QJEEM23Ph8e9d: JEE Main 2023 Physics - Class 12 JEEM23Ph8e9d Solution: A point source of emits light with efficiency. At a distance of …

A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency. At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows:

The intensity at a distance of $5$ meters can be calculated as:

Intensity at 5 m $$=\frac{5}{4\pi \times 5^2}\left(\frac{W}{m^2}\right)$$

$$=\frac{1}{20\pi }\left(\frac{W}{m^2}\right)$$

Since the intensity due to the electric field component is half of the total intensity:

Intensity due to electric field $$=\frac{1}{40\pi }\left(\frac{W}{m^2}\right)$$

Thus, we arrive at our result:

$$=\frac{1}{40\pi }\left(\frac{W}{m^2}\right)$$