In Young's double slit experiment, if the widths of the slit are in the ratio $4 : 9$ , ratio of intensity of maxima to intensity of minima will be

25 : 1

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9 : 4

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3 : 2

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81 : 16

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Solution

The correct option is B $25 : 1$
Width of slit (W) $\propto$ Light intensity (I), $\frac{W_{1}}{W_{2}} = \frac{I_{1}}{I_{2}} = \frac{4}{9} = \frac{4 k}{9 k} . . . . (k i s a c o n s t a n t)$
$I_{m a x} = 4 k + 9 k + 2 \sqrt{4 k 9 k} = 25 k$
$I_{m i n} = 4 k + 9 k - 2 \sqrt{4 k 9 k} = k$
$\frac{I_{m a x}}{I_{m i n}} = 25 : 1$

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