Sunlight of intensity $1.3 kW m^{-2}$ is incident normally on a thin convex lens of focal length $20 cm$ . Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light, in $kW m^{-2}$ , at a distance $22 cm$ from the lens on the other side is __________.

Solution

$\frac{r}{R}=\frac{2}{20}=\frac{1}{10}$
$\therefore$ Ratio of area $=\frac{1}{100}$
Let energy incident on lens be E
$\therefore$ Given $\frac{E}{A}=1.3$
$E=A\times 1.30$
Also $\frac{a}{A}=\frac{1}{100}$
$\therefore$ Average intensity of light at $22 cm=\frac{E}{a}=\frac{A\times 1.3}{a}=100\times 1.3=130k W/m^2$

Hence, the correct answer is 130.