Question

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

Answer 1

Correct option is D. $$20\times 10^{-8}$$N
Force on the surface ($$25\%$$ reflecting and rest absorbing)
$$F=\dfrac{25}{100}\left(\dfrac{2I}{C}\right)+\dfrac{75}{100}\left(\dfrac{I}{C}\right)=\dfrac{125}{100}\left(\dfrac{I}{C}\right)$$
$$=\dfrac{125}{100}\times \left(\dfrac{50}{3\times 10^{8}}\right)=20.83\times 10^{-8}$$N.