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(C) The intensity of light $I$ is inversely proportional to the square of the distance $d$ from the source,i.e.,$I \propto \frac{1}{d^2}$.Since the nu

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Increase by a factor of $4$.

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(C) The intensity of light $I$ is inversely proportional to the square of the distance $d$ from the source,i.e.,$I \propto \frac{1}{d^2}$.Since the number of photoelectrons emitted is directly proportional to the intensity of incident light,we have $N \propto I$.Therefore,the ratio of the number of electrons is given by $\frac{N_2}{N_1} = \frac{I_2}{I_1} = \left( \frac{d_1}{d_2} \right)^2$.Given $d_1 = 1 \ m$ and $d_2 = \frac{1}{2} \ m$,we get:$\frac{N_2}{N_1} = \left( \frac{1}{1/2} \right)^2 = (2)^2 = 4$.Thus,the number of emitted electrons increases by a factor of $4$.

$A$ photocell is illuminated by a small bright light source placed at a distance of $1 \ m$. When the same light source is placed at a distance of $\frac{1}{2} \ m$,the number of electrons emitted by the photocathode will be:

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