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 is: ($c$ = speed of light in vacuum and $g$ = acceleration due to gravity)
Answer: (D) $\theta = \dfrac{2E}{mc\sqrt{gl}}$
Light of energy $E$ carries momentum $E/c$. On reflection from the mirror its momentum reverses, so the mirror receives an impulse $\dfrac{2E}{c}$:
$$mv = \frac{2E}{c} \Rightarrow v = \frac{2E}{mc}$$
The mirror then swings up like a pendulum bob: $\tfrac12mv^2 = mgl(1 - \cos\theta) \approx mgl\dfrac{\theta^2}{2}$ for small $\theta$, so
$$v = \theta\sqrt{gl} \Rightarrow \theta = \frac{2E}{mc\sqrt{gl}}$$
Solution by Sreeraj P, M.Sc Physics