A small mirror of mass m is suspended by a massless thread of length . 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 du

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Solution

Q.
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) [2025]

1 $θ = \frac{3 E}{4 m c \sqrt{g l}}$

2 $θ = \frac{E}{m c \sqrt{g l}}$

3 $θ = \frac{E}{2 m c \sqrt{g l}}$

4 $θ = \frac{2 E}{m c \sqrt{g l}}$

Ans.
(4)

Force due to beam assuming complete reflection

$F = \frac{2 P}{c} = \frac{2}{c} \frac{d E}{d t}$ ; P is power

So change in momentum of mirror.

$m (v - 0) = \int F d t = \frac{2}{c} \int d E = \frac{2 E}{c}$ ... (i)

Now using work energy theorem

$W_{g} = △ k$

$- m g l (1 - \cos θ) = 0 - \frac{1}{2} m v^{2}$

$g l (2 \sin^{2} \frac{θ}{2}) = \frac{v^{2}}{2}$

as $θ$ is small

$g l 2 {(\frac{θ}{2})}^{2} = \frac{1}{2} \frac{4 E^{2}}{m^{2} c^{2}}$ from eq. (i)

$g l θ^{2} = \frac{4 E^{2}}{m^{2} c^{2}} \Rightarrow θ = \frac{2 E}{m c \sqrt{g l}}$

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