Question

Energy density of sunlight, $50W{m}^{-2}$, is normally incident on the surface of a solar panel. Some part of incident energy $\left(25\text{ percent}\right)$ is reflected from the surface and the rest is being absorbed. The force exerted on $1m^2$ surface area will be close to? $(C=3\times {10}^{8}m{s}^{-1})$
$$\begin{array}{rl}& A.15\times {10}^{-8}N\\ & B.35\times {10}^{-8}N\\ & C.10\times {10}^{-8}N\\ & D.20\times {10}^{-8}N\end{array}$$

Answer 1

Hint: The force exerted by the sunlight will be the ratio of the product of the twice the intensity and the energy part of the light involved.to the velocity of light. This is the case when the incident energy is being absorbed. When the energy gets reflected the value will become twice of this. Find the sum of both the situations. And then substitute the values in it. This will help you in solving this question.

Complete answer:
It has been mentioned in the question that $$25$$of the energy incident on the solar panel is getting reflected. And the rest of the energy, that is, $$75$$ of the energy is getting absorbed.
Force exerted by the reflected ray will be twice that of absorbed ray. Therefore we can write that, the force exerted by the reflected ray will be,
$$F_r={\displaystyle \frac{25}{100}}\left({\displaystyle \frac{2I}{C}}\right)$$
The force exerted by the absorbed ray can be written as,
$$F_a={\displaystyle \frac{75}{100}}\left({\displaystyle \frac{I}{C}}\right)$$
Therefore the net force can be written as,
$$F=F_r+F_a$$
Substituting the equations of both in this will give,
$$\begin{array}{rl}& F={\displaystyle \frac{25}{100}}\left({\displaystyle \frac{2I}{C}}\right)+{\displaystyle \frac{75}{100}}\left({\displaystyle \frac{I}{C}}\right)\\ & \Rightarrow F={\displaystyle \frac{125}{100}}\left({\displaystyle \frac{I}{C}}\right)\end{array}$$
The energy density of the light is given as,
$$I=50W{m}^{-2}$$
As we all know, the velocity of the light is found to be as,
$$C=3\times {10}^{8}m{s}^{-1}$$
Substituting this in the equation will give,
$$F={\displaystyle \frac{125}{100}}\left({\displaystyle \frac{50}{3\times {10}^8}}\right)=20.83\times {10}^{-8}N$$
Therefore the value of the force exerted has been obtained.

So, the correct answer is “Option D”.

Note:
Energy density is defined as the amount of energy which has been stored in a specific system or region of space per unit volume. It can be otherwise referred to as specific energy or the energy per unit mass. Energy density is a scalar quantity.