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 ( \(\mathrm{c}=\) speed of light in vacuum and \(\mathrm{g}=\) acceleration due to gravity)

Force due to beam assuming complete reflection \(\mathrm{F}=\frac{2 \mathrm{P}}{\mathrm{C}}=\frac{2}{\mathrm{C}} \frac{\mathrm{dE}}{\mathrm{dt}} ; \mathrm{P}\) is power
So change in momentum of mirror.
\(\mathrm{m}(\mathrm{~V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}} \int \mathrm{dE}=\frac{2 \mathrm{E}}{\mathrm{C}}\)
Now using work energy theorem ...(1)

\(\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \\
& -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \\
& \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2}
\end{aligned}\)
as \(\theta\) is small
\(\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} \ell}}\end{aligned}\)