**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°**