$(a)\;1\;mm \\ (b)\;2\;mm \\ (c)\;2.1\;cm \\ (d)\;1.2\;cm $

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When a phase difference of exists between the two interfering waves.

$I= 4 I_0 \cos^2 \large\frac{\phi}{2}$

$I_0 =4 I_0 \cos^2 \large\frac{\phi}{2}$

$\cos \large\frac{\phi}{2} =\large\frac{1}{2}$

$\large\frac{\phi}{2}=\large\frac{\pi}{3}$

$\phi =\large\frac{2 \pi}{3}$

$\therefore $ The path difference $\Delta x$ is given by

$\phi =\large\frac{2 \pi}{3} =\large\frac{ 2 \pi }{\lambda}$$ (\Delta x)$

$\Delta x= \large\frac{\lambda}{3}$

Also If y is distance from central maxima

$\Delta x =\large\frac{yd}{D}$

$\large\frac{\lambda}{3} =y . \large\frac{d}{D}$

$y= \large\frac{\lambda}{3 \times \Large\frac{10^{-4}}{1}}$

$y= \large\frac{6 \times 10^{-7}}{3 \times 10^{-4}}$

$\qquad= 2 \times 10^{-3} m$

$\qquad= 2\;mm$

Hence b is the correct answer.

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