= Solution
The slide adds <optical path length> $\Delta=(n-1)t=0.5(0.1\,\mathrm{mm})=50\,\mu\mathrm m$ to the covered beam. Put the covered slit at transverse coordinate $+s/2$. For positive screen coordinate $x$ its geometrical path is shorter by approximately $sx/d$, so the total covered-minus-uncovered <optical path difference> is
$$
\Delta_{\rm tot}=(n-1)t-\frac{sx}{d}.
$$
The zero-order fringe follows equal total <optical path lengths>, $\Delta_{\rm tot}=0$. Hence
$$
\boxed{x_0=\frac{d(n-1)t}{s}=0.050\,\mathrm m=5.0\,\mathrm{cm},\quad\text{towards the covered slit}.}
$$
The covered beam needs a shorter geometric path to compensate its extra <optical path length>. This is the paraxial estimate; exact point-to-point distances give $x_0\simeq5.006\,\mathrm{cm}$. “Central” here follows the zero-order fringe continuously as the slide is introduced. At one fixed <wavelength>, the final <monochromatic wave> <interference> pattern is periodic, so intensity alone identifies a fringe position only modulo the fringe spacing. Slide reflections and absorption have been neglected.
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