$(a)\;5.67\;mm \\ (b)\;4.67\;mm \\ (c)\;6.67 \\ (d)\;3.67\;mm $

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

here $I= \large\frac{3}{4} $$(4 I_0) =3I_0$

$\therefore \cos \large\frac{\phi}{2} =\large\frac{\sqrt 3}{2}$

Thus $\large\frac{\phi}{2} $$=n \pi \pm \large\frac{\pi}{6}$

or $ \phi = 2 n \pi +\pm \large\frac{\pi}{3}$

Since$\phi =\large\frac{2 \pi}{\lambda}$$P$

$\therefore \large\frac{2 \pi}{\lambda} \frac{dy_n}{D} $$= 2n\pi \pm \large\frac{n}{3}$

or $y_n = \bigg( n \pm \large\frac{1}{6}\bigg) \large\frac{\lambda}{d}$

For point lying b/w third minima and third maxima,

$n=3$ and $y_3= \bigg(3 -\large\frac{1}{6} \bigg)\frac{\lambda}{d}$

or $y_3= \large\frac{17}{6} \frac{\lambda D}{d}$

Putting $\lambda, p,d$

$y_3=5.67\;mm$

Hence a is the correct answer.

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