Question

A light of intensity \(12\;Wm^{-2}\) incidents on a black surface of area \(4\;cm^2\). The radiation pressure on the surface is

• A. \(1\times10^{-8}\;Pa\)
• B. \(4\times10^{-8}\;Pa\)
• C. \(1.6\times10^{-7}\;Pa\)
• D. \(4.8\times10^{-7}\;Pa\)

Hint:

For radiation pressure: \[ P=\frac{I}{c} \] for a perfectly absorbing surface, and \[ P=\frac{2I}{c} \] for a perfectly reflecting surface.

Answer 1

The Correct Option is B

Step 1: Recall radiation pressure on a black surface.
A perfectly absorbing (black) surface feels a radiation pressure \[ P = \frac{I}{c} \] where $I$ is the intensity and $c$ the speed of light.
Step 2: Note why the area does not enter.
Pressure is force per unit area, and although a larger area receives more force, the pressure depends only on the intensity, so the given $4\ \text{cm}^2$ is not needed for the pressure itself.
Step 3: List the data.
Here $I = 12\ \text{W m}^{-2}$ and $c = 3 \times 10^{8}\ \text{m s}^{-1}$.
Step 4: Substitute into the formula.
\[ P = \frac{12}{3 \times 10^{8}} \]
Step 5: Simplify the arithmetic.
Dividing, \[ P = 4 \times 10^{-8}\ \text{Pa} \]
Step 6: State the answer.
Hence the radiation pressure on the surface is \[ \boxed{4 \times 10^{-8}\ \text{Pa}} \]