In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Correct Option: , 3
Amplitude $\propto$ Width of slit
$\Rightarrow A_{2}=3 A_{1}$
$\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\left(\frac{\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}}{\left|\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right|}\right)^{2}$
$\because$ Intensity $\mathrm{I} \propto \mathrm{A}^{2}$
$\Rightarrow \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\left(\frac{\mathrm{A}_{1}+\mathrm{A}_{2}}{\left|\mathrm{~A}_{1}-\mathrm{A}_{2}\right|}\right)^{2}$
$=\left(\frac{A_{1}+3 A_{1}}{\left|A_{1}-3 A_{1}\right|}\right)^{2}$
$=\left(\frac{4 A_{1}}{2 A_{1}}\right)^{2}=4: 1$