When a thin transparent sheet is placed in front of a Young's...

The fringe width in Young's double slit experiment is given by $\beta = \dfrac{\lambda D}{d}$.

When a thin transparent sheet of thickness $t$ and refractive index $\mu$ is placed in front of one of the slits, an additional path difference of $(\mu - 1)t$ is introduced. This causes the entire fringe pattern to shift by a distance $\Delta y = \dfrac{D}{d}(\mu - 1)t$.

The fringe width $\beta$ depends only on the wavelength $\lambda$, the distance between the slits and the screen $D$, and the distance between the slits $d$. Since these parameters remain unchanged, the fringe width remains the same.