Light of uniform intensity shines perpendicularly on a totally absorbing surface fully illuminating the surface . If the area of the surface is decreased

Options

A. The radiation pressure increases and the radiation force increases
B. The radiation pressure increases and the radiation force decreases
C. The radiation pressure stays the same and the radiation force increases
D. The radiation pressure stays the same and the radiation force decreases

Answer

To solve the problem, we need to analyze the effects of decreasing the area of a totally absorbing surface on the radiation force and radiation pressure when illuminated by light of uniform intensity.

### Step-by-Step Solution:

1. **Understanding Radiation Force**:
The radiation force \( F \) acting on a totally absorbing surface can be expressed using the formula:
\[ F = \frac{2IA}{c} \]
where:
- \( I \) is the intensity of the light,
- \( A \) is the area of the surface,
- \( c \) is the speed of light.

2. **Effect of Decreasing Area**:
If the area \( A \) of the surface is decreased, the radiation force \( F \) will also decrease because \( F \) is directly proportional to \( A \). Thus, if \( A \) decreases, \( F \) will decrease.

3. **Understanding Radiation Pressure**:
Radiation pressure \( P \) is defined as the force per unit area:
\[ P = \frac{F}{A} \]
Substituting the expression for \( F \):
\[ P = \frac{2IA}{cA} = \frac{2I}{c} \]
This shows that radiation pressure \( P \) is independent of the area \( A \) of the surface.

4. **Conclusion**:
- When the area of the surface is decreased, the radiation force \( F \) decreases.
- However, the radiation pressure \( P \) remains constant regardless of the change in area.

### Summary:
- Decreasing the area of the surface decreases the radiation force.
- The radiation pressure remains constant when the area is changed.