In a double slit experiment utilizing light of wavelength λ , a...

When a transparent plate of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits, the additional path difference introduced at the centre of the screen is $\Delta x = (\mu - 1)t$.

For the intensity at the centre of the fringe pattern to be zero, the path difference must satisfy the condition for destructive interference.

The condition for a minimum is $\Delta x = (2n - 1)\dfrac{\lambda}{2}$, where $n = 1, 2, 3, \dots$

For the minimum thickness $t$, we take $n = 1$, which gives $\Delta x = \dfrac{\lambda}{2}$.

Equating the expressions for path difference, we get:

$(\mu - 1)t = \dfrac{\lambda}{2}$

$t = \dfrac{\lambda}{2(\mu - 1)}$