= Solution
The slide adds an <optical path length> $\Delta=(n-1)t=50\,\mu\mathrm m$ in the covered path. If the covered slit is at positive transverse position, its geometric path to $y>0$ is shorter, giving total path difference $\Delta-by/D$. The zero-difference fringe therefore moves toward that slit:
$$
\boxed{y_0=\frac{D(n-1)t}{b}=5.0\times10^{-2}\,\mathrm m=50\,\mathrm{mm}.}
$$
This is about $94$ fringe periods. The leading grating-angle correction, $b\sin\theta=\Delta$ and $y=D\tan\theta$, gives $50.06$ mm, consistent with the paraxial result. In a strictly monochromatic periodic pattern, identifying the continuously tracked central order is essential: the final intensity pattern determines displacement only modulo one fringe spacing.
Back to article page
