First slide

Young's double slit Experiment

Question

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the $ratio \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$ will be

Easy

A

$\frac{\sqrt{n}}{n + 1}$
B

$\frac{2 \sqrt{n}}{n + 1}$
C

$\frac{\sqrt{n}}{{(n + 1)}^{2}}$
D

$\frac{2 \sqrt{n}}{{(n + 1)}^{2}}$

Solution

$Here, \frac{I_{1}}{I_{2}} = n$
$\begin{array}{l} \frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2} = {(\frac{\sqrt{I_{1} / I_{2}} + 1}{\sqrt{I_{1} / I_{2}} - 1})}^{2} = {(\frac{\sqrt{n} + 1}{\sqrt{n} - 1})}^{2} \\ \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{\frac{I_{\max}}{I_{\min}} - 1}{\frac{I_{\max}}{I_{\min}} + 1} \\ = \frac{{(\frac{\sqrt{n} + 1}{\sqrt{n} - 1})}^{2} - 1}{{(\frac{\sqrt{n} + 1}{\sqrt{n} - 1})}^{2} + 1} = \frac{(\sqrt{n} + 1)^{2} - (\sqrt{n} - 1)^{2}}{(\sqrt{n} + 1)^{2} + (\sqrt{n} - 1)^{2}} = \frac{4 \sqrt{n}}{2 (n + 1)} = \frac{2 \sqrt{n}}{n + 1} \end{array}$

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