Nodal Lines & Accessible Orders

Fix the two sources S₁ and S₂ a distance d apart and ask for every point whose distances satisfy r₂ − r₁ = k, a constant. That is the definition of a hyperbola with foci S₁ and S₂, so each fringe is a conic, not a line. Put the midpoint at the origin, the source line along y and the central axis along x. The branch obeys y²/a² − x²/b² = 1 with semi-transverse axis a = |k|/2, focal distance c = d/2, and b = √(c² − a²). Antinodal lines take k = mλ, so a = |m|λ/2; nodal lines take k = (m + ½)λ. In a ripple tank with d = 50.0 mm and λ = 15.0 mm the m = 2 antinodal branch has a = 15.0 mm, c = 25.0 mm and b = √(625 − 225) = 20.0 mm. The case m = 0 is the degenerate hyperbola a = 0: the perpendicular bisector, a straight line, and the only fringe that ever is one.

Walk along the segment joining the sources — that is the y axis of the previous section, not the x axis. At a point y from the midpoint, r₂ = d/2 + y and r₁ = d/2 − y, so r₂ − r₁ = 2y exactly, with no approximation anywhere. The m-th antinodal branch therefore cuts the baseline at y = mλ/2, and that crossing is its vertex. A vertex can sit on the segment only while |m|λ/2 ≤ d/2, that is |m| ≤ d/λ; at equality it lands on the source itself, b = 0, and the branch has collapsed onto the line beyond the source, surviving only as the grazing direction θ = 90°. The count follows: N = 2⌊d/λ⌋ + 1 antinodal branches, including the central one. Two loudspeakers 1.60 m apart at 686 Hz in air (λ = 0.500 m) give d/λ = 3.20, so m runs 0, ±1, ±2, ±3 — seven loud loci, and the m = 3 vertex sits 0.750 m from the midpoint, only 50 mm short of the speaker itself. Nodal branches are counted the same way with half-integers: |m + ½| ≤ 3.20 admits ±0.5, ±1.5 and ±2.5, so six of them.

A hyperbola's asymptotes pass through its centre with direction vector (b, a) in these coordinates, and that vector has length √(a² + b²) = c. So the asymptote makes an angle θ with the central axis where sin θ = a/c = (mλ/2)/(d/2) = mλ/d. The familiar d sin θ = mλ is therefore not an approximation to the fringe: it is the exact direction the fringe approaches, and the only approximation in far-field work is the claim that the observer already sits on the asymptote. Read the result the way it is written and one more fact drops out. Consecutive orders are evenly spaced in sin θ, by Δ(sin θ) = λ/d, never in θ. At d/λ = 3.20 the orders sit at θ = 0°, 18.2°, 38.7° and 69.6°: the first gap is 18°, the last is 31°, so a pattern that looks regular near the axis crowds towards the source line.

Put a microphone on a circular track of radius R ≫ d centred on the pair and log the angle ψ from the central axis. The far-field path difference is d sin ψ, which rises from 0 at ψ = 0 to d at ψ = 90°, the source-line direction, and falls back. Each branch has two asymptotic directions, so every branch is met twice per circuit. With d/λ = 3.20 that gives 7 × 2 = 14 loud directions and 6 × 2 = 12 silent ones — and 14 ≠ 12, which looks impossible if maxima and minima must alternate. They need not. The path difference is not monotonic in ψ: it turns around at 3.20λ, short of the 3.5λ needed for the next null. So the two m = 3 maxima sit either side of ψ = 90° with no silence between them, only a dip to cos²(0.2π) = 65.5% of the maximum intensity — and the m = −3 pair does the same either side of ψ = 270°. Two turning points, two unpaired maxima, and the difference 14 − 12 = 2 is accounted for. Counting the nulls on a circle is how you measure d/λ without a ruler.

Measure R and θ from the midpoint. Then r₂ and r₁ are √(R² ± R d sin θ + d²/4), the upper sign belonging to r₂, and expanding in d/R gives r₂ − r₁ = d sin θ − d³ sin θ cos²θ/(8R²) + … . The quadratic Fresnel term appears in each range but is identical for both sources, so it cancels in the difference — the leading error is cubic. Its largest value over angle is 0.385d³/(8R²), so keeping the phase error below π/10 needs only R ≳ d√(d/λ). For d = 0.500 mm and λ = 589 nm that is 14.6 mm, not the 42.4 cm that the usual d²/λ rule would demand — a factor of 29 more generous. The second approximation is separate and is the one that really bites: replacing sin θ by y/L on a flat screen. That small-angle step fails at large angle however far away the screen is, and no increase in L repairs it.

Let source 1 lead source 2 by Δα. The phase difference at P is Δφ = 2π(r₂ − r₁)/λ + Δα, so antinodal loci become r₂ − r₁ = (m − Δα/2π)λ. The geometry is unchanged — still confocal hyperbolae, still vertices at half the path difference — but every branch is relabelled, and the family slides towards the lagging source. Two speakers 0.700 m apart at λ = 0.250 m give d/λ = 2.80 and five antinodal directions when driven in phase. Add a quarter-cycle lead, Δα = π/2, and the condition becomes sin θ = (m − 0.25)λ/d: the central lobe tips 5.12° towards the lagging speaker, and orders now run m = −2 to +3, six of them, because the accessible window |m − 0.25| ≤ 2.80 is no longer symmetric. At Δα = π the nodal and antinodal families swap places exactly.