Question 73: EM waves

Question

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:

Accepted Answer

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 $$= \frac{5}{4\pi 	imes 5^2}\left(\frac{W}{m^2}ight)$$
$$= \frac{1}{20\pi}\left(\frac{W}{m^2}ight)$$
Since the intensity due to the electric field component is half of the total intensity:
Intensity due to electric field $$= \frac{1}{40\pi}\left(\frac{W}{m^2}ight)$$
Thus, we arrive at our result:
$$= \frac{1}{40\pi}\left(\frac{W}{m^2}ight)$$

Answer

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

Answer

$\frac{1}{10 \pi} \frac{W}{m^2}$

Answer

$\frac{1}{20 \pi} \frac{W}{m^2}$

Answer

$\frac{1}{2 \pi} \frac{W}{m^2}$

EM waves

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:

Choose the correct answer: