Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section B / B7 / d / Solution

A relative phase $\pi$ changes the sign of one slit field. Their sum becomesThe central interference fringe is now dark. The maxima move to $D\sin \theta =(m+1/2)\lambda$, while the minima occur at $D\sin \theta =m\lambda$. Thus the pattern shifts by half a fringe, $2.00\mathrm{mrad}$ near the axis, but its angular fringe separation remains $\boxed{4.00\mathrm{mrad}}$. This assumes the glass changes phase uniformly and introduces negligible absorption or beam deflection.