Question

An electromagnetic wave of intensity $I$ is incident on a non-reflecting surface. If $C$ is the speed of light in free space, then, the ratio $\frac{I}{C}$ is same as

• A. momentum
• B. force
• C. pressure
• D. pressure per unit area
• E. force \(\times\) area

Hint:

For a perfectly reflecting surface, the radiation pressure is doubled: \(P = \frac{2I}{c}\). This is because the change in momentum is doubled when the wave bounces back.

Answer 1

The Correct Option is C

Concept: Radiation pressure \(P\) is the pressure exerted upon any surface exposed to electromagnetic radiation. For a non-reflecting (perfectly absorbing) surface, the radiation pressure is related to the intensity \(I\) of the wave and the speed of light \(c\).

Step 1: Define Intensity and Momentum.
Intensity \(I\) is the energy (\(U\)) incident per unit area (\(A\)) per unit time (\(t\)): \[ I = \frac{U}{A \cdot t} \] The momentum \(p\) carried by electromagnetic radiation of energy \(U\) is: \[ p = \frac{U}{c} \]

Step 2: Derive the ratio \(I/c\).
Pressure is defined as Force per unit Area (\(F/A\)), and Force is the rate of change of momentum (\(dp/dt\)). For a perfectly absorbing surface: \[ \text{Pressure } (P) = \frac{F}{A} = \frac{1}{A} \cdot \frac{p}{t} = \frac{1}{A} \cdot \frac{U/c}{t} \]

Step 3: Relate to Intensity.
Rearranging the terms: \[ P = \frac{1}{c} \left( \frac{U}{A \cdot t} \right) \] Since the term in parentheses is the Intensity \(I\): \[ P = \frac{I}{c} \] Therefore, the ratio \(I/c\) represents radiation pressure.