f f = +  25 1 40 1 1 − + + = F  F cm 3 200 = − , hence D f cm P 1.5 200 / 3 100 ( ) 100 = − − = = 19. (b) Initially F = 60 cm (Focal length of combination) Hence by using 1 2 1 1 1 F f f = +  60 1 1 1 1 2 + = f f  1 2 1 2 f f f f + ......(i) Finally by using 1 2 1 2 1 1 1 f f d F f f = + −  where F = 30 cm and d = 10 cm  1 2 1 2 1 1 10 30 1 f f f f = + − ......(ii) From equations (i) and (ii) 600 . f1 f2 = − From equation (i) 10 f1 + f2 = − .....(iii) Also, difference of focal lengths can written as 1 2 2 f1 − f2 = (f1 + f2 ) − 4 f f  f1 − f2 = 50 .....(iv) From (iii)  (iv) f1 = 20 and 30 f2 = − 20. (b) By using 2( − 1) =  R f  ( ) 30.7 31 . 2 1.65 1 40 f = cm  cm − = 21. (a) By using v u R  2 1  2 − 1 − = Where 1, 1 = 1.5,  2 = u = – OP, v = OQ Hence OQ OP (+ R) − = − − 1.5 1 1.5 1  OP OP R 1.5 1 0.5 + = P' P x 12cm 6 cm 4 cm Side 1 Side 2 Air bubble t x 45° 45° A B C B A  C  60 Air o Air Glass
 OP = 5 R 22. (b) By using D D x f 4 2 2 − =  f 21 cm 4 100 100 40 2 2 =  − = Hence power ( ) D F cm P 5 21 100 100 = =  + 23. (d) Here ( ) 100 3 20 1 1 1.6 1 1 1 −  =      −  = − f .......(i) ( ) 20 1 20 1 20 1 1.5 1 1 2  =      − = − − f .......(ii) ( ) 100 1 3 20 1 1.6 1 1 3 −  =       − − = − f .......(iii) By using 1 2 3 1 1 1 1 F f f f = + +  100 3 20 1 100 1 3 + − − = F  F = −100 cm 24. (a) By using F f l fm 1 2 1 = + Since fm =   cm f F l 10 2 20 2 = = = (After silvering concave lens behave as convex mirror) 25. (a) In concave lens, image is always formed on the same side of the object. Hence the given lens is a convex lens for which u = – 25 cm, v = 75 cm. By using f v u 1 1 1 = −  ( ) ( 25 ) 1 75 1 1 − − + = f  f = + 18.75 cm. 26. (c) By using 1 2 O = I I  O = 8  2 = 4 cm 27. (a) By using f u f m + = here ( ) ( f) u f m + + + − =  f u f f u m = + + − = 1 1  f m m u . 1       + = − 28. (a) By using v u R  2 1  2 − 1 − = where u = ? , v = – 1 cm, 1.5 1 = , 1  2 = , R = – 2 cm. ( 2) 1.5 1 1.5 1 1 − − − = − u  1.2 . 5 6 u = − = − cm 29. (c) From figure 2 10 11 = d D  2.8 . 10 2 1.4 10 11 9 d = cm   = 30. (a) The given condition will be satisfied only if one source (S1) placed on one side such that u < f (i.e. it lies under the focus). The other source (S2) is placed on the other side of the lens such that u > f (i.e. it lies beyond the focus). If 1 S is the object for lens then f y − x − − = 1 1 1  y x f 1 1 1 = − ........(i) If S 2 is the object for lens then (24 ) 1 1 1 f y − − x − + = Sun (D) (d) Image   f 1011 m 1=1.5 2=1 C u R = 2cm v =1cm  + F Fe Fm  + + f1 f2 f3 F P Q O
 (24 ) 1 1 1 y f − x = − ........(ii) From equation (i) and (ii) (24 ) 1 1 1 1 x f f − x − = −  9 2 2 (24 ) 1 1 = = − + x x f  24 108 0 2 x − x + = On solving the equation x = 18 cm , 6 cm 31. (c) Consider the refraction of the first surface i.e. refraction from rarer medium to denser medium 1 2 1 1 2 R u v     + − = −  v R R v 9 2 3 3 4 3 4 2 3 1 1 +  =  =        −      Now consider the refraction at the second surface of the lens i.e. refraction from denser medium to rarer medium v R R R v       = − +  = − − 2 1 3 9 2 3 2 3 1 2 2 The image will be formed at a distance do R 2 3 . This is equal to the focal length of the lens. 32. (d) 2 sin sin A i  =  2 sin sin 45 2 A =  2 1 2 2 1 2 sin = = A o o A A 30 60 2  =  = 33. (c) Given  m = A , then by using 2 sin 2 sin A A  m  + = 2 sin 2 sin A A + A   = 2 2 cos 2 sin sin A A A = =       = 2 cos 2 sin 2 sin A A A  o o A A A A 82 2 41 2 0.75 cos 2 1.5 = 2 cos  =  =  = 34. (c) Incident ray and emergent ray are symmetrical in the cure, when prism is in minimum deviation position. Hence in this condition 2 sin sin A i  =        = 2 sin sin A i   o o i i 45 2 1 sin = 1.414  sin 30 =  = 35. (a) By using sin 0.75 48 36' sin 30 sin 1.5 sin sin o i i i A i  =  =  =  = Also from 48 36' 30 18 36' o o o  = i − A   = − = 36. (c) This is the case when light ray is falling normally an second surface. Hence by using A i sin sin  =  o o i i i 45 2 1 sin 2 sin 30 sin 2 =  =   = 37. (d) Given that o A = 60 and o i e A 60 45 4 3 4 3 = = =  = By using i + e = A +   o 45 + 45 = 60 +    = 30 38. (a)  = ( − 1)A as A is halved, so  is also halves 39. (c) o o ism ( 1)A (1.5 1)4 2  Pr =  − = − =   Total =  Prism +  Mirror o o = ( − 1)A + (180 − 2i) = 2 + (180 − 2  2) = 178 40. (d) If  = maximum value of vase angle for which light is totally reflected from hypotenuse. (90 −) = C = minimum value of angle of incidence an hypotenuse for TIR   1 sin(90 − ) = sin C =          = −   1 cos 1 I1 I Water Air I1 I2 S1 S2 x (24 – 4) 24 cm y