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Interference Pattern

Key idea: State the conditions for a stable interference pattern and use phase/path difference to identify maxima and minima (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Explain single-aperture diffraction and apply first-minimum and Rayleigh criteria.
  • Relate phase difference to separations in time and position.
  • Explain coherent two-source interference using phase and path difference.
  • Analyse Young double-slit interference and its small-angle assumptions.

1. Definitions (Must Know)

Two-source interference and path differenceTwo coherent point sources produce circular wavefronts. Curves mark antinodal and nodal loci, and a sample point shows the two path lengths used to find path difference.Two coherent sources, constant phase differenceS₁S₂antinode: Δr = nλnode: Δr = (n + ½)λPr₁r₂At P: path difference Δr = |r₂ − r₁|, then phase difference Δφ = 2πΔr/λ.
Scroll diagram horizontally to read all labels.
For in-phase sources, an antinodal line joins points with path difference nλ; a nodal line joins points with path difference (n + ½)λ. Similar amplitudes make the minima more complete.

A. Interference pattern

An interference pattern is a stable pattern of maxima (bright/loud) and minima (dark/quiet) formed when waves superpose.

B. Coherent sources

Coherent sources produce waves with a constant phase difference.

C. Path difference and phase difference

At a point, the path difference Δ x between two waves produces a phase difference: Δφ = 2π(Δ x)/λ

2. Key Ideas (What Earns Marks)

  • Conditions for a clear, stable interference pattern:
    • coherent sources (constant phase difference)
    • similar amplitudes (good fringe contrast)
    • same polarisation plane (or both unpolarised) for light
  • For two in-phase sources:
    • maxima (constructive): Δ x = nλ
    • minima (destructive): Δ x = (n + 1/2)λ
Link: superposition and phase difference

See: Principle of Superposition and Phase Difference.

3. Detailed Explanations

A. Why coherence is required

If the phase difference between the two sources changes randomly, the positions of maxima/minima move around. The average intensity becomes uniform and the fringes wash out.

Coherent sources keep a constant phase difference, so the pattern is steady.

B. Why similar amplitudes matter (contrast)

At a minimum, the waves are out of phase and oppose each other.

If the amplitudes are equal, you can get almost complete cancellation (very dark/quiet minima). If one amplitude is much larger, cancellation is incomplete and the pattern has poor contrast.

C. Why polarisation must match (light)

For light, complete cancellation requires the electric fields to oppose along the same line.

If the waves are polarised in perpendicular planes, they do not cancel completely even if they are π out of phase.

4. Common Mistakes

  • Saying “same frequency” is enough (you need constant phase difference, not just same f).
  • Using Δ x = nλ for destructive interference (it is for constructive, for in-phase sources).
  • Confusing the interference pattern with the diffraction envelope (in double-slit patterns).

5. Exam Tips

  • Start by stating the conditions for fringes (coherence + comparable amplitude + matching polarisation for light).
  • If the question gives a path difference, classify the point as maximum/minimum by comparing with nλ or (n + 1/2)λ.
  • Use n = 0 for the central maximum (if sources are in phase).

6. Worked Examples

Modelled example 1

Identify maximum or minimum from path difference

Core

Problem

Two coherent sources are in phase. The wavelength is λ = 0.40 m. At a point, the path difference is Δ x = 0.60 m. Is this point a maximum or a minimum?
Study the worked solution

  1. Express the path difference in wavelengths

    Method

    The path difference is 1.5λ.

    Reason

    Interference conditions are recognised by comparing Δ x with one wavelength.

    Working

    (Δ x)/λ = 0.60/0.40 = 1.5 = 1 + 1/2

  2. Match the interference condition

    Method

    The point is a minimum.

    Reason

    For in-phase sources, a half-integer number of wavelengths gives antiphase arrival and destructive interference.

    Working

    Δ x = (n + 1/2)λ with n = 1

Guided practice 2

Small change in path difference

About 3 min

Problem

Two coherent in-phase sources have wavelength λ. A point on the screen changes from a maximum to the next minimum. By what multiple of λ did the path difference change?

Try this before viewing the solution

Unit: λ

Hints

Hint 1: write neighbouring conditions
Write one maximum as nλ and the next minimum as (n + 1/2)λ.
View solution step by step

  1. Write the two neighbouring conditions

    Method

    Use nλ for the maximum and (n + 1/2)λ for the next minimum.

    Reason

    Constructive and destructive points alternate as path difference increases.

    Working

    nλ → (n + 1/2)λ

  2. Find the change

    Method

    The path difference changes by half a wavelength.

    Reason

    Subtract the initial condition from the final condition.

    Working

    Δ(Δ x) = (n + 1/2)λ-nλ = (1/2)λ

Common misconception 3

Identify order from path difference

Find and correct the mistake

Learner claim

Two coherent in-phase sources have λ = 0.25 m and Δ x = 0.50 m at a point. A learner calls it the “second minimum” because the ratio is 2. Diagnose the claim and identify the fringe correctly.

Try this before viewing the solution

Correct classification

View solution step by step

  1. Calculate the wavelength ratio

    Method

    The path difference is two wavelengths.

    Reason

    The ratio determines whether the multiplier is an integer or half-integer.

    Working

    (Δ x)/λ = 0.50/0.25 = 2

  2. Separate type from order

    Method

    This is a constructive maximum with order n = 2.

    Reason

    For in-phase sources, integer multiples nλ are maxima; the integer gives the order, not the fringe type.

    Working

    Δ x = 2λ

Examiner practice 4

Using phase difference directly

3 marks

Examination question

Two coherent in-phase sources produce waves of wavelength λ = 0.80 m. At a point, the phase difference is Δφ = π. Find the path difference and state whether the point is a maximum or minimum. [3 marks]

Try this before viewing the solution

Unit: m

View solution step by step

  1. Relate phase and path

    1 mark

    Method

    Use Δφ = 2πΔ x/λ.

    Reason

    A full wavelength of path difference corresponds to 2π radians.

    Working

    Δ x = (Δφ/2π)λ

  2. Calculate the path difference

    1 mark

    Method

    Δ x = 0.40 m.

    Reason

    A phase difference of π is half a cycle.

    Working

    Δ x = (π/2π)(0.80) = 0.40 m = (1/2)λ

  3. Classify the point

    1 mark

    Method

    The point is a minimum.

    Reason

    For in-phase sources, a half-wavelength path difference makes the waves arrive in antiphase.

    Working

    destructive interference

Challenge 5

From one bright fringe to the next

Minimal support

Independent transfer

For two coherent in-phase sources, a second-order maximum occurs where the path difference is 0.90 m. Find the path difference at the adjacent third-order maximum.

Try this before viewing the solution

Unit: m

Hints

Hint 1: infer the wavelength
At the second-order maximum, 0.90 = 2λ.
View solution step by step

  1. Infer the wavelength

    Method

    λ = 0.45 m.

    Reason

    The second-order maximum has path difference 2λ.

    Working

    λ = 0.90/2 = 0.45 m

  2. Advance to the next maximum

    Method

    The third-order path difference is 1.35 m.

    Reason

    Adjacent maxima differ in order by one and therefore in path difference by one wavelength.

    Working

    Δ x₃ = 3λ = 3(0.45) = 1.35 m

7. Mind Stretchers

Mind stretcher 1: Fringe contrast explanationExtension

Two coherent waves have amplitudes A and 0.3A. They arrive π out of phase at a point.

Explain why the point is not completely dark/quiet.

Show Answer

Out of phase means the displacements oppose, but because the amplitudes are not equal there is incomplete cancellation. The resultant amplitude is A-0.3A = 0.7A, not zero, so the intensity is not zero.

Mind stretcher 2: Why matching polarisation matters (light)Extension

Two light waves of the same frequency are coherent, but they are polarised in perpendicular planes.

Will you get complete destructive interference? Explain.

Show Answer

No. Complete cancellation requires the electric fields to be along the same line so they can add algebraically to zero.

If the polarisations are perpendicular, the fields are in different directions and do not cancel fully, so the fringe contrast is reduced or the interference pattern may not be observed.

Mind stretcher 3: Optional (Enrichment)Extension

A. Sources initially out of phase

If the two sources start π out of phase, the roles of maxima/minima swap:

  • maxima: Δ x = (n + 1/2)λ
  • minima: Δ x = nλ

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027