Question

A light of wavelength $650\text{nm}$ falls on a surface at an angle ${60}^{\circ }$ and bounces back at the same angle from the normal. If the reflectivity of the surface is $50\%$ and intensity of light is $100{\text{W/m}}^2$, find the pressure experienced by the surface due to the reflected radiation.

• A. $1.67\times {10}^{-9}{\text{N/m}}^2$
• B. $1.67\times {10}^{-2}{\text{N/m}}^2$
• C. $1.67\times {10}^{-5}{\text{N/m}}^2$
• D. $1.67\times {10}^{-7}{\text{N/m}}^2$

Answer 1

The correct option is D $1.67\times {10}^{-7}{\text{N/m}}^2$

$\lambda =650\text{nm}$
$\theta ={60}^{\circ }$
$r=50\%=0.5$
$I=100{\text{W/m}}^2$

Pressure experienced by the surface is,

$P=\frac{F}{A}=\frac{\left(2p\cos \theta \right)}{A}$

As, $p=\frac{E}{c}=\frac{IA}{c}$

$\therefore P=\frac{0.5\times I}{c}\times 2\cos \theta$

$=\frac{0.5\times 100}{3\times {10}^8}\times 2\times \cos {60}^{\circ }$

$=1.67\times {10}^{-7}{\text{N/m}}^2$

Hence, $\left(A\right)$ is the correct answer.