Question
Question: Energy density of sunlight, \(50W{{m}^{-2}}\), is normally incident on the surface of a solar panel....
Energy density of sunlight, 50Wm−2, is normally incident on the surface of a solar panel. Some part of incident energy (25 percent) is reflected from the surface and the rest is being absorbed. The force exerted on 1m2 surface area will be close to? (C=3×108ms−1)
& 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{aligned}$$Solution
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 25of 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,
Fr=10025(C2I)
The force exerted by the absorbed ray can be written as,
Fa=10075(CI)
Therefore the net force can be written as,
F=Fr+Fa
Substituting the equations of both in this will give,