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 $60 \, W$ 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
$=p_{2}-\left(- p_{1}\right)=p_{2}+p_{1}=\left(2 + 1 .2\right)\times \left(10\right)^{- 7}=3.2\times \left(10\right)^{- 7} \, N$