If a 100 watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3m is

Options

A. `0.002Wm^-2`
B. `0.22Wm^-2`
C. `4.4Wm^-2`
D. `0.033Wm^-2`

Explanation

To find the intensity of a 100-watt bulb acting as a point source with 2.5% efficiency at a distance of 3 meters, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Effective Power Output**:
The bulb has a power rating of 100 watts and operates at 2.5% efficiency. Therefore, the effective power output (useful power) can be calculated as:
\[ \text{Effective Power} = \text{Total Power} \times \text{Efficiency} = 100 \, \text{W} \times \frac{2.5}{100} = 2.5 \, \text{W} \]

2. **Calculate the Area of the Sphere**:
Since the bulb acts as a point source, the intensity is distributed over the surface area of a sphere. The formula for the surface area \(A\) of a sphere is:
\[ A = 4 \pi r^2 \]
where \(r\) is the radius (distance from the source). Given \(r = 3 \, \text{m}\):
\[ A = 4 \pi (3)^2 = 4 \pi \times 9 = 36 \pi \, \text{m}^2 \]

3. **Calculate the Intensity**:
Intensity \(I\) is defined as the power per unit area:
\[ I = \frac{\text{Power}}{\text{Area}} = \frac{2.5 \, \text{W}}{36 \pi \, \text{m}^2} \]
Now substituting the value of \(\pi \approx 3.14\):
\[ I = \frac{2.5}{36 \times 3.14} = \frac{2.5}{113.04} \approx 0.0221 \, \text{W/m}^2 \]

4. **Final Result**:
Therefore, the intensity of the bulb at a distance of 3 meters is approximately:
\[ I \approx 0.0221 \, \text{W/m}^2 \]