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In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, the minimum distance of the point from the central maximum is ________ $\mu \mathrm{m}$.
(Given : $\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$ )
(Given : $\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$ )
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The correct answer is:
200
$\begin{aligned} & \mathrm{I}=\mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \frac{\mathrm{I}_0}{4}=\cos ^2\left(\frac{\Delta \phi}{2}\right) \\ & \Delta \phi=\frac{2 \pi}{3} \\ & \frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3} \\ & \mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}\end{aligned}$
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