Monochromatic (single-wavelength), coherent light passes through two narrow slits separated by distance d. The light reaching the screen at distance L from the slits is the superposition of two waves, one from each slit.

What is coherence requirement?

The two sources must have a fixed phase relationship. In practice, both slits are illuminated by the same monochromatic source (e.g., a laser, or a single slit illuminated first), ensuring coherence.

A regular pattern of bright and dark fringes appears on the screen. Bright fringes correspond to constructive interference (the waves arrive in phase); dark fringes correspond to destructive interference (the waves arrive out of phase by π).

What is width of the central maximum?

From the first dark fringes on either side: sin _1 ≈ λ / w, so the angular half-width is _1 ≈ λ / w. The angular full-width is 2 λ / w.

Refer to the Melbourne University HeNe demonstration. (a) Calculate fringe spacing for λ = 633 nm, d = 0.20 mm, L = 2.0 m. (b) Determine the wavelength if a green laser produces 5.32 mm fringes.

VICPhysicsSyllabus dot point

How has understanding of the physical world changed?

Investigate the wave model of light, including diffraction and constructive and destructive interference (Young's double-slit experiment), and apply $\Delta x = \lambda L / d$ for fringe spacing in the small-angle limit

A focused answer to the VCE Physics Unit 4 dot point on the wave model of light. Covers Young's double-slit experiment, the path-difference condition for constructive and destructive interference, the fringe-spacing formula $\\Delta x = \\lambda L / d$ in the small-angle limit, and single-slit diffraction.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-25

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

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What this dot point is asking
The wave model of light
Coherent light and the double-slit experiment
Path difference conditions
Fringe spacing in the small-angle limit
Single-slit diffraction
Diffraction grating
What the interference and diffraction patterns tell us
Examples in context
Try this

What this dot point is asking

VCAA wants you to describe the wave model of light through its key signatures (interference and diffraction), explain Young's double-slit experiment, apply the path-difference conditions for constructive and destructive interference, and use the small-angle fringe-spacing formula $\Delta x = \lambda L / d$ . The dot point is the wave-side of the wave-particle duality story in Unit 4.

The wave model of light

Light is a transverse electromagnetic wave. Like all waves, it can:

Reflect off surfaces.
Refract when passing between media of different refractive index.
Diffract around obstacles or through narrow openings.
Interfere with other waves, producing constructive and destructive patterns.

The interference and diffraction phenomena are the diagnostic wave behaviours. They cannot be explained by a purely particle model of light, so their observation is direct evidence for the wave nature of light.

The wave equation $v = f \lambda$ connects wave speed, frequency and wavelength. For light in vacuum, $v = c \approx 3.0 \times 10^8$ m/s, so:

c = f \lambda

Cross-link: see the wavelength-frequency calculator for fast conversions.

Coherent light and the double-slit experiment

Young's double-slit experiment (1801) is the canonical demonstration of light interference.

Setup: Monochromatic (single-wavelength), coherent light passes through two narrow slits separated by distance $d$ . The light reaching the screen at distance $L$ from the slits is the superposition of two waves, one from each slit.
Coherence requirement: The two sources must have a fixed phase relationship. In practice, both slits are illuminated by the same monochromatic source (e.g., a laser, or a single slit illuminated first), ensuring coherence.
Observation: A regular pattern of bright and dark fringes appears on the screen. Bright fringes correspond to constructive interference (the waves arrive in phase); dark fringes correspond to destructive interference (the waves arrive out of phase by $\pi$ ).

Path difference conditions

The phase relationship between the two waves at a point on the screen depends on the path difference: the difference in the distance each wave has travelled.

\text{path difference} = d \sin \theta

where $\theta$ is the angle from the centreline to the point on the screen.

Constructive interference (bright fringe):

d \sin \theta = m \lambda \quad \text{for } m = 0, \pm 1, \pm 2, \ldots

Destructive interference (dark fringe):

d \sin \theta = (m + \tfrac{1}{2}) \lambda \quad \text{for } m = 0, \pm 1, \pm 2, \ldots

The integer $m$ is the order of the fringe. $m = 0$ is the central bright fringe (directly opposite the midpoint of the slits, equal path lengths from both slits). $m = \pm 1$ are the first-order bright fringes on either side, and so on.

Fringe spacing in the small-angle limit

For small $\theta$ (the typical Young's setup where $L \gg d$ ), $\sin \theta \approx \tan \theta = y / L$ where $y$ is the distance from the centreline on the screen.

Substituting into the constructive condition: $d \cdot y / L = m \lambda$ , so $y = m \lambda L / d$ .

The fringe spacing $\Delta x$ (distance between adjacent bright fringes, or between adjacent dark fringes) is:

\Delta x = \frac{\lambda L}{d}

This formula is the working tool for Young's double-slit problems. Three takeaways:

Longer wavelength gives wider spacing. Red light (700 nm) makes wider fringes than blue (450 nm).
Greater screen distance gives wider spacing. Moving the screen further from the slits spreads the pattern.
Wider slit separation gives narrower spacing. Slits further apart cram the fringes closer.

The formula is approximate; it assumes the small-angle limit (typically valid when $\theta < 5$ degrees) and that the slits are very narrow compared to the wavelength (diffraction from each slit is wide compared to the separation).

Single-slit diffraction

When light passes through a single narrow slit of width $w$ , it spreads out and produces a diffraction pattern on a distant screen.

Pattern. A central bright band, flanked by progressively dimmer side bands separated by dark fringes. The central band is twice as wide as the side bands.

Dark fringe condition. Dark minima occur at angles where:

w \sin \theta = m \lambda \quad \text{for } m = \pm 1, \pm 2, \ldots

(Note: $m = 0$ corresponds to the centre of the bright central maximum, not a dark fringe.)

Width of the central maximum. From the first dark fringes on either side: $\sin \theta_1 \approx \lambda / w$ , so the angular half-width is $\theta_1 \approx \lambda / w$ . The angular full-width is $2 \lambda / w$ .

This means narrow slits diffract more (wide pattern) and wide slits diffract less (narrow pattern). The same principle explains why low-frequency (long-wavelength) sound waves diffract around corners more than high-frequency (short-wavelength) sound waves.

Diffraction grating

A diffraction grating is many parallel slits (typically thousands per mm). The constructive interference condition is the same as for two slits:

d \sin \theta = m \lambda

where $d$ is the spacing between adjacent slits. The diffraction pattern produced by a grating has very sharp bright maxima (because of constructive interference between many sources) separated by dark regions.

Diffraction gratings are used to disperse light into its spectrum (different wavelengths diffract at different angles) and underpin spectroscopy.

What the interference and diffraction patterns tell us

The wave model predicts both the position and the spacing of the fringes correctly. Specifically:

Existence of the pattern. A purely particle (corpuscular) model of light predicts a single bright band where the particles pass through, not a fringe pattern.
Wavelength dependence of fringe spacing. Different wavelengths give different spacings, exactly as $\Delta x = \lambda L / d$ predicts.
Wavelength dependence of single-slit diffraction. Narrower slits and longer wavelengths give wider diffraction, again consistent with the wave model.

The wave model is therefore the working model for low-intensity classical optics: refraction, reflection, interference, diffraction, polarisation. The photon (particle) model takes over for high-energy interactions with matter (photoelectric effect, atomic transitions), and the matter-wave model unifies both.

Worked example

Example 1. Calculate fringe spacing

Wavelength 600 nm, slit separation 0.20 mm, screen distance 1.5 m.

$\Delta x = (600 \times 10^{-9}) \times 1.5 / (0.20 \times 10^{-3}) = 9.0 \times 10^{-7} / 0.20 \times 10^{-3} = 4.5 \times 10^{-3}$ m = 4.5 mm.

Example 2. Wavelength from fringe spacing

Fringe spacing 2.0 mm, slit separation 0.50 mm, screen distance 1.2 m. Find $\lambda$ .

Rearrange: $\lambda = \Delta x \cdot d / L = (2.0 \times 10^{-3}) \times (0.50 \times 10^{-3}) / 1.2 = 1.0 \times 10^{-6} / 1.2 \approx 8.3 \times 10^{-7}$ m $= 830$ nm (in the infrared / red region).

Example 3. Path difference and fringe identification

In a setup with $\lambda = 500$ nm, the path difference at a point on the screen is 1500 nm.

Express in wavelengths: $1500 / 500 = 3$ . The path difference is exactly $3 \lambda$ (integer multiple), so constructive interference: this is the third-order bright fringe ( $m = 3$ ).

Common errors

Unit conversion forgotten: Wavelengths in nm, slit separation in mm, screen distance in m. Convert all to metres before substituting.
Confusing $d$ and $w$: $d$ is the slit separation in Young's double-slit. $w$ is the slit width in single-slit diffraction. Different quantities, different formulas.
Applying small-angle formula at large angles: $\Delta x = \lambda L / d$ assumes small $\theta$ . For large $\theta$ (close to the slits, or first-order maxima with very short $L$ ), use $d \sin \theta = m \lambda$ directly.

Treating the $m = 0$ position as a dark fringe in single-slit diffraction. The single-slit minimum condition has $m \neq 0$ ; $m = 0$ is the bright centre.

Forgetting coherence. Two independent light sources are not coherent, so they do not produce a stable interference pattern. Young used one source illuminating both slits to ensure coherence.

Confusing path difference with phase difference. Path difference (in metres) becomes phase difference (in radians) via $\phi = 2 \pi \Delta / \lambda$ . Constructive: $\Delta = m \lambda$ (so $\phi = 2 \pi m$ ). Destructive: $\Delta = (m + 0.5) \lambda$ (so $\phi = (2m + 1) \pi$ ).

Examples in context

Example 1. School laboratory double-slit with HeNe laser in Carlton. A Melbourne University physics demonstration uses a $632.8$ nm helium-neon laser with double-slit separation $d = 0.20$ mm projected on a screen $L = 2.00$ m away. Fringe spacing is $\Delta x = \lambda L/d = 632.8 \times 10^{-9} \times 2.00/(0.20 \times 10^{-3}) = 6.33 \times 10^{-3}$ m or $6.33$ mm, easily measured with a ruler. Doubling slit separation halves fringe spacing to $3.17$ mm. Using a green $532$ nm laser instead would give $\Delta x = 532/633 \times 6.33 = 5.32$ mm, demonstrating the wavelength dependence and confirming the wave model.

Example 2. Square Kilometre Array interferometry at Murchison. SKA-Low achieves angular resolution far beyond any single dish by combining signals from antennas separated by up to $74$ km. For $300$ MHz observations ( $\lambda = 1.0$ m), the diffraction-limited angular resolution is $\theta \approx 1.22 \lambda/D = 1.22 \times 1/74000 = 1.65 \times 10^{-5}$ rad = $3.4$ arc seconds. Each antenna pair creates an interference pattern on the sky; combining many baselines via Fourier inversion produces an image with the resolution of a single $74$ km dish, despite using small dipole elements. This is the same physics as Young's double slit, scaled up.

Try this

Q1. State the formula for fringe spacing in a Young's double-slit experiment, with each symbol defined. [2 marks]

Cue. $\Delta x = \lambda L/d$ . $\lambda$ is wavelength, $L$ slit-to-screen distance, $d$ slit separation.

Q2. A double-slit experiment with $d = 0.50$ mm and $L = 1.5$ m produces fringes $1.9$ mm apart. Calculate the wavelength. [3 marks]

Cue. $\lambda = d\Delta x/L = 0.50 \times 10^{-3} \times 1.9 \times 10^{-3}/1.5 = 6.33 \times 10^{-7}$ m = $633$ nm.

Q3. Refer to the Melbourne University HeNe demonstration. (a) Calculate fringe spacing for $\lambda = 633$ nm, $d = 0.20$ mm, $L = 2.0$ m. (b) Determine the wavelength if a green laser produces $5.32$ mm fringes. (c) Explain how the experiment supports the wave model of light. [2+2+2 marks]

Cue. (a) $6.33$ mm. (b) $\lambda = 5.32 \times 0.20/2.0 \times 10^{-3} = 532$ nm. (c) Interference requires phase-coherent superposition, a wave property.

Exam-style practice questions

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

2024 VCAA4 marksCoherent monochromatic light of wavelength 600 nm passes through two slits separated by 0.40 mm. A screen is placed 2.0 m beyond the slits. (a) Calculate the fringe spacing on the screen. (b) State two observations that support a wave model of light from this experiment.

Show worked answer →

(a) Fringe spacing. Use $\Delta x = \frac{\lambda L}{d}$ .

Convert units. $\lambda = 600 \times 10^{-9}$ m; $L = 2.0$ m; $d = 0.40 \times 10^{-3}$ m.

$\Delta x = \frac{600 \times 10^{-9} \times 2.0}{0.40 \times 10^{-3}} = \frac{1.2 \times 10^{-6}}{0.40 \times 10^{-3}} = 3.0 \times 10^{-3}$ m $= 3.0$ mm.

(b) Two observations supporting the wave model.

A regular bright-dark fringe pattern appears on the screen, which is the signature of constructive and destructive interference (only waves interfere).
The fringe spacing depends on wavelength: changing $\lambda$ shifts the spacing, consistent with the $\Delta x = \lambda L / d$ wave prediction.

Markers reward correct formula and unit handling, and two distinct observations linked to wave behaviour rather than just the existence of a pattern.

2023 VCAA3 marksIn a Young's double-slit experiment, the path difference from the two slits to a point on the screen is found to be

1.5 \lambda

. State and justify whether constructive or destructive interference occurs at this point.

Show worked answer →

Path difference $= 1.5 \lambda = \frac{3 \lambda}{2}$ is a half-integer multiple of the wavelength.

Destructive interference occurs when path difference $= (m + \frac{1}{2}) \lambda$ for integer $m$ .

Here, $m = 1$ gives $1.5 \lambda$ , so destructive interference occurs at this point. The two waves arrive out of phase by $\pi$ radians (half a wavelength), so they cancel.

Markers reward identifying the path-difference condition for destructive interference (half-integer wavelengths) and an explicit phase-difference statement.