Phase-reversed double-slit diffraction

Two equal rectangular apertures of width $a$, centre spacing $D$, and relative phase $\pi$ have Fraunhofer diffraction amplitude proportional to $2ia\mathrm{sinc}(qa/2)\sin (qD/2)$, with $q=(2\pi /\lambda )\sin \theta$ and $\mathrm{sinc}z=\sin z/z$. HenceThe interference zeros satisfy $D\sin \theta =m\lambda$, including the central dark fringe. The single-aperture envelope vanishes at $a\sin \theta =n\lambda$ for nonzero integer $n$. A uniform phase reversal translates the narrow-slit fringes by half a fringe without changing their spacing.

Two adjacent half-slits of width h with opposite field transmission have centres separated by h. The aperture transform is $h\mathrm{sinc}(qh/2)[e^{-iqh/2}-e^{iqh/2}]$. Thus the phase-reversed pair factor multiplies the finite-half-slit envelope, giving the displayed normalized intensity with $s=qh/2$ and $\mathrm{sinc}s=\sin s/s$. The on-axis field cancels. Positive side maxima satisfy $\tan s=2s$, and total integrated transmission is unchanged by a lossless phase plate even though the angular distribution changes.