Question

If electromagnetic waves of power 600 W incident on a non-reflecting surface, then the total force acting on the surface is

• A. $12 \times 10^{-6}$ N
• B. $9 \times 10^{-9}$ N
• C. $6 \times 10^{-6}$ N
• D. $2 \times 10^{-6}$ N

Hint:

For total absorption: $F = P/c$. For total reflection: $F = 2P/c$.

Answer 1

The Correct Option is D

Step 1: Formula for Radiation Pressure Force:
When radiation is absorbed (non-reflecting surface), the momentum transfer per unit time is $P/c$. The force $F$ exerted is: \[ F = \frac{\text{Power}}{c} \] where $c$ is the speed of light ($3 \times 10^8$ m/s).
Step 2: Calculation:
Power $P = 600$ W. \[ F = \frac{600}{3 \times 10^8} = 200 \times 10^{-8} = 2 \times 10^{-6} \text{ N} \]