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How does Young's double-slit experiment confirm the wave nature of light and let us measure its wavelength?

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

Young's double-slit experiment, why it produces evenly spaced bright and dark fringes, and the relationship linking fringe spacing, wavelength, slit separation and screen distance.

Generated by Claude Opus 4.89 min answer

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

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  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

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).

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.

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:

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.

Exam-style practice questions

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

SACE 20233 marksIn 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 20253 marksLaser 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.

Openly AI-built from the official SACE Board syllabus, past papers and marking guides. Not individually reviewed; always check SACE Boardand your school's materials for current rules.

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