From adding waves to single detection events

Two openings in front of a screen might suggest two bright marks. Instead, light can make many alternating bright and dark bands. The dark places are particularly puzzling: how can adding another opening leave some places darker? To approach that question, we look at two waves arriving at the same position. Their changes can reinforce or cancel, depending on how the waves line up; this is interference. We then connect that alignment to the distances travelled through the two openings. Keep the dark bands in mind when individual detections begin to build the screen pattern: that is the observation the explanation must account for.

1 · What is wave interference?

Take a small section of a rope. At any instant its displacement can be above or below its resting position, so displacement has a sign. If two small waves overlap, their displacements add at the same place and time. This is the superposition principle for linear waves.

Phase says how far a repeating oscillation has progressed through its cycle. One complete cycle is 360°, or $2\pi$ radians. A phase difference compares two oscillations; 180° means half a cycle apart. Wavelength $\lambda$ is the distance between successive crests of the same travelling wave.

The first two traces below show $y_1$ and $y_2$ ; the third is their sum. At 0° phase difference, crests meet crests and the displacement doubles. At 180°, a crest meets an equal trough and the total displacement is zero. Use the slider and pause the motion to compare the three traces at the same horizontal position. Interference means this addition, including reinforcement and cancellation.

Relative phase 0 °

Crests arrive together: constructive interference.

y₁(0) = 0.00, y₂(0) = 0.00, y(0) = 0.00

Light is an electromagnetic wave, not a vibrating rope, but its electric-field amplitudes also add in a linear medium. For monochromatic light of a fixed frequency, average intensity is proportional to the squared amplitude. Doubling amplitude gives four times the intensity; a zero combined amplitude gives a dark region. The rope is a model for signed addition, not a claim that light needs a material rope.

y=y_1+y_2,\qquad I\propto A_{\rm total}^2

2 · Two paths give different arrival phases

Illuminate two narrow slits with the same coherent light source. “Coherent” means that their relative phase remains stable enough for a stable pattern. Beyond each slit the wave spreads. A point on the screen receives contributions from both slits.

At the centre, the distances are equal, so the waves arrive in phase. Move to one side of the screen and one distance becomes longer. One extra wavelength means one full cycle, or 2π radians, so it still reinforces. Half an extra wavelength means opposite phases and cancellation. Different screen positions therefore alternate between bright and dark.

Choose one point and follow the extra distance

Keep both slits coherent. The marked screen point in the diagram moves with this control. Above the centre is positive $y$ . The drawing enlarges the small vertical distances; the calculation uses metres. We compare two possible wave contributions to this same point.

Screen position

y=0.000\,\mathrm{mm}

The two slits are $d/2$ above and below the centre. Each path is the hypotenuse of a right triangle: squared length equals the sum of the squared horizontal and vertical distances. Call the upper path $r_1$ and the lower one $r_2$ .
$r_1=\sqrt{L^2+(y-d/2)^2}=1.500000003\,\mathrm m$
$r_2=\sqrt{L^2+(y+d/2)^2}=1.500000003\,\mathrm m$
Subtract the paths. The difference is tiny compared with either whole path, so express it in nanometres, the unit used for wavelength.
$\Delta r=r_2-r_1=0.00\,\mathrm{nm},\quad 1\,\mathrm m=10^9\,\mathrm{nm}$
Divide by one wavelength to find the extra number of cycles. One full cycle is $360^\circ$ ; half a cycle puts a crest against a trough.
$\Delta r/\lambda=0.00/550=0.0000,\quad\Delta\phi=0.00^\circ$
Whole extra cycles reinforce; half-integer extra cycles cancel for equal amplitudes. In between, there is partial reinforcement. The value below compares this interference factor with its maximum of $1$ . The finite slit width still supplies the broader envelope shown in the full experiment.
$\cos^2(\pi\Delta r/\lambda)=1.000$

The dashed lines compare path lengths to one point; they do not assign a photon a hidden route. The arcs indicate spreading waves and are schematic. $d$ is slit separation, $L$ is screen distance, $y$ is displacement along the screen, and $\theta$ is the angle from the central direction. In the far field the path difference is $d \sin \theta$ ; close to the centre, $\sin \theta \approx y/L$ .

Wavelength λ 550 nmSlit separation d 0.18 mm

Preparing the model…

\Delta r=d\sin\theta\approx\frac{dy}{L},\qquad\Delta\phi=\frac{2\pi\Delta r}{\lambda}

3 · What changes when detections arrive one at a time?

At sufficiently low light intensity, a sensitive detector records separated, localised events. Each event marks one position. Repeating identical preparations builds a distribution with interference fringes. One event is not a whole stripe, and a stripe is not the path of one photon.

Quantum theory assigns amplitudes to the alternatives. With both slits open and no distinguishing path record, the probability density is proportional to $\lvert \psi_1 + \psi_2\rvert^2$ . We add amplitudes first, then square. Simply adding $\lvert \psi_1\rvert^2$ and $\lvert \psi_2\rvert^2$ removes the interference term and predicts a different distribution. The simulation below samples these probabilities; it does not solve the measurement problem or display a definite trajectory between source and detector.

4 · Build a fresh pattern, one point at a time

First use “One detection” several times. Each click produces exactly one simulated detected event. Then start slow accumulation and finally switch to faster accumulation. Both rates use the same probability distribution. Blue dots are the individual events. Below them, blue bars count events in equal-width screen bins; the amber curve is the expected count for the same total sample size.

The screen is now shown facing us. Its horizontal coordinate runs from −25 to +25 mm; vertical dot positions are spread uniformly only to keep the individual marks visible. The physical intensity calculation is one-dimensional. Changing wavelength, slit separation or the path condition clears the screen so results from different experiments cannot get mixed together.

Wavelength λ 550 nmSlit separation d 0.18 mmPath condition

Two-slit near-axis spacing

\lambda L/d

4.58 mm

Accumulation rateN = 0

Counts are conditioned on detection inside the displayed screen window. Time is slowed for teaching. Dots stay after each event; a brief ring marks the newest one. After 6,000 detections the run stops; clear the screen for a new run.

With coherent alternatives, the distribution has repeated minima and maxima under a wider diffraction envelope. A small sample need not look like this smooth curve; agreement is statistical and improves with more events.

Compare 550 nm with 750 nm: the fringes spread farther apart. Reset the screen and compare $d = 0.18\ \mathrm{mm}$ with $d = 0.30\ \mathrm{mm}$ : the fringes move closer together. The single-slit width stays at 0.04 mm. Notice that changing accumulation speed changes waiting time, not fringe spacing.

Check: can one detection reveal the interference pattern?

No. It gives one position. A distribution is a statement about many repeated trials. A point in a relatively dim region is not a contradiction: dim means lower probability, while a true zero of the ideal density is a special limiting case.

5 · Where the stripe formula comes from

Now give the earlier addition a mathematical form. At one screen point, take equal amplitudes $A$ , common angular frequency $\omega$ , and a phase difference $\Delta \varphi$ . The cosine addition identity combines the two oscillations into one with a new amplitude.

y_1=A\cos\omega t,\quad y_2=A\cos(\omega t-\Delta\phi)

The factor multiplying the time-dependent cosine is 2A $\cos (\Delta \varphi /2)$ . Squaring that amplitude gives the interference factor $\cos^2(\Delta \varphi /2)$ . Substitute $\Delta \varphi \approx 2\pi dy/(\lambda L)$ . A finite-width slit also diffracts, so this factor is multiplied by the single-slit envelope.

y_1+y_2=2A\cos(\Delta\phi/2)\cos(\omega t-\Delta\phi/2)

\begin{aligned}I(y)&\propto\operatorname{sinc}^2\!\left(\frac{\pi ay}{\lambda L}\right)\\ &\quad{}\times\cos^2\!\left(\frac{\pi dy}{\lambda L}\right)\\ \operatorname{sinc}(u)&=\frac{\sin u}{u}\end{aligned}

Here $a = 0.04\ \mathrm{mm}$ is one slit’s width. The sinc envelope comes from adding contributions across that width; it is a far-field aperture result used here, not derived by treating a slit as one point. At $u = 0$ , $\operatorname{sinc}(u)$ is defined by its limit, 1. For interference maxima, the path difference is an integer $m$ times $\lambda$ . Adjacent integer values then give the spacing below.

\begin{gathered}d\sin\theta=m\lambda\\ \Longrightarrow\quad y_m\approx\frac{m\lambda L}{d}\\ \Delta y\approx\frac{\lambda L}{d}\end{gathered}

The model assumes equal rectangular slits, monochromatic illumination and paraxial far-field propagation. The envelope can suppress a predicted interference maximum. The illustrated wavefronts and detector timing are schematic; the probability sampling uses the stated aperture model. The histogram normalises over the finite displayed window, so it compares like with like rather than pretending to detect every emitted photon.