Wave model of light and coherence

Light is a transverse electromagnetic wave, and it is the electric field that drives every detector you will use — an eye, a photodiode, a photographic grain. Along one direction the field of a single plane wave is E(x, t) = E₀ cos(kx − ωt + φ₀), with k = 2π/λ counting radians per metre and ω = 2πf radians per second. At λ₀ = 550 nm the frequency is c/λ₀ = 5.45 × 10¹⁴ Hz and one period lasts 1.83 fs. Nothing follows that. The fastest photodiode integrates for nanoseconds and the eye for tens of milliseconds, so every instrument reports the time average of the energy flux: I = ⟨ε₀cE²⟩ = ½ε₀cE₀², the ½ being the average of cos² over a cycle. That consequence sets up the rest of the topic. The observable is a squared amplitude, the oscillation itself is invisible, and any claim about phase has to be cashed out as a claim about a time-averaged intensity. For scale, a 1.0 mW laser filling 1.0 mm² gives I = 1.0 × 10³ W m⁻², so E₀ = √(2I/ε₀c) = 8.7 × 10² V m⁻¹.

Superposition is a statement about fields: where two beams overlap, E = E₁ + E₂ at every instant. Take E₁ = E₀₁ cos(ωt) and E₂ = E₀₂ cos(ωt + δ) at one point. Squaring and time-averaging gives ⟨E²⟩ = ⟨E₁²⟩ + ⟨E₂²⟩ + 2⟨E₁E₂⟩, and the product averages to ⟨E₀₁E₀₂ cos(ωt) cos(ωt + δ)⟩ = ½E₀₁E₀₂ cos δ. Multiply through by ε₀c and the result is I = I₁ + I₂ + 2√(I₁I₂) cos δ. The first two terms are what you would get by adding brightnesses like sand; the third is the interference, and it is the only term that knows the beams are related. For equal beams, I₁ = I₂ = I₀, it collapses to I = 2I₀(1 + cos δ) = 4I₀ cos²(δ/2), which runs from 4I₀ at δ = 0 down to exactly zero at δ = π. Nothing is created or destroyed: cos δ averages to zero across a full pattern, so the mean intensity is still I₁ + I₂. Interference moves energy out of the dark fringes and into the bright ones.

Everything now rests on δ, and δ has exactly three sources. The beams may cover different geometric distances; they may pass through media of different refractive index; and they may have left their sources with different phases. Inside a medium the frequency is unchanged but the wavelength shrinks to λ = λ₀/n, so a distance r holds nr/λ₀ cycles rather than r/λ₀. Counting cycles and multiplying by 2π, δ = (2π/λ₀)(n₂r₂ − n₁r₁) + (φ₂ − φ₁). The quantity nr is the optical path length, and Δ = n₂r₂ − n₁r₁ is the optical path difference, so δ = 2πΔ/λ₀ whenever the two sources start in phase. Worth a number: sliding a 12.0 μm sheet of glass of n = 1.52 into one beam, in place of the same thickness of air, adds (1.52 − 1.00)(12.0 μm) = 6.24 μm of optical path — 10.6 wavelengths at λ₀ = 589 nm, so the pattern shifts by 10.6 fringes for a sheet thinner than a hair.

Write the observed intensity honestly, with the average taken over the detector's integration time T: I = I₁ + I₂ + 2√(I₁I₂)⟨cos δ⟩. Two limits follow. If δ is genuinely fixed over T then ⟨cos δ⟩ = cos δ, the full cross term appears, and the beams are coherent. If δ wanders across the whole range 0 to 2π during T then ⟨cos δ⟩ = 0, the beams add like sand, and they are incoherent. The general case is written 2√(I₁I₂)|γ| cos δ, where |γ| between 0 and 1 is the degree of coherence, and what a meter actually returns is the fringe visibility V = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ) = 2√(I₁I₂)|γ|/(I₁ + I₂). Two independent LEDs are hopeless: their relative phase is redrawn roughly every 10⁻¹³ s, so a 1/60 s exposure averages about 1.7 × 10¹¹ independent draws and ⟨cos δ⟩ sinks to some 2 × 10⁻⁶. The practical response is never to use two sources. Split one beam — by wavefront at a pair of slits, or by amplitude at a half-silvered mirror — so that both arms inherit the same phase accidents.

A perfectly monochromatic wave would be one frequency extending over all time. Real light is emitted in bursts, and a burst lasting τ_c cannot be a single frequency: Fourier analysis spreads it over a bandwidth Δν ≈ 1/τ_c. Read backwards, a measured spectral width fixes how long the phase is remembered, and multiplying by c turns that memory into a distance, the coherence length L_c = cτ_c ≈ λ₀²/Δλ. This is the tolerance on path difference. Recombine two arms differing by Δ and the visibility stays high while Δ ≪ L_c, collapses near Δ ≈ L_c, and is gone past it; the number of usable fringes is L_c/λ₀ ≈ λ₀/Δλ. The spread across sources is enormous. White light at λ₀ = 550 nm with Δλ ≈ 300 nm gives L_c ≈ 1.0 μm, about two fringes, which is why a soap film shows colour only where it is submicron thin. A sodium line filtered to Δλ = 0.60 nm gives L_c = 0.58 mm. A helium-neon laser of Δν = 1.5 GHz gives L_c = 0.20 m, and a stabilised laser reaches kilometres.

Temporal coherence compares one beam with itself at a later time. Spatial coherence compares two points across the same wavefront at the same time, which is the question that matters when two slits or pinholes have to be illuminated. A point source is perfect: one emitter drives both points, so their phase difference is fixed by geometry alone. Widen the source to a width a at distance R and every element of it drives its own pattern, shifted slightly from its neighbours because it sees a slightly different path difference to the two points. Summing over the source washes the fringes out once that shift reaches one fringe, which happens when the point separation d reaches the transverse coherence width wₜ ≈ λ₀R/a = λ₀/θₛ, with θₛ = a/R the angular width of the source. Numbers make the rule vivid: the Sun subtends θₛ = 9.3 mrad, so sunlight at 550 nm is coherent across only 59 μm, while a 0.20 mm filament at 1.0 m manages 2.8 mm. That is why Young set a single narrow slit ahead of his double slit, and why a laser, spatially coherent across its whole beam, needs no such trick.