A point source of light emits isotopically with a power of 200 W . What is the force due to the light on a totally absorbing sphere of radius 2.0 cm at a distance of 20 mfrom the source?

The radius of a sphere,$r=0.02m$

The distance,$d=20m$

The power, $P=200W$

The intensity can be calculated from the following equation.

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{\pi }{\mathbf{d}}^{\mathbf{2}}}$

By using the value of intensity in the above equation, you can find the force due to the light on a totally absorbing sphere.

$\mathit{F}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}\mathbf{c}}$

Here, $\mathit{I}$ is the intensity, $\mathit{P}$ is the power, $\mathit{d}$ is the distance, $\mathit{A}$ is the area, and $\mathit{c}$ is the speed of light having a value $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$.

First, calculate the intensity at distance $r$ by using the formula below.

$$\begin{gathered} I=\frac{P}{4\pi d^2} \\ =\frac{200W}{4\times 3.14\times {(20m)}^2} \\ =0.040W/m^2 \end{gathered}$$

Now, if the radiation is totally absorbed by the surface, the force is given by,

$F=\frac{IA}{c}$ ….. (1)

Where, $A$ is the area of the sphere, and it can be calculated as,

$$\begin{gathered} A=4\pi r^2 \\ =4\times 3.14\times {\left(0.02\text{m}\right)}^2 \\ =5.024\times {10}^{-3}{m}^2 \end{gathered}$$

Substitute known values into equation (1), and you get

$$\begin{gathered} F=\frac{2\times (0.040w/m^2)\times (5.024\times {10}^{-3}{m}^2)}{3\times {10}^{8}m/s} \\ =1.33\times {10}^{-12}N \end{gathered}$$

Hence, the radiation force on the surface from the light rays is $1.33\times {10}^{-12}N$.