Wave Interference / Two-Source

constructive vs destructive interference, visualized live

Worked example — Wave Interference / Two-Source

d/lambda	4

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Why 1+1 can become 0 🖖

When two waves meet, their amplitudes add point-by-point — this is the superposition principle. For path difference Δr = mλ the crests align and intensity quadruples: I = 4I₀. For Δr = (m+½)λ a crest meets a trough and intensity drops to zero. The general result is Aₛᵤₘ = 2A₀ cos(δ/2), so intensity I ∝ cos²(δ/2). This is exploited in noise-canceling headphones (inject δ = π), optical anti-reflection coatings (thin-film thickness λ/4), and radio phased-array antennas.

What the bright and dark bands mean 🖖

Two sources send out overlapping ripples. Pick any point on the screen: usually one wave has traveled slightly farther than the other, and that extra distance — the path difference — decides everything. When it equals a whole number of wavelengths, crests line up and you get a bright fringe; when it's off by half a wavelength, a crest meets a trough and the spot goes dark. Widen the separation d and the tool packs more fringes in, following sin θ = nλ/d.

A single electron interferes with itself 🖖

You might assume interference needs a whole crowd of waves crossing at once, but it doesn't. Fire electrons through a double slit one at a time — so lonely that only one is ever in the apparatus — and each lands as a single dot, seemingly at random. Let thousands accumulate and the same striped pattern emerges. Each particle somehow samples both paths and interferes with itself, a result readers voted "the most beautiful experiment in physics."

Problem solved in full

Two sources 2 units apart with a wavelength of 0.5 5 steps
Two sources 2 units apart, wavelength 0.5. Find the second-order bright and dark directions, then count how many bright fringes exist at all.
1. Everything follows from one geometric fact: at an angle θ the extra distance the far source's wave must travel is d sin θ. Whether that arrives in step or out of step is the whole subject.
  
  $Δ r = d sin θ, bright : Δ r = nλ, dark : Δ r = (n + \frac{1}{2}) λ$
2. A whole number of wavelengths of extra path means the crests coincide. Order 2 needs one full unit of path difference, and this separation supplies it at a sine of one half.
  
  $sin θ_{bright} = \frac{nλ}{d} = \frac{2 \times 0.5}{2} = \frac{1}{2} ⟹ θ = 3 0^{\circ}$
3. Half a wavelength out and the crests meet troughs. The 2.5 is not a typo — dark fringe n sits between bright n and bright n+1.
  
  $sin θ_{dark} = \frac{2.5 \times 0.5}{2} = 0.625 ⟹ θ = 38.6 8^{\circ}$
4. Now bound it. The sine cannot exceed 1, so the ratio d/λ caps the order, and the fringes exist on both sides of the centre plus the centre itself.
  
  $sin θ \leq 1 ⟹ n \leq \frac{d}{λ} = 4, bright orders = 2 n_{m a x} + 1 = 9$
5. Take that limit seriously and it becomes a design rule. Below one wavelength of separation there is nothing but the central lobe, and at half a wavelength there is exactly one.
  
  $d < λ ⟹ n_{m a x} = 0 : one lobe only. d = \frac{λ}{2} is the array spacing that has no grating lobe.$
Answer

30° and 38.68°, with nine bright fringes in total. The count is the interesting one, and it comes from a constraint rather than a calculation: sin θ can never exceed 1, so no order beyond d/λ = 4 has anywhere to go. Shrink the separation below one wavelength and every order but the central one disappears — the pattern collapses to a single lobe. That single fact explains a hole and a nuisance in two different fields at once. It is why an aperture narrower than a wavelength radiates almost evenly in all directions, and why a phased array is built at half-wavelength spacing: at d = λ/2 the first side order is pushed to sin θ = 2, which does not exist, and the array steers without spraying copies of its beam into the sky.

$θ_{bright} = 3 0^{\circ}, θ_{dark} = 38.6 8^{\circ}, n_{m a x} = ⌊ d / λ ⌋ = 4$

Learning path

From one spring to a standing wave

Leads to Standing waves superposition you can see.

References (2)

Superposition, the cos²(δ/2) intensity and thin-film coatings: Eugene Hecht, Optics, 5th edition, ch. 7 and 9. Pearson, 2017. ISBN 978-0-13-397722-6.
The single-electron double slit in the third block: A. Tonomura, J. Endo, T. Matsuda, T. Kawasaki and H. Ezawa, "Demonstration of single-electron buildup of an interference pattern." American Journal of Physics 57(2), 117–120, 1989.

By the 42Math team · Updated 2026-05-15

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

wide spacing - wide spacing d=4, lambda=0.5 -> many narrow bright/dark fringes
narrow spacing - narrow spacing d=1.2, lambda=0.5 -> fewer, wider fringes
long wavelength - longer wavelength lambda=1.4 spreads fringes farther apart
sound-like - sound-like long wavelength lambda=2.6 shows broad interference regions