Derive the expression for radiation pressure on a perfectly absorbing surface illuminated by an electromagnetic wave of intensity $I$, and calculate the pressure when $I=500{\text{W m}}^{-2}$.

Step-by-step solution

1. Radiation pressure $p$ is defined as the force per unit area exerted by the wave.
2. For a perfectly absorbing surface, the momentum delivered per unit time per unit area equals the incident energy flux divided by the speed of light: $p=\frac{I}{c}$.
3. Insert $c=3.0\times 10^8{\text{m s}}^{-1}$ and $I=500{\text{W m}}^{-2}$: $p=\frac{500}{3.0\times 10^8}$.
4. Perform the division: $p=1.666\times 10^{-6}{\text{N m}}^{-2}$, which we round to $1.67\times 10^{-6}{\text{N m}}^{-2}$.
5. Thus the radiation pressure on the absorbing surface is $1.67\times 10^{-6}{\text{N m}}^{-2}$.

Concept tested: Radiation pressure of EM wave

Published 9 September 2026 · How this solution was prepared and checked