 Solution

For the narrow pair, the amplitude is $e^{-iqd/2}-e^{iqd/2}=-2i\sin (qd/2)$. Hence phase-reversed double-slit diffraction givesThere is a central dark fringe. Minima obey $d\sin \theta =m\lambda$, maxima $d\sin \theta =(m+1/2)\lambda$, and the near-axis angular fringe period is $\lambda /d=0.005\mathrm{rad}$. Finite slit widths multiply this interference profile by their single-slit envelope. The parent plot includes the narrow-pair profile.