Q. 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) $5.12 \times 10^{- 8}N$
Explanation
(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
$= p_{2} - \left( - p_{1} \right) = p_{2} + p_{1} = (2 + 1.2) \times 10^{- 7} = 3.2 \times 10^{- 7}$N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.