Light of uniform intensity shines perpendicularly on a totally absorbing surface, fully illuminating the surface. If the area of the surface is decreased, what is the effect on radiation pressure?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Remains the same
Light carries momentum, so it exerts a force when it is absorbed. For a totally absorbing surface at normal incidence, the momentum delivered per second per unit area is \( I/c \), so the radiation pressure is
\( \displaystyle P = \frac{I}{c} \)
This depends only on the intensity and the speed of light. The force \( F = IA/c \) falls when the area is reduced, but the force per unit area stays the same, so the radiation pressure remains the same.
Why the other options are incorrect
Option (1) Increases — Incorrect. A smaller area receives less light, so the force also falls in the same ratio; the pressure does not rise.
Option (2) Decreases — Incorrect. It is the total force that decreases; pressure is force per unit area and does not change.
Option (4) Cannot be said — Incorrect. The result follows directly from \( P = I/c \), so it can be stated with certainty.