Question

$5\%$ of the power of $100W$ bulb is converted to visible radiation. Average intensity of visible radiation at a distance of $10m$ from the bulb is

• A. $\frac{5}{2\pi (10{)}^2}watt/m^2$
• B. $\frac{5}{4\pi (10{)}^2}watt/m^2$
• C. $\frac{5}{\pi (10{)}^2}watt/m^2$
• D. $\frac{5}{8\pi (10{)}^2}watt/m^2$

Answer 1

The correct option is B $\frac{5}{4\pi (10{)}^2}watt/m^2$

Given,

$P=100W,d=10m$

Power of visible radiation, $P\text{'}=\frac{5\times 100}{100}=5W$

Hence, the intensity of radiation, $I=\frac{P\text{'}}{4\pi d^2}=\frac{5}{4\pi (10{)}^2}watt/m^2$