Q. A perfectly reflecting mirror has an area of 1 cm$^{2}$ Light energy is allowed to fall on it for 1h at the rate of 10 Wcm$^{-2}$. The force that acts on the mirror is
A) $3.35\ \times \ 10^{- 8}N$
B) $6.7\ \times \ 10^{- 8}N$
C) $1.34\ \times \ 10^{- 7}N$
D) $2.4\ \times 10^{- 4}N$
Explanation
(b) Let $E =$ Energy falling on the surface per second = 10 J
Momentum Of photons
$p = \frac{h}{\lambda}\ \ \ = \frac{h}{\left( c_{1}v \right)}$
$= \ \frac{hv}{c}\ = \ \frac{E}{C}$
On reflection,
Change in momentum per second $= 2p = \frac{2E}{c}$
We know that,
Change in momentum per second = force
$F = \frac{2E}{c}\ = \ \frac{2\ \times \ 10}{3\ \times \ 10^{8}}$
$= \ 6.7\ \times \ 10^{- 8}\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.