First slide

Young's double slit Experiment

Question

Two slits in Youngs experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern, $\frac{I_{\max}}{I_{\min}} is$

Moderate

A

$\frac{49}{121}$
B

$\frac{4}{9}$
C

$\frac{9}{4}$
D

$\frac{121}{49}$

Solution

$As, intensity I \propto width of slit W$
$Also, intensity I \propto square of amplitude A$
$∴ \frac{I_{1}}{I_{2}} = \frac{W_{1}}{W_{2}} = \frac{A_{1}^{2}}{A_{2}^{2}}$
$But \frac{W_{1}}{W_{2}} = \frac{1}{25} (given)$
$∴ \frac{A_{1}^{2}}{A_{2}^{2}} = \frac{1}{25} or \frac{A_{1}}{A_{2}} = \sqrt{\frac{1}{25}} = \frac{1}{5}$
$∴ \frac{I_{\max}}{I_{\min}} = \frac{{(A_{1} + A_{2})}^{2}}{{(A_{1} - A_{2})}^{2}} = \frac{{(\frac{A_{1}}{A_{2}} + 1)}^{2}}{{(\frac{A_{1}}{A_{2}} - 1)}^{2}}$
$= \frac{{(\frac{1}{5} + 1)}^{2}}{{(\frac{1}{5} - 1)}^{2}} = \frac{{(\frac{6}{5})}^{2}}{{(- \frac{4}{5})}^{2}} = \frac{36}{16} = \frac{9}{4}$

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