If the distance of surface from light source is doubled then the illuminance will become

\frac{1}{2}

times

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2

times

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\frac{1}{4}

times

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4

times

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Solution

The correct option is C $\frac{1}{4}$ times
We have illuminance,
$I \propto \frac{1}{R^{2}}$
where $R$ is the distance
If the distance is doubled then
$I^{'} \propto \frac{I}{4}$

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