A lamp is placed at a distance of 16 cm, from a photocell and the current observed is 1 mA. If the distance from the lamp is reduced to 8 cm, the photocurrent now will be

Options

A. 4 mA
B. 0.25 mA
C. 1 mA
D. 0.125 mA

Answer

A

Explanation

We know, that
Intensity `prop(1)/(("distance")^(2))` and photocurrent, (i)`prop`Intensity (I)
Given, distance `r_(1)=16cm`
distance, `r_(2)=8cm implies`current, `I_(1)=1mA`
`therefore(I_(1))/(I_(2))=(r_(2)^(2))/(r_(1)^(2))=(8^(2))/(16^(2))=(1)/(4)`
`thereforeI_(2)=4I_(1)=4xx1=4mA`