Question

In a photocell 4 unit photoelectric current is flowing, the distance between source and cathode is 4 units. Now the distance between source and cathode becomes 1 unit. What will be photoelectric current now?

Answer 1

Hint: Intensity is defined as the power transferred by the light source per unit area where the area is measured in the plane perpendicular to the direction of propagation of the light.
$I = \dfrac{P}{A}$

Complete step by step solution:
One of the results of the photoelectric effect experiment is –
The number of photoelectrons emitted per second also called photoelectric current, is directly proportional to the intensity of the incident light.
$Current \propto I$
In order to calculate the intensity, we have the definition of intensity –
$I = \dfrac{P}{A}$
If we consider a point source of light, the light emanates in all directions in the form of a sphere. Thus, the area here is considered as the surface area of the sphere.

Thus,
$Area,A = 4\pi {r^2}$
Substituting,
$
  I = \dfrac{P}{A} \\
  I = \dfrac{P}{{4\pi {r^2}}} \\
 $
Therefore, from this relation, we get to know that –
$I \propto \dfrac{1}{{{r^2}}}$
and also,
$
  Current \propto I \\
   \Rightarrow Current \propto \dfrac{1}{{{r^2}}} \\
 $
In this problem,
At 4 units of photocurrent $ \Rightarrow $ Distance between source and cathode, $r = 4 $units
At how many (?) units of photocurrent $ \Rightarrow $ Distance between source and cathode, $r = 1$units
To calculate the units of photocurrent,
$
  Current \propto \dfrac{1}{{{r^2}}} \\
  \dfrac{{Curren{t_2}}}{{Curren{t_1}}} = \dfrac{{r_1^2}}{{r_2^2}} \\
  Curren{t_2} = Curren{t_1}\left( {\dfrac{{{4^2}}}{{{1^2}}}} \right) \\
  Curren{t_2} = 4 \times \left( {\dfrac{{{4^2}}}{{{1^2}}}} \right) \\
  Solving, \\
  \therefore Current = {4^3} = 64 \\
 $
Thus, the new photocurrent = 64 units.

Note:
This principle of variation of photoelectric current with the intensity of light is applicable in the device photocell. This device is used as a sensor in automatic lighting systems. When there is excess light in the surroundings, the photocell detects the high change in intensity, and the photoelectric current increases. This increase in current triggers the lights to turn dim.