Two slits in Young’s experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, is [2015]

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Solution

Q.
Two slits in Young’s experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, $\frac{I_{max}}{I_{min}}$ is [2015]

1 $\frac{49}{121}$

2 $\frac{4}{9}$

3 $\frac{9}{4}$

4 $\frac{121}{49}$

Ans.
(3)

As, intensity $I \propto$ width of slit W

Also, intensity $I \propto$ square of amplitude A

$∴$    $\frac{I_{1}}{I_{2}} = \frac{W_{1}}{W_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}}$

But   $\frac{W_{1}}{W_{2}} = \frac{1}{25}$ (given) $∴$    $\frac{A_{1}^{2}}{A_{2}^{2}} = \frac{1}{25} or \frac{A_{1}}{A_{2}} = \sqrt{\frac{1}{25}} = \frac{1}{5}$

$∴$ $\frac{I_{max}}{I_{min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}} = \frac{{(\frac{A_{1}}{A_{2}} + 1)}^{2}}{{(\frac{A_{1}}{A_{2}} - 1)}^{2}}$

   $= \frac{{(\frac{1}{5} + 1)}^{2}}{{(\frac{1}{5} - 1)}^{2}} = \frac{{(\frac{6}{5})}^{2}}{{(- \frac{4}{5})}^{2}} = \frac{36}{16} = \frac{9}{4}$

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