In Young's double slit experiment, the ratio of slit widths is 4 : 1. The intensity ratio in the interference pattern would be:
Answer & explanation
Correct answer: option 3
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
$∴\frac{I_1}{I_2}=\frac{4}{1}$
$∴\frac{I_{max}}{I_{min}}=\frac{(\sqrt{I_1}+\sqrt{I_2})^2}{(\sqrt{I_1}-\sqrt{I_2})^2}=\frac{(\sqrt{\frac{I_1}{I_2}}+1)^2}{(\sqrt{\frac{I_1}{I_2}}-1)^2}$
$=\frac{(\sqrt{4}+1)^2}{(\sqrt{4}-1)^2}=\frac{9}{1}$