Question

In a photo cell 4 unit photo electric current is flowing, the distance between source and cathode is 4 unit. Now distance between source and cathode becomes 1 unit. What will be photo electric current now?

Answer 1

Since Current $\propto$ Intensity

And intensity is given by $I=\frac{P}{4\pi r^2}$ where P is the power of point source and $r$ is the distance between source and cathode.

So,

$\frac{I_1}{I_2}={\left(\frac{r_2}{r_1}\right)}^2$

$\frac{4}{I_2}={\left(\frac{1}{4}\right)}^2$

$I_2=64units$