= Solution
Put the plate behind the slit at transverse coordinate $+d/2$, and let positive screen coordinate $x$ point toward that slit. The extra <optical path length> is
$$
\Delta=(n_1-1)t=0.52(200\,\mu\mathrm m)=104\,\mu\mathrm m.
$$
In the <paraxial approximation>, the covered path minus the uncovered path has total <optical path difference> $\Delta-dx/L$. Its zero determines the shifted central <interference fringe>:
$$
\boxed{x_0=\frac{L(n_1-1)t}{d}=1.04\times10^{-2}\,\mathrm m=10.4\,\mathrm{mm},}
$$
\b[toward the slit covered by the plate]. Moving toward that slit shortens its geometric path and compensates the extra delay in the plate.
More generally, the <phase difference> is $(2\pi/\lambda)(\Delta-dx/L)$, so every <interference fringe> translates by the same $x_0$ in this approximation. The spacing $\lambda L/d$ is unchanged, and the shift corresponds to $\Delta/\lambda$ fringe periods. For an ideal uniform transparent plate that changes only the <wave phase>, the envelope and <interferometric visibility> are unchanged. Reflection losses or unequal transmission through a real plate could additionally reduce the <interferometric visibility>; those are separate amplitude effects. No numerical <wavelength> is needed for the physical displacement of the zero-order fringe.
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