Electrostatics Question 14
Question 14 - 30 January - Shift 2
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 :
(1) $\frac{1}{2 \pi} \frac{W}{m^{2}}$
(2) $\frac{1}{40 \pi} \frac{W}{m^{2}}$
(3) $\frac{1}{10 \pi} \frac{W}{m^{2}}$
(4) $\frac{1}{20 \pi} \frac{W}{m^{2}}$
Show Answer
Answer: (2)
Solution:
$I _{EF}=\frac{1}{2} \times \frac{5}{4 \pi \times 5^{2}}$
$=\frac{1}{40 \pi} W / m^{2}$
