Past exam of the electromagnetism and optics course of the University of Oxford / 2016 / 6 / 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 isIn 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: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.