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Double-slit diffraction (SACE Stage 2 Physics)

Syllabus dot point

“Analyse the double-slit interference pattern and use the fringe-spacing relationship to determine wavelength.”

9 min read

Quick answer

In Young's double-slit experiment, light passing through two narrow, close slits produces an evenly spaced pattern of bright and dark fringes on a screen - proof of wave interference. The fringe spacing is Δy=λLd\Delta y = \dfrac{\lambda L}{d}, where λ\lambda is the wavelength, LL the slit-to-screen distance, and dd the slit separation. This relationship lets you measure the wavelength of light.

Jump to a section
  1. What this dot point is asking
  2. Why a pattern forms
  3. The fringe-spacing relationship
  4. Why it confirms the wave model
  5. How SACE assesses this
  6. Exam-style questions

What this dot point is asking

You need to explain why the double-slit experiment produces an interference pattern and use the fringe-spacing equation to find a wavelength (or another quantity).

Finding the wavelength of light

Two slits 0.25 mm apart give fringes 2.4 mm apart on a screen 1.0 m away

Find the wavelength.

Rearrange the fringe-spacing equation:

Δy=λLd  ⟹  λ=Δy dL.\Delta y = \frac{\lambda L}{d} \implies \lambda = \frac{\Delta y\, d}{L}.

Substitute (Δy=2.4×10−3\Delta y = 2.4\times10^{-3} m, d=0.25×10−3d = 0.25\times10^{-3} m, L=1.0L = 1.0 m):

λ=(2.4×10−3)(0.25×10−3)1.0=6.0×10−7 m=600 nm.\lambda = \frac{(2.4\times10^{-3})(0.25\times10^{-3})}{1.0} = 6.0\times10^{-7}\ \text{m} = 600\ \text{nm}.

This is in the orange-red part of the visible spectrum.

Why a pattern forms

Monochromatic light is shone on two narrow slits a small distance dd apart. Each slit diffracts the light, so the two slits act as coherent sources (they come from the same wavefront, so they keep a constant phase relationship). The light from the two slits overlaps on a screen a distance LL away.

Where the path difference is a whole number of wavelengths, the waves arrive in phase and interfere constructively - a bright fringe. Where it is an odd number of half-wavelengths, they cancel - a dark fringe. The result is a series of evenly spaced bright and dark bands.

Bright fringe condition

A bright fringe occurs where the path difference from the two slits is a whole number of wavelengths:

dsin⁡θ=nλn=0,1,2,…d\sin\theta = n\lambda \qquad n = 0, 1, 2, \dots

n=0n = 0 is the central bright fringe (zero path difference), with higher-order fringes either side.

The fringe-spacing relationship

For small angles (slits close together, screen far away), the bright fringes are evenly spaced, and the spacing between adjacent fringes is:

Fringe spacing

Δy=λLd\Delta y = \frac{\lambda L}{d}

Δy\Delta y is the distance between adjacent bright (or dark) fringes (m), λ\lambda the wavelength (m), LL the slit-to-screen distance (m), and dd the slit separation (m).

This tells you the pattern spreads out (larger Δy\Delta y) for:

  • longer wavelength (red spreads more than blue),
  • a larger screen distance LL,
  • a smaller slit separation dd.

Why it confirms the wave model

Only waves interfere. The double-slit pattern - alternating bright and dark fringes that depend on wavelength exactly as the wave model predicts - cannot be explained by treating light purely as particles travelling in straight lines. Young's experiment (1801) was decisive evidence for the wave nature of light. Remarkably, the same pattern appears even when single photons or electrons pass through one at a time, a key clue to wave-particle duality.

How SACE assesses this

SACE Stage 2 double-slit questions give three of the four quantities in Δy=λLd\Delta y = \dfrac{\lambda L}{d} and ask for the fourth, most often the wavelength from a measured fringe spacing using λ=Δy dL\lambda = \dfrac{\Delta y\, d}{L}. The single biggest source of lost marks is inconsistent units: the slit separation is usually in micrometres or millimetres, the wavelength in nanometres and the fringe spacing in centimetres, so convert everything to metres before substituting. Keep dd (the tiny slit separation) and Δy\Delta y (the larger fringe spacing) clearly distinct. Several "show that" parts give the target wavelength, so rearrange the formula first, substitute in SI units, and confirm the printed value, finishing with the colour or region the wavelength corresponds to.

Common errors
Confusing dd and Δy\Delta y
dd is the (tiny) slit separation; Δy\Delta y is the (larger) fringe spacing on the screen. Mixing them gives a wildly wrong wavelength.
Unit slips
Convert everything to metres. A slit separation in millimetres and a wavelength in nanometres must be in consistent units before substituting.
Thinking only one slit is needed
A single slit diffracts but does not give the two-source interference pattern. Two coherent slits are needed for the evenly spaced fringes.
Forgetting the small-angle assumption
Δy=λL/d\Delta y = \lambda L/d assumes small angles (close slits, distant screen). For wide angles use dsin⁡θ=nλd\sin\theta = n\lambda directly.

Exam-style questions

Questions in the style of SACE Board 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.

SACE 2023-style3 marks
In a two-slit experiment with a sodium lamp, the slit-to-screen distance was 1.40 m1.40\ \text{m} and the slits were separated by 40.0 μm40.0\ \mu\text{m}. The lamp emitted light of wavelength 589 nm589\ \text{nm}. Determine the distance between adjacent bright fringes.
Show worked answer →

The fringe-spacing relationship is Δy=λLd\Delta y = \dfrac{\lambda L}{d}, with LL the slit-to-screen distance and dd the slit separation.

Convert units: λ=589×10−9 m\lambda = 589\times10^{-9}\ \text{m}, d=40.0×10−6 md = 40.0\times10^{-6}\ \text{m}, L=1.40 mL = 1.40\ \text{m}.

Δy=(589×10−9)(1.40)40.0×10−6=8.246×10−74.00×10−5=2.06×10−2 m.\Delta y = \dfrac{(589\times10^{-9})(1.40)}{40.0\times10^{-6}} = \dfrac{8.246\times10^{-7}}{4.00\times10^{-5}} = 2.06\times10^{-2}\ \text{m}.

The adjacent bright fringes are about 2.06×10−2 m2.06\times10^{-2}\ \text{m} (2.06 cm2.06\ \text{cm}) apart. 1 mark for the equation, 1 mark for consistent unit conversion, 1 mark for the answer.

SACE 2025-style3 marks
Laser light passes through two slits separated by 1.50×10−4 m1.50\times10^{-4}\ \text{m} onto a screen 2.50 m2.50\ \text{m} away. The average distance between bright fringes is 1.09 cm1.09\ \text{cm}. Calculate the experimental value for the wavelength of the laser light.
Show worked answer →

Rearrange the fringe-spacing relationship for wavelength: λ=Δy dL\lambda = \dfrac{\Delta y\, d}{L}.

Convert the fringe spacing: Δy=1.09 cm=1.09×10−2 m\Delta y = 1.09\ \text{cm} = 1.09\times10^{-2}\ \text{m}. Then d=1.50×10−4 md = 1.50\times10^{-4}\ \text{m}, L=2.50 mL = 2.50\ \text{m}.

λ=(1.09×10−2)(1.50×10−4)2.50=1.635×10−62.50=6.54×10−7 m.\lambda = \dfrac{(1.09\times10^{-2})(1.50\times10^{-4})}{2.50} = \dfrac{1.635\times10^{-6}}{2.50} = 6.54\times10^{-7}\ \text{m}.

The wavelength is about 6.5×10−7 m6.5\times10^{-7}\ \text{m} (654 nm654\ \text{nm}). 1 mark for the rearranged equation, 1 mark for converting the fringe spacing to metres, 1 mark for the answer.

Practise this

Sources & how we know this

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

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