Question

Light with an average flux of $20\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$c{m}^2$}\right.$ falls on a non-reflecting surface at normal incidence having surface area $20c{m}^2$. The energy received by the surface during the time span of $1minute$ is

Answer 1

Step 1. Given data:

Average flux $\left(I\right)=20\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$c{m}^2$}\right.$

Area $\left(A\right)=20c{m}^2$

Time $t=1min=1\times 60sec=60sec$

Step 2. Calculation:

The intensity of the light incident on the surface is given by:
$I=\frac{E}{At}$, where $E$= Energy received by the surface, $I,A,t$ are their usual meaning as described above.

Substituting the given values, we get

$E=20\times 20\times 60=24\times {10}^{3}J=24kJ$

Hence, the energy received by the surface during the time span of $1min$ is $24kJ$.