Preview
Wave Interference
Superposition and two-source interference patterns
Drop two dippers into a ripple tank and let them bob in phase. The surface redraws itself into a map: radiating still lines where the water never moves, and between them lanes where the waves rear up twice as tall as either source could make alone. Nothing about that map is decorative — it is arithmetic made visible. Every point on the surface listens to both dippers, and the only question that matters is how much farther the point is from one than from the other. Where that path difference is a whole number of wavelengths, crest lands on crest: the bright lanes. Where it is a whole number plus one half, crest lands on trough: the still lines. And here is the twist this lab is built around — the map is drawn in units of wavelength, not meters. The sources sit exactly where they sat, you turn only the frequency knob, and the geometry redraws itself: shorter wavelength, more still lines sliding in from the edges. One more thing the still lines are not: they are not places where waves died. The two waves pass through each other untouched — what cancels at a node is the displacement, and the energy it seems to be missing is exactly the energy piled up on the bright lanes (the same bookkeeping your noise-canceling headphones run in reverse). Last lesson you watched one slit bend light through a continuum of tiny wavelet sources; today there are just two sources and one path difference, the cleanest interference there is. You have three live controls — frequency, separation, amplitude — and a prediction to make before you touch the first one.
What you'll be able to do
- Apply the path-difference conditions quantitatively on the panel's own rows: at defaults (f = 1.20 Hz → λ = 4.17 m, d = 4.0 m) destructive lines sit where Δr = (n+½)λ and constructive where Δr = nλ (OpenStax Eq. 3.1/3.2), and because the largest path difference anywhere on the plane is the baseline itself (|Δr| ≤ d, reached on the axis beyond the sources), the default state admits exactly one nodal pair (½λ = 2.08 m ≤ 4.0 m; 3λ/2 = 6.25 m > 4.0 m) and no off-axis antinodal order (λ = 4.17 m > d)
- Price the whole pattern with one dimensionless ratio, d/λ: raising frequency 1.20 → 3.00 Hz shrinks λ 4.17 → 1.67 m and lifts d/λ from 0.96 to 2.40, so a second nodal pair (Δr = 0.83 m, 2.50 m) and antinodal orders m = 1, 2 (1.67 m, 3.33 m) slide in under the same geometric ceiling; sweeping separation 1.0 → 8.0 m at fixed f moves the same ratio 0.24 → 1.92 — below d = ½λ = 2.08 m even the first nodal pair cannot exist (the Close Sources preset's all-bright regime, d/λ = 0.48); 'these angles depend on wavelength and the distance between the slits' (OpenStax §3.1 verbatim)
- Keep two-source interference and single-slit diffraction apart as mechanisms: this pattern is the path-difference ladder between TWO coherent point sources (Δl = mλ / (m+½)λ, Eq. 3.1/3.2 — the ripple-tank two-plunger situation), while the sister unit's single slit is pair-wise cancellation of a continuum of Huygens wavelets across ONE aperture (a·sinθ = mλ, Eq. 4.1) whose central maximum is twice as wide as any side maximum ('the second maximum is only about half as wide as the central maximum', §4.1 verbatim) — and confirm that amplitude is nowhere in either geometry: dragging Amplitude 0.2 → 2.0 scales the surface tenfold while the nodal tubes never move
Formulas
Make a prediction
Default state: f = 1.20 Hz (the panel reads λ = 4.17 m), source separation d = 4.0 m — exactly one pair of red nodal lines floats over the surface, where Δr = ½λ = 2.08 m. Now drag Frequency to 3.00 Hz (separation and amplitude untouched). What happens to the interference pattern?
No grading here — commit to a guess, then scroll down and test it yourself.
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Your prediction
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Answer: A second nodal pair slides in — the pattern densifies: λ shrinks to 1.67 m, so Δr = (n+½)λ = 0.83 m and 2.50 m both fit under the geometric ceiling |Δr| ≤ d = 4.0 m, where at λ = 4.17 m only the 2.08 m line fit
λ = v/f = 5.0/3.0 = 1.67 m, and nodal lines live wherever the path difference equals a half-integer number of wavelengths: Δr = (n+½)λ (OpenStax Eq. 3.2). On this plane the largest possible path difference is the baseline itself — |Δr| ≤ d = 4.0 m, reached on the axis beyond the sources. At λ = 4.17 m the half-integer ladder (2.08, 6.25, … m) fits only its first rung under that ceiling; at λ = 1.67 m the ladder (0.83, 2.50, 4.17, … m) fits two rungs (4.17 > 4.0 misses by a hair), and the antinodal orders m = 1, 2 (1.67 m, 3.33 m) come along too. Same geometry, new pattern — because the pattern is drawn in units of λ, not meters: 'regions of constructive and destructive interference move out from the slits at well-defined angles… These angles depend on wavelength and the distance between the slits' (OpenStax §3.1, verified). Option B is the geometry-only instinct — the sources make the waves, so their layout should rule the map — and the HTML's own quiz Q2 carries the correction in-scene: 'Fringe count depends on d/λ — changing d does change the count'; frequency is simply the same ratio's other lever. Option C inverts the law: a shorter λ packs MORE path-difference rungs under the same ceiling, so lines arrive closer together, never wider. The lab's machine gate counts nodal pairs at both frequencies and brackets the ratio at exactly 2.
Quiz (0/5)
Two sources 1.5 m apart emit waves of λ = 0.5 m. At what angles do the first two constructive interference maxima occur?
A standing wave on a 0.6 m string vibrates in its 3rd harmonic. What is the wavelength? How many nodes are there?
If you double the wavelength of both sources, how does the spacing of maxima change?
Two sources emit waves in phase. A point P is 3.25λ from source 1 and 1.75λ from source 2. Is P a node or antinode?
A string fixed at both ends has v = 120 m/s and L = 0.8 m. Find all resonant frequencies up to 500 Hz.
You can now
- Apply the path-difference conditions quantitatively on the panel's own rows: at defaults (f = 1.20 Hz → λ = 4.17 m, d = 4.0 m) destructive lines sit where Δr = (n+½)λ and constructive where Δr = nλ (OpenStax Eq. 3.1/3.2), and because the largest path difference anywhere on the plane is the baseline itself (|Δr| ≤ d, reached on the axis beyond the sources), the default state admits exactly one nodal pair (½λ = 2.08 m ≤ 4.0 m; 3λ/2 = 6.25 m > 4.0 m) and no off-axis antinodal order (λ = 4.17 m > d)
- Price the whole pattern with one dimensionless ratio, d/λ: raising frequency 1.20 → 3.00 Hz shrinks λ 4.17 → 1.67 m and lifts d/λ from 0.96 to 2.40, so a second nodal pair (Δr = 0.83 m, 2.50 m) and antinodal orders m = 1, 2 (1.67 m, 3.33 m) slide in under the same geometric ceiling; sweeping separation 1.0 → 8.0 m at fixed f moves the same ratio 0.24 → 1.92 — below d = ½λ = 2.08 m even the first nodal pair cannot exist (the Close Sources preset's all-bright regime, d/λ = 0.48); 'these angles depend on wavelength and the distance between the slits' (OpenStax §3.1 verbatim)
- Keep two-source interference and single-slit diffraction apart as mechanisms: this pattern is the path-difference ladder between TWO coherent point sources (Δl = mλ / (m+½)λ, Eq. 3.1/3.2 — the ripple-tank two-plunger situation), while the sister unit's single slit is pair-wise cancellation of a continuum of Huygens wavelets across ONE aperture (a·sinθ = mλ, Eq. 4.1) whose central maximum is twice as wide as any side maximum ('the second maximum is only about half as wide as the central maximum', §4.1 verbatim) — and confirm that amplitude is nowhere in either geometry: dragging Amplitude 0.2 → 2.0 scales the surface tenfold while the nodal tubes never move