In Young's double slit experiment, if the ratio of intensities of bright and dark fringes is $9 / 4$ , then the ratio of intensities of interfering waves is?

16 : 1

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4 : 1

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25 : 1

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8 : 9

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Solution

The correct option is D $25 : 1$
Given, $\frac{I_{m a x}}{I_{m i n}} = \frac{9}{4}$
So, $\frac{I_{m a x}}{I_{m i n}} = {(\frac{a_{1} + a_{2}}{a_{1} - a_{2}})}^{2} = \frac{9}{4}$
$\Rightarrow \frac{a_{1} + a_{2}}{a_{1} - a_{2}} = \frac{3}{2}$
$\Rightarrow 2 a_{1} + 2 a_{2} = 3 a_{1} - 3 a_{2}$
$\Rightarrow 5 a_{2} = a_{1} \Rightarrow \frac{a_{1}}{a_{2}} = 5$
As $I \propto a^{2}$
So $\frac{I_{1}}{I_{2}} = {(\frac{a_{1}}{a_{2}})}^{2} = \frac{25}{1}$ .

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