Tiêu đề 28. Ray Optics Med Ans.pdf ✅

1. (a): The light is an electromagnetic wave of wavelength range 390 nm – 700 nm belonging to the visible part of the spectrum. 2. (c): Let f be focal length of the convex mirror. According to new Cartesian sign convention object distance, u f =− , focal length = + f According to mirror formula 1 1 1 u v f + = 1 1 1 f v f + = − of 1 1 1 2 v f f f = + = of f v 2 = The image is formed at a distance f 2 behind the mirror It is a virtual image. Magnification, , v (f / 2) 1 m u ( f ) 2 = − = − = − Also, 1 O Height of image(h ) m Height of object(h ) = I O 1 h mh (1m) 0.5m 2  = = = 3. (c): For spherical mirror, R f 2 = Here, R 35.0cm = 35 f 17.5cm 2  = = Now, 1 1 1 f v u = + Also, magnification v m u − = ro v = - mu 1 1 1 f u mu  = − or 1 u f 1 m   = −     1 17.5 1 17.5 (06) 10.5cm 2.5   = − =  =     4. (b): According to mirror formula, 1 1 1 u v f + = Differentiating with respect to t, we get 2 2 1 du 1 dv 0 ( f iscostan t) u v dt dt − − = or 2 dv v du dt u dt   = −    2 i 0 i o v dv du v v v and v u dt dt     = − = =         Substituting thegiven values, we get 2 1 i 10 v 9 1ms 30   − = −  = −     1 1 v 1ms− = 5. (a):10 cm 6. (a): Here, h 2cm,u 16cm 1 = = − 2 h 3cm = − (Since image is real and inverted 2 1 h v m h u −  = = 2 1 h 3 v u ( 16) 24cm h 2 −  = =  − = − Now, 1 1 1 1 1 2 3 5 f v u 24 16 48 48 − − − = + = − − = = 48 f 9.6cm 5 − = = − 7. (d): Here, f = - 10 cm For end A, Au 20cm = − Image position of end A,, A A 1 1 1 v u f + = A 1 1 1 v ( 20) ( 10) + = − − or A 1 1 1 1 v ( 10) (20) 20 = + = − − For end B,u 30cm B = − Image position of end B, B B 1 1 1 v u f + = B 1 1 1 v ( 30) ( 10) + = − − or B 1 1 1 2 v 10 30 30 = + = − − v 15cm B = − Length of the image A B = − = − = v v 20cm 15cm 5cm 8. (a): As shown in figure glass slab will form the image of bottom i.e., mirror MM’ at a depth   d      from its front face. So the distance of object O from virtual mirror mm’ will be d h   +      Now as a plane mirror forms image behind the mirror at the same distance as the object is in front of it, the distance of image I from mm’ will b d h   +      and as the distance of virtual mirror from the front face of slab is
  d      the distance of image I from front face as seen by observer will be d d 2d h h   = + + = +        9. (c): Since the refractive index is less at the beam boundary, the ray at the edges of the beam move faster compared to the axis of beam. Hence, the beam converges. 10. (b): o a g o sin 60 sin 35  = ...(i) o a w o sin 60 sin 41  = ...(ii) o a g sin 41 sin  =  ...(iii) a a a    =  g g g o o o o o sin 60 sin 41 sin 60 sin 41 sin 35 sin  =  (Using (i), (ii) and (iii)) o sin sin35  =o = 35 11. (a): As refractive index, Realdepth Apparent depth  =  Apparent depth of the vessel when viewed from above is apparent 1 1 2 x x x 1 1 d 2 2 2   = + = +         2 1 1 2 1 2 1 2 x x( ) 2 2    +   +  = =         12. (d): Let 1 x be apparent depth of the dot when seen from air. 1 1 h / 3 x  = (Here, h/3 is real depth of the dot under liquid of density 1 d ) 1 1 1 x 3  =  Similarly, apparent depths of the dot when seen from air through two other liquids are 2 2 h x 3 =  and 3 3 h x 3 =   Apparent depth of the dot 1 2 3 = + + x x x 2 3 1 2 3 1 1 h h 1 1 1 3 3 3 3   = + + = + +           13. (c): Here, o i 45 = Applying snell’s law at air – glass surface, we get a g  =  sini sinr' 1sini 2 sinr' = 1 sin r ' sin i 2 = 1 o sin 45 2 = 1 sin r ' 2  =1 o 1 r ' sin 30 2 −   = =     From figure, o r r' 180 +  + =o o i 30 180 ( i r) +  + = = o o o 45 30 180 +  + = o o o  = − = 180 75 105 Hence, the angle between reflected and refracted rays is o 105 14. (c): From figure, in right angled CDB  = − CBD (i r)CD d sin(i r) BC BC  − = = or d = BC sin (i – r) ...(i) Also, in right angled CNB BN t cosr BC BC = =
t BC cosr = ...(ii) Substitute equation (ii) in equation (i), we get t d sin(i r) cos r = − For small angles sin (i r) i r;cosr 1 −  −  r d t(i r),d it 1 i   = − = −     15. (c): Actual depth of the needle in water h 15.5cm 1 = Apparent depth of needle in water h 8.5cm 2 = 1 water 2 h 15.5 1.82 h 8.5  = = = Hence,  = water 1.82, when water replaced by a liquid of refractive index ' = 1.94. The actual depth remains the same, but its apparent depth changes. Let H be the new apparent depth of the needle 1 h ' H  = or 1 h 15.5 H 7.98cm ' 1.9 = = =  Here, H is less than 2 h Thus to focus the needle again, the microscope should be moved up. distance by which the microscope should be moved up = − = 8.5 7.98 0.52cm 16. (c): Refraction at P, o 1 sin 60 3 sin r = 1 1 sin r 2 = or o 1 r 30 = Since, 2 1 r r = o 2  = r 30 Refraction at 2 2 sin r 1 Q, sin i 3 = or o 2 sin 30 1 sin i 3 = Or o 2 i 60 = At point ' o Q,r r 30 2 2 = = o ' o o o o 2 2  = − + = − + = 180 (r i ) 180 (30 60 ) 90 17. (c): Total internal reflection of light 18. (c): Greater than 2 19. (d): Here, 8 1 Av 1.8 10 ms− =  8 1 Bv 2.4 10 ms− =  Light travels slower in denser medium. Hence medium A is a denser medium and medium B is a rarer medium, Here, light travels from medium A to medium B. Let C be the critical angle between them. A B B A 1  =  = sin C  Refractive index of medium B w.r.t. to medium A is A A B B Velocityof light in medium A v Velocityof light in medium B v  = = 8 A 8 B v 1.8 10 3 sin C v 4 2.4 10   = =  or 1 3 C sin 4 −   =     20. (a): The figure shows incidence from water at critical angle c for the limiting case. Now, sin c 1  =  So that c 2 1/2 1 tan ( 1)  =  − From figure, c r tan h  = where r is the radius of disc. Therefore, diameter of the disc is c 2 1/2 2h 2r 2h tan ( 1) =   − 21. (b): 2 1 v sin v'   =  where v and v’ are the speeds of light in medium (i) and medium (ii) respectively. v v' sin =  22. (b): In total internal reflection, light travel from denser to rarer medium. 23. (b): Since refraction occurs from denser to rerer medium, Therefore, 2 1 1 2 u v R    −  −+=
3 1 3 1 1 2 2( 3) v 5 10 − − + = = − − or 1 1 1 4 v 10 2 10 = − = −  = − v 2.5cm 24. (a): Here, f 0.2m,v 0.3m = − = + The lens formula, 1 1 1 v u f − = 1 1 1 1 1 0.5 u v f 0.3 0.2 0.06  = − = + = 0.06 u 0.12m 0.5 = = 25. (b): The focal length of the plano – convex lens is 1 1 1 1 (1.5 1) f 10 20   = − − =     +  Focal length of plano –concave lens is 1 1 1 1 (1.5 1) f 10 20   − = − − =      Since parallel beams are incident on the lens, its image from plano – concave lens will be formed at + 20 cm from it (at the focus) and will act as an object for the plano – concave lens. Since the two lens are at a distance of 10 cm from each other, therefore, for the next lens u = + 10 cm. uf 10 20 v 20cm u f 10 20   = = = + − 26. (d): Here, P 10D 1 = and P 5D 2 =− Therefore, power of the combined lens is P P P 10 ( 5) 5D = + = + + − = + 1 2 Now, magnification, f m u f = + Here, m = 2 and 1 1 f 0.2m 20cm P 5 = = = = 20 2 u 20 10 u 20  =  + = +  = − u 10cm 27. (b) Using, 1 2 2 1 u v R    −  − + = Here, 1 2 4 R 20cm, , 1.60 3 =  =  = u 200cm = − 4 / 3 1.60 1.60 (4 / 3) 200 v 20 − − + = −  = v 240m 28. (a): Using 1 2 1 1 1 ( 1) f R R   =  − −     Here, R 20cm.R 20cm,u 10cm 1 2 = = − = − and =1.56 1 1 1 (1.56 1) f 20 20   = − +     1 0.56 2 20 f 17.86cm f 20 0.56 2  =  = =  Now, from lens equation, uf 10 17.86 v 22.86cm u f 10 17.86 −  = = = − + − + Since v is negative, the image will be formed on the same side as that of object. 29. (d): The far point of 6.0 m tell us that the focal length of the lens is f 6.0m,u 18m = − = − and h 2m = Using, 1 1 1 1 1 1 1 1 f v u v f u 6.0 18.0 = −  = + = − −  = − v 4.5m  The image size, h’ = h v 4.5 2 0.50m u 18.0     − − =  − =         30. (b): Power of the lens combination is 1 2 1 2 1 1 1 1 P P P f (in m) f (in m) 0.40m 0.25m = + = + + + + − 1 1.5m 1.5D − = − = − 31. (b): Here, u 25cm,v 50cm = − = − 1 1 1 1 1 1 f 50cm f v u 50 25 50 = − = + =  = + − 32. (d): Magnification for f m f u − = − Here, 2 f 10cm 10 10 m, − = =  11 u 1.5 10 m =  2 13 2 11 10 10 m 6.67 10 10 10 1.5 10 − − − −   = =   −   Diameter of the image 9 d m 1.39 10 =  13 9 4 6.67 10 1.39 10 9.27 10 m − − =    =  33. (a): Area of a square card = 1 mm ×1 mm = 1 2 mm Focal length of magnifying lens (converging lens), F = + 10 cm Object distance, u = - 9 cm According to thin lens formula