SQA SQA AH style-style practice: In a Young's double-slit experiment the slits are 0.50\ mm apart and the screen is 2.0\ m away. The fringe spacing is 2.4\ mm. Calculate the wavelength of the light.

The fringe spacing in Young's experiment is x = λ Dd, where d is the slit separation and D the slit-to-screen distance. Rearrange for the wavelength: λ = x \, dD. Substitute (in metres): λ = 2.4 × 10^-3 × 0.50 × 10^-32.0. Evaluate: λ = 1.2 × 10^-62.0 = 6.0 × 10^-7\ m, that is 600\ nm. Markers reward the correct fringe relationship, consistent SI units, and the wavelength with unit (in the visible range).

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How do coherent waves interfere by division of amplitude and division of wavefront?

Coherence and path difference, constructive and destructive interference, division of amplitude in thin films and wedges, and division of wavefront in Young's double-slit experiment.

An SQA Advanced Higher Physics answer on interference, covering coherence and path difference, the conditions for constructive and destructive interference, division of amplitude in thin films and wedges, and division of wavefront in Young's double-slit experiment.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Interference is the superposition of coherent waves. Two sources are coherent if they keep a constant phase relationship, which is needed for a stable pattern. Where the path difference is a whole number of wavelengths, $\Delta L = m\lambda$ , the waves arrive in phase and interfere constructively (bright); where it is an odd number of half-wavelengths, $\Delta L = (m + \tfrac{1}{2})\lambda$ , they interfere destructively (dark). Division of amplitude splits one beam at a surface, as in thin-film colours and wedge fringes, where a $\pi$ phase change on reflection from a denser medium must be included. Division of wavefront splits a single wavefront through two slits, as in Young's experiment, where the fringe spacing is $\Delta x = \frac{\lambda D}{d}$ .

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What this key area is asking
Coherence and path difference
Constructive and destructive interference
Division of amplitude
Division of wavefront: Young's slits
Examples in context
Try this

What this key area is asking

The SQA wants you to understand coherence and path difference, state the conditions for constructive and destructive interference, solve problems on division of amplitude (thin films and wedge fringes, including the phase change on reflection), and solve problems on division of wavefront (Young's double-slit experiment).

Coherence and path difference

Coherence is essential: only coherent sources produce a steady pattern of bright and dark fringes, because the phase relationship at each point stays fixed. Incoherent sources (two ordinary lamps) have a randomly varying phase difference, so any interference averages out and is unobservable. This is why interference experiments use a single source split into two, or a laser, to guarantee coherence.

Constructive and destructive interference

A whole number of wavelengths of path difference means the waves arrive in phase and reinforce; an odd number of half-wavelengths means they arrive in antiphase and cancel. This single rule underlies every interference pattern, whether the waves are split by amplitude or by wavefront.

Division of amplitude

This explains the colours of thin films such as oil on water and soap bubbles: light reflected from the top and bottom surfaces of the film interferes, and the path difference (set by the film thickness) decides which colours reinforce. Wedge fringes, formed by a thin air wedge between two glass plates, produce equally spaced dark and bright fringes used to test surface flatness, with the half-wavelength phase change on reflection making the contact edge dark.

Division of wavefront: Young's slits

Each slit samples the same incoming wavefront, so the two are automatically coherent. The fringe spacing increases with wavelength and with screen distance, and decreases as the slits are brought closer; measuring it gives the wavelength of light. Young's experiment was historically decisive evidence that light is a wave.

Examples in context

The iridescent colours of soap bubbles, oil slicks and butterfly wings are thin-film interference by division of amplitude. Anti-reflection coatings on lenses and solar panels use a quarter-wavelength film so that reflected light interferes destructively, increasing transmission. Optical flats and Newton's rings use wedge fringes to test surfaces to a fraction of a wavelength. Young's double slit remains the standard demonstration of wave interference and underlies modern quantum-interference experiments.

Try this

Q1. State what is meant by two coherent sources. [1 mark]

Cue. Same frequency and a constant phase relationship.

Q2. State the path-difference condition for destructive interference. [1 mark]

Cue. $\Delta L = (m + \tfrac{1}{2})\lambda$ (an odd number of half-wavelengths).

Q3. State whether Young's double-slit experiment is division of amplitude or division of wavefront. [1 mark]

Cue. Division of wavefront.

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style5 marksIn a Young's double-slit experiment the slits are

0.50\ \text{mm}

apart and the screen is

2.0\ \text{m}

away. The fringe spacing is

2.4\ \text{mm}

. Calculate the wavelength of the light.

Show worked answer →

The fringe spacing in Young's experiment is $\Delta x = \dfrac{\lambda D}{d}$ , where $d$ is the slit separation and $D$ the slit-to-screen distance.

Rearrange for the wavelength: $\lambda = \dfrac{\Delta x \, d}{D}$ .

Substitute (in metres): $\lambda = \dfrac{2.4 \times 10^{-3} \times 0.50 \times 10^{-3}}{2.0}$ .

Evaluate: $\lambda = \dfrac{1.2 \times 10^{-6}}{2.0} = 6.0 \times 10^{-7}\ \text{m}$ , that is $600\ \text{nm}$ .

Markers reward the correct fringe relationship, consistent SI units, and the wavelength with unit (in the visible range).

SQA AH style4 marksState the condition on path difference for constructive interference, and explain why the two sources must be coherent.

Show worked answer →

Constructive interference occurs where the path difference is a whole number of wavelengths, $\Delta L = m\lambda$ , so the waves arrive in phase and reinforce.

The sources must be coherent, meaning they have a constant phase relationship (and the same frequency), so that the interference pattern is stable.

If the phase relationship varied, the bright and dark fringes would shift randomly and average out, leaving no visible pattern.

Markers reward the path-difference condition for constructive interference and the explanation that coherence keeps the pattern stable.

Related dot points

Sources & how we know this

SQA Advanced Higher Physics Course Specification — SQA (2019)