Amplitude, Intensity & Fringe Visibility

Superposition is a statement about fields: where two waves overlap the displacements add, y = y₁ + y₂. No detector reads that. A diaphragm, a photodiode or an ear reports intensity, the energy crossing unit area per second, and for a harmonic wave in a medium I = ½ρvω²A², quadratic in amplitude. So the order of operations is fixed: add first, square second. Take y₁ = A₁cos(ωt) and y₂ = A₂cos(ωt − δ). Square the sum and average over one period, using (1/T)∫₀^T cos(ωt) cos(ωt − δ) dt = ½cos δ, and you get ⟨y²⟩ = ½A₁² + ½A₂² + A₁A₂cos δ. The first two terms are what each wave delivers on its own. The third exists only because the square of a sum carries a cross product, and it is the interference term. Square each wave first and then add, and that term is gone — which is exactly the mistake the rest of this lesson exists to prevent.

Two waves of the same frequency but different amplitude and phase are added most cheaply as phasors. Draw A₁ along a reference direction, draw A₂ at angle δ to it, and the resultant is the closing side of the triangle, so the cosine rule gives A_R² = A₁² + A₂² + 2A₁A₂cos δ at once; the resultant's own phase relative to wave 1 follows from tan α = A₂ sin δ / (A₁ + A₂ cos δ). With A₁ = 3.0 mm, A₂ = 2.0 mm and δ = 60°, the cross term is 2(3.0)(2.0)(0.500) = 6.0 mm², so A_R = √(9.0 + 4.0 + 6.0) = 4.36 mm, an intensity 2.11 times that of source 1 alone. Two things are worth reading straight off the triangle. The resultant is confined between |A₁ − A₂| and A₁ + A₂, so complete cancellation demands equal amplitudes and nothing else will do. And δ enters only through its cosine, so δ, −δ and δ + 2π all give one answer: only the phase difference modulo 2π matters.

When the amplitudes match, the algebra simplifies sharply. Sum-to-product gives A cos(ωt) + A cos(ωt − δ) = 2A cos(δ/2) cos(ωt − δ/2): one wave at the original frequency, lagging by half the phase difference, carrying an amplitude 2A|cos(δ/2)| that δ alone controls. Squaring it produces the pattern this topic is named for, I = 4I₁cos²(δ/2), with I₁ the intensity one source delivers alone at that point. Read its scale carefully. At δ = 0 the intensity is 4I₁ — four times one source, not twice, because the amplitude doubled and intensity follows the square. At δ = π it is exactly zero. And because cos² is never negative, the pattern is not a sinusoid about zero but a raised one: it swings between 0 and 4I₁ about a mean of 2I₁, and it passes through that mean at δ = π/2, where the two phasors stand at right angles and the cross term vanishes — a perfectly coherent pair handing back the incoherent answer.

Absolute intensities are awkward to measure and easy to miscalibrate; a ratio is neither. Fringe visibility, V = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ), divides out whatever gain the detector applies, so two observers with different instruments still agree on it. For two coherent beams the extremes come from adding and subtracting amplitudes, Iₘₐₓ = (√I₁ + √I₂)² and Iₘᵢₙ = (√I₁ − √I₂)², so V = 2√(I₁I₂)/(I₁ + I₂), equal to 1 only when the beams are balanced. What surprises students is how slowly it falls. Make one beam 25 times the intensity of the other and V is still 2 × 5/26 = 0.38, with Iₘₐₓ = 36 and Iₘᵢₙ = 16 in units of the weaker beam — fringes anyone can see, because visibility tracks the amplitude ratio 5:1, not the intensity ratio 25:1. So a low visibility measured with beams you know to be balanced is evidence about coherence, not about amplitudes.

The cross term survives only if δ holds still. Two independent sources — two lamps, two loudspeakers on separate oscillators — drift in relative phase on a timescale far shorter than any detector's integration time, so what the meter reports is the average of I₁ + I₂ + 2√(I₁I₂) cos δ over a δ that is uniformly distributed. Since (1/2π)∫₀²π cos δ dδ = 0, the cross term vanishes and I = I₁ + I₂: intensities simply add, there are no fringes, and V = 0. Between the two limits sits partial coherence, handled by multiplying the cross term by a factor |γ| lying between 0 and 1, which makes V = 2√(I₁I₂)|γ|/(I₁ + I₂). This is why visibility is the experimenter's coherence meter. Block each source in turn to get I₁ and I₂, predict the fully coherent visibility from them, and the ratio of the measured visibility to that prediction is |γ|.

Anyone who has just written I = 0 at a dark fringe should feel uneasy about the energy. Integrate the pattern instead of sampling it. Averaging the equal-amplitude result over a full cycle of phase gives ⟨4I₁cos²(δ/2)⟩ = 4I₁ × ½ = 2I₁, which is precisely I₁ + I₂ for equal sources: the incoherent answer. The unequal case behaves identically, because ⟨cos δ⟩ = 0 leaves ⟨I⟩ = I₁ + I₂. Interference redistributes power across the pattern without altering the total, and every watt missing from a minimum reappears in a maximum — which is why the maxima climb to 4I₁ rather than 2I₁. Two loudspeakers several wavelengths apart and driven in phase radiate very nearly the power they would radiate alone; they merely refuse to spread it evenly. Bring them within a wavelength and each begins to work against the other's pressure field, the radiated power itself changes, and this simple audit stops applying.