A small mirror of mass m is suspended by a ma | vedclass

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(D) The force exerted by the laser pulse on the mirror,assuming complete reflection,is given by $F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt}$,where $

Vedclass · Accepted Answer

$	heta=\frac{2 E }{ mc \sqrt{ gl }}$

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(D) The force exerted by the laser pulse on the mirror,assuming complete reflection,is given by $F = \frac{2P}{c} = \frac{2}{c} \frac{dE}{dt}$,where $P$ is the power.Integrating this over the duration of the pulse,the change in momentum of the mirror is:$mv = \int F dt = \frac{2}{c} \int dE = \frac{2E}{c} \implies v = \frac{2E}{mc}$.Now,using the work-energy theorem for the pendulum:$W_g = \Delta K$$-mg l(1 - \cos 	heta) = 0 - \frac{1}{2}mv^2$$mg l(2 \sin^2 \frac{	heta}{2}) = \frac{1}{2}mv^2$Since $	heta$ is small,$\sin 	heta \approx 	heta$,so $2 \sin^2 \frac{	heta}{2} \approx 2(\frac{	heta}{2})^2 = \frac{	heta^2}{2}$.Substituting this into the energy equation:$mg l \frac{	heta^2}{2} = \frac{1}{2} m \left(\frac{2E}{mc}ight)^2$$mg l 	heta^2 = \frac{4E^2}{m c^2}$$	heta^2 = \frac{4E^2}{m^2 c^2 g l}$$	heta = \frac{2E}{mc \sqrt{gl}}$

$A$ small mirror of mass $m$ is suspended by a massless thread of length $l$. Find the small angle through which the thread will be deflected when a short pulse of laser of energy $E$ falls normally on the mirror. ($c=$ speed of light in vacuum and $g=$ acceleration due to gravity)

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