Solution:
Given,
Radius of the curvature (R) = 200 m,
Rotation fomirror (f) = 20 rev/s,
Velocity of the light (c) = 3 x 10^8 m/s,
Angle (θ) = ?
We have,
\(\theta=\frac{4\pi fR}c=\frac{4\pi\times20\times200}{3\times10^8}=1.67\times10^{-4}\) radian
= \(\left(9.57\times10^{-3}\right)^\circ\)
Now the angle between the incident and the reflected ray is
\(=2\theta=2\times9.5\times10^{-3}=0.0192\) Degree