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A source of light is placed at a distance $x$ from a photocell and cut off potential is $V$. If the distance of the same source from the photocell is doubled, then cut off potential will be :

$\begin {array} {1 1} (a)\;V/4 & \quad (b)\;V/2 \\ (c)\;V & \quad (d)\;2V \end {array}$


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Ans : (c) V
$V$ depends on frequency and not on the intensity of radiations.


answered Dec 24, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1

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