Question

A small photocell is placed at a distance of a 4 m from photosensitive surface. When light falls on the surface the current is 5 mA. If the distance of cell is decreased to 1 m , the current will become:

• A. $1.25mA$
• B. $\left(\frac{5}{16}\right)mA$
• C. $20mA$
• D. $80mA$

Answer 1

Answer: (D)

$80mA$

Assuming that the energy is emitted equally in all the directions,
$E=I\left(4\pi r^2\right)$
where E is the energy emitted at the center of the sphere, and I is the intensity at a radial distance $r$ from center,
Intensity is directly related to the no. of photons/electrons and hence the current.
$I\left(r\right)=\frac{E}{4\pi r^2}$
$I\left(r\right)\propto \frac{1}{r^2}$
$I\left(r\right)=\frac{K}{r^2}$
$\frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}$
$I_2=I_1(\frac{4}{1}{)}^2$
$I_2=16I_1=80mA$