A plane light wave of intensity I = 50 Wcm^-2 falls on a plane mirror surface with reflection coefficient $\mathrm{\rho }=0.8$. The angle of incidence is 45^o. In terms of corpuscular theory, find the magnitude of the normal pressure exerted on that surface (in $\times {10}^{-3} {\mathrm{Nm}}^{-2}$). 

Momentum incident per second corresponding to incident photons, normal to the every infinitesimal area $dA$ of the surface, 

           ${\left(\frac{dp}{dt}\right)}_{\text{incident }}=\frac{I\left(dA\right)}{c}co{s}^2\theta$,  where $\theta =45°$ is the angle of incidence.

Since refection coefficient is $\mathrm{\rho }$, so the momentum of the reflected photons per second normal to surface, for every infinitesimal area $dA$,

           ${\left(\frac{dp}{dt}\right)}_{\text{reflected }}=\rho \frac{I\left(dA\right)}{c}co{s}^2\theta$

Hence, rate of change of momentum of the photons,

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

From Newton's third law, $F={\left(\frac{dp}{dt}\right)}_{\text{surface }}=\left(1+\rho \right)\frac{I\left(dA\right)}{c}co{s}^2\theta$

Hence, pressure exerted on surface, $P=\frac{dF}{dA}=\left(1+\rho \right)\frac{I}{c}co{s}^2\theta$

On substituting values, we get $P=(1+0.8)\frac{\left(50\times {10}^4\right)}{\left(3\times {10}^8\right)}\times \frac{1}{2}=1.5\times {10}^{-3} {\mathrm{Nm}}^{-2}$