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Young's double-slit interference: WACE Year 12 Physics Unit 4

Syllabus dot point

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

8 min read

Quick answer

Light through two narrow, closely spaced slits produces a pattern of bright and dark fringes because the waves from the slits interfere. Bright fringes occur where the path difference is a whole number of wavelengths (dsin⁡θ=mλd\sin\theta=m\lambda) and dark fringes where it is a half-integer number. The fringe spacing on a screen is Δy=λLd\Delta y=\dfrac{\lambda L}{d}. This pattern is decisive evidence that light is a wave.

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 dd 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,

dsin⁡θ=mλ,m=0,1,2,…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,

dsin⁡θ=(m+12)λ.d\sin\theta=\left(m+\tfrac{1}{2}\right)\lambda.

The central bright fringe (m=0m=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 LL from the slits, the bright fringes are evenly spaced by

Δy=λLd.\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 Δy\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.

Wavelength from a fringe pattern

Finding wavelength from fringe spacing

Light passes through two slits 0.25 mm0.25\ \text{mm} apart onto a screen 1.8 m1.8\ \text{m} away, producing bright fringes 4.3 mm4.3\ \text{mm} apart. Find the wavelength.

Rearrange the fringe-spacing equation:

λ=Δy dL=(4.3×10−3)(0.25×10−3)1.8.\lambda=\frac{\Delta y\,d}{L}=\frac{(4.3\times10^{-3})(0.25\times10^{-3})}{1.8}.

λ=1.075×10−61.8=6.0×10−7 m=600 nm.\lambda=\frac{1.075\times10^{-6}}{1.8}=6.0\times10^{-7}\ \text{m}=600\ \text{nm}.

This is orange-red light, consistent with the visible range of about 400400 to 700 nm700\ \text{nm}.

What the fringe-spacing equation tells you

The relation Δy=λLd\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 LL, so moving the screen back spreads the fringes. It is inversely proportional to the slit separation dd, 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 dsin⁡θ=mλd\sin\theta=m\lambda for bright fringes and the half-integer version for dark, and use Δy=λL/d\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 mm counts fringes out from the centre.

Swapping the bright and dark fringe conditions

A frequent slip is using the half-integer condition for bright fringes. Bright (constructive) fringes need a whole number of wavelengths of path difference; dark (destructive) fringes need a half-integer number. Mixing them up shifts every predicted position by half a fringe. Anchor your reasoning on the central bright fringe at zero path difference.

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 2022-style7 marks
In a double-slit experiment, slits separated by 0.20 mm0.20\ \text{mm} are illuminated by laser light of wavelength 650 nm650\ \text{nm}. The interference pattern is observed on a screen 2.5 m2.5\ \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.
Show worked answer →

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

(a) Fringe spacing
Δy=λLd=(650×10−9)(2.5)0.20×10−3=1.625×10−62.0×10−4=8.1×10−3 m\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 3Δy=3×8.1×10−3=2.4×10−2 m3\Delta y=3\times8.1\times10^{-3}=2.4\times10^{-2}\ \text{m} from the centre.
(c) Halving the slit separation
Since Δy=λLd\Delta y=\dfrac{\lambda L}{d}, halving dd doubles the fringe spacing to about 1.6×10−2 m1.6\times10^{-2}\ \text{m}.

Markers reward Δy=λL/d\Delta y=\lambda L/d near 8.1×10−3 m8.1\times10^{-3}\ \text{m}, the third-order distance of 2.4×10−2 m2.4\times10^{-2}\ \text{m} and the inverse relation doubling the spacing.

WACE 2020-style5 marks
Explain 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.
Show worked answer →

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, dsin⁡θ=mλd\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.

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.

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