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

Let $I_s$ be the intensity that either narrow slit would produce alone, and let $y$ measure screen displacement from the perpendicular bisector. The optical path difference is approximately $\delta =b\sin \theta \simeq by/D$. Equal coherent field amplitudes then giveThe maxima are at $y_m=m\lambda D/b$; the minima are halfway between them. HereThe expression uses the paraxial approximation and neglects the slowly varying single-slit diffraction envelope because the slit widths are unspecified.

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.