To solve this problem, we need to calculate the intensity of light produced by the electric field component at a distance of 5 meters from a point source, given the power emitted by the source and its efficiency.
- First, determine the power that is actually converted to light. The source emits light with an efficiency of \(5\%\). Therefore, the light power emitted is
\(P_{\text{light}} = 0.05 \times 100 \, \text{W} = 5 \, \text{W}\)
- Next, calculate the intensity of light at a distance from a point source. The intensity \((I)\) of light from a point source is given by the formula:
\(I = \frac{P_{\text{light}}}{4 \pi r^2}\)
where \(r\) is the distance from the source. Substituting the given values,
\(I = \frac{5 \, \text{W}}{4 \pi (5 \, \text{m})^2}\)
This simplifies to
\(I = \frac{5}{4 \pi \times 25} \, \text{W/m}^2 = \frac{5}{100 \pi} \, \text{W/m}^2 = \frac{1}{20 \pi} \, \text{W/m}^2\)
- Therefore, the calculated intensity is \(\frac{1}{20 \pi} \, \text{W/m}^2\). However, let's revisit the options given:
- \(\frac{1}{40 \pi} \, \text{W/m}^2\)
- \(\frac{1}{10 \pi} \, \text{W/m}^2\)
- \(\frac{1}{20 \pi} \, \text{W/m}^2\)
- \(\frac{1}{2 \pi} \, \text{W/m}^2\)
- Notice that the calculated intensity matches one of the answer choices — specifically, \(\frac{1}{20 \pi} \, \text{W/m}^2\) is the correct calculation result but not matching the listed correct answer. Therefore, I believe there might be an error in the listed correct answer choice. The calculated \(\frac{1}{20 \pi} \, \text{W/m}^2\) aligns with one of the given answer options, indicating that this should be the intended correction.
The selected and calculated correct choice based on the logical assessment and step-by-step analysis is:
\(\frac{1}{20 \pi} \, \text{W/m}^2\)