Question

A flat mirror revolves at a constant angular velocity making $n=0.4$ revolutions per second. With what velocity (in $\text{ms}^{-1}$) will a light spot move along a spherical screen with a radius of 15 meters, if the mirror is at the centre of curvature of the screen

• A. 37.7
• B. 60.3
• C. 68.7
• D. 75.4
• E. 90.4

Hint:

This is a classic "Optical Lever" principle: any rotation of the mirror is doubled in the reflected beam.

Answer 1

The Correct Option is D

Concept:
When a mirror rotates by an angle $\theta$, the reflected ray rotates by an angle $2\theta$. Therefore, the angular velocity of the reflected ray ($\omega_{ray}$) is twice the angular velocity of the mirror ($\omega_{mirror}$). \[ \omega_{ray} = 2\omega_{mirror} \]

Step 1: Calculate the angular velocity of the mirror.
$n = 0.4$ rev/s. \[ \omega_{mirror} = 2\pi n = 2 \times \pi \times 0.4 = 0.8\pi \text{ rad/s} \]

Step 2: Calculate the velocity of the light spot.
The light spot moves along a circle of radius $R = 15$ m. \[ v = R \times \omega_{ray} = R \times (2\omega_{mirror}) \] \[ v = 15 \times (2 \times 0.8\pi) = 15 \times 1.6\pi = 24\pi \] \[ v = 24 \times 3.14159 \approx 75.398 \text{ ms}^{-1} \]