Question

A beam of light of intensity $I$ is normally incident on a plate of area $A$ . If plate absorbs fraction $\eta$ of incident light and reflects rest of it back, then the force exerted by the incident light on the plate is given by [ $c$ =speed of light]

• A. $\frac{IA}{c}\left(1 - \eta\right)$
• B. $\frac{IA}{c}\left(\eta + 1\right)$
• C. $\frac{IA}{c}\left(2 \eta - 1\right)$
• D. $\frac{IA}{c}\left(2 - \eta\right)$

Answer 1

Answer: (D)

Force exerted by the fraction of ,ight that is absorbed is given as, $F_{1}=\frac{ηIA}{c}$ .
Force exerted by the fraction of light that is totally reflected is given as, $F_{2}=\frac{\left(1 - \eta\right) \times 2 IA}{c}$ .
Hence, total force will be given as, $F=F_{1}+F_{2}=\frac{\left(2 - \eta\right) IA}{c}$ .