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

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$\frac{1}{20 π} \frac{W}{m^{2}}$

$\frac{1}{10 π} \frac{W}{m^{2}}$

$\frac{1}{2 π} \frac{W}{m^{2}}$

$\frac{1}{40 π} \frac{W}{m^{2}}$

answer is B.

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Detailed Solution

Complete Solution:

Calculate Total Power Emitted as Light:
Total Power = Input Power × Efficiency
Given input power = 100 W and efficiency = 5%, we find:
Total Power Emitted = 100 W × (5/100) = 5 W
Determine Total Intensity (I):
Intensity I at distance r from a point source is calculated by:
I = Power / (4 π r²) = 5 W / (4 π × (5 m)²) = 5 / (100 π) = 1 / (20 π) W/m²
Intensity of Electric Field Component:
The electric field component intensity is half of the total intensity:
I_electric = (1/2) × (1 / 20 π) = 1 / (40 π) W/m²

Final Answer:

The intensity produced by the electric field component at a distance of 5 m from the source is:

1 / (40 π) W/m²

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