A perfectly reflecting mirror has an area of 1 cm2. Light energy is allowed to fall on it for 1 h at the rate of 10 W cm-2. The force that acts on the mirror is:
Answer & explanation
Correct answer: option 2
Let, E = energy falling on the surface per second = 10 J
Momentum of photons, $p=\frac{h}{\lambda}=\frac{h}{(c / v)}=\frac{h v}{c}=\frac{E}{c}$
On reflection, change in momentum per second = force
= 2 p = $\frac{2 E}{c}=\frac{2 \times 10}{3 \times 10^8}=6.7 \times 10^{-8} N$