SOLUTIONS TO CONCEPTS CHAPTER 21 1.
In the given Fizeau’ apparatus, 3 D = 12 km = 12 × 10 m n = 180 8 c = 3 × 10 m/sec 2Dn We know, c = c c 180 = rad/sec = deg/sec 2Dn 2Dn =
2.
180 3 10 8
4
= 1.25 × 10 deg/sec 24 10 3 180 In the given Focault experiment, R = Distance between fixed and rotating mirror = 16m = Angular speed = 356 rev/ = 356 × 2 rad/sec b = Distance between lens and rotating mirror = 6m a = Distance between source and lens = 2m –3 s = shift in image = 0.7 cm = 0.7 × 10 m So, speed of light is given by,
4 16 2 356 2 2 4R 2 a 8 = = 2.975 × 10 m/s s(R b) 0.7 10 3 (16 6) In the given Michelson experiment, 3 D = 4.8 km = 4.8 × 10 m N=8 D N We know, c = 2 C=
3.
=
2c c 3 10 8 3 rad/sec = rev/sec = = 7.8 × 10 rev/sec DN DN 4.8 10 3 8
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21.1