Solution:
Given,
Radius of curvature (R) = 40 m,
Frequency (f) = 2600 rev/s,
Angle (2θ) = ?,
Velocity of light (c) = 3 x 10^8 m/s,
We have,
\(\theta=\frac{4\pi fR}c=\frac{4\pi\times2600\times40}{3\times10^8}=4.35\times10^{-3}\) rad
= \(\left(4.35\times10^{-3}\times\frac{180}{\mathrm\pi}\right)^\circ=0.25^\circ\)
Then the angle between the incident and the reflected ray is
2 x θ = 0.25 x 2 = 0.5°