A perfectly reflecting mirror has an area of $1c{m}^2$. Light energy is allowed to fall on it for $1h$ at the rate of $10Wc{m}^{-2}$. The force that acts on the mirror is

The correct option is B

$6.7\times {10}^{-8}N$

The explanation for the correct answer is,

Step 1. Given Data:

Area $=1c{m}^2$

Light rays are quanta of energy called photons.
Let,$E=$ Energy fall on the surface per second $=10J$

Step 2. We know that:

The momentum of the photons,$p=\frac{E}{c}$

Change in momentum per second = Force

$$\begin{gathered} F=\frac{2E}{c} \\ \Rightarrow \frac{2E}{c}=\frac{2x10}{(3x{10}^8)} \\ \Rightarrow \frac{2E}{c}=6.7x{10}^{-8}N \end{gathered}$$

All photons emitted from a source, travel through space at the same speed (equal to the speed of light).

Hence, the correct option is (B).