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.

=

2c 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