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How does Young's double-slit experiment provide evidence for the wave model of light?

Apply path difference and the double-slit equation to analyse two-source interference of light

A focused answer to the WACE Year 12 Physics Unit 4 content point on two-slit interference. Coherent sources, path difference conditions for bright and dark fringes, the fringe-spacing equation, and why the experiment supports the wave model of light.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking

WACE wants you to apply the interference conditions, use the fringe-spacing equation, and explain why two-slit interference cannot be accounted for by treating light as particles. Coherence is the key requirement: the two sources must keep a fixed phase relationship.

Coherent sources and path difference

When light passes through two slits a distance $d$ apart, each slit acts as a source. Because they come from the same original beam, they are coherent. At a point on the screen the two waves have travelled slightly different distances; this path difference determines whether they arrive in step or out of step.

Bright and dark fringes

Constructive interference (a bright fringe) occurs when the path difference is a whole number of wavelengths,

d\sin\theta=m\lambda,\qquad m=0,1,2,\dots

Destructive interference (a dark fringe) occurs when the path difference is a half-integer number of wavelengths,

d\sin\theta=\left(m+\tfrac{1}{2}\right)\lambda.

The central bright fringe ( $m=0$ ) sits directly opposite the midpoint of the slits, where the path difference is zero.

Fringe spacing on the screen

For small angles and a screen a distance $L$ from the slits, the bright fringes are evenly spaced by

\Delta y=\frac{\lambda L}{d}.

This shows that closer slits or a longer screen distance spread the fringes further apart, and that longer-wavelength light gives wider fringes. Measuring $\Delta y$ is a standard way to determine the wavelength of light.

Why this proves light is a wave

Particles cannot cancel each other, but waves can: the dark fringes are places where light plus light gives darkness through destructive interference. Only a wave model explains alternating bright and dark bands, so Young's experiment was the historic confirmation of the wave nature of light.

What the fringe-spacing equation tells you

The relation $\Delta y=\dfrac{\lambda L}{d}$ is worth interpreting, not just substituting, because examiners ask how the pattern changes when a variable is altered. The fringe spacing is proportional to the wavelength, so red light gives wider fringes than blue, and white light produces a central white fringe flanked by coloured ones. It is proportional to the screen distance $L$ , so moving the screen back spreads the fringes. It is inversely proportional to the slit separation $d$ , so bringing the slits closer together widens the spacing. This last point seems counter-intuitive but follows directly from the equation, and it is the reverse of the order spacing in a diffraction grating, where many finely spaced slits give very sharp, well-separated maxima. Reasoning from the equation in this way handles most "what if" parts without any arithmetic.

Keeping the conditions straight

Use $d\sin\theta=m\lambda$ for bright fringes and the half-integer version for dark, and use $\Delta y=\lambda L/d$ for spacing on the screen. Convert slit separations from millimetres to metres and keep the screen distance in metres. The order number $m$ counts fringes out from the centre.

Exam-style practice questions

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

WACE 20227 marksIn a double-slit experiment, slits separated by

0.20\ \text{mm}

are illuminated by laser light of wavelength

650\ \text{nm}

. The interference pattern is observed on a screen

2.5\ \text{m}

from the slits. (a) Calculate the spacing between adjacent bright fringes. (b) Calculate the distance from the central bright fringe to the third-order bright fringe. (c) State what happens to the fringe spacing if the slit separation is halved.

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A 7 mark calculation rewards the fringe spacing, a higher-order position and a scaling statement.

(a) Fringe spacing: $\Delta y=\dfrac{\lambda L}{d}=\dfrac{(650\times10^{-9})(2.5)}{0.20\times10^{-3}}=\dfrac{1.625\times10^{-6}}{2.0\times10^{-4}}=8.1\times10^{-3}\ \text{m}$ .
(b) Third-order position: The bright fringes are evenly spaced, so the third-order fringe is $3\Delta y=3\times8.1\times10^{-3}=2.4\times10^{-2}\ \text{m}$ from the centre.
(c) Halving the slit separation: Since $\Delta y=\dfrac{\lambda L}{d}$ , halving $d$ doubles the fringe spacing to about $1.6\times10^{-2}\ \text{m}$ .

Markers reward $\Delta y=\lambda L/d$ near $8.1\times10^{-3}\ \text{m}$ , the third-order distance of $2.4\times10^{-2}\ \text{m}$ and the inverse relation doubling the spacing.

WACE 20205 marksExplain why coherent light sources are necessary to observe a stable two-slit interference pattern, and explain why a bright fringe forms where the path difference is a whole number of wavelengths.

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A 5 mark explanation needs the coherence requirement and the constructive-interference condition.

Coherence. A stable pattern requires the two sources to maintain a constant phase relationship (be coherent). If the phase difference between the sources varied randomly, the positions of constructive and destructive interference would shift rapidly and the pattern would wash out into uniform brightness. Using a single source split by two slits guarantees the slits stay in step.

Whole-number path difference. A bright fringe forms where the waves from the two slits arrive in phase. This happens when their path difference is a whole number of wavelengths, $d\sin\theta=m\lambda$ , because then crest meets crest and trough meets trough, giving constructive interference and maximum brightness.

Markers reward the constant-phase (coherence) requirement to avoid washout, and the in-phase, crest-on-crest condition at whole-wavelength path differences.