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  • In 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 minimum intensity in the int

In 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 minimum intensity in the interference pattern.
Option: 1 2:1
Option: 2 4:1
Option: 3 3:1
Option: 4 1:4

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\begin{aligned} &\text { Amplitude } \propto \text { Width of slit }\\ &\begin{array}{l} \Rightarrow \mathrm{A}_{2}=3 \mathrm{~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} \end{array} \\ &\because \text { 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}\\ \\ &\begin{array}{l} \Rightarrow \frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=\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 \end{array} \end{aligned}

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