A point source of light is placed at the centre of curvature of a hemispherical surface.The source emits a power of `24W`.The radius of curvature of hemisphere is `10cm` and the inner surface is completely reflecting.The force on the hemisphere due to the light falling on it is_____ `times10^(-8)N`.

Answer

To solve the problem, we need to determine the force on the hemispherical surface due to the light emitted from a point source located at its center of curvature. Here’s a step-by-step solution:

### Step 1: Understand the parameters given
- Power of the light source, \( P = 24 \, \text{W} \)
- Radius of curvature of the hemisphere, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- The inner surface of the hemisphere is completely reflecting.

### Step 2: Calculate the intensity of light at the surface of the hemisphere
The intensity \( I \) of light at a distance \( R \) from a point source is given by the formula:
\[ I = \frac{P}{4\pi R^2} \]
Substituting the values:
\[ I = \frac{24 \, \text{W}}{4\pi (0.1 \, \text{m})^2} \]
Calculating \( 4\pi (0.1)^2 \):
\[ 4\pi (0.1)^2 = 4\pi (0.01) = 0.04\pi \, \text{m}^2 \]
Now substituting this back into the intensity formula:
\[ I = \frac{24}{0.04\pi} = \frac{24}{0.1256} \approx 191.0 \, \text{W/m}^2 \]

### Step 3: Calculate the power received by the hemispherical surface
The area of the hemispherical surface is given by:
\[ A = 2\pi R^2 \]
Substituting \( R = 0.1 \, \text{m} \):
\[ A = 2\pi (0.1)^2 = 2\pi (0.01) = 0.02\pi \, \text{m}^2 \]
The power \( P' \) received by the hemispherical surface is:
\[ P' = I \cdot A = 191.0 \cdot 0.02\pi \approx 12.0 \, \text{W} \]

### Step 4: Calculate the force on the hemisphere due to the light
The force \( F \) on the surface due to the light can be calculated using the formula:
\[ F = \frac{2P'}{c} \]
Where \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)). Substituting the values:
\[ F = \frac{2 \cdot 12.0}{3 \times 10^8} = \frac{24.0}{3 \times 10^8} = 8 \times 10^{-8} \, \text{N} \]

### Final Answer
The force on the hemisphere due to the light falling on it is \( 8 \times 10^{-8} \, \text{N} \).