$(a)\;\frac{1}{2 \sqrt 3}\;m \\ (b)\;\frac{1}{\sqrt 3} m \\ (c)\;\frac{1}{\sqrt 2}m \\ (d)\;\frac{2}{\sqrt 3}\;m $

$D >> d$

Hence the path difference $\Delta P= d \sin \theta$

$\large\frac{d}{\lambda}$$=2$

Clearly $, n << \large\frac{d}{\lambda}$$=2$

It is not possible for any values of n .

Hence $\Delta P=\large\frac{dy}{D} $ can not be used

for 1st maxima,

$\Delta P= d \sin \theta=\lambda$

$=> \sin \theta =\large\frac{\lambda}{D} =\frac{1}{2}$

$=> \theta= 30^{\circ}$

Hence $y= D \tan \theta =\large\frac{1}{\sqrt 3}$$m$

Hence b is the correct answer.

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