IB Physics SL Topic C.3 — How Waves Behave Paper 1 & 2 path difference = nλ ~8 min read

Interference

When two waves overlap, superposition adds their displacements — and the pattern we observe from that adding is interference. In some spots the waves reinforce into a big signal; in others they wipe each other out completely. Which happens where comes down to one thing: path difference.

📘 What you need to know

Constructive and Destructive Interference

Superposition can play out in two extreme ways. When two waves meet in phase — crest lined up with crest — their displacements add and you get a wave of bigger amplitude: constructive interference. When they meet in antiphase — crest lined up with trough — they cancel out to leave nothing: destructive interference.

in phase → constructive (2A) antiphase → destructive (0)
Left: two in-phase waves (dashed) build a resultant of double amplitude (purple). Right: two antiphase waves (dashed) cancel to a flat line (purple).

Coherence

You only get a stable, observable interference pattern if the two sources are coherent. That means they must have:

If the phase difference keeps jumping around, the bright and dark spots wander and smear out, so no steady pattern forms. This is why monochromatic laser light and two speakers driven at the same frequency are the classic coherent sources. At points that are neither fully in phase nor fully in antiphase, the resultant amplitude sits somewhere in between the two extremes.

Path Difference

What decides whether two waves arrive in phase or antiphase at a given point is the path difference — how much farther one wave has travelled than the other to reach that point.

S₁ S₂ P r₁ r₂ path difference = r₂ − r₁
Two coherent sources reach point P by different distances. The path difference (r₂ − r₁) sets whether they arrive in phase or antiphase.

Compare the path difference with the wavelength and you know instantly what you’ll see:

Constructive interference path difference = nλ
Destructive interference path difference = (n + ½)λ

where n = 0, 1, 2, 3… A path difference of a whole number of wavelengths brings the waves back into step (constructive); an odd number of half-wavelengths puts them exactly out of step (destructive). On a wavefront diagram you can just count wavelengths from each source to the point and subtract.

🧭 Finding the interference at a point

  1. Count the wavelengths (or measure the distance) from each source to the point
  2. Subtract to get the path difference
  3. A whole number of wavelengths (nλ) → constructive
  4. A half-integer number of wavelengths ((n + ½)λ) → destructive
  5. Check the sources are coherent first, or no steady pattern exists
Quick recap: constructive interference (in phase, bigger) happens at path difference nλ; destructive interference (antiphase, cancels) at (n + ½)λ — and only coherent sources give a steady pattern.
WE 1

Two coherent sources S₁ and S₂ emit waves of wavelength λ. At point P the waves travel 6λ and 6.5λ from the sources. At point Q they travel 7λ and 6λ.

State the type of interference at P and at Q.

Point P Path difference = 6.5λ − 6λ = 0.5λ = (0 + ½)λ destructive Point Q Path difference = 7λ − 6λ = λ = 1λ constructive Half-integer wavelengths cancel; whole wavelengths reinforce.
WE 2

On a wavefront diagram from two coherent sources, the distances to three points are: X → 5.5λ and 4.5λ; Y → 3.5λ and 3.5λ; Z → 4λ and 3.5λ.

Identify the interference at X, Y and Z.

Point X 5.5λ − 4.5λ = λ → constructive Point Y 3.5λ − 3.5λ = 0 → constructive Point Z 4λ − 3.5λ = 0.5λ → destructive Equal distances (path difference 0) are constructive — it’s still a whole number of wavelengths, just n = 0.

💡 Top tips

⚠ Common mistakes

Up next: we put all of this to work in the most famous interference experiment of all — Young’s Double-Slit Experiment — and derive the equation for the fringe spacing.

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