$(a)\;6 \times 10^{-7} \;m \\ (b)\;3 \times 10^{-7} \;m \\ (c)\;4 \times 10^{-6}\;m \\ (d)\;3 \times 10^{-6} $

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Fringe width $= \large\frac{D \lambda}{d}$

Where d - is separation between two sources.

Since $S_2 $ is image of $S_1$

$S_1 S_2 =2 \times 1 m =2m$

$\therefore \beta =\large\frac{600 \times 10^{-9} \times 1}{2}$

$\qquad= 300 \times 10^{-9} \;m$

$\qquad= 3 \times 10^{-7} \;m$

Hence b is the correct answer.

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