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The intensity of light emerging from one of the slits in a Young's double slit experiment is found to be $1.5$ limes the intensity of light emerging from the other slit. What will be the approximate ratio of intensity of an interference maximum to that of an interference minimum
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98
Hint:
$I_{1}=1.51_{2}$
$\frac{\left.\right|_{1}}{I_{2}}=\frac{3}{2} \cdot \quad \frac{l_{\max }}{I_{\min }}=\left(\frac{\sqrt{L_{1}}+\sqrt{L_{2}}}{\sqrt{I_{1}}-\sqrt{L_{2}}}\right)^{2}=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^{2}=98$
$I_{1}=1.51_{2}$
$\frac{\left.\right|_{1}}{I_{2}}=\frac{3}{2} \cdot \quad \frac{l_{\max }}{I_{\min }}=\left(\frac{\sqrt{L_{1}}+\sqrt{L_{2}}}{\sqrt{I_{1}}-\sqrt{L_{2}}}\right)^{2}=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^{2}=98$
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