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 mir

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

TestHub · Accepted Answer

Force due to beam assuming complete reflection $F = \frac{2 P}{C} = \frac{2}{C} \frac{dE}{dt}; P$ is power
So change in momentum of mirror.
$m (V - 0) = \int Fdt = \frac{2}{C} \int dE = \frac{2 E}{C}$
Now using work energy theorem ...(1)
$W_{g} = Δ k - mg ℓ (1 - cos θ) = 0 - \frac{1}{2} mv^{2} g ℓ (2 sin^{2} \frac{θ}{2}) = \frac{v ^{2}}{2}$
as $θ$ is small
$g ℓ 2 (\frac{θ}{2})^{2} = \frac{1}{2} \frac{4 E ^{2}}{m ^{2} c ^{2}} g ℓ θ^{2} = \frac{4 E ^{2}}{m ^{2} c ^{2}} θ = \frac{2 E}{mc g ℓ}$

Physics - Modern Physics Question with Solution | TestHub

Options:

Doubts & Discussion