Question

A square, perfectly reflecting surface is oriented in space to be perpendicular to the light rays from the Sun. The surface has anedge lengthof $\mathit{l}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{m}$andislocated$\mathit{r}\mathbf{=}\mathbf{3}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{11}}\mathbf{\text{\hspace{0.17em}m}}$fromthe Sun’s center. What is the radiation force on the surface from the light rays?

Short answer

The radiation force on the surface of the square from the light rays is$F=9.2\mu N$

Step-by-step solution

Listing the given quantities

$l=2.0m$

$r=3.0\times {10}^{11}\text{\hspace{0.17em}m}$

Understanding the concepts of intensity 

First, we have to find the intensity by usingequation. Then, we can calculate the radiation force on the surface of the square by using the equation

Formula:

$I=\frac{p_s}{4\pi r^2}$

Calculations of theradiation force on the surface of the square from the light rays 

$I=\frac{P_s}{4\pi r^2}$

where powerfor the sun from Appendix$C$is

$P_s=3.90\times {10}^{26}W$

So,

$\begin{array}{c}I=\frac{3.90\times {10}^{26}}{4\pi \times {(3.0\times {10}^{11})}^2}\\ =344.84W/m^2\end{array}$

Now, if the radiationis totally reflected back along its original path, the force is given as

$F=\frac{2IA}{c}$

Where$A$is the area of the reflecting surface and it can be calculated as

$\begin{array}{c}A={\left(l\right)}^2\\ =4m^2\end{array}$

So,

$\begin{array}{c}F=\frac{2\times 344.84\times 4}{3\times {10}^8}\\ =9.2\times {10}^{-6}N\\ =9.2\mu N\end{array}$

Hence, the radiation force on the surface of the square from the light rays is$9.2\mu N$