Question

A parallel beam of light is incident normally on a plane surface absorbing $40\%$ of the light and reflecting the rest. If the incident beam carries $60W$ of power, the force exerted by it on the surface is

• A. $3.2\times 10^{-8}N$
• B. $3.2\times 10^{-7}N$
• C. $5.12\times 10^{-7}N$
• D. $5.12\times 10^{-8}N$

Answer 1

Answer: (B)

Momentum of incident light per second
$p_1=\frac{E}{c}=\frac{60}{3\times 10^8}=2\times 10^{-7}$
Momentum of reflected light per second
$p_2=\frac{60}{100}\times \frac{E}{c}=\frac{60}{3\times 10^8}=1.2\times 10^{-7}$
Force on the surface = change in momentum per second
$=p_2-\left(-p_1\right)=p_2+p_1=\left(2+1.2\right)\times {\left(10\right)}^{-7}=3.2\times {\left(10\right)}^{-7}N$