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# In a Young's double slit experiment 'd' distance between slit is $1 \times 10^{-4}\;m$ and $D=1\; m$. ( distance of screen from slits ) At a point P on the screen the resulting intensity is equal to the intensity due to individual slit $I_0$ The distance of P from central maxima when wavelength of $x=6000\;A^{\circ}$ is

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

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A)
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.