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Wave Interference / Two-Source

constructive vs destructive interference, visualized live

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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

  1. 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.

    S₂ S₁ d θ = 30° θ Δr = 2λ

    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.

      Δr=dsin⁡θ,bright:Δr=nλ,dark:Δr=(n+12)λ\Delta r=d\sin\theta,\qquad \text{bright}: \Delta r=n\lambda,\qquad \text{dark}: \Delta r=\left(n+\tfrac{1}{2}\right)\lambda

    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⁡θbright=nλd=2×0.52=12⟹θ=30∘\sin\theta_{\text{bright}}=\frac{n\lambda}{d}=\frac{2\times0.5}{2}=\frac{1}{2}\quad\Longrightarrow\quad \theta=30^{\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⁡θdark=2.5×0.52=0.625⟹θ=38.68∘\sin\theta_{\text{dark}}=\frac{2.5\times0.5}{2}=0.625\quad\Longrightarrow\quad \theta=38.68^{\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≤1 ⟹ n≤dλ=4,bright orders=2nmax⁡+1=9\sin\theta\le1\ \Longrightarrow\ n\le\frac{d}{\lambda}=4,\qquad \text{bright orders}=2n_{\max}+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 ⟹ nmax⁡=0: one lobe only. d=λ2 is the array spacing that has no grating lobe.d<\lambda\ \Longrightarrow\ n_{\max}=0:\ \text{one lobe only. } d=\tfrac{\lambda}{2}\ \text{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.

    θbright=30∘, θdark=38.68∘, nmax⁡=⌊d/λ⌋=4\theta_{\text{bright}}=30^{\circ},\ \theta_{\text{dark}}=38.68^{\circ},\ n_{\max}=\left\lfloor d/\lambda\right\rfloor=4

Learning path

From one spring to a standing wave

References (2)

· Reviewed against published sources. Corrections welcome.