Question

A light of intensity $8{\text{kWm}}^{-2}$ falls on a plane mirror with reflection coefficient $r=0.95$. The angle of incidence is ${60}^{\circ }.$ The pressure exerted by the light on the mirror is

• A. $1.3\times {10}^{-5}{\text{Nm}}^{-2}$
• B. $0.3\times {10}^{-5}{\text{Nm}}^{-2}$
• C. $30\times {10}^{-5}{\text{Nm}}^{-2}$
• D. $1.3\times {10}^{-6}{\text{Nm}}^{-2}$

Answer 1

The correct option is A $1.3\times {10}^{-5}{\text{Nm}}^{-2}$


Given,

$I=8{\text{kW m}}^{-2};\theta ={60}^{\circ };r=0.95$

Due to absorption and reflection, the rate of change of momentum of the surface of the mirror is,

$F=(\frac{I}{c}\cos \theta +r\frac{I}{c}\cos \theta )A$

$A=\text{Area of the of beam.}c=\text{Speed of light in free space}$

The pressure exerted on the surface of mirror is,

$P=\frac{(1+r)\frac{I}{c}A\cos \theta }{\frac{A}{\cos \theta }}=(1+r)\frac{I}{c}{\cos }^2\theta$

$=\frac{(1+0.95)\times 8\times {10}^3}{3\times {10}^8}\times {\cos }^2{60}^{\circ }$

$=\frac{1.95\times 8\times {10}^3}{3\times {10}^8}\times {\left(\frac{1}{2}\right)}^2$

$=1.3\times {10}^{-5}{\text{Nm}}^{-2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question? Key Concept: Pressure applied is always calculated normal to the plane. So, always use the normal component of the area of beam incidence on the surface.