Optics-09-Answersheet

1. (c) 2. (d) 3. (d) 4. (b) 5. (b) 6. (d) 7. (c) 8. (b) 9. (b) 10. (a) 11. (d) 12. (a) 13. (a) 14. (c) 15. (d) 1. (c) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a) 7. (a) 8. (c) 9. (c) 10. (c) 11. (c) 12. (a) 13. (d) 14. (a) 15. (c) 1. (i) The image is –30 cm from mirror, -3 , magnified, real and inverted. (ii) The image is 15 cm at the block of the mirror, 3, magnified, virtual and errect. 2. 54 cm is fornt of the mirror, 54, inverted and real, - 5cm, screen should move away from mirror upto, when object is at –18 cm from mirror. 3. V = 6.66 cm; 0.55 cm, image is errect and virtual; progressively diminished in size. 4. 2.8 cm2. 5. 5 cm; the location of slab will not effect the answer. 6. sin1 and sin1 0.375 7. 4.89 cm from the bottom 8. 2.58 cm2 9. i  sin1 (0.35) 10. R   11. 5000 Å, 6 1014 Hz, 45 12. 400 nm, 51014 Hz, 2 108 ms1 13. 600 nm 14. (a) Reflected light (wavelength, frequency, speed same as incident light), λ  589nm, f  5.09 1014 Hz, c  3108 ms1 (b) Reracted light (frequency same as incident frequency) f  5.09 1014 Hz, f  2.26 108 ms1 λ  444 nm 1. 2 cos 15º 2. 60 cm 3. R 2 (μn  μ  1) f  x , R  x1 x2 , μ  x1 4. 1 x  x x  x 1 2 1 2 x  d  (d  f1 ) f2 , d  f1  f2 y  (d  f1 ) d  f1  f2 7. Final image will formed at optical center of L1 8. μ  1.382 9. (a) 0.1 m (b) 0.2 mm (c) 51 (d) 5001 (e) 2501 UNSOLVED SUBJECTIVE LEVEL - III 1. 52 cm 9 2. Image will coincide with object f  n2 R1R2 , f  n1R1R2 f1  f2  1 3. 2 n(R  R )  (n R  n R ) 1 n(R  R )  (n R  n R ) a b 4. 15 cm 13 5. 100 cm 19 1 2 1 2 2 1 1 2 1 2 2 1 6. 17.1cm 8. (a)0.3276 mm (b) 1.2 cm (c) 0.7 mm 3 10. 4 PROBLEMS 1. (a) AB  1.8cm , M = – 1.5 (b) tmin  90 nm 2. (a) 4 .33 mm (b) 0.75 (c) 650 nm and 433.3 nm. 3. ˆr  2 (3ˆi  4ˆj  5kˆ) 10 4. (a) 0  60 nm (b) sin1 1.5 i  2  42.8o 5. (a) 4 minima at positions 0.258 m and 1.13m (b) y  & 6. t  9.3106 m 7. (a) m   sin1  1    (b) K1/2x  4y1/4 (c) P(x1y1)  P(4, 1) (d)   90o .