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Question
In a Young's double slit experiment the intensity at a point where the path difference is ${\lambda \over 6}$ ( $\lambda $ being the wavelength of light used ) is $I$. If ${I_0}$ denotes the maximum intensity, ${I \over {{I_0}}}$ is equal to
${3 \over 4}$
${1 \over {\sqrt 2 }}$
${{\sqrt 3 } \over 2}$
${1 \over 2}$

Solution

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