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In a Young's double slit experiment the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $\mathrm{I}$. If $\mathrm{I}_0$ denotes the maximum intensity, $\frac{\mathrm{I}}{\mathrm{I}_0}$ is equal to
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$3 / 4$
$3 / 4$
$\frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)$
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