Double-slit experiment

The double-slit experiment compares propagation through two openings. Coherent illumination gives double-slit interference; independent incoherent source patches contribute light intensities whose displaced patterns can reduce the observed interferometric visibility.

Evacuating a transparent cell of optical length $l$ in one coherent beam changes its optical path by $l(n-1)$ relative to an unchanged cell in the other beam. With vacuum wavelength ${\lambda }_0$, the number of completed fringe periods is $N=l(n-1)/{\lambda }_0$. The initial maximum is not an additional completed period. Counting only integer fringes leaves a phase uncertainty of order one fringe unless the final fractional phase is measured; the corresponding index scale is ${\lambda }_0/l$. Fixed equal window contributions cancel in the path difference.

In oblique-incidence two-slit interference, the incident wave contributes a constant phase offset while changing observation angle changes the outgoing optical path difference by $a\Delta \sin \theta$. Subtracting the phase conditions of an adjacent minimum and maximum cancels the unknown incidence angle and order and gives the displayed wavelength. Nonadjacent extrema need their intervening phase-order difference; angular spacing should not be replaced by a small-angle formula unless justified.

For two narrow slits separated by $a$, oblique coherent illumination contributes a fixed relative incident wave phase. With a consistent angular convention, the observation-dependent phase is $\Delta \varphi =(2\pi a/\lambda )(\sin \theta -\sin \alpha )$, where $\alpha$ is the unknown incidence angle.

The term containing $\alpha$ shifts the whole fringe sequence without changing the phase difference between two observed directions. Subtracting the conditions at adjacent bright and dark extrema eliminates this offset and leaves a phase change of $\pi$. This is the basis of incidence-independent adjacent-fringe wavelength; using the absolute position of one maximum without knowing $\alpha$ would confuse source tilt with wavelength or fringe order.

A sequence of individual localized electron detections can accumulate a two-path interference distribution when the apparatus maintains coherence. The 1989 experiment led by Akira Tonomura used a field-emission source, an electron biprism, electron lenses and a position-sensitive electron counter. It realized two coherent paths without literal machined double slits. Low occupation of the instrument excludes interactions between different electrons as the origin of the fringes.^[ref]

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.