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Standing Waves Visualization

A monochromatic light wave at normal incidence meets one reflector or is trapped between two mirrors. The coherent sum of the counterpropagating waves forms a standing pattern whose nodes, antinodes and standing-wave ratio follow from the complex reflection coefficient.

Physical units: x in µm, frequency in THz, optical time in fs • Settings are kept in the URL • The animation is slowed by an explicit factor
ν 500 THz Frequency
λ 600 nm Medium wavelength
r 1.00 ∠180° Reflection
S ∞ SWR
⚫ 0 Nodes (E = 0)
t 0.00 fs Optical time
⏱ 1.00×10¹⁵ Slow-down
📊 Field, energy and spectrum

Instantaneous field E(x, t) 1D plane wave

Signed field in units of the incident amplitude E₀. Dashed curves: analytic envelope ±|Ẽ(x)|. Circles on the axis mark E minima (nodes when |r| = 1), diamonds mark maxima. x is to scale. Drag or use ←/→ on a plot to move the probe xp.

Time-averaged energy density ⟨u⟩ ∝ |Ẽ|²

|Ẽ|²/E₀² (electric, solid) and η²|H̃|²/E₀² (magnetic, dashed); ⟨uE⟩ = ε|Ẽ|²/4. Their sum is constant, 2(1 + |r|²). An incident wave alone gives 1 for each.

Probe time trace at xp

E(xp, t) and ηH(xp, t) over two optical periods; the white line is the current optical time. In a pure standing wave E and H are 90° apart; in a travelling wave they are in phase.

💡 How to Use

Learn with this tool

Learning objectives

  • Predict where the nodes of a standing pattern fall, and its standing-wave ratio, from the complex reflection coefficient r. Then measure both with the probe.
  • Derive the allowed frequencies of a two-mirror resonator from the round-trip condition, and relate the mode number m to the node count and the free spectral range.
  • Explain why partially transmitting mirrors turn discrete modes into Airy resonances of width FSR/F, and estimate the finesse, Q and photon lifetime.

Prerequisites

Model

A monochromatic plane wave at normal incidence in a lossless medium of index n. The tool uses the phasor convention E(x, t) = Re[Ẽ(x) e−i(ωt − φ)]. One boundary: the reflector is at x = L, and s = L − x is the distance in front of it. Resonator: mirrors sit at x = 0 and x = L, and s = x. In both cases Ẽ = E₀[e−iks + r eiks] and ηH̃ = ±E₀[e−iks − r eiks]. Two mirrors allow only frequencies with r₁r₂e2ikL = 1. With partial mirrors the same round trip is summed as a geometric series, which gives the Airy transmission.

ν, ω = 2πν
optical frequency (THz); the same in every medium
λ₀ = c/ν, λ = λ₀/n
vacuum and medium wavelength
k = 2πnν/c
wavenumber in the medium
r = |r|eiθ
amplitude reflection coefficient, Eref/Einc at the boundary
SWR
standing-wave ratio |E|max/|E|min = (1 + |r|)/(1 − |r|)
η = η₀/n
wave impedance of the medium (η₀ ≈ 376.73 Ω)
FSR = c/(2nL)
free spectral range, the spacing between adjacent modes
R = |r|², F
mirror power reflectance; finesse F = π(R₁R₂)1/4/(1 − √(R₁R₂))
τ, Q
photon lifetime τ = trt/(−ln R₁R₂) with trt = 2nL/c; quality factor Q = 2πντ

Assumptions and validity: scalar, one-dimensional, infinitely wide plane waves; lossless, non-dispersive real n; ideal reflectors with frequency-independent r; steady state (every transient has decayed); perfect temporal coherence. The field plot of the resonator uses ideal mirrors (|r| = 1). Partial mirrors enter only the spectrum preview.

Derivation: envelope, extrema, modes and Airy function

Envelope. |Ẽ|² = E₀²|e−iks + |r|ei(ks+θ)|² = E₀²[1 + |r|² + 2|r| cos(2ks + θ)]. Maxima E₀(1 + |r|) occur where 2ks + θ = 2πm, and minima E₀(1 − |r|) occur where 2ks + θ = (2m + 1)π. Adjacent minima are therefore Δs = π/k = λ/2 apart. Replacing |r| by −|r| gives η²|H̃|², so |Ẽ|² + η²|H̃|² = 2E₀²(1 + |r|²) everywhere.

Global phase. Changing φ multiplies Ẽ(x) by eiφ at every x. |Ẽ(x)|, and hence every node, is unchanged.

Time-averaged flux. ⟨S⟩ = Re(Ẽ H̃*)/2 = (E₀²/2η)(1 − |r|²). The cross terms are purely imaginary and carry no net power. They describe energy sloshing between the electric and magnetic forms.

Resonator. Start at x = 0 with the wave leaving mirror 1. After a round trip it has gained e2ikL and been multiplied by r₂ and r₁. A steady field without a source needs r₁r₂e2ikL = 1, so 2kL + θ₁ + θ₂ = 2πq. This gives νq = (q − (θ₁ + θ₂)/2π)·c/(2nL). For two PEC mirrors θ₁ = θ₂ = π, so νm = m c/(2nL) and Ẽ ∝ sin(mπx/L), with m antinodes and m + 1 nodes.

Airy function. Let an external wave enter through mirror 1 (amplitude transmission t₁) and leave through mirror 2 (t₂). The transmitted amplitude sums t₁t₂eikL(1 + ρ + ρ² + …) with ρ = r₁r₂e2ikL. For |ρ| < 1 the series gives t₁t₂eikL/(1 − ρ). For lossless mirrors |tj|² = 1 − Rj (the index factors cancel for equal outer media), so T = (1 − R₁)(1 − R₂)/[(1 − √(R₁R₂))² + 4√(R₁R₂) sin²(δ/2)], with δ = 2kL + θ₁ + θ₂. Setting the denominator to twice its minimum gives the half-width. In the high-finesse limit it is FSR/F.

Exercise 1: Wiener's node at a mirror

  1. Light at ν = 500 THz in air reflects from a perfect conductor. Where is the first electric-field node, and how far away is the second? Where is the first antinode?
  2. Press PEC mirror. Turn on ηH and watch the animation.
  3. Move the probe to the two E minima nearest the mirror and read their distance from the boundary. Then read the probe values of |Ẽ|²/E₀² and η²|H̃|²/E₀² at the mirror.
  4. Why did Wiener's tilted photographic film show a dark fringe exactly at the silvered surface?
Show answer

λ = c/ν = 599.6 nm. Tangential E must vanish on a conductor, so the first node lies at the surface. The next is λ/2 = 299.8 nm in front of it, and the first antinode is λ/4 = 149.9 nm away. At the mirror |Ẽ|² = 0 while η²|H̃|² = 4, the magnetic antinode. The emulsion responds to E, so it stays dark at the surface. This showed that the electric field drives the photochemistry.

Exercise 2: a weak standing wave at glass

  1. For air → glass (n₂ = 1.5), predict r, the SWR, the ratio of maximum to minimum energy density, and whether the interface sits at an E maximum or minimum. What changes for glass → air?
  2. Press Air → glass, then Glass → air.
  3. With the probe, read |Ẽ|²/E₀² at an adjacent maximum and minimum and form their ratio. Read the net power fraction from the readouts.
  4. Why is a dielectric interface not a "free end", and where does the missing power go?
Show answer

r = (1 − 1.5)/(1 + 1.5) = −0.2, so SWR = 1.2/0.8 = 1.5 and the energy-density ratio is SWR² = 2.25 (1.44/0.64). With r < 0 the interface sits at an E minimum of 0.8E₀. For glass → air r = +0.2, and the interface is a maximum of 1.2E₀. In both cases 1 − |r|² = 96 % of the incident power flows on through the interface. The field is continuous there, with t = 1 + r, so the boundary neither pins nor frees the field.

Exercise 3 (limiting case): from standing to travelling wave

  1. Choose Custom complex r with arg r = 180°. As |r| goes from 1 to 0, predict the SWR, the envelope, and the largest phase lag between E and ηH. It occurs λ/8 from an E minimum.
  2. Type |r| = 1, 0.5, 0.1 and 0 into the number box.
  3. Record the SWR. Put the probe λ/8 from a minimum and read the E–H phase lag in the readouts; compare with the time trace.
  4. Which quantity tells you that energy is flowing, and why is it zero for |r| = 1?
Show answer

SWR = ∞, 3, 1.22 and 1. The envelope flattens to |Ẽ| = E₀ at |r| = 0, the Absorber preset. From the model, ẼηH̃* = E₀²[1 − |r|² + 2i|r| sin(2ks + θ)]. The real part is the net flux, ⟨S⟩ ∝ 1 − |r|². The imaginary part is reactive energy that sloshes back and forth. The largest lag, atan[2|r|/(1 − |r|²)], is 90°, 53.1°, 11.4° and 0° for the four values. At |r| = 1, E and H are in quadrature at every point and no net energy flows. At |r| = 0 they are in phase, as in a travelling wave. At the extrema themselves the lag is always 0.

Exercise 4: mode number, finesse and linewidth

  1. Two PEC mirrors, L = 1.5 µm, n = 1. Predict ν₄, the number of nodes of mode 4, and the FSR. If the mirrors had R = 0.9, what would the finesse and linewidth be? What happens as R → 1?
  2. Press Cavity m = 4. Then slide R towards 0.995 and down to 0.3.
  3. Count nodes in the field plot. Read the FSR, F and δν in the readouts, and check the half-maximum points with the spectrum cursor.
  4. Why do the peak positions not move when R changes, while their widths do?
Show answer

FSR = c/(2L) = 99.93 THz, so ν₄ = 399.7 THz, with 5 nodes including both mirrors. For R = 0.9, F = π√0.9/0.1 = 29.8 and δν = FSR/F ≈ 3.35 THz (exact Airy FWHM 3.355 THz), so Q = ν/δν ≈ 119. As R → 1, F → ∞ and δν → 0, which recovers the discrete ideal modes of the field plot. The peak condition 2kL + θ₁ + θ₂ = 2πq depends only on the mirror phases. |r| sets how fast light leaks out, and hence the width.

Worked example: a semiconductor microcavity

A GaAs layer (n ≈ 3.5, dispersion neglected) of thickness L = 1.5 µm sits between two mirrors. Model them as conductors with R = 0.99. Which longitudinal mode lies nearest the 1.31 µm telecom band, and what are its linewidth, Q and photon lifetime? Reproduce it with the Microcavity preset.

  1. FSR = c/(2nL) = 2.998×10⁸/(2 · 3.5 · 1.5×10⁻⁶) = 28.55 THz.
  2. At λ₀ = 1.31 µm, ν = 228.9 THz, so m = ν/FSR ≈ 8.02. The nearest mode is m = 8, with ν₈ = 228.41 THz (λ₀ = 1312.5 nm).
  3. Inside, λ/2 = λ₀/(2n) = 187.5 nm. Mode 8 has 9 E-field nodes, including both mirrors, spaced 187.5 nm apart.
  4. F = π√0.99/(1 − 0.99) = 312.6, so δν = FSR/F = 91.3 GHz and Q = ν₈/δν ≈ 2.50×10³.
  5. trt = 2nL/c = 35.0 fs. Each round trip keeps R² = 0.9801 of the energy, so τ = trt/(−ln 0.9801) = 1.74 ps. Check: 1/(2πτ) = 91.3 GHz, the same linewidth.

Real vertical-cavity lasers use Bragg mirrors with R > 0.99. Their reflection phase varies with frequency and the field penetrates the stack, so the effective length exceeds L. The thin-film and passive resonator tools handle that; this ideal-mirror model does not.

When the model fails

  • Real metals: at optical frequencies r is complex with |r| < 1 (silver ≈ 0.95–0.98). The field penetrates a skin depth, so the node sits slightly behind the surface. Use a custom r to mimic this.
  • Oblique incidence and polarization: r then depends on angle and on s/p polarization, and the standing pattern has period λ/(2 cos θ). See the Fresnel tool.
  • Finite beams: a Gaussian beam gains a Gouy phase that shifts the resonances and adds transverse modes. Curved mirrors also decide stability. See the Gaussian beam and passive resonator tools.
  • Finite coherence: a source of bandwidth Δν washes the pattern out beyond a distance of about c/(2nΔν) from the mirror. This model assumes a single frequency.
  • Dispersion, absorption and gain: n(ν) moves the modes and makes the FSR use the group index, and absorption damps the pattern. Gain is a laser problem (laser cavity).
  • Driven partial-mirror field: the field plot always shows the ideal-mirror mode. Only the Airy spectrum includes finite R.

References

  • E. Hecht, Optics, 5th ed. (Pearson, 2017): superposition and standing waves (ch. 7) and the Fabry–Pérot interferometer (ch. 9).
  • B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics (Wiley): the chapter "Resonator Optics" covers planar-mirror resonators, finesse and photon lifetime.
  • D. J. Griffiths, Introduction to Electrodynamics, 4th ed.: reflection at a conductor and at dielectric interfaces (ch. 9).
  • O. Wiener, "Stehende Lichtwellen und die Schwingungsrichtung polarisirten Lichtes", Annalen der Physik 40, 203 (1890).
  • A. E. Siegman, Lasers (University Science Books, 1986): Fabry–Pérot resonances and cavity lifetime.
  • MIT OCW 6.974 Fundamentals of Photonics: lecture notes.

📚 Physics Background

〰️ Standing patterns versus resonances

Two coherent waves of the same frequency travelling in opposite directions add to a standing pattern: the intensity envelope is fixed in space while the field oscillates in time. A single reflector is enough. Light reflected from one mirror forms a standing pattern at any frequency, with nodes spaced by λ/2.

Discrete frequencies appear only when a second boundary also constrains the field. Then the wave must reproduce itself after a round trip, and only a set of resonant modes survives. Resonance is a property of the two-boundary resonator, not a condition for a standing pattern.

📐 Model and coordinates (identical to the code)

Each field is written as a complex phasor with time dependence e−iωt. The plotted field is E(x, t) = Re[Ẽ(x) e−i(ωt − φ)], with k = 2πnν/c and ω = 2πν.

One boundary. The medium (index n = n₁) fills 0 ≤ x ≤ L, the reflector sits at x = L, and s = L − x is the distance in front of it. With r = |r|eiθ defined at the boundary:

Incident and reflected fields

E~(x)=E0(e−iks+reiks)\tilde E(x)=E_0\left(e^{-iks}+r e^{iks}\right)
ηH~(x)=E0(e−iks−reiks),η=η0n\eta\tilde H(x)=E_0\left(e^{-iks}-r e^{iks}\right),\qquad\eta=\frac{\eta_0}{n}
∣E~(x)∣=E01+∣r∣2+2∣r∣cos⁡(2ks+θ)|\tilde E(x)|=E_0\sqrt{1+|r|^2+2|r|\cos(2ks+\theta)}

The envelope is fixed by r alone, so the extrema follow analytically:

Maxima and minima

Emax⁡=E0(1+∣r∣),2ks+θ=2πmE_{\max}=E_0(1+|r|),\qquad 2ks+\theta=2\pi m
Emin⁡=E0(1−∣r∣),2ks+θ=(2m+1)πE_{\min}=E_0(1-|r|),\qquad 2ks+\theta=(2m+1)\pi

Standing-wave ratio

SWR=∣E∣max⁡∣E∣min⁡=1+∣r∣1−∣r∣\mathrm{SWR}=\frac{|E|_{\max}}{|E|_{\min}}=\frac{1+|r|}{1-|r|}

Adjacent minima are λ/2 = c/(2nν) apart, and the reflection phase θ sets where they fall relative to the boundary. The global phase φ multiplies every point by the same factor eiφ, so it cannot move a node.

🧱 Reflectors

Perfect electric conductor, r = −1

Tangential E must vanish at a perfect conductor, so θ = π places an E-field node exactly on the surface (s = 0) and further nodes every λ/2 in front of it. The magnetic field has its maximum 2E₀/η there. Real metals at optical frequencies are close to this, but they absorb, and |r| is slightly below 1.

Dielectric interface (Fresnel, normal incidence)

Interface coefficients

r=n1−n2n1+n2,t=2n1n1+n2=1+rr=\frac{n_1-n_2}{n_1+n_2},\qquad t=\frac{2n_1}{n_1+n_2}=1+r

At normal incidence s and p polarization give the same r. Going into a denser medium (n₂ > n₁) gives r < 0 and an E-field minimum at the interface. Going into a rarer medium gives r > 0 and a maximum. In both cases |r| is small (0.2 for air/glass). The minima are therefore not zero and most power is transmitted. A dielectric interface is not a "free end". The field continues into medium 2 with amplitude t, drawn to the right of the boundary.

Ideal magnetic conductor (r = +1) and matched absorber (r = 0)

A perfect magnetic wall is the dual idealization. Tangential H vanishes and E has an antinode at the surface. It is approximated by some metamaterial or high-impedance surfaces, not by ordinary optics. A matched absorber reflects nothing and leaves a pure travelling wave with SWR = 1.

🎵 Two-boundary resonator

With mirrors r₁ at x = 0 and r₂ at x = L, the field in between is Ẽ(x) = E₀ [ e−ikx + r₁ e+ikx ]. This is the same expression as above with s = x. The wave reflected at x = L must reproduce it after one round trip:

Resonance condition

r1r2e2ikL=1r_1r_2e^{2ikL}=1
ν=(q−θ1+θ22π)c2nL\nu=\left(q-\frac{\theta_1+\theta_2}{2\pi}\right)\frac{c}{2nL}

Two perfect conductors

νm=mc2nL,E~∝sin⁡ ⁣(mπxL)\nu_m=\frac{mc}{2nL},\qquad\tilde E\propto\sin\!\left(\frac{m\pi x}{L}\right)

Mode indices m = 1, 2, 3, …

  • Mode m has m antinodes and m + 1 E-field nodes, including both mirror surfaces.
  • The mode spacing is the free spectral range c/(2nL).
  • PMC | PMC gives the same frequencies with cos(mπx/L) profiles.
  • PEC | PMC gives quarter-wave modes ν = (2m − 1) c/(4nL).

The field plot uses ideal, lossless mirrors. With partially reflecting mirrors (power reflectance R) the discrete modes broaden into Airy resonances of width FSR/F, with finesse F = π√R/(1 − R). The spectrum panel previews this at the same mode frequencies. The passive resonator tool models the driven, partially transmitting cavity field itself.

⚡ Electric and magnetic fields, energy flow

In a standing wave E and H are complementary: |E|² + η²|H|² = 2E₀²(1 + |r|²) at every x. E-field nodes are H-field antinodes, and in time the two oscillate 90° apart, so energy sloshes between electric and magnetic form. The time-averaged Poynting flux toward the boundary is ⟨S⟩ = (1 − |r|²) Sinc. It is zero for a perfect reflector and in a lossless cavity mode, even though the instantaneous flux is not.

⏱ Time scale and limits of the model

Visible light oscillates at about 5×10¹⁴ Hz, so one optical period lasts about 2 fs. The animation advances the optical time t (shown in fs) by (wall time)/(slow-down factor). The slow-down factor is the chosen wall time per period multiplied by ν. The time origin t = 0 is the instant when the field at the first antinode peaks. This is only a fixed choice of clock zero.

  • One-dimensional monochromatic plane wave at normal incidence; the field shown is the tangential E (E ⊥ x).
  • Real, non-dispersive, lossless indices; the amplitude E₀ is normalized to 1.
  • The pixel sampling is only for drawing. Nodes, antinodes, SWR and mode frequencies are computed in closed form.

🌟 Real-World Examples

  • Wiener's experiment (1890): a photographic film tilted in front of a mirror recorded dark fringes λ/2 apart, with the first at the mirror surface. This showed that the E field, not H, exposes the emulsion.
  • Laser and Fabry–Pérot cavities: two mirrors select longitudinal modes spaced by c/(2nL); gain and losses decide which of them lase.
  • Optical lattices: counterpropagating laser beams form standing waves that trap cold atoms at nodes or antinodes.
  • Antireflection and dielectric mirrors: the reflection phase of each interface decides where standing-wave maxima fall inside thin-film stacks.
  • Microwave ovens: standing waves in the metal cavity create hot and cold spots λ/2 apart.