# About 5% of the power of a $100\: W$ light bulb is converted to visible radiation. What is the average intensity of visible radiation at a distance of $10\: m$ from the bulb?

$\begin {array} {1 1} (a)\;0.398\: W/m^2 & \quad (b)\;0.3098 \: W/m^2 \\ (c)\;0.0398\: W/m^2 & \quad (d)\;0.00398\: W/m^2 \end {array}$

$P = 100\: W$
Power of visible radiation, $P’ = \large\frac{5 \times 100}{100}$$= 5W d = 10m Hence, intensity of radiation, I = \large\frac{P’}{4 \pi d^2} = \large\frac{5}{4 \pi (10)^2}$$ = 0.00398\: W/m^2$
Ans : (d)