Skip to main content
ExamExplained

Light as a wave: WACE Year 12 Physics Unit 4

Syllabus dot point

“Explain diffraction, two-slit interference and the electromagnetic spectrum as evidence for the wave model of light”

8 min read

Quick answer

Light spreads (diffracts) when it passes through gaps comparable to its wavelength, and two coherent slits produce an interference pattern of bright and dark fringes. Bright fringes occur where the path difference is a whole number of wavelengths, giving the fringe spacing Δy=λLd\Delta y=\frac{\lambda L}{d}. This wave behaviour, plus the electromagnetic spectrum, is the evidence light is a wave.

What this dot point is asking

WACE wants you to use diffraction and two-slit interference as the historical and quantitative evidence that light is a wave, work with the path-difference conditions and the fringe-spacing formula, and place visible light within the broader electromagnetic spectrum. The skill being tested is connecting the geometry of the slits to the observed pattern.

Diffraction

Diffraction is the spreading of a wave as it passes an edge or through a gap. The effect is significant when the gap width is similar to or smaller than the wavelength: a wide gap gives little spreading, a narrow gap gives a lot. Because visible light has a tiny wavelength (hundreds of nanometres), noticeable diffraction needs very narrow slits, which is why everyday light appears to travel in straight lines.

Two-slit interference

In Young's experiment, monochromatic light illuminates two narrow, closely spaced slits that act as coherent sources. The diffracted light from each slit overlaps and superposes, producing alternating bright and dark fringes on a distant screen. Bright fringes (constructive interference) occur where the path difference is a whole number of wavelengths,

dsin⁡θ=mλ,m=0,1,2,…d\sin\theta=m\lambda,\quad m=0,1,2,\dots

Dark fringes (destructive interference) occur at half-integer multiples, dsin⁡θ=(m+12)λd\sin\theta=(m+\tfrac{1}{2})\lambda. Here dd is the slit separation and θ\theta the angle to the fringe.

Fringe spacing

For small angles and a screen a distance LL away, the spacing between adjacent bright fringes is

Δy=λLd.\Delta y=\frac{\lambda L}{d}.

So fringes spread out for longer wavelengths (red), a more distant screen, or closer slits. Measuring Δy\Delta y, LL and dd lets you calculate the wavelength of light, which is how the experiment first pinned down λ\lambda for visible light.

Coherence and monochromatic light

A clear pattern needs coherent sources: a constant phase relationship and a single wavelength. This is why a single source illuminating two slits (or a laser) works, while two independent lamps do not. White light produces overlapping coloured fringes because each wavelength has its own spacing.

The electromagnetic spectrum

Light is one part of a continuous spectrum of electromagnetic waves, all travelling in vacuum at c=3.0×108 m s−1c=3.0\times10^{8}\ \text{m s}^{-1} and all obeying c=fλc=f\lambda. In order of increasing frequency (decreasing wavelength) the spectrum runs radio, microwave, infrared, visible, ultraviolet, X-ray and gamma. Visible light occupies only a narrow band from about 400 nm400\ \text{nm} (violet) to 700 nm700\ \text{nm} (red). All these waves are transverse oscillations of electric and magnetic fields.

Wavelength from a double-slit pattern

Finding lambda from fringe spacing

Light passes through two slits separated by d=0.25 mmd=0.25\ \text{mm} and forms fringes on a screen L=2.0 mL=2.0\ \text{m} away. Adjacent bright fringes are Δy=4.8 mm\Delta y=4.8\ \text{mm} apart.

Rearranging the fringe-spacing formula,

λ=Δy dL=(4.8×10−3)(0.25×10−3)2.0.\lambda=\frac{\Delta y\,d}{L}=\frac{(4.8\times10^{-3})(0.25\times10^{-3})}{2.0}.

The numerator is 1.2×10−61.2\times10^{-6}, so

λ=1.2×10−62.0=6.0×10−7 m=600 nm,\lambda=\frac{1.2\times10^{-6}}{2.0}=6.0\times10^{-7}\ \text{m}=600\ \text{nm},

an orange wavelength, consistent with visible light.

Keeping the units straight

Slit separations are millimetres, fringe spacings millimetres, wavelengths nanometres, and the screen distance metres. Convert everything to metres before substituting, and decide whether a question is asking for a fringe position (use dsin⁡θ=mλd\sin\theta=m\lambda) or a spacing (use Δy=λLd\Delta y=\frac{\lambda L}{d}).

Confusing slit separation with slit width

Students mix up dd, the distance between the two slits, with the width of each slit. The two-slit interference formulae use the separation dd between slit centres, not the width of an individual slit (the width controls the overall diffraction envelope). Mislabelling these, or forgetting to convert millimetres and nanometres to metres, throws the wavelength out by orders of magnitude. Label the diagram carefully and convert all lengths to metres first.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2023-style6 marks
In a Young's double-slit experiment, two slits 0.25 mm0.25\ \text{mm} apart are illuminated by light of wavelength 633 nm633\ \text{nm}, and fringes are observed on a screen 1.8 m1.8\ \text{m} away. (a) Calculate the spacing between adjacent bright fringes. (b) Explain why a bright fringe forms at the centre of the pattern.
Show worked answer →

A 6 mark answer rewards the fringe-spacing calculation and a path-difference explanation.

(a) Fringe spacing. Using Δy=λLd\Delta y=\dfrac{\lambda L}{d}:

Δy=(633×10−9)(1.8)0.25×10−3=1.14×10−62.5×10−4=4.6×10−3 m.\Delta y=\frac{(633\times10^{-9})(1.8)}{0.25\times10^{-3}}=\frac{1.14\times10^{-6}}{2.5\times10^{-4}}=4.6\times10^{-3}\ \text{m}.

(b) Central bright fringe. At the centre the two paths from the slits are equal in length, so the path difference is zero. Zero path difference means the waves arrive in phase and interfere constructively, producing a bright fringe.

Markers reward Δy=λL/d\Delta y=\lambda L/d, the value near 4.6×10−3 m4.6\times10^{-3}\ \text{m} and the zero-path-difference, in-phase explanation for the central maximum.

WACE 2020-style5 marks
Explain how the observation of diffraction and two-slit interference of light supports the wave model rather than a particle model of light.
Show worked answer →

A 5 mark explanation needs both phenomena tied to wave behaviour.

Diffraction
Light spreads out when it passes through a narrow gap or around an edge, bending into the geometric shadow. Spreading at an aperture comparable to the wavelength is characteristic of waves; a stream of particles travelling in straight lines would not bend this way.
Interference
When light passes through two slits it produces a pattern of alternating bright and dark fringes. Dark fringes occur where waves from the two slits arrive out of phase and cancel (destructive interference), which only makes sense if light is a wave that can superpose. Particles arriving independently could only add up, never cancel to give darkness where light is shone.
Conclusion
Both bending at apertures and the cancellation producing dark fringes are wave properties, so they support the wave model.

Markers reward diffraction as wave spreading and, crucially, the dark-fringe cancellation that a particle model cannot explain.

Practise this

Sources & how we know this

Based on the official SCSA syllabus. Not yet individually reviewed by a person; always check SCSA and your school's materials for current rules.

ExamExplained