Question 100 of 200

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)

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

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Answer: (D) $\theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} l}}$

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
$\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} \ell}}\end{aligned}$