A parallel beam of monochromatic light of wavelength $663nm$ is incident on a totally reflecting plane mirror. The angle of incidence is ${60}^o$ and the number of photons striking the mirror per second is $1.0\times {10}^{19}$. Calculate the force exerted by light beam on the mirror.

Force exerted = rate of change of momentum

Since light ray is incident at the angle 60 degree so there is two component of the momentum

Initial momentum of horizontal direction$=P\sin \theta$

Initial momentum of vertical direction$=P\cos \theta$

Final momentum of horizontal direction$=P\sin \theta$

Final momentum of vertical direction$=-P\cos \theta$

We include negative sign because direction of the final momentum is opposite

Change in momentum in horizontal direction $=P\sin \theta -P\sin \theta =0$

Change in momentum in vertical direction $=P\cos \theta -(-P\sin \theta )=2P\cos \theta$

We know relation between wave length and momentum is given as

$\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

$Force=\frac{Changeinmomentum}{Timetaken}$

$Force=Numberofphotonspersecond\times Changeinmomentum$

$Force={10}^{19}\times \frac{6.6\times {10}^{-34}}{6.63\times {10}^{-9}}\cos 60$

$Force={10}^{-8}N$