Past exam of the physics 2 paper of the University of Oxford / 2021 / 9 / c

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 isThe zero-order fringe follows equal total optical path lengths, ${\Delta }_{\mathrm{tot}}=0$. HenceThe 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 $x_0\simeq 5.006\mathrm{cm}$. “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.