Question

Two slits in Young's experiment have widths in the ratio of $1:25$. The ratio of intensity at the maxima and minima in the interference pattern, $I_{max}/I_{min}$ is?

• A. $4/9$
• B. $9/4$
• C. $121/49$
• D. $49/121$

Answer 1

The correct option is B $9/4$

Given,

$\frac{I_1}{I_2}=\frac{W_1}{W_2}=\frac{1}{25}$

Therefore,

$\frac{I_{max}}{I_{min}}=\frac{{(\sqrt{I_2}+\sqrt{I_1})}^2}{{(\sqrt{I_2}-\sqrt{I_1})}^2}={\left[\frac{\sqrt{\frac{I_2}{I_1}}+1}{\sqrt{\frac{I_2}{I_1}}-1}\right]}^2$

Substitute all value in above equation

$\frac{I_{max}}{I_{min}}={\left[\frac{\sqrt{\frac{25}{1}}+1}{\sqrt{\frac{25}{1}}-1}\right]}^2={\left[\frac{5+1}{5-1}\right]}^2={\left[\frac{6}{4}\right]}^2=\frac{36}{16}=\frac{9}{4}$

Option B is correct option