In a double slit experiment, when a thin film of thickness t having refractive index μ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ($$\lambda$$ is the wavelength of the light used):

In a Young’s double-slit experiment, the condition for constructive interference at any point on the screen is that the path difference between the two interfering waves must be an integral multiple of the wavelength. Expressed mathematically, the path difference $$\Delta$$ required for a bright fringe is

$$\Delta = m\lambda \quad (m = 0,1,2,\dots)$$

Normally, at the centre of the screen the geometrical path difference between the two slits is zero, so the central point corresponds to $$m = 0$$ and is therefore bright.

Now we introduce a thin transparent film of thickness $$t$$ and refractive index $$\mu$$ just in front of one of the slits. Whenever a wave travels a distance $$t$$ inside a medium of refractive index $$\mu$$, its optical path length becomes $$\mu t$$. In the absence of the film the same geometrical distance would count as simply $$t$$ (because it would be travelled in air or vacuum, whose refractive index is $$1$$). Therefore, the additional optical path introduced by the film is

$$\text{Extra path difference} = \mu t - t = (\mu - 1)t.$$

This extra path difference shifts the entire interference pattern. If the shift of the central maximum equals one complete fringe width, it means that the point which was originally the central maximum now satisfies the bright-fringe condition for the next integral order. In other words, the optical path difference has effectively increased by exactly one wavelength $$\lambda$$:

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

We now solve this simple algebraic equation for the thickness $$t$$. Dividing both sides by $$(\mu - 1)$$ we get

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

This is precisely the required thickness that shifts the central bright fringe by one fringe width when the film is inserted in front of one slit.

Hence, the correct answer is Option D.