Question 65: Dual Nature of Matter and Radiation

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

Accepted Answer

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)
$$\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \
& -\mathrm{mg} \ell(1-\cos 	heta)=0-\frac{1}{2} \mathrm{mv}^2 \
& \mathrm{~g} \ell\left(2 \sin ^2 \frac{	heta}{2}ight)=\frac{\mathrm{v}^2}{2}
\end{aligned}$$
as $	heta$ is small
$$\begin{aligned}
& \mathrm{g} \ell 2\left(\frac{	heta}{2}ight)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \quad 	ext{(from eq. (1))}\
& \mathrm{~g} \ell 	heta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \
& 	heta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{~g} \ell}}
\end{aligned}$$

Answer

$	heta=\frac{E}{m c \sqrt{g l}}$

Answer

$	heta=\frac{E}{2 m c \sqrt{g l}}$

Answer

$	heta=\frac{3 E}{4 m c \sqrt{g l}}$

Answer

$	heta=\frac{2 E}{m c \sqrt{g l}}$

Dual Nature of Matter and Radiation

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 g= acceleration due to gravity)

Choose the correct answer: