JEE Main 2025 Apr 4 Shift 1, Physics Q2: Introduction and Photoelectric Effect

JEE Main2025Apr 4, Shift 1Physics

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

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

A
$θ = \frac{3 E}{4 m c g l}$
B
$θ = \frac{E}{m c g l}$
C
$θ = \frac{E}{2 m c g l}$
D
$θ = \frac{2 E}{m c g l}$

Solution

Step 1 — momentum delivered to the mirror. For a photon pulse of energy $E$ reflecting normally, the momentum change of the radiation is $2 E / c$ , so by Newton's third law the impulse delivered to the mirror is also $2 E / c$ . The mirror acquires initial speed $v$ with $m v = \frac{2 E}{c}$ , hence $v = \frac{2 E}{m c}$ .

Step 2 — energy conservation for the pendulum swing. The kinetic energy $\frac{1}{2} m v^{2}$ converts to gravitational PE at the maximum deflection: $\frac{1}{2} m v^{2} = m g l (1 - cos θ)$ .

For small $θ$ , $1 - cos θ \approx \frac{θ ^{2}}{2}$ , giving $v^{2} = g l θ^{2}$ .

Step 3 — solve for $θ$ . $θ = \frac{v}{g l} = \frac{2 E}{m c g l}$ .

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JEE Main 2025 Apr 4 Shift 1, Physics Q2: Introduction and Photoelectric Effect

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