JEE Physics 2023 Question with Solution

Given: Source power is $100 \, \text{W}$, efficiency is $5\%$, and distance is $5 \, \text{m}$.

Find: The intensity produced by the electric field component.

Total light power emitted is

$$P = 100 \times \frac{5}{100} = 5 \, \text{W}$$

The total intensity at distance $r = 5 \, \text{m}$ is

$$I = \frac{P}{4\pi r^2} = \frac{5}{4\pi (5)^2} = \frac{1}{20\pi} \, \text{W/m}^2$$

This intensity is due to both electric and magnetic field components. Therefore, the intensity due to the electric field component alone is half of the total intensity:

$$I_{EF} = \frac{1}{2} I = \frac{1}{2} \times \frac{1}{20\pi} = \frac{1}{40\pi} \, \text{W/m}^2$$

Therefore, the intensity produced by the electric field component is $\frac{1}{40\pi} \, \text{W/m}^2$. The solution concludes this value but labels it as option A; among the given options, this value matches option B.

The solution explicitly computes the final value as

$$\frac{1}{40\pi} \, \text{W/m}^2$$

Both approaches on the page state this same numerical result. However, the page incorrectly says "The Correct Option is A". Comparing with the listed options shows that $\frac{1}{40\pi} \, \text{W/m}^2$ is actually option B. Hence the defensible answer is B.