The wave model and interference (SACE Stage 2 Physics)

Syllabus dot point

“Describe the wave model of light and apply the principle of superposition to constructive and destructive interference.”

9 min readUpdated 2 Jun 2026

Quick answer

The wave model treats light as a transverse wave with a wavelength, frequency and speed linked by $v = f\lambda$ . When two waves overlap, their displacements add (superposition). They interfere constructively (bright) where they arrive in phase - a path difference of a whole number of wavelengths - and destructively (dark) where they arrive out of phase - a path difference of an odd number of half-wavelengths.

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What this dot point is asking
The wave model of light
The principle of superposition
Coherence and path difference
Why interference is evidence for waves
How SACE assesses this
Exam-style questions

What this dot point is asking

You need to describe the wave model of light and apply the principle of superposition to predict where constructive and destructive interference occur.

Constructive or destructive?

Two coherent speakers emit sound of wavelength 0.50 m; a listener is 4.00 m from one and 5.25 m from the other

Decide whether the listener hears a loud (maximum) or quiet (minimum) point.

Path difference: $5.25 - 4.00 = 1.25\ \text{m}$ .

Compare with the wavelength: $\dfrac{1.25}{0.50} = 2.5$ wavelengths.

This is an odd number of half-wavelengths ( $2.5\lambda = \tfrac{5}{2}\lambda$ ), so the waves arrive exactly out of phase. The interference is destructive and the listener is at a minimum (quiet point). The same path-difference reasoning applies to light and to radio waves.

The wave model of light

The wave model treats light as a transverse electromagnetic wave. Its key quantities are wavelength $\lambda$ , frequency $f$ and speed $v$ (in a vacuum, $c = 3.00\times10^8\ \text{m s}^{-1}$ ), linked by:

v = f\lambda.

Different colours correspond to different wavelengths; the visible range runs from about

400\ \text{nm}

(violet) to

700\ \text{nm}

(red). The wave model successfully explains reflection, refraction, diffraction and interference.

The principle of superposition

Superposition

When two or more waves overlap at a point, the resultant displacement is the sum of the individual displacements at that instant. Where two crests (or two troughs) coincide, the waves reinforce; where a crest meets a trough, they cancel.

Superposition is what makes interference possible: the combined wave can be larger or smaller than either wave alone, depending on the phase relationship at that point.

Coherence and path difference

For a stable interference pattern the two sources must be coherent, meaning they have the same frequency and a constant phase relationship. The outcome at any point is then decided by the path difference, the difference in the distances travelled by the two waves to that point.

Constructive and destructive interference

Constructive interference (maximum): the waves arrive in phase. The path difference is a whole number of wavelengths, $n\lambda$ ( $n = 0, 1, 2, \dots$ ).
Destructive interference (minimum): the waves arrive exactly out of phase. The path difference is an odd number of half-wavelengths, $\left(n + \tfrac{1}{2}\right)\lambda$ .

So you classify any point by dividing its path difference by the wavelength: a whole number gives a maximum, a half-integer gives a minimum.

Why interference is evidence for waves

Interference - the alternating reinforcement and cancellation of two overlapping disturbances - is a uniquely wave phenomenon. Particles travelling in straight lines cannot cancel one another out, yet two coherent light beams produce dark fringes where they overlap. The double-slit pattern (next dot point) is the classic demonstration and was historically decisive evidence for the wave nature of light.

How SACE assesses this

SACE Stage 2 questions on superposition give the path difference to a point (often from two antennas or two slits) and ask whether the interference is constructive or destructive. The reliable method is to compute the path difference, divide it by the wavelength, and classify the result: a whole number of wavelengths gives a maximum (constructive), an odd number of half-wavelengths gives a minimum (destructive). A frequent explanation part asks you to account for the dark fringes in a two-slit pattern, where the marking points are superposition of the two coherent waves, destructive interference (a crest meeting a trough), and the path-difference condition of an odd number of half-wavelengths. State your reasoning explicitly and quote the path-difference condition you are using.

Common errors

Forgetting the half-wavelength condition: Minima occur at odd numbers of half-wavelengths, not at half-integer wavelengths quoted carelessly. A path difference of $\tfrac{5}{2}\lambda$ is a minimum.
Ignoring coherence: A stable pattern needs coherent sources (same frequency, fixed phase). Two ordinary light bulbs do not interfere visibly.
Using distance instead of path difference: It is the difference in the two path lengths, compared with the wavelength, that decides constructive or destructive, not either distance alone.
Confusing amplitude and intensity: Cancellation refers to displacement (amplitude); the observed brightness or loudness depends on the resultant amplitude at that point.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2024-style3 marks

Two vertical transmitting antennas A and B emit coherent electromagnetic waves of wavelength

62.5\ \text{m}

. A receiver at point P is

30.0\ \text{km}

from A and

40.0\ \text{km}

from B. Determine whether the waves at P are at maximum or minimum amplitude.

Show worked answer →

Whether the interference is constructive (maximum) or destructive (minimum) depends on the path difference compared with the wavelength.

Path difference $= 40.0 - 30.0 = 10.0\ \text{km} = 10\,000\ \text{m}$ .

Divide by the wavelength: $\dfrac{10\,000}{62.5} = 160$ , a whole number of wavelengths.

A path difference of a whole number of wavelengths means the waves arrive in phase, so they interfere constructively and the amplitude at P is a maximum. 1 mark for the path difference, 1 mark for dividing by the wavelength to get a whole number, 1 mark for concluding maximum amplitude.

SACE 2023-style3 marks

In a two-slit experiment with a sodium lamp, a pattern of dark and light fringes appears on a screen. Explain the production of the dark fringes.

Show worked answer →

Light from the two slits spreads out and overlaps on the screen. Because the light is coherent, the two sets of waves combine by the principle of superposition.

A dark fringe forms where the two waves arrive exactly out of phase, so a crest from one slit meets a trough from the other. This is destructive interference, and the waves cancel to give minimum (zero) amplitude.

This happens where the path difference is an odd number of half-wavelengths ( $\tfrac{1}{2}\lambda$ , $\tfrac{3}{2}\lambda$ , and so on). 1 mark for superposition of the two coherent waves, 1 mark for destructive interference (crest meets trough), 1 mark for the odd-number-of-half-wavelengths condition.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Stage 2 Physics subject outline (2025 onwards) — SACE Board
Stage 2 Physics: external assessment — SACE Board