Question

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 __________.

Answer 1

$\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 $\frac{E}{A}=1.3$
So final, $\frac{E}{a}=? ? $
$E=A \times 1.30$
Also $\frac{a}{A}=\frac{1}{100}$
$\therefore $ Average intensity of light at
$22\, cm =\frac{E}{a}=\frac{A \times 1.3}{a}$
$=100 \times 1.3=130 \,kW / m ^{2}$