In Young's double slit experiment, the ratio of slit widths is 4 : 1. The intensity ratio in the interference pattern would be:

9 : 1

The correct answer is Option (3) → 9 : 1

Ratio of the slit width, $\frac{W_1}{W_2}=\frac{4}{1}$

let $l_1,l_2$ are intensity of light coming from two slits -

As, $I∝W$

I = intensity

$∴\frac{I_1}{I_2}=\frac{4}{1}$

$∴\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}=\frac{(\sqrt{\frac{I_1}{I_2}}+1)^2}{(\sqrt{\frac{I_1}{I_2}}-1)^2}$

$=\frac{(\sqrt{4}+1)^2}{(\sqrt{4}-1)^2}=\frac{9}{1}$