When a monochromatic point source of light is | vedclass

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(C) The intensity of light $I$ from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$.When the distance $r$ is doubled (from $0.

Vedclass · Accepted Answer

$0.8\, V$ and $6\, mA$

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(C) The intensity of light $I$ from a point source follows the inverse square law,$I \propto \frac{1}{r^2}$.When the distance $r$ is doubled (from $0.1\, m$ to $0.2\, m$),the intensity becomes $\frac{1}{4}$ of the original value.The saturation current is directly proportional to the intensity of incident light. Therefore,the new saturation current will be $\frac{24\, mA}{4} = 6\, mA$.The cut-off voltage (stopping potential) depends only on the frequency of the incident light and the work function of the material,not on the intensity. Since the source remains the same,the frequency is unchanged,so the cut-off voltage remains $0.8\, V$.

When a monochromatic point source of light is placed at $0.1\, m$ from a photocell,the cut-off voltage and saturation current are $0.8\, V$ and $24\, mA$ respectively. If the same source is placed at $0.2\, m$ away from the photocell,then the cut-off voltage and saturation current will be:

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