Question

A small mirror of mass $m$ is suspended by a massless thread of length $l$. Then the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normal on the mirror ( $c=$ speed of light in vacuum and $g=$ acceleration due to gravity).

• A. $\theta=\frac{3 \mathrm{E}}{4 \mathrm{mc} \sqrt{g l}}$
• B. $\theta=\frac{\mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$
• C. $\theta=\frac{\mathrm{E}}{2 \mathrm{mc} \sqrt{\mathrm{gl}}}$
• D. $\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}$

Hint:

Use the work-energy theorem to find the deflection angle.

Answer 1

The Correct Option is D

1. Force due to the laser beam: \[ F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt} \]
2. Change in momentum of the mirror: \[ m(V - 0) = \int F dt = \frac{2}{c} \int dE = \frac{2E}{c} \]
3. Using work-energy theorem: \[ W_g = \Delta K \] \[ -\frac{mg l (1 - \cos \theta)}{l} = \frac{1}{2} m V^2 \] \[ \frac{g l (1 - \cos \theta)}{l} = \frac{1}{2} \left( \frac{4E^2}{m^2 c^2} \right) \] \[ \frac{g l \theta^2}{2} = \frac{2E^2}{m^2 c^2} \] \[ \theta = \frac{2E}{mc \sqrt{gl}} \] Therefore, the correct answer is (4) $\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{gl}}}$.