Question

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

• A. 0.4 W / $m^2$
• B. 0.04 W / $m^2$
• C. 0.004 W / $m^2$
• D. 0.0004 W / $m^2$

Answer 1

The correct option is C 0.004 W / $m^2$

Given:- Power of light bulb=$P_o=100W$

Powerof visible radiation=$P=5$ of $P_o$

$P=\frac{5}{100}\times 100$

$P=5W$

$Distance=d=10m$

solution:-By considering light bulb as a point source.

Since radiation trable by spherical wave fronts.

Intensity of visible radiation at a distance$d=10m$

is given by :-

$I=\frac{P}{4\Pi d^2}=\frac{5}{4\times 3.14\times {10}^2}$

$I=0.00398W/m^2$

$I\approx 0.004W/m^2$