Question

A thin lens of focal length $f$ and its aperture diameter $d$ , forms a real image of intensity I. Now the central part of the aperture upto diameter $\left(\frac{d}{2}\right)$ is blocked by an opaque paper. If the new focal length is $xf$ and new image intensity is $\frac{yI}{4}$,then $xy$ is

Answer 1

Initially energy/sec $=I\times \pi {\left(\frac{d}{2}\right)}^2=\frac{\pi d^{2}I}{4}$

Now energy/sec $=I[\pi {\left(\frac{d}{2}\right)}^2-\pi {\left(\frac{d}{4}\right)}^2]$

$=I\pi d^2\left[\frac{3}{16}\right]$

So, Now $\frac{\text{Intensity final}}{\text{intensity initial}}=\frac{\text{I}\pi d^{2}3/16}{I\pi d^2/4}=\frac{3}{4}$

So, new intensity would be $\frac{3I}{4}$.

Focus will not change.
$\therefore x=1,y=3\Rightarrow xy=3$