Question

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

• A. $\frac{49}{121}$
• B. $\frac{4}{9}$
• C. $\frac{9}{4}$
• D. $\frac{121}{49}$

Answer 1

The Correct Option is C

(c): As, intensity $ I \propto $ width of slit W
Also, intensity $ I \propto $ square of amplitude A
$ \therefore \, \, \, \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)
$ \therefore \frac{ {A_1}^2}{{A_2}^2} = \frac{1}{25} \, \, or \, \, \frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$
$\therefore \, \, \, \frac{I_{max}}{I_{min}} = \frac{ (A_1+A_2)^2}{(A_1-A_2)^2} = \frac{ \bigg( \frac{A_1}{A_2}+1\bigg)^2 }{ \bigg( \frac{A_1}{A_2}-1 \bigg)^2}$
$ = \frac{\bigg( \frac{1}{5}+1 \bigg)^2}{ \bigg(\frac{1}{5}-1 \bigg)^2} = \frac{ \bigg(\frac{6}{5}\bigg)^2}{\bigg(-\frac{4}{5}\bigg)^2} = \frac{36}{16} =\frac{9}{4}$