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

169 : 25

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81 : 16

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25 : 1

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9 : 4

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Solution

The correct option is C $25 : 1$
Slit width ratio $= 4 : 9$ , hence $I_{1} : I_{2} = 4 : 9$

$∴ \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{4}{9} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{2}{3}$

$∴ \frac{I_{m a x}}{I_{m i n}} = \frac{(a_{1} + a_{2})^{2}}{(a_{1} - a_{2})^{2}} = \frac{25}{1}$

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