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In Young's double slit experiment with a monochromatic light, maximum intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?
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The correct answer is:
$1 / 9$
$\left(\sqrt{I_1}+\sqrt{I_2}\right)^2=4\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
$\sqrt{l_1}+\sqrt{I_2}=2\left(\sqrt{l_1}-\sqrt{I_2}\right), 3 \sqrt{I_2}=\sqrt{I_1}, \frac{l_2}{I_1}=\frac{1}{9}$
$\sqrt{l_1}+\sqrt{I_2}=2\left(\sqrt{l_1}-\sqrt{I_2}\right), 3 \sqrt{I_2}=\sqrt{I_1}, \frac{l_2}{I_1}=\frac{1}{9}$
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