
Physics III course sequence · Interactive lesson
Coherence, phase & superposition of light
Everything in physical optics rests on one operation: add two fields, square the sum, average. This lesson makes it quantitative — how the phase difference is built from geometry and refractive index, why a real source holds its phase only for a coherence length, and how fringe visibility measures how much interference survived.
UPIII-05.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Build the model
Fields always add. Whether you see it depends on the phase staying put.
The key idea
Two light waves are coherent at a detector when their phase difference stays fixed for longer than the detector averages; their fields then add to make fringes, and the fringe visibility measures how completely they do so.
Example. A red LED at 630 nm with a 20 nm linewidth has ℓc ≈ λ₀²/Δλ = 630²/20 nm ≈ 20 μm: split its light along two paths differing by 5 μm and the fringes are clear, by 20 μm and they are gone.
The complete physical model and assumptions
Maxwell's equations are linear, so where two light waves overlap the total field is E₁ + E₂, always. A detector, though, cannot follow a field oscillating at 5 × 10¹⁴ Hz; it records the time average of E², and that average contains a cross term E₁E₂⟨cos δ⟩ that depends on the phase difference δ between the two waves. The whole subject turns on that one term. If δ is fixed over the detector's integration time the cross term survives and the intensity swings between (√I₁ + √I₂)² and (√I₁ − √I₂)²: fringes. If δ wanders randomly, cos δ averages to zero and the intensities add. Two things set δ. Geometry and medium fix its deterministic part — each metre of path in a medium of index n advances the phase by 2πn/λ₀ — while the source fixes how long δ stays put: a source of bandwidth Δν emits wave trains whose phase is predictable only for τc ≈ 1/Δν, so two paths differing by more than ℓc = cτc compare unrelated trains and lose their fringes. An extended source adds a second limit, since each of its points makes its own shifted pattern, and the patterns cancel once the apertures sit further apart than λ₀/θₛ. The fully coherent, monochromatic, scalar picture is therefore a limit: it holds when the path difference is small against ℓc, the aperture separation small against ℓₜ, and both beams share a polarisation. Outside it the fields still add — only the fringes have gone.
1Add the fields, then square and average
Where two beams overlap the field is E = E₁ cos(ωt) + E₂ cos(ωt + δ); nothing in Maxwell's equations lets the intensities add instead. A detector integrates over millions of optical cycles, so what it records is ⟨E²⟩, and with ⟨cos² ωt⟩ = ½ and ⟨cos ωt cos(ωt + δ)⟩ = ½ cos δ the average is ½E₁² + ½E₂² + E₁E₂ cos δ. Since I ∝ ⟨E²⟩, that is I = I₁ + I₂ + 2√(I₁I₂) cos δ. The phasor picture says the same thing: two arrows of length E₁ and E₂ set δ apart, and the cosine rule gives the resultant. For equal beams of intensity I₀ the total swings from 4I₀ at δ = 0 to zero at δ = π, averaging 2I₀ across the fringes — energy is not created, only moved. The cross term is the entire content of interference, and every later result in this section is a statement about what happens to cos δ.
2Count the phase in vacuum wavelengths of optical path
A wave of vacuum wavelength λ₀ travelling a distance s through a medium of index n has wavelength λ₀/n there, so it accumulates phase 2πns/λ₀. The product ns is the optical path length, and the phase difference between two routes is δ = (2π/λ₀) × (difference in optical path). Inserting a 1.00 mm slab of glass (n = 1.52) into one route replaces 1.00 mm of air by 1.52 mm of optical path, an excess of 0.52 mm — at 550 nm that is 945.45 wavelengths, a phase of 5940 rad. What the detector sees depends only on the fractional part, here 0.45 of a wavelength, so the slab shifts the fringes by 0.45 of a fringe; but thinning the slab by just 1 μm changes the excess by 0.52 μm and moves them by another 0.95 of a fringe. This is the double edge of interferometry: phase is exquisitely sensitive to optical path, so fringes measure thickness or index to nanometres, and predicting a fringe position from three-figure data is hopeless. Reflection adds its own bookkeeping — a shift of π when the wave reflects off a higher-index medium — which the thin-film lesson takes up.
3A real source emits finite wave trains
No source is monochromatic. An atom radiates for a limited time before a collision or re-excitation resets its phase, so light arrives as a sequence of wave trains whose phases are unrelated, and the Fourier transform of a train of duration τc has bandwidth Δν ≈ 1/τc. Read the other way, that is the useful statement: a source of bandwidth Δν has coherence time τc ≈ 1/Δν and coherence length ℓc = cτc ≈ c/Δν. Because ν = c/λ, a small spread in ν maps to Δλ = λ₀²Δν/c, giving the form spectroscopists use, ℓc ≈ λ₀²/Δλ. The range is enormous. White light spanning 400 to 700 nm has ℓc ≈ 550²/300 nm ≈ 1 μm, about two wavelengths. A red LED with Δλ = 20 nm at 630 nm has ℓc ≈ 20 μm. A multimode He–Ne laser with Δν ≈ 1.5 GHz has ℓc ≈ 0.20 m; a frequency-stabilised one at 1 MHz reaches 300 m. Coherence length is not a property of laser light as against lamp light but a number every source has, and each interference lesson that follows quietly assumes its path difference sits well inside it.
4Visibility fades as the path difference approaches ℓc
Split light of bandwidth Δν between two paths that differ by Δ. Each frequency ν in the spectrum makes its own fringe term cos(2πνΔ/c), and terms at different frequencies do not interfere with one another — their cross terms oscillate at the difference frequency and average away — so the total is the intensity-weighted sum I(Δ) ∝ ∫ S(ν)[1 + cos(2πνΔ/c)] dν. For a flat-topped spectrum of width Δν centred on ν₀ the integral is elementary and gives 1 + sinc(πΔνΔ/c) cos(2πν₀Δ/c): the fast cosine is the fringe pattern and the sinc is its envelope. Writing ℓc = c/Δν, the visibility is |γ(Δ)| = |sin(πΔ/ℓc)/(πΔ/ℓc)|. It is 1 at Δ = 0, 0.64 at Δ = ℓc/2, and zero at Δ = ℓc, where the fringes of the reddest and bluest components are a whole fringe out of step, so every component in the band is cancelled by the one half a bandwidth away. The weak revivals beyond that belong to the sharp-edged model; a Gaussian line gives a Gaussian envelope with none. In general the envelope is the Fourier transform of the spectrum, which is the principle behind Fourier-transform spectroscopy and the Michelson lesson later in this section.
5An extended source blurs the phase across a wavefront
Temporal coherence compares a beam with a delayed copy of itself. Spatial coherence asks instead whether two points across a wavefront share a phase. A point source answers yes for any pair, because both points lie on the same spherical wave. Widen the source to an angular size θₛ and every source point sends its own wave, reaching two apertures a distance d apart with its own path difference. Points at opposite edges of the source differ in that path by about dθₛ, so their fringe patterns are displaced; once dθₛ reaches λ₀ the shifts span a whole fringe and the sum is a uniform wash. So the transverse coherence width is ℓₜ ≈ λ₀/θₛ, and for a uniform strip source the visibility falls as |sinc(πdθₛ/λ₀)|, the same shape as the temporal case with dθₛ playing the role of Δ; a uniform disc gives an Airy factor with its first zero at 1.22λ₀/θₛ. The Sun subtends 9.3 mrad, so at 550 nm sunlight is coherent across only about 59 μm. That is why Young first passed sunlight through a pinhole: a 0.5 mm hole seen from 2 m subtends 0.25 mrad and widens ℓₜ to 2.2 mm, well beyond his slit spacing.
6Where the coherent, scalar picture stops
Three approximations sit under every fringe formula, and each fails its own way. Full coherence fails gradually rather than suddenly: the cross term is really 2√(I₁I₂)|γ| cos δ, with |γ| the degree of coherence, and V simply tracks |γ| — partial coherence is the general case, and visibility is how you measure it. Monochromaticity fails when Δ becomes comparable to ℓc, as the last two moves showed. The scalar treatment fails with polarisation: the cross term is 2⟨E₁ ⋅ E₂⟩, a dot product, so two beams polarised at right angles superpose perfectly and show no fringes at all — the Fresnel–Arago result the polarisation lesson uses. Two independent sources are not a failure of superposition but of the average: their fields add at every instant, and a detector fast enough to follow the phase sees beats — two lasers a few megahertz apart on a fast photodiode show exactly that — but over any ordinary integration time the phase difference of two lamps wanders through many cycles and ⟨cos δ⟩ is zero. Coherence is a comparison between how fast δ wanders and how long the detector looks.
Key equations
Read the units, assumptions and limits before substituting.
Superposition and the interference term
I = I₁ + I₂ + 2√(I₁I₂) cos δ
Add the fields, square, average over a cycle: the cross term is the whole of interference, and it lives or dies with cos δ.
I, I₁, I₂ time-averaged intensities in W m⁻²; δ phase difference at the detector in rad; same frequency and polarisation assumedPhase from optical path
δ = (2π/λ₀) ΔOPL, where OPL = Σ nᵢ sᵢ
Counts the difference in vacuum wavelengths along the two routes — a slab of thickness t adds (n − 1)t to one of them.
λ₀ vacuum wavelength in m; nᵢ dimensionless; sᵢ geometric lengths in m; add π for a reflection off a denser mediumFringe visibility
V = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ) = 2√(I₁I₂) |γ| / (I₁ + I₂)
Splits contrast into two independent factors: how well balanced the beams are, and how well their phases stay locked.
V and |γ| dimensionless, from 0 to 1; |γ| is the degree of coherence between the two beamsCoherence time and coherence length
ℓc = c τc ≈ c/Δν ≈ λ₀²/Δλ
How far along a beam the phase stays predictable — fringes need the path difference to sit well inside it.
Δν bandwidth in Hz, Δλ linewidth in m, c = 3.00 × 10⁸ m s⁻¹; the last step uses Δν/ν₀ = Δλ/λ₀Visibility against path difference, flat spectrum
|γ(Δ)| = |sin(πΔ/ℓc) / (πΔ/ℓc)|
Fringes fade smoothly and first vanish at Δ = ℓc; the envelope is the Fourier transform of the spectrum.
Δ path difference in m, ℓc in m; holds for a flat-topped spectrum of width Δν = c/ℓcSpatial coherence width
ℓₜ ≈ λ₀/θₛ = λ₀ R/w
The widest aperture separation that still sees fringes from an extended source — why Young needed a pinhole.
θₛ angular width of the source in rad; w source width in m seen from distance R in mInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Leave the linewidth at 20 nm and drag the path difference from 0 towards 18 μm: the fringes flatten and vanish exactly at ℓc = 18 μm, then weakly revive. Then set Δ back to 4 μm and pull the intensity ratio down to 0.1: |γ| stays at 0.92 while V drops to 0.53 — contrast has two independent factors.
Live results
- COHERENCE LENGTH ℓc = λ₀²/Δλ
- 18.0 μm
- BALANCE 2√(I₁I₂)/(I₁+I₂)
- 1.000
- COHERENCE FACTOR |γ|
- 0.921
- VISIBILITY V
- 0.921
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Path difference Δ: 4.0 μm; Source linewidth Δλ: 20 nm; Intensity ratio I₂/I₁: 1.00
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Coherence is not a switch that lasers have on and lamps have off, and incoherence does not mean the waves fail to superpose. Every source has a coherence length: a laser's fringes disappear once the path difference passes its ℓc, and sunlight makes fringes wherever the path difference is within a micrometre or so and the apertures sit closer than λ₀/θₛ — which is how Young saw them in 1801 and why Newton's rings appear in white light. When two independent lamps overlap, their fields still add at every instant; what vanishes is the time average of cos δ, because the phase difference drifts through many cycles while the detector integrates. Coherence compares two times — how long δ stays put against how long the detector looks — and it is measured by the visibility, not assumed from the label on the source.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A green LED emits at λ₀ = 525 nm with a linewidth Δλ = 35 nm. Find its bandwidth in hertz, its coherence time and its coherence length, and express the coherence length in wavelengths. Then find the coherence time and length of a He–Ne laser at 632.8 nm whose bandwidth is 1.5 GHz.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Convert the linewidth to a bandwidth with Δν = cΔλ/λ₀², which follows from ν = c/λ: Δν = (3.00 × 10⁸ m s⁻¹)(35 × 10⁻⁹ m) / (525 × 10⁻⁹ m)² = 10.5 / 2.756 × 10⁻¹³ = 3.8 × 10¹³ Hz.
- The coherence time is the reciprocal of the bandwidth: τc ≈ 1/Δν = 2.63 × 10⁻¹⁴ s, about 26 femtoseconds — the length of one wave train.
- The coherence length is the distance light travels in that time: ℓc = cτc = (3.00 × 10⁸)(2.63 × 10⁻¹⁴) = 7.9 × 10⁻⁶ m = 7.9 μm. Check by the shortcut λ₀²/Δλ = 525²/35 nm = 7875 nm ✓, which is λ₀/Δλ = 15 wavelengths.
- For the laser the bandwidth is given directly: τc = 1/(1.5 × 10⁹ Hz) = 6.7 × 10⁻¹⁰ s and ℓc = (3.00 × 10⁸)(6.7 × 10⁻¹⁰) = 0.20 m. Its linewidth is Δλ = λ₀²Δν/c = (632.8 × 10⁻⁹)²(1.5 × 10⁹)/(3.00 × 10⁸) = 2.0 × 10⁻¹² m, two picometres.
- Compare: 0.20 m against 7.9 μm is a factor of 2.5 × 10⁴. An interferometer fed by the LED must hold its path difference to a few micrometres; the laser tolerates centimetres.
Answer. LED: Δν ≈ 3.8 × 10¹³ Hz, τc ≈ 26 fs, ℓc ≈ 7.9 μm (15 wavelengths). He–Ne laser: τc ≈ 0.67 ns, ℓc ≈ 0.20 m, some 25 000 times longer.
MediumExample 2
Light from an LED with coherence length 7.9 μm and a flat-topped spectrum is split into two beams that meet again with I₁ = 4.00 W m⁻² and I₂ = 1.00 W m⁻² after paths differing by 3.0 μm. Find the fringe visibility, the maximum and minimum intensities, the path difference at which the fringes first vanish, and which factor is worth fixing.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Visibility has two independent factors. The balance factor comes from the intensities alone: 2√(I₁I₂)/(I₁ + I₂) = 2√(4.00 × 1.00)/5.00 = 0.800.
- The coherence factor comes from the path difference against ℓc: the argument is πΔ/ℓc = π × 3.0/7.9 = 1.19 rad, so |γ| = |sin(1.19)/1.19| = 0.929/1.19 = 0.78.
- Multiply: V = 0.800 × 0.78 = 0.62. The mean intensity is unaffected by coherence, I₁ + I₂ = 5.00 W m⁻².
- Fringe extremes: Iₘₐₓ = (I₁ + I₂)(1 + V) = 5.00 × 1.62 = 8.1 W m⁻² and Iₘᵢₙ = 5.00 × 0.38 = 1.9 W m⁻². Check the limiting case: with perfect coherence and balance the extremes would be 4 × 2.5 = 10 and 0, so the fringes have lost contrast from both ends.
- The fringes first vanish where the sinc has its first zero, Δ = ℓc = 7.9 μm.
- Which factor to fix: balancing the beams perfectly would raise V only to |γ| = 0.78, while shortening Δ to 1.0 μm raises |γ| to 0.97 and V to 0.78 as well — each fix alone buys the same; only doing both reaches V ≈ 0.97.
Answer. V ≈ 0.62; Iₘₐₓ ≈ 8.1 W m⁻², Iₘᵢₙ ≈ 1.9 W m⁻²; the fringes vanish at Δ = 7.9 μm. Balancing the beams and shortening the path difference are worth about the same, and both are needed for V above 0.9.
HardExample 3
The Sun subtends 9.3 mrad. Two pinholes d apart are lit by direct sunlight through a narrow 550 nm filter, so temporal coherence is not the limit. (a) Find the transverse coherence width. (b) Using the uniform-strip model |γ| = |sinc(πdθₛ/λ₀)|, find the visibility for d = 20 μm and for d = 40 μm. (c) Young instead let the sunlight through a 0.50 mm hole and set the pinholes 2.0 m behind it: find the new θₛ and ℓₜ, and the largest d that keeps V ≥ 0.90.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- (a) Two apertures see fringes while the extra path from opposite edges of the source, dθₛ, is small against a wavelength: ℓₜ ≈ λ₀/θₛ = (550 × 10⁻⁹ m)/(9.3 × 10⁻³ rad) = 5.9 × 10⁻⁵ m = 59 μm.
- (b) At d = 20 μm the argument is πdθₛ/λ₀ = π × 20/59.1 = 1.06 rad, so |γ| = sin(1.06)/1.06 = 0.874/1.06 = 0.82: clear fringes.
- At d = 40 μm the argument is 2.12 rad and |γ| = 0.850/2.12 = 0.40. Doubling the separation more than halved the contrast, and at d = 59 μm it reaches zero — a uniform disc would put that zero at 1.22 × 59 = 72 μm instead, so the strip model is pessimistic by a fifth.
- (c) A 0.50 mm hole seen from 2.0 m subtends θₛ = 0.50 × 10⁻³/2.0 = 2.5 × 10⁻⁴ rad, so ℓₜ = 550 × 10⁻⁹/2.5 × 10⁻⁴ = 2.2 × 10⁻³ m = 2.2 mm, 37 times wider than in direct sunlight.
- V ≥ 0.90 needs sinc(u) ≥ 0.90, and sin(u)/u = 0.90 at u = 0.79 rad, so πdθₛ/λ₀ ≤ 0.79 gives d ≤ 0.25 ℓₜ = 0.55 mm. The pinhole costs almost all the light, which is the price of a phase that is the same across both apertures.
Answer. (a) ℓₜ ≈ 59 μm. (b) V ≈ 0.82 at d = 20 μm and V ≈ 0.40 at d = 40 μm. (c) θₛ = 0.25 mrad, ℓₜ ≈ 2.2 mm, and pinholes up to 0.55 mm apart keep V ≥ 0.90.
Quick check
Select an answer and explain it before checking.
Review & continue
Two light waves are coherent at a detector when their phase difference stays fixed for longer than the detector averages; their fields then add to make fringes, and the fringe visibility measures how completely they do so.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.