Q 12-08-062JEE MainJEE Main 2024 (8 Apr, Shift 1)Medium
Average force exerted on a non-reflecting surface at normal incidence is $2.4\times10^{-4}\ \text{N}$. If $360\ \text{W/cm}^2$ is the light energy flux during a span of 1 hour 30 minutes, then the area of the surface is:
Answer: (D) $0.02\ \text{m}^2$
For a perfectly absorbing surface, the force is $F = \dfrac{IA}{c}$ (the duration does not matter for the average force).
$I = 360\ \text{W cm}^{-2} = 3.6\times10^6\ \text{W m}^{-2}$
$$A = \frac{Fc}{I} = \frac{2.4\times10^{-4}\times3\times10^8}{3.6\times10^6} = 0.02\ \text{m}^2$$
Solution by Sreeraj P, M.Sc Physics