In young’s double slit interference experiment, using two coherent waves of different amplitudes, the intensities ratio between bright and dark fringes is 3. Then, the value of the ratio of the amplitudes of the wave that arrive there is:

\frac{\sqrt{3} + 1}{\sqrt{3} - 1}

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\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

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\sqrt{3} : 1

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1 : \sqrt{3}

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Solution

The correct option is A $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$
$\frac{I_{m a x}}{I_{m i n}} = \frac{3}{1} \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}} = \frac{3}{1} \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} = \frac{\sqrt{3}}{1} \sqrt{3} \sqrt{I_{1}} - \sqrt{3} \sqrt{I_{2}} = \sqrt{I_{1}} + \sqrt{I_{2}} \sqrt{3} \sqrt{I_{1}} - \sqrt{I_{1}} = \sqrt{I_{2}} + \sqrt{3} \sqrt{I_{2}} (\sqrt{3} - 1) \sqrt{I_{1}} = \sqrt{I_{2}} (\sqrt{3} + 1) \frac{\sqrt{I_{1}}}{\sqrt{I_{2}}} = \frac{A_{1}}{A_{2}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

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