Diffraction and the Double Slit

Beginner

Step 1, light can turn corners, a little. We line things up by sighting along them because light normally goes straight. But squeeze light through a very narrow gap and it fans out on the other side. This spreading is called diffraction. Sound does the same through an open door, which is why you hear people around a corner; for light the gap must be tiny, because light waves are so short. The narrower the gap, the wider the spread. Tiny bullets would fly straight through; only waves spread like this.

Step 2, two gaps. Now give the light two narrow slits close together. Each sends out its own circular ripples, and the two sets overlap. Where a crest from one slit meets a crest from the other, they team up and the screen is bright. Where a crest meets a trough, they cancel and the screen stays dark. This adding up of waves is interference.

In the animation, straight waves arrive from the left, spread out from both slits, and the screen fills with bright and dark stripes; the orange bars show how bright each spot is.

Step 3, why stripes. The middle stripe is bright because it is equally far from both slits, so the waves arrive in step. Move up a little and one path gets longer: half a wave longer gives darkness, a whole wave longer gives the next bright stripe.

Light added to light can make darkness. Thomas Young found this with these slits around 1801, and it convinced physicists that light is a wave. Cover one slit and the stripes vanish, leaving one smooth patch.

Check yourself: at a spot on the screen one path is exactly two wavelengths longer than the other. Bright or dark? Answer: bright, because the crests arrive together again.

Intermediate

Diffraction: waves spread through openings

Light seems to travel in straight lines, which is why shadows have sharp edges. Yet every wave spreads out after passing through an opening: this is diffraction. By Huygens' principle each point of a wavefront acts as a source of new wavelets, and the wave further on is their sum. A wide opening rebuilds a straight beam; a narrow one acts almost as a single source, and the wave fans out in circles.

For a wavelength $\lambda$ and an opening of width $a$, the spread is about $\lambda /a$ radians. Sound, with wavelengths near 1 m, pours through a doorway into the whole room; green light, with $\lambda \approx 500\mathrm{nm}$, spreads from a hole 1 mm wide by only half a millimetre per metre. To show light diffracting, a slit must be a fraction of a millimetre wide.

Two slits lit by one wave

In the animation straight crests reach two narrow slits a distance $d$ apart, with a screen a distance $L$ behind them. The same crest arrives at both slits at once, so they send out wavelets in step: they are coherent sources. A point on the screen at distances $r_1$ and $r_2$ from the slits is bright or dark depending on the path difference $\Delta r=r_2-r_1$.

Where one path is longer by a whole number of wavelengths, crest arrives with crest; where it is longer by an odd number of half wavelengths, crest arrives with trough:

• $\Delta r=m\lambda \text{(bright)}$
• $\Delta r=\left(m+\frac{1}{2}\right)\lambda \text{(dark)}$

Here $m=0,\pm 1,\pm 2,\dots$ is the order of the fringe. The central fringe, $m=0$, is bright because the axis is equally far from both slits; the orders are negative below the axis and positive above it, as on the animation's readout.

From path difference to angle

Seen from a distant screen ($L\gg d$), the two paths to a point are practically parallel, at an angle $\theta$ from the central axis. A perpendicular dropped from the upper slit onto the lower path cuts off a small right triangle whose hypotenuse is $d$ and whose side opposite the angle $\theta$ is the extra path,
$\Delta r=d\sin \theta$

The conditions for bright and dark fringes therefore become conditions on the angle:
$d\sin \theta =m\lambda \text{(bright)}$
$d\sin \theta =\left(m+\frac{1}{2}\right)\lambda \text{(dark)}$
Since $\sin \theta \le 1$, at most $d/\lambda$ orders fit on each side.

Near the axis the small-angle approximation $\sin \theta \approx \tan \theta =y/L$ holds, where $y$ is the height on the screen, so the bright fringe of order $m$ sits at $y_m=m\lambda L/d$ and the fringes are equally spaced:
$\Delta y=\frac{\lambda L}{d}$
Red light, slits closer together and a more distant screen all give wider fringes.

Worked example: the pattern in the animation

The drawing is not to scale. In pixels it has $\lambda =8$, $d=50$ and $L=220$, so the slits are only 6.25 wavelengths apart, where real ones are hundreds. The small-angle formula gives
$\Delta y=\frac{\lambda L}{d}=\frac{8\cdot 220}{50}=35.2\mathrm{px}$

The first bright fringe has $\sin {\theta }_1=\lambda /d=0.16$, so ${\theta }_1=9.2^{\circ }$. Its exact position, where $r_2-r_1=\lambda$, is 35.9 px above the centre, the length of the $\Delta y$ mark beside the screen: at 9° the small-angle formula is only 2 % short.

Follow the probe dot up the screen, trailing the green path $r_1$ from the upper slit and the purple path $r_2$ from the lower one. Its readout runs from $-1.76\lambda$ at the bottom through 0 on the central fringe, 0.50 on the first dark fringe, 1.00 on the first bright one and 1.50 on the second dark one, to 1.76 at the top. So the screen holds three bright fringes ($m=0,\pm 1$) and four dark ones; the second order, at 18.7°, lies just beyond the edge at 16.5°.

The orange bars show the side fringes at only about three quarters of the central height: the width of each slit, treated at the expert level, dims them.

Worked example: measuring the wavelength of light

Light with $\lambda =600\mathrm{nm}$ falls on slits $d=0.25\mathrm{mm}$ apart, and the screen is $L=1.0\mathrm{m}$ away. How far apart are the fringes? With every length in metres,
$\Delta y=\frac{600\times 10^{-9}\cdot 1.0}{0.25\times 10^{-3}}=2.4\times 10^{-3}\mathrm{m}=2.4\mathrm{mm}$
The first fringe lies at ${\theta }_1=\lambda /d=2.4\times 10^{-3}\mathrm{rad}$, about 0.14°, where the small-angle approximation is excellent. Fringes this fine are one reason the effect went unnoticed for so long.

Turned around, $\lambda =d\Delta y/L$ measures wavelengths. With the same slits and screen, ten spacings of a laser pointer's fringes span 26 mm (measuring many spacings at once divides the error), so $\Delta y=2.6\mathrm{mm}$ and
$\lambda =\frac{d\Delta y}{L}=\frac{(0.25\times 10^{-3})\cdot (2.6\times 10^{-3})}{1.0}=6.5\times 10^{-7}\mathrm{m}=650\mathrm{nm}$

White light holds all wavelengths at once. The central fringe is white, but violet (400 nm) gives fringes 1.6 mm apart and red (700 nm) 2.8 mm apart, so each first-order fringe is a small spectrum, violet inside and red outside; after two or three orders the colours overlap into a whitish glow.

Reading the fringes on the screen

The stripes and the orange bars show the intensity $I$, the power arriving on each square metre of the screen. For two narrow, equal slits it follows a smooth ${\cos }^2$ curve, bright bands and dark gaps of equal width, with each dark fringe halfway between two bright ones: in the animation the first dark fringe is 17.8 px from the centre and the first bright one 35.9 px.

Amplitudes add, and intensity goes as the square of the amplitude: at a bright fringe the amplitude doubles and the intensity is $2^2=4$ times one slit's, while at a dark fringe it is zero. Averaged over many fringes it is exactly twice one slit's, because the light missing from the dark fringes has gone into the bright ones. Cover one slit and the fringes vanish into a broad, smooth patch.

Common mistakes

• Dark fringes are not shadows of the wall between the slits: both slits light the first dark fringe, and covering either one makes that spot a quarter as bright as the central peak.
• Moving the slits apart does not spread the fringes; it squeezes them. The spacing goes as $1/d$, so halving $d$ from 0.25 mm to 0.125 mm doubles $\Delta y$ from 2.4 mm to 4.8 mm.
• The formula $\Delta y=\lambda L/d$ holds only at small angles: for slits two wavelengths apart the first order is at 30°, where $y=L\tan 30^{\circ }=0.58L$, not $0.5L$.
• Two separate lamps, or even two lasers, make no visible fringes: their phases wander far faster than the eye can follow, so only the sum of their intensities remains. Both slits must be lit by the same wave.

Where you meet it

Gratings with hundreds of lines per millimetre measure wavelengths in most spectrometers, and the rainbow on a CD comes from its tracks, 1.6 µm apart. Diffraction also limits sharpness: light passing through a telescope or the pupil of an eye, of diameter $D$, spreads by about $\lambda /D$, as The Eye, Cameras and Telescopes explains.

X-rays, with wavelengths near 0.1 nm, diffract from the rows of atoms in a crystal and reveal where the atoms sit. Electrons and even whole molecules build up the same fringes one at a time, the starting point of Wave–Particle Duality. Next, Polarization of Light turns to another property of light waves: the direction in which they vibrate.

Advanced

Adding the two fields

Two-slit fringes come from adding the waves themselves, not their brightnesses, so the first job is to add the two fields correctly. Let the slits be a distance $d$ apart, and look at a distant point in the direction $\theta$, measured from straight ahead, at distances $r_1$ and $r_2$ from the two slits. Each slit delivers a field of the same amplitude $E_0$ and angular frequency $\omega$: $E_0\cos (\omega t-k{r}_1)$ from the first and $E_0\cos (\omega t-k{r}_2)$ from the second, where $k=2\pi /\lambda$ is the wavenumber. The term $\omega t$ is common to both, so only the phase difference matters:
$\phi =k(r_2-r_1)=kd\sin \theta$
The last step uses the far-field path difference $r_2-r_1=d\sin \theta$; the next section checks how accurate it is.

Legend

$E_0$	Amplitude of the field of one slit; both slits give the same.	$V/m$
$\phi$	Phase difference of the two fields, $\phi =kd\sin \theta$: the turn from the dashed extension of the first arrow to the second.	$\mathrm{rad}$
$\phi /2$	Half the phase difference: the angle between the resultant and the first phasor.	$\mathrm{rad}$
$E$	Resultant (total field): the sum of the two phasors, closing the triangle.	$V/m$

Figure 1 adds the two fields as phasors. A phasor is an arrow whose length is the amplitude of an oscillation and whose angle is its phase; adding two oscillations of the same frequency is then the same as adding two arrows head to tail. The blue arrow is the field of the first slit, of length $E_0$. The green arrow starts at its head and is the field of the second slit: the same length $E_0$, but turned by $\phi$ away from the dashed extension of the blue arrow. The purple arrow $E$ joins the start of the blue arrow to the tip of the green one, so it is the total field.

Now look at the shape. Two equal sides make an isosceles triangle, so its two base angles are equal, and together they equal the outside angle $\phi$ at the joint: each is $\phi /2$. The dashed altitude from the joint meets $E$ at a right angle and cuts the triangle into two equal right triangles, each with hypotenuse $E_0$ and angle $\phi /2$ at the tail of the blue arrow. Half of the resultant is therefore $E_0\cos (\phi /2)$, and the whole is
$E=2E_0|\cos (\phi /2)|$
The modulus keeps $E$ positive when $\phi$ passes $\pi$ and the cosine turns negative. Two checks: at $\phi =0$ the arrows lie in one line and $E=2E_0$; at $\phi =\pi$ they point in opposite directions and cancel.

A detector cannot follow the rapid oscillation of the field. It measures the intensity, which is proportional to the time average of the field squared, so $I\propto E^2$. Squaring $E$ and writing $\phi /2=\pi d\sin \theta /\lambda$ gives
$I=I_0{\cos }^2\left(\frac{\pi d\sin \theta }{\lambda }\right)$

Here $I_0$ is the intensity at the centre, where the two fields arrive in step. The amplitude there is doubled, so $I_0$ is four times the intensity $I_1$ of one slit alone. In Interference and Superposition $I_0$ meant one source; here it means the central peak. No energy is created: the mean of ${\cos }^2$ over many fringes is $\frac{1}{2}$, so the average intensity is $I_0/2=2I_1$, exactly what the two slits deliver together. Interference only moves light out of the dark fringes into the bright ones.

How accurate is the path difference?

The result above used $r_2-r_1=d\sin \theta$, which treats the two rays to the point as parallel. That is exact only infinitely far away, so it is worth knowing the error at a real screen.

Measure the distance $R$ and the angle $\theta$ from the midpoint of the slits. The law of cosines gives each exact distance, $r^2=R^2+d^2/4\pm Rd\sin \theta$, with the plus sign for the farther slit. Expand each square root in powers of $d/R$. Both distances contain the same second-order term $d^2{\cos }^2\theta /(8R)$, which cancels in the difference; the first surviving correction is of third order:
$\Delta r=d\sin \theta \left(1-\frac{d^2{\cos }^2\theta }{8R^2}\right)$
So the relative error is at most $(d/R{)}^2/8$: it shrinks with the square of $d/R$.

In the animation $d/R\approx 0.22$ at the first bright fringe, so the correction is only 0.6 %. On the screen, the exact curve $r_2-r_1=\lambda$ (a hyperbola) meets it at 35.9 px, the far-field formula with $y=L\tan \theta$ gives 35.7 px, and the small-angle formula $y=\lambda L/d$ gives 35.2 px.

One slit of finite width

So far each slit was a point that sends the same amplitude $E_0$ in every direction. A real slit has a width $a$, and every part of it acts as a source, so its brightness depends on direction. Adding those parts gives the slow envelope under the fringes.

Legend

$E_0$	Length of the arc: the amplitude straight ahead, where all strips are in phase.	$V/m$
$E$	Resultant of the strips: the chord from the start of the arc to its end.	$V/m$
$\rho$	Radius of the arc (dashed), in field units.	$V/m$
$\beta$	Half the angle the arc subtends at its centre; $2\beta =ka\sin \theta$ is the phase difference between the edges of the slit.	$\mathrm{rad}$

Figure 2 adds up the parts of one slit with phasors. Cut the slit into many narrow strips of equal width; each strip sends a tiny phasor, shown by the dots along the blue curve. In the direction $\theta$ each strip's path is longer than its neighbour's by the same small amount, so each phasor is turned from the one before by the same small angle. A chain of equal arrows, each turned by the same angle, bends evenly: it follows an arc of a circle, here of radius $\rho$ (the dashed radii).

How far does the arc turn? The two edges of the slit differ in path by $a\sin \theta$, so the first and last phasors differ in phase by $ka\sin \theta$. The direction along an arc turns by the same angle as the arc subtends at the centre, so that central angle is $ka\sin \theta$. Call it $2\beta$; Figure 2 draws it as two halves $\beta$ on either side of the dashed bisector. With $k=2\pi /\lambda$,
$\beta =\frac{\pi a\sin \theta }{\lambda }$

Now compare the arc with the chord. Straight ahead, at $\theta =0$, all strips are in phase, the chain is a straight line, and its length is the full amplitude $E_0$. In any other direction the tiny phasors are just as long, so the arc still has length $E_0$, and since arc length is radius times angle, $E_0=2\rho \beta$. The resultant is the purple chord $E$ from the start of the arc to its end. The dashed bisector cuts it into two halves of $\rho \sin \beta$, so $E=2\rho \sin \beta$. Dividing the two removes $\rho$:
$E=E_0\frac{\sin \beta }{\beta }$

The chord is always shorter than the arc, and it vanishes when the arc closes. At $2\beta =2\pi$ the chain curls into the full circle on the right of Figure 2, its end returns to its start (the purple dot), and $E=0$. This happens when the edges differ in path by one wavelength:
$a\sin \theta =\lambda \text{(first envelope zero)}$

A pairing argument gives the same zero without phasors. Split the slit into an upper and a lower half. Each strip in the upper half has a partner $a/2$ lower, whose path is then longer by $\lambda /2$, so the two cancel, and so does every pair.

In a double slit each slit sends the amplitude $E_0\sin \beta /\beta$ in the direction $\theta$, the same for both. Putting this in place of $E_0$ in the result of Figure 1 makes the two-slit pattern a product of fast fringes and a slow envelope:
$I=I_0{\cos }^2\left(\frac{\pi d\sin \theta }{\lambda }\right){\left(\frac{\sin \beta }{\beta }\right)}^2$

This is the graph in the animation's expert layer, whose formula writes $\sin \theta$ as $y/L$. There $a=14\mathrm{px}=1.75\lambda$ and $d=6.25\lambda$. The first side peaks of the gold curve reach 0.77 of the central height, where they touch the dashed envelope; the envelope has fallen to 0.37 at the edges of the screen, and its first zero, at 34.8°, lies far off the screen.

A bright fringe that falls exactly on a zero of the envelope is missing. With $d=3a$ the third order sits at $\sin \theta =3\lambda /d=\lambda /a$, the first envelope zero, and vanishes. In general, whenever $d/a$ is a whole number $n$, the orders $n$, $2n$, … are missing.

The pattern as a Fourier transform

The phasor sums of Figures 1 and 2 are two cases of one rule: the far-field amplitude is the Fourier transform of the aperture. This rule explains at once why the two factors multiply and why small features give wide patterns.

Legend

$A$	Transmission of the aperture: 1 inside a slit, 0 on the wall.
$x$	Position across the aperture, along the line joining the slits.	$\mathrm{m}$
$a$	Width of each slit.	$\mathrm{m}$
$d$	Distance between the centres of the two slits.	$\mathrm{m}$
$I$	Intensity of the pattern in the direction $\theta$: gold fringes under the dashed envelope.	$W/{\mathrm{m}}^2$
$\sin \theta$	Sine of the angle $\theta$ of the outgoing light, measured from straight ahead; the horizontal axis of the pattern.
$\lambda /d$	Fringe period: the spacing of neighbouring bright fringes in $\sin \theta$.
$\lambda /a$	Value of $\sin \theta$ at the first zeros of the envelope, marked by rings where a fringe is missing.

Figure 3 puts an aperture above its far-field pattern. The top graph is the transmission $A$ against the position $x$ across the aperture: $A=1$ inside each slit and $A=0$ on the wall between them. The slits have width $a$ (purple bar) and centres a distance $d$ apart (gold bar). The bottom graph is the intensity $I$ against $\sin \theta$: gold fringes under a dashed purple envelope. Match the colours. The gold spacing $d$ sets the gold fringe period $\lambda /d$, and the purple width $a$ sets the purple envelope, whose first zeros are at $\sin \theta =\pm \lambda /a$. Here $d=3a$, so the third fringe on each side lands on an envelope zero and is missing, at the rings.

Here is the rule. Far from the aperture, a strip of width $dx$ at the position $x$ has a path that differs by $x\sin \theta$ from that of the strip at $x=0$, by the same geometry as for the two slits, so its phase differs by $kx\sin \theta$. Write each contribution as a complex exponential, weight it by the transmission $A(x)$ and add over all strips:
$E(\theta )\propto \int A(x)e^{-iqx}dx,q=k\sin \theta$
This integral is the Fourier transform of $A(x)$, with $q$ in the role of the spatial frequency.

It reproduces both earlier results. Two very narrow slits at $x=\pm d/2$ give $e^{-iqd/2}+e^{iqd/2}=2\cos (qd/2)$, and since $qd/2=\pi d\sin \theta /\lambda$, its square is the ${\cos }^2$ fringes. One slit of width $a$, from $-a/2$ to $a/2$, gives $a\sin (qa/2)/(qa/2)=a\sin \beta /\beta$, because $qa/2=\beta$: the envelope.

Two wide slits are one wide slit copied to two places. In the language of transforms, the aperture is the single slit convolved with the pair of narrow slits, and the transform of a convolution is the product of the transforms. That is why the fringe factor and the envelope multiply.

The transform also shows the reciprocity behind all diffraction: a feature of size $s$ in the aperture makes structure on the angular scale $\lambda /s$. In Figure 3 the large spacing $d$ makes the fine fringes and the small width $a$ the broad envelope. A lens forms the far-field pattern in its focal plane, at a finite distance, and the X-ray patterns that revealed the structures of crystals are the squared magnitudes of the transform of the crystal's electron density.

Many slits: the diffraction grating

Adding more slits at the same spacing keeps the bright fringes in the same places but makes them far sharper and brighter. That sharpness is what makes a grating a precise tool for measuring wavelengths.

Legend

$E_0$	Amplitude of the field of one slit: the length of each phasor.	$V/m$
$\phi$	Phase step between neighbouring slits, $\phi =kd\sin \theta$: the turn from one phasor to the next.	$\mathrm{rad}$
$E$	Resultant of all the phasors: the closing side of the polygon.	$V/m$

The left half of Figure 4 adds the fields of $N=5$ equal slits with spacing $d$. Each blue arrow has length $E_0$ and is turned by the same angle $\phi =kd\sin \theta$ from the one before; the angle is marked at one joint. Equal arrows with equal turns trace part of a regular polygon, so all the corners lie on one circle (dashed). The purple resultant $E$ is the closing side of the polygon, from the first tail to the last head.

The circle gives the length of $E$. A chord that subtends an angle $\alpha$ at the centre of a circle of radius $\rho$ has length $2\rho \sin (\alpha /2)$. Each side turns by $\phi$, so, as in Figure 2, it subtends $\phi$ at the centre and $E_0=2\rho \sin (\phi /2)$. The closing side spans all $N$ sides and subtends $N\phi$, so $E=2\rho |\sin (N\phi /2)|$. Dividing removes $\rho$, and squaring gives
$I=I_1{\left(\frac{\sin (N\phi /2)}{\sin (\phi /2)}\right)}^2$
where $I_1$ is the intensity from one slit. For $N=2$ the ratio is $\sin \phi /\sin (\phi /2)=2\cos (\phi /2)$, so this is $4I_1{\cos }^2(\phi /2)$ again, the two-slit result.

Whenever $\phi$ is a whole multiple of $2\pi$, all the arrows point the same way and line up: $E=N{E}_0$, and the intensity reaches $N^2{I}_1$. These principal maxima stay where $d\sin \theta =m\lambda$, just as for two slits.

The same chain of arrows also sets how narrow each maximum is. Moving away from a maximum, the chain curls up and first closes at $\phi =2\pi /N$: that is the right half of Figure 4, a closed pentagon with $E=0$ (the purple dot). So the half-width of each maximum, from its peak to its first zero, is $1/N$ of the spacing $2\pi$ between maxima, $N/2$ times narrower than with two slits. Between neighbouring maxima lie $N-1$ zeros and $N-2$ weak secondary peaks. The light is concentrated, not created: the mean intensity is still $N{I}_1$.

A grating with 500 lines per millimetre has a spacing of 2.0 µm. For 600 nm light $\lambda /d=0.3$, so $\sin \theta =m\lambda /d$ is 0.3, 0.6 and 0.9 in the first three orders: they leave at 17.5°, 36.9° and 64.2°. A fourth order would need $\sin \theta =1.2$ and does not exist.

Two nearby wavelengths count as separated when the maximum of one falls on the first zero of the other. Take the angle of the order-$m$ maximum of $\lambda +\Delta \lambda$, where $d\sin \theta =m(\lambda +\Delta \lambda )$. There the phase step for $\lambda$ is $2\pi m+2\pi m\Delta \lambda /\lambda$, and it must exceed $2\pi m$ by the first-zero step $2\pi /N$. Setting $m\Delta \lambda /\lambda =1/N$ gives the resolving power
$\lambda /\Delta \lambda =mN$
The yellow sodium lines, 589.0 and 589.6 nm, need about 1000 lines in the first order. 2 cm of this grating holds 10,000 lines and, in the first order, separates lines 0.06 nm apart at 600 nm.

Worked problem: slit sizes from a pattern

Laser light of 600 nm passes a double slit, and a screen 1.0 m away shows fringes 2.4 mm apart. The fringes fade towards the sides, and the fifth on each side, at ±12 mm, is missing, leaving nine in the central band. Find the slit separation $d$, the slit width $a$, and the brightness of each fringe.

The fringe spacing gives the separation. Neighbouring bright fringes differ in path by one wavelength, so for small angles $\Delta y=\lambda L/d$, and
$d=\lambda L/\Delta y=(600\times 10^{-9}\cdot 1.0)/(2.4\times 10^{-3})=0.25\mathrm{mm}$

The gap gives the width. The missing fringe sits on the envelope's first zero, $a\sin \theta =\lambda$, which for small angles is at $y=\lambda L/a=12\mathrm{mm}$, so
$a=(600\times 10^{-9}\cdot 1.0)/(12\times 10^{-3})=0.050\mathrm{mm}$
Thus $d=5a$, which is why the fifth order is the one missing. A check on the far field: $a^2/\lambda =4.2\mathrm{mm}$, far below 1.0 m, so the far-field formulas apply.

At the fringe of order $m$, $\sin \theta =m\lambda /d$, so $\beta =\pi a\sin \theta /\lambda =\pi ma/d=\pi m/5$. Every bright fringe has ${\cos }^2=1$, so its brightness is set by the envelope alone: $(\sin \beta /\beta {)}^2$ gives 1, 0.88, 0.57, 0.25 and 0.05 of the centre for $m=0$ to 4, then 0 for $m=5$. Beyond the gap the fringes return in a side band whose brightest fringe stays below 5 % of the centre.

The animation has $d/a=50/14=3.57$, not a whole number, so no order is lost completely. Its first orders reach 0.77, the second 0.31 and the third only 0.033, just inside the envelope's zero at $m=3.57$; a wide enough screen would show seven fringes in the central band. Slits widened to $a=d/3\approx 16.7\mathrm{px}$ would remove the third order exactly.

One quantum at a time

The same pattern survives when the light is so dim that it arrives one photon at a time, and it appears for electrons too. Here the rule for adding fields becomes the central rule of quantum mechanics.

Dim the light until photons arrive one at a time: the screen records single flashes at random points, yet after many thousands of flashes they pile up into the same ${\cos }^2$ fringes. Electrons do the same, with the de Broglie wavelength $\lambda =h/p$, where $p$ is the momentum and $h$ Planck's constant. Accelerated through 100 V, an electron has $\lambda =0.123\mathrm{nm}$, so slits 1 µm apart would give fringes 0.12 mm apart on a screen 1 m away.

The rule is the one for fields, applied to probability amplitudes. Each way of reaching a point has a complex amplitude, ${\varphi }_1$ through one slit and ${\varphi }_2$ through the other, and the probability of arriving there is the squared magnitude of their sum:
$P=|{\varphi }_1+{\varphi }_2{|}^2=|{\varphi }_1{|}^2+|{\varphi }_2{|}^2+2|{\varphi }_1||{\varphi }_2|\cos (k\Delta r)$
The first two terms are the single-slit patterns. The last term is the interference, with $k=p/\hslash$ and $\hslash =h/(2\pi )$; without it the slits would simply add their single-slit patterns, with no fringes.

Recording which slit each particle passes removes the cross term, and the fringes go with it. The reason is the kick the measurement gives. Light that can tell the slits apart must have a wavelength below about $d$, so each of its photons carries a momentum of at least about $h/d$. Scattering one turns the particle's path by up to about $(h/d)/p=\lambda /d$, which is one fringe spacing in angle, and kicks of random size smear the fringes away, as Wave–Particle Duality explains.

Limits of validity

The scalar theory used here treats light as a single wave amplitude, ignoring the direction of its electric field, and treats the slits as holes in a perfectly black screen. That works for slits several wavelengths wide. A slit narrower than one wavelength has no envelope zero ($a\sin \theta =\lambda$ needs $a\ge \lambda$), spreads light almost evenly behind it, and transmits differently for fields along it and across it, a polarization effect the scalar picture leaves out.

Fringes also need coherence: the waves must keep a steady phase relation. A spread of wavelengths $\Delta \lambda$ limits the coherence length, the path difference over which the phase stays steady, to about ${\lambda }^2/\Delta \lambda$. For white light that is about 1 µm, only two or so wavelengths, so only the first two or so orders show colours, while a laser, coherent over centimetres to metres, gives fringes far out.

A source of angular size $\alpha$ blurs the pattern too. Each point of the source makes its own copy of the pattern, shifted in angle by its own direction, and the copies wash out once $\alpha$ approaches the fringe spacing $\lambda /d$. For the 0.25 mm slits of the worked problem that is 2.4 mrad, but the Sun spans 9.3 mrad, so direct sunlight makes fringes only through slits closer than about 0.06 mm: hence the pinhole or single slit in front of classic set-ups. Partial coherence lowers the visibility $V=(I_{\max }-I_{\min })/(I_{\max }+I_{\min })$ below 1.

Finally, the envelope formula assumes the far field, $L\gg a^2/\lambda$; nearer, in the Fresnel regime, the pattern changes shape with distance. The two-slit fringe positions need only $d\ll L$, as the section on the path difference showed. That is why the animation's fringes obey $d\sin \theta =m\lambda$ although $d^2/\lambda =312\mathrm{px}$ exceeds $L=220\mathrm{px}$.

Where this leads

Many slits make the grating at the heart of most spectrometers, and the line spectra they record lead to Atomic Models: the Bohr Atom. A round aperture of diameter $D$ spreads light into a spot of angular radius $1.22\lambda /D$, the resolution limit of The Eye, Cameras and Telescopes. The fields that add here are those of Electromagnetic Waves, and their direction of oscillation is the subject of Polarization of Light.

The amplitude rule $P=|{\varphi }_1+{\varphi }_2{|}^2$ is the core of quantum mechanics: Wave–Particle Duality builds it up from single electrons, and Schrödinger's Equation and Tunnelling turns it into a wave equation for matter.