
Physics III course sequence · Interactive lesson
Phase difference, path difference, and coherence
Interference is bookkeeping before it is anything else. Every extra half-wavelength of path, every reversed pair of leads, every random restart inside the source adds a term to one running phase difference. Get the ledger right and the pattern follows; leave the last term out and the fringes quietly vanish.
UPIII-03.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Build the model
Interference needs a phase difference that holds still.
The key idea
The phase difference is the single number δ saying how far one arriving wave's cycle sits behind another's at the same point, built from the path difference and from whatever offset the sources already carried.
Example. Two speakers driven in step by one amplifier, λ = 0.500 m, one of them 0.75 m further away: δ = (2π/0.500 m)(0.75 m) = 3π rad, an odd multiple of π, so the two arrivals cancel.
The complete physical model and assumptions
Two waves arriving at one point can differ in phase for exactly three reasons, and the reasons add. The sources may already have been out of step by Δφ₀ when the waves left; the paths may differ in length; and one path may have crossed a region where the wave travelled more slowly. Because a sinusoid advances 2π of phase in every wavelength it covers, a path difference converts at the fixed rate 2π/λ, and the whole account collapses to one number, δ = (2π/λ)Δr + Δφ₀, with a slow segment entering the first term as the extra 2π f L (1/v − 1/v₀) it inserts. That single number decides whether the arrivals reinforce or cancel. The model is powerful because δ can be computed from geometry and wiring before anything is measured. It stops working the moment Δφ₀ refuses to hold still. Real sources radiate in bursts: an atom, a laser mode, an oscillator driving a speaker each reset their phase on some timescale τc, the coherence time, and once the two paths differ by more than Lc = c τc the arrivals no longer remember a shared phase. Then δ wanders through 2π during any realistic exposure, the time average of cos δ collapses to zero, and the intensities merely add. Coherence is therefore not a property of one wave but of the pair: it measures how long, and across how wide a patch, the phase difference stays predictable.
1Delay is the primitive; wavelength is the shortcut
Two waves leaving one oscillator reach a point at different times, and phase is what a delay costs. A sinusoid advances ω radians every second, so an arrival-time difference Δt appears as δ = ω Δt = 2π f Δt. In a uniform medium of speed v the extra travel time for extra distance Δr is Δr/v, and since v = fλ the frequency cancels: δ = 2π f Δr/v = 2π Δr/λ. That is the familiar conversion, and it carries a warning inside it — λ must be the wavelength in the medium the wave actually crossed, not in air or vacuum. A 1.00 kHz tone in air at 343 m s⁻¹ has λ = 0.343 m, so an extra 0.100 m of path is Δt = 2.92 × 10⁻⁴ s and δ = 2π(1000 Hz)(2.92 × 10⁻⁴ s) = 1.83 rad, which is 0.58π: noticeably out of step, but well short of the π that would cancel.
2The ledger: three entries and a sign convention
Write δ = (2π/λ)Δr + Δφ₀ and fix what every symbol means before substituting anything. Both entries have to count the same thing — how far wave 2 sits behind wave 1 — or adding them is meaningless. Choose an order for the sources and keep it: let Δr = r₂ − r₁, so a positive Δr means wave 2 travelled further and therefore lags. Then Δφ₀ must be the lag wave 2 already carried when it left, Δφ₀ = φ₀₁ − φ₀₂, and not the other subtraction: write φ₀₂ − φ₀₁ into this formula and every source offset enters with the wrong sign. Two speakers fed from one amplifier have Δφ₀ = 0; swap the leads on the second and Δφ₀ = π, permanently and everywhere in the room. A wave reflected at a boundary with a denser medium comes away π behind, so that π enters the ledger with a plus sign when it is wave 2 that reflected and a minus sign when it is wave 1. The convention is not decoration. Relabel the sources half way through a problem and δ changes sign, which cos δ forgives but which ruins any statement about which arrival is ahead.
3Coherence time: how long the phase stays predictable
The ledger assumes Δφ₀ is a constant. No real source obliges indefinitely. An atom radiates for a finite time and restarts with a fresh phase; a laser's phase diffuses; even a quartz oscillator drifts. Call τc the coherence time, the interval over which the phase stays predictable to within about a radian. It is not a separate property to be measured on its own: a wave train of duration τc cannot have a sharp frequency, and Fourier analysis ties the two by Δν τc ≈ 1. Narrow the spectrum and the phase holds longer, in strict inverse proportion to the bandwidth. A quartz-referenced 1.00 kHz tone stable to 0.01 Hz has τc ≈ 100 s. A green line from a discharge lamp, Doppler-broadened to Δν ≈ 1.5 GHz, has τc ≈ 0.67 ns. Sunlight through a 10 nm filter at 550 nm has Δν ≈ 9.9 × 10¹² Hz, so τc ≈ 0.10 ps.
4Coherence length is the largest path difference that works
Multiply the coherence time by the wave speed and the result is a length, Lc = c τc. Substituting Δν = c Δλ/λ² turns it into the working form Lc ≈ λ²/Δλ, and the ratio λ/Δλ counts the fringes available before the contrast dies. The rule of use is blunt: fringes need |Δr| ≪ Lc, they weaken as |Δr| approaches Lc, and beyond it nothing stable survives. White light, λ ≈ 550 nm with Δλ ≈ 300 nm, has Lc ≈ 1.0 µm — about two fringes, which is exactly why a white-light pattern is a few coloured bands and no more. The same lamp behind a 10 nm filter gives Lc = 30 µm and roughly 55 fringes. An isolated Doppler-broadened line at Δν = 1.5 GHz gives Lc = 0.20 m, and a frequency-stabilised laser hundreds of metres.
5Two independent sources, and the term that averages away
Now the headline result. Let two sources at the same nominal frequency run from separate oscillators. Their offset Δφ₀(t) is not fixed: it wanders on the scale of τc and, over any longer observation, takes every value in [0, 2π) with equal probability. The interference term carries cos δ, so a slow detector records the average (1/T)∫₀ᵀ cos δ(t) dt, and a uniformly distributed phase makes that integral zero. The intensities simply add, everywhere, and no fringes form. Model the wander instead as Gaussian about a mean δ with standard deviation σ, and the average of the cosine is the cosine of that mean reduced by the factor e(−σ²/2) — the full term at σ = 0, about half of it at σ = 1.2 rad, and 1% of it by σ = 3 rad. None of this is about mismatched frequencies: two generators agreeing to 0.002 Hz at 440 Hz still sweep δ through 2π every 500 s, so a one-second exposure sees a pattern and a ten-minute one sees none. Detect fast enough and even two separate lasers interfere.
6Spatial coherence: the source has width
Temporal coherence asks how far apart in time two samples of one beam may be taken; spatial coherence asks how far apart across it. An extended source is a crowd of independent emitters, and each one presents its own Δφ₀ to a pair of slits. Two points of the source separated by a, viewed from distance R, reach slits separated by d with path differences that differ by roughly ad/R. Once that difference reaches λ the patterns from the two edges of the source sit a whole fringe apart, the emitters between them fill that interval evenly, and the sum of all of them is flat — so fringes survive only while d ≲ λR/a. That length, ℓₜ ≈ λR/a, is the transverse coherence width, equivalently λ/θₛ with θₛ the angular width of the source. The Sun subtends 9.3 × 10⁻³ rad, so at 550 nm its coherence width at the ground is about 59 µm: pinholes closer together than that make fringes in raw sunlight. Read backwards, the same argument turns an interferometer into a ruler for θₛ, which is how stellar diameters are measured.
Key equations
Read the units, assumptions and limits before substituting.
The phase-difference ledger
δ = (2π/λ) Δr + Δφ₀
One number settles the outcome: cos δ = +1 reinforces, cos δ = −1 cancels.
δ in rad; Δr = r₂ − r₁ the path difference in m; λ the wavelength in that medium in m; Δφ₀ the lag wave 2 already carried behind wave 1 at emission, in rad.Delay is the primitive
δ = ω Δt = 2π f Δt
Use it whenever the two paths cross different media, where counting Δr/λ alone would mislead.
ω in rad s⁻¹, f in Hz, Δt the arrival-time difference in s. In one uniform medium Δt = Δr/v, with v in m s⁻¹.Phase bought by a slower segment
Δφ = 2π f L (1/v − 1/v₀)
Counts the phase a glass slide or a layer of warm air inserts without changing any distance.
L the altered segment in m; v its speed and v₀ the speed elsewhere, both m s⁻¹; f in Hz, unchanged at the boundary.Coherence time from bandwidth
Δν τc ≈ 1
Phase predictability and spectral purity are one property counted two ways: narrow the spectrum and the phase holds longer.
Δν the spectral width in Hz, τc the coherence time in s. The product is of order unity; its value depends on line shape.Coherence length
Lc = c τc ≈ λ²/Δλ
The largest path difference that still gives a stable pattern — the practical ceiling on any interferometer.
Lc in m; c the wave speed in m s⁻¹; λ the mean wavelength and Δλ the spectral width, both in m. λ/Δλ counts the fringes.Transverse coherence width
ℓₜ ≈ λR/a = λ/θₛ
The widest slit separation an extended source can still illuminate coherently.
ℓₜ in m; a the source width in m; R the source-to-slit distance in m; θₛ ≈ a/R the source's angular width in rad.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set the source offset to zero, hold the jitter at zero, and slide the path difference to 0.50 λ: half a wavelength of extra path is a π lag and the curves cancel. Then open the jitter and watch the surviving term collapse — that is two independent sources, with no stable pattern left to see.
Live results
- PHASE DIFFERENCE δ
- 3.77 rad
- COS δ
- -0.81
- COHERENCE FACTOR
- 0.88
- SURVIVING COS TERM
- -0.71
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 Δr / λ: 0.35 λ; Source phase offset Δφ₀: 0.50 π rad; Phase jitter σ: 0.50 rad
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The error is to hear same frequency and conclude coherent. Two signal generators both dialled to 440 Hz, or two lamps filtered to the same line, give no stable interference pattern at all: the intensities simply add. Coherence is a statement about the difference of the phases, not about the frequencies. What must hold still is Δφ₀, and nothing couples two independent oscillators to make it do so. Their frequencies also differ in the digits they do not display, so δ creeps through 2π unaided, and each source restarts its phase at random on the scale of τc. Averaged over any exposure long compared with τc, the cos δ term integrates to zero. That is why interference demonstrations split one source into two paths, or drive both speakers from one amplifier: the split manufactures the common phase that two separate sources can never supply.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Two loudspeakers are driven from a single amplifier and emit a steady 686 Hz tone; the speed of sound is 343 m s⁻¹. A listener stands 4.25 m from one speaker and 5.50 m from the other. Find the phase difference at the listener and say what is heard. The leads to the second speaker are then reversed. What changes?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Wavelength: λ = v/f = 343 m s⁻¹ ÷ 686 Hz = 0.500 m.
- Path difference: Δr = 5.50 m − 4.25 m = 1.25 m, so Δr/λ = 2.50.
- One amplifier feeding both speakers means Δφ₀ = 0, so δ = 2π(2.50) = 5π rad = 15.7 rad. That is an odd multiple of π, so cos δ = −1 and the two arrivals cancel: the listener is at a minimum.
- Reversing the leads sets Δφ₀ = π, giving δ = 5π + π = 6π rad = 18.8 rad, an even multiple of π, so cos δ = +1 and the listener now hears a maximum. Nothing about the geometry moved; only the second column of the ledger changed.
Answer. δ = 5π rad ≈ 15.7 rad, a minimum. After the leads are reversed, δ = 6π rad ≈ 18.8 rad, a maximum.
MediumExample 2
A red LED emits at a mean wavelength of 630 nm with a spectral width of 20 nm. Find its bandwidth, its coherence time and its coherence length, and estimate how many fringes it could produce in a two-path interferometer. Take c = 3.00 × 10⁸ m s⁻¹.
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 spectral width to bandwidth: Δν = c Δλ/λ² = (3.00 × 10⁸ m s⁻¹)(20 × 10⁻⁹ m) ÷ (630 × 10⁻⁹ m)² = 1.51 × 10¹³ Hz.
- Coherence time: τc ≈ 1/Δν = 1 ÷ (1.51 × 10¹³ Hz) = 6.6 × 10⁻¹⁴ s, that is 66 fs.
- Coherence length: Lc = c τc = (3.00 × 10⁸ m s⁻¹)(6.6 × 10⁻¹⁴ s) = 2.0 × 10⁻⁵ m, or 20 µm. The shortcut λ²/Δλ = (630 nm)² ÷ 20 nm returns the same 20 µm, as it must.
- Fringes need |Δr| ≲ Lc, and successive fringes are one wavelength apart in Δr, so the count is Lc/λ = λ/Δλ = 630 ÷ 20 ≈ 32.
- Limiting check: as Δλ → 0 the coherence length diverges and the fringe count is unlimited, which is the monochromatic idealisation the bare ledger quietly assumes.
Answer. Δν = 1.5 × 10¹³ Hz, τc = 6.6 × 10⁻¹⁴ s, Lc = 2.0 × 10⁻⁵ m (20 µm), giving about 32 fringes.
HardExample 3
A sodium lamp filtered to λ = 589.0 nm with Δλ = 0.60 nm illuminates a source slit of width a = 0.10 mm. The slit sits R = 0.80 m in front of a double slit of separation d = 0.50 mm, and fringes are viewed on a screen D = 2.0 m further on, out to 30 mm either side of the centre. Decide whether temporal coherence, spatial coherence, or neither limits the pattern, then find the source-slit width at which the fringes disappear.
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.
- Temporal limit: Lc = λ²/Δλ = (589.0 × 10⁻⁹ m)² ÷ (0.60 × 10⁻⁹ m) = 5.8 × 10⁻⁴ m, that is 0.58 mm.
- The largest path difference actually used is at the screen edge: Δr ≈ d y/D = (0.50 × 10⁻³ m)(0.030 m) ÷ (2.0 m) = 7.5 × 10⁻⁶ m, only 1.3% of Lc. Temporal coherence is nowhere near the limit.
- Spatial limit: ℓₜ ≈ λR/a = (589.0 × 10⁻⁹ m)(0.80 m) ÷ (0.10 × 10⁻³ m) = 4.7 × 10⁻³ m, that is 4.7 mm. The double slit uses d = 0.50 mm, about one ninth of ℓₜ, so this condition is comfortably met as well.
- Neither limit bites, so good fringes appear. To find where spatial coherence fails, set λR/a = d: aₘₐₓ = λR/d = (589.0 × 10⁻⁹ m)(0.80 m) ÷ (0.50 × 10⁻³ m) = 9.4 × 10⁻⁴ m, about 0.94 mm.
- Sanity check: opening the source slit by a factor of 9.4 destroys the pattern while leaving the lamp's spectrum, and therefore Lc, untouched. The two coherences are independent limits, and here it is the spatial one the experimenter must respect.
Answer. Lc = 0.58 mm against a used path difference of 7.5 µm, and ℓₜ = 4.7 mm against d = 0.50 mm: neither limits the fringes. They vanish once the source slit reaches a ≈ 0.94 mm.
Quick check
Select an answer and explain it before checking.
Review & continue
The phase difference is the single number δ saying how far one arriving wave's cycle sits behind another's at the same point, built from the path difference and from whatever offset the sources already carried.
- 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.