JEE Main 2025 Apr 3 Shift 1, Physics Q4: Electromagnetic Waves

For a perfectly reflecting surface, radiation pressure is $P_{\text{rad}}=\frac{2I}{c}$, where $I$ is the intensity at the surface and $c$ is the speed of light.

Assuming isotropic emission, intensity at distance $r=2$ m:

$I=\frac{P}{4\pi r^2}=\frac{450}{4\pi (2{)}^2}=\frac{450}{16\pi }{\text{W m}}^{-2}.$

Then

$P_{\text{rad}}=\frac{2\times 450}{16\pi \times 3\times 10^8}=\frac{150}{8\pi \times 10^8}\approx 5.97\times 10^{-8}\text{Pa}\approx 6\times 10^{-8}\text{Pa}.$