Question

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:

• A. $\frac{1}{40 \pi} \frac{W}{m^2}$
• B. $\frac{1}{10 \pi} \frac{W}{m^2}$
• C. $\frac{1}{20 \pi} \frac{W}{m^2}$
• D. $\frac{1}{2 \pi} \frac{W}{m^2}$

Answer 1

The Correct Option is A

The correct option is(A): \(\frac{1}{40 \pi} \frac{W}{m^2}\)

\(I_{EF}​=\frac{1}{2}​×\frac{5}{4π×5^2}​ \)

\(=\frac{1}{40π}1​W/m^2\)

Answer 2

The correct option is (A): \(\frac{1}{40\pi} \frac{w}{m^2}\)

Total power emitted = \(100w \times 5\%  = 100 \times \frac{5}{100} = 5W\)  

\(\text{Intensity} = \frac{\text{Power}}{\text{area}}\)
\(= \frac{5}{4\pi(r^2)}\)
\(= \frac{5}{4\pi (5)^2}\)
\(= \frac{1}{20π} \frac{W}{m^2}\)

This intensity is due to both electric and magnetic fields,
So, due to electric field \(= \frac{1}{2} \times I\)
\(= \frac{1}{2} \times \frac{1}{20\pi}\)
\(= \frac{1}{40\pi} \frac{W}{m^2}\)

Therefore, the intensity produced by the electric field component is \(= \frac{1}{40\pi} \frac{W}{m^2}\)