Question

When a source is at $1.5\,m$ from a photocell the photocurrent is $1\,A$. If the source is at $2\,m$, the current is :
A. $\dfrac{4}{3}$A
B. $\dfrac{9}{16}$A
C. $\dfrac{3}{4}$A
D. $\dfrac{9}{4}$A

Answer 1

A photocell is a light detection and measurement unit. The dusk to dawn photocell sensor switch switches on lights and fans at night and turns them off throughout the day. A photocell sensor can be thought of as a transducer that detects the strength of light.

Complete answer:
The electric current that flows through a photosensitive system, such as a photodiode, as a result of exposure to radiant power is known as photocurrent. Photocurrent may be caused by the photoelectric, photoemissive, or photovoltaic effects. Internal gain induced by contact between ions and photons under the control of applied fields, as in an avalanche photodiode, will boost the photocurrent (APD).

As an appropriate radiation is used, the photoelectric current is directly proportional to the strength of the radiation and increases with increasing accelerating potential until it reaches a point where it reaches its limit and does not increase with increasing accelerating potential. Saturation current is the peak (maximum) value of the photo-current. Cut-off voltage or stopping potential for a given incident ray frequency is the value of retarding potential at which photo-current becomes zero.

Initially $text{Current}=1\,A$ and distance =$1.5\,m$
We know that
Intensity $\propto$ Photo current --(1)
Also for a point source
${\rm{ Intensity }} \propto \dfrac{1}{{{{\bf{r}}^2}}}$-
Given $r’ = 2\,m$
$\dfrac{{r'}}{r} = \dfrac{2}{{1.5}}$
$\Rightarrow {\bf{r}} = \dfrac{4}{{\bf{3}}}{\bf{r}}$
From (2)
${\bf{I}} \propto \dfrac{1}{{{{({\bf{r}}')}^2}}}$
$\Rightarrow I \propto \dfrac{1}{{{{\left( {\dfrac{4}{3}r} \right)}^2}}}$
$\Rightarrow {\bf{I}} \approx \dfrac{{\bf{I}}}{{\dfrac{{{\bf{16}}}}{{\bf{9}}}{{\bf{r}}^2}}} = \dfrac{{\bf{9}}}{{{\bf{16}}}}{\bf{I}}$ -(3)
Now from 1 and 3
$({\rm{ Current' }}) = \dfrac{9}{{16}} \times {\rm{ Current }}$
$\Rightarrow ({\rm{ Current' }})= \dfrac{9}{{16}} \times 1$
$\therefore ({\rm{ Current' }}) = \dfrac{9}{{16}}\,A$

Note: Photodiodes are similar to standard semiconductor diodes, with the exception that they may be exposed (to detect vacuum UV or X-rays) or packaged with a window or optical fibre link that allows light to penetrate the sensitive part of the chip. To maximise the speed of reaction, many diodes engineered specifically for use as a photodiode use a PIN junction rather than a p–n junction. A photodiode is made to work in reverse bias mode.