8.0 Force exerted by a light beam on a surface

8.1 Surface exposed to normal radiation
8.2 Surface exposed to radiation inclined at an angle

Let us consider a perfectly absorbing surface. A light beam of wavelength $\lambda $ cross-sectional area $S$ and power $P$ watt is incident on it.

$$\begin{equation} \begin{aligned} {\text{Number of photons emitted per second ,}}\quad N = \frac{{P\lambda }}{{hc}} \\ \\\end{aligned} \end{equation} $$

This number of photons strike the surface and get absorbed by it. Hence, there is a sudden momentum change of $$\Delta P = 0 - ( - P\widehat i) = P\widehat i$$

So, force exerted on the body = total momentum change per second.

$$\begin{equation} \begin{aligned} F = \frac{{P\lambda }}{{hc}} \times \frac{h}{\lambda } = \frac{P}{c} \\ \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} {\text{Similarly, for a perfectly reflecting body}}\quad F = \frac{{2P}}{c} \\ \\\end{aligned} \end{equation} $$

Now, let us study some cases.

Before that remember these notations.

$a=$ absorption coefficient and $r=$ reflection coefficient.

Assume no transmission. Let a surface of area $A$ exposed to intensity $I$.