Question

$50{\text{W/m}}^2$ energy density of sunlight is normally incident on the surface of a solar panel. $25\text{\%}$ of the incident energy is reflected from the surface and the rest is absorbed. The force exerted on $1{\text{m}}^2$ surface area will be close to
$(c=3\times {10}^8\text{m/s})$

• A. $20\times {10}^{-8}N$
• B. $10\times {10}^{-8}N$
• C. $35\times {10}^{-8}N$
• D. $15\times {10}^{-8}N$

Answer 1

The correct option is A $20\times {10}^{-8}N$

Given, $I=50{\text{W/m}}^2$
Here, change in momentum is,
$\Delta P=\Delta P_r+\Delta P_a...\left(1\right)$

Since, $25\%$ of incident energy is reflected back and rest is absorbed, equation $\left(1\right)$ can be written as,

$\Delta P=(2\times \frac{P}{4})+\frac{3}{4}\times P=\frac{5P}{4}$

Now, $F=\frac{\Delta P}{\Delta t}=\frac{5IA}{4c}$

$F=\frac{5}{4}\times \frac{50\times 1}{3\times {10}^8}\approx 20\times {10}^{-8}\text{N}$

Hence, $\left(A\right)$ is the correct answer.