An isotropic point source having radiation of light of power $P$ is located on the axis of an ideal mirror plate. The distance between the source and the plate exceeds the radius of the plate $\eta$-fold. Find the force that the light radiation exerts on the plate.

Intensity of photons at distance of $r$ from source,

$I=\frac{P}{4{\mathrm{\pi r}}^2}$

If radius of the mirror is $R$ and distance of the point source from the mirror is $l$, then according to the question, $\frac{l}{R}=n$,

Momentum per second of photons striking the mirror in normal direction,

${\left(\frac{dp}{dt}\right)}_{n,incident}=\frac{I}{c}dA{\cos }^2\theta$

Momentum per second of reflected photons,

${\left(\frac{dp}{dt}\right)}_{n,reflected}=-\rho \frac{I}{c}dA{\cos }^2\theta$

where $\rho$ is reflection coefficient.

Hence rate of change of momentum of photons,

${\left(\frac{dp}{dt}\right)}_{photons}=-\left(1+\rho \right)\frac{I}{c}dA{\cos }^2\theta$

From Newton's third law, rate of change of momentum of mirror,

${\left(\frac{dp}{dt}\right)}_{mirror}=\left(1+\rho \right)\frac{I}{c}dA{\cos }^2\theta$

Since mirror plate is perfect reflector, $\rho =1$ and force on its surface is

$dF=\frac{2I}{c}2\pi \left(ltan\theta \right)\left(lse{c}^2\theta d\theta \right)co{s}^2\theta$

[as $y=l\tan \theta , dy=lse{c}^2\theta d\theta$]

$=\frac{2}{c}\left(\frac{P}{4\pi r^2}\right)2\pi l^{2}tan\theta d\theta$      [as $I=\frac{P}{4{\mathrm{\pi r}}^2}$]

$=\frac{P}{c}\sin \theta \cos \theta d\theta$      [as $\frac{l^2}{r^2}={\cos }^2\theta$]

$\therefore F={\int }_0^{{\theta }_0}\frac{P}{c}\sin \theta \cos \theta d\theta$

$=\frac{P}{2c}{\sin }^2{\theta }_0$

$=\frac{P}{2c}\times \left(\frac{R^2}{R^2+l^2}\right)=\frac{P}{2c}\left(\frac{1}{1+n^2}\right)$