Physics · Atomic Physics

JEE Main 2025 — 4 April, Morning Shift — Question 52

A small mirror of mass $m$ is suspended by a massless thread of length $I$ . 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)

Answer: A

Momentum of photon $=\frac{h}{\lambda}=\frac{E}{C}$ $P-\frac{E}{C}=\frac{E}{C}$

$\mathrm{KE}_{i}=\mathrm{PE}_{f} \quad P=\frac{2 E}{c}$

$\frac{\left(\frac{2 E}{c}\right)^{2}}{2 m}=m g l(1-\cos \theta)$

$\frac{4 E^{2}}{2 c m}=m g l \frac{\theta^{2}}{2}$

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

Answer key and solution verified before publishing.

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