Past exam of the physics 2 paper of the University of Oxford / 2014 / 7 / a / 2 / 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: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.