1. Effect of introducing a thin film in one path

When a thin film of thickness $t$ and refractive index $\mu$ is placed in front of one slit, the optical path in that arm increases by

$\Delta = (\mu - 1)t$

This is the extra optical path difference introduced between the two waves.

2. Relation between fringe shift and path difference

In Young’s double slit experiment, if an extra path difference $\Delta$ is introduced, the fringe pattern shifts by

$n = \frac{\Delta}{\lambda}$

fringe widths.

Given that the central maximum shifts by one fringe width, we have

$\frac{\Delta}{\lambda} = 1$

So,

$\Delta = \lambda$

3. Substitute the expression for $\Delta$

$(\mu - 1)t = \lambda$

Therefore,

$t = \frac{\lambda}{\mu - 1}$

4. Match with the options

This corresponds to:

$\boxed{\frac{\lambda}{\mu - 1}}$

So the correct option is D.

5. Comparison with stored correct answer

Stored correct answer: D

Derived answer: D

Hence, the derived answer agrees with the stored correct answer.