A point source of `100W` emits light with `5%` efficiency.At a distance of `5m` from the source,the intensity produced by the electric field component is

Options

A. `(1)/(40 pi)(W)/(m^(2)),`
B. `(1)/(10 pi)(W)/(m^(2)),`
C. `(1)/(20 pi)(W)/(m^(2))`
D. `(1)/(2pi)(W)/(m^(2)),`

Answer

To solve the problem, we need to find the intensity produced by the electric field component at a distance of 5 meters from a point source of light that emits 100 W with an efficiency of 5%.

### Step-by-Step Solution:

1. **Calculate the Total Power Emitted:**
The total power emitted by the source can be calculated using the efficiency:
\[ \text{Total Power Emitted} = \text{Input Power} \times \text{Efficiency} \]
Given that the input power is 100 W and the efficiency is 5%, we can write:
\[ \text{Total Power Emitted} = 100 \, \text{W} \times \frac{5}{100} = 5 \, \text{W} \]

2. **Determine the Intensity:**
Intensity (I) is defined as power per unit area. For a point source, the area over which the power is distributed is the surface area of a sphere, which is given by \(4\pi r^2\), where \(r\) is the distance from the source.
\[ I = \frac{\text{Power}}{\text{Area}} = \frac{P}{4\pi r^2} \]
Substituting the values we have:
\[ I = \frac{5 \, \text{W}}{4\pi (5 \, \text{m})^2} \]
\[ I = \frac{5}{4\pi \times 25} = \frac{5}{100\pi} = \frac{1}{20\pi} \, \text{W/m}^2 \]

3. **Calculate the Intensity Due to the Electric Field:**
The intensity of the electromagnetic wave is equally divided between the electric and magnetic fields. Therefore, the intensity due to the electric field component is half of the total intensity:
\[ I_{\text{electric}} = \frac{1}{2} I = \frac{1}{2} \times \frac{1}{20\pi} = \frac{1}{40\pi} \, \text{W/m}^2 \]

### Final Answer:
The intensity produced by the electric field component at a distance of 5 m from the source is:
\[ \frac{1}{40\pi} \, \text{W/m}^2 \]