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 equally illuminated slits of different widths have field amplitudes $A_j(\theta )\propto w_j\mathrm{sinc}(\pi w_j\sin \theta /\lambda )$. Their interference intensity is $A_1^2+A_2^2+2A_1{A}_2\cos (2\pi d\sin \theta /\lambda +\delta )$. Near the centre, $A_2/A_1=w_2/w_1$, so the local maximum-to-minimum ratio is $(w_1+w_2{)}^2/(w_2-w_1{)}^2$. At one slit envelope's zero the interference visibility vanishes even if the other slit still transmits.

The ratio of an envelope width to a fringe period counts spatial intervals. It need not equal the number of bright fringe centres strictly inside the envelope's zeros. For a symmetric overlap from phase index $-25$ to $25$, the width spans 50 periods, but the strictly interior integer centres are $-24$ through $24$, totaling 49. Endpoints where one aperture amplitude vanishes have no two-beam contrast. State the count convention rather than treating these as inconsistent answers.

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.