$(a)\;\lambda \\ (b)\;3 \lambda \\ (c)\;4 \lambda \\ (d)\;2 \lambda $

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Path difference due to slab should be integral multiple of $\lambda$ or $\Delta x = n \lambda$

or $(\mu-1)t=n \lambda$$ \qquad n=1,2,3...$

or $t= \large\frac{n \lambda}{\mu-1}$

For minimum value of t , we have n =1

$\therefore t= \large\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}$

$\qquad= 2 \lambda$

Hence d is the correct answer.

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