Double-slit interference ($I=I_1+I_2+2\sqrt{I_1{I}_2}|{\gamma }_{12}|\cos \delta$)

Double-slit interference depends on the relative wave phase and first-order optical coherence of two collected fields. For equal separately measured intensities, the interferometric visibility is $|{\gamma }_{12}|$. A small source direction change ${\vartheta }_x$ introduces phase difference $2\pi D{\vartheta }_x/\lambda$ for projected interferometric baseline $D$.

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.