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(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.

Vedclass · Accepted Answer

$4$

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(C) The intensity $I$ of light from a point source follows the inverse square law,$I \propto \frac{1}{d^2}$,where $d$ is the distance from the source.Since the number of photoelectrons emitted is directly proportional to the intensity of incident light,we have $N \propto I \propto \frac{1}{d^2}$.Let $N_1$ be the number of photons at distance $d_1 = 1 \ m$ and $N_2$ be the number of photons at distance $d_2 = 2 \ m$.Then,$\frac{N_1}{N_2} = \left( \frac{d_2}{d_1} \right)^2$.Substituting the values: $\frac{N_1}{N_2} = \left( \frac{2}{1} \right)^2 = 4$.Therefore,$N_1 = 4N_2$,which means the number of photons emitted at $1 \ m$ is $4$ times the number of photons emitted at $2 \ m$.

How many times more photons are emitted from a photocell by a source placed at a distance of $1 \ m$ compared to a source placed at a distance of $2 \ m$?

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