Solution
Given,
Speed of light (c) = 3 x 10^8 m/s,
No. of face of mirror (m) = 8,
No . of rotation per sec (n) = ?,
Distance traveled (D) = 4.8 km
∴ The distance between two mirrors, \(d=\frac D2=\frac{4800}2=2400\) m
We have,
c = 2mnd
Or, \(3\times10^8=2\times8\times n\times2400\)
Or, n = 7812.5 rev/s
Hence the frequency of rotation of the mirror where the image is formed is 7812.5 rev/s