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Thomas Young's double slit experiment (1801) provided the first conclusive evidence that light behaves as a wave. It remains one of the most important experiments in physics and is a required part of the AQA A-Level specification.
Spec mapping: This lesson sits under AQA 7408 section 3.3.2 and covers the apparatus, theory and derivation for Young's double-slit experiment, including the fringe-spacing relation w = λD/s with the small-angle approximation, the use of monochromatic vs white light, and the experimental technique forming Required Practical 4 (interference of two coherent sources). (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Required Practical 4: the laser double-slit measurement of wavelength is the explicit practical here; students are expected to identify uncertainties, suggest improvements, and analyse the resulting graph.
- Section 3.3.2 (diffraction gratings): the grating equation d sin θ = nλ is the wide-angle generalisation of the double-slit small-angle relation; the conceptual prerequisite is exactly this lesson.
- Section 3.2.2 (wave–particle duality): electron-diffraction experiments through analogous slit geometries reproduce the same fringe pattern, demonstrating that the wave model extends to matter — a key synoptic theme for Paper 2.
The apparatus consists of:
Light from the single source passes through the double slit. Each slit acts as a separate coherent source (because they originate from the same wavefront). The two sets of diffracted waves overlap and interfere on the screen.
When using a non-laser source, a single slit is placed before the double slit to ensure spatial coherence. The single slit produces a single diffraction pattern, illuminating both slits of the double slit with light from the same wavefront. This guarantees a constant phase relationship between the two slits.
A laser already produces spatially coherent light, so the single slit can be omitted.
On the screen, an alternating pattern of bright fringes (maxima) and dark fringes (minima) is observed:
The fringes are equally spaced (for small angles) and parallel to the slits. The central bright fringe (zeroth order, n = 0) is on the axis of symmetry, where the path difference is zero.
The distance between adjacent bright fringes (or adjacent dark fringes) is called the fringe spacing (or fringe width), w:
w = λD/s
where:
Consider the nth bright fringe at a position y on the screen, measured from the central maximum. For a bright fringe, the path difference from the two slits must equal nλ.
Using geometry, for small angles θ:
The spacing between the nth and (n+1)th maxima is: w = y(n+1) − y(n) = (n+1)λD/s − nλD/s = λD/s
Exam Tip: This derivation requires the small angle approximation (sin θ ≈ tan θ ≈ θ for small θ in radians). This is valid when D >> s, which is always the case in practice.
Worked Example 1 — Red laser light of wavelength 635 nm passes through a double slit with slit separation 0.40 mm. The screen is 2.5 m from the slits. Calculate the fringe spacing.
λ = 635 nm = 635 × 10⁻⁹ m = 6.35 × 10⁻⁷ m s = 0.40 mm = 4.0 × 10⁻⁴ m D = 2.5 m
w = λD/s = (6.35 × 10⁻⁷ × 2.5)/(4.0 × 10⁻⁴)
w = (1.5875 × 10⁻⁶)/(4.0 × 10⁻⁴) = 3.97 × 10⁻³ m
w = 4.0 × 10⁻³ m = 4.0 mm
Worked Example 2 — In a Young's double slit experiment, the fringe spacing is measured as 2.8 mm when the slit-to-screen distance is 1.80 m and the slit separation is 0.35 mm. Calculate the wavelength of the light.
w = 2.8 mm = 2.8 × 10⁻³ m D = 1.80 m s = 0.35 mm = 3.5 × 10⁻⁴ m
Rearranging w = λD/s:
λ = ws/D = (2.8 × 10⁻³ × 3.5 × 10⁻⁴)/1.80
λ = (9.80 × 10⁻⁷)/1.80 = 5.44 × 10⁻⁷ m
λ = 544 nm (green light)
Worked Example 3 — A Young's double slit experiment uses light of wavelength 590 nm. The slit separation is 0.50 mm and the screen is 2.0 m away. How far from the central maximum is the 3rd dark fringe?
Dark fringes occur at path differences of (n + ½)λ, so the mth dark fringe from the centre corresponds to m = 0, 1, 2, ... The 3rd dark fringe corresponds to m = 2 (since we count from m = 0).
Position: y = (m + ½)λD/s = (2 + ½) × (590 × 10⁻⁹ × 2.0)/(5.0 × 10⁻⁴)
y = 2.5 × (1.18 × 10⁻⁶)/(5.0 × 10⁻⁴) = 2.5 × 2.36 × 10⁻³ = 5.9 × 10⁻³ m
y = 5.9 mm from the central maximum
When white light (containing all visible wavelengths) is used instead of monochromatic light:
Exam Tip: A common 6-mark question asks you to describe and explain the white light fringe pattern. State clearly: central white fringe, coloured higher-order fringes with blue/violet nearer the centre and red further out, pattern becomes indistinct after a few orders due to overlapping.
Young's experiment was historically crucial because:
Only waves can produce interference patterns. If light were purely a stream of particles (as Newton proposed), you would not observe the alternating bright and dark fringes — particles passing through two slits would produce two bright bands, not an interference pattern.
The experiment allowed the wavelength of light to be measured for the first time.
Combined with later experiments (photoelectric effect), it led to the understanding of wave-particle duality.
Worked Example 4 — A student measures the distance across 8 bright fringes as 25.6 mm. Calculate the fringe spacing.
8 bright fringes span 7 fringe spacings (from the 1st to the 8th bright fringe, there are 7 gaps).
w = 25.6/7 = 3.66 mm (to 3 s.f.)
Common Misconception: Students often divide by 8 instead of 7. If you count 8 fringes, there are only 7 spaces between them. Always count the number of gaps, not the number of fringes.
| Change | Effect on Fringe Spacing w |
|---|---|
| Increase wavelength λ | w increases (w ∝ λ) |
| Increase slit separation s | w decreases (w ∝ 1/s) |
| Increase screen distance D | w increases (w ∝ D) |
| Use white light instead of monochromatic | Central white fringe; coloured fringes spread out, then overlap |
The double-slit experiment produces not only a fringe position pattern but also a characteristic intensity profile. For two slits of negligible width (idealised pure interference), the intensity at angle θ from the central axis is:
I(θ) = 4I₀ cos²(πs sin θ / λ)
where I₀ is the intensity from one slit alone. At θ = 0 the cosine equals 1, giving I = 4I₀ (maximum). At the dark fringes (path difference (n + ½)λ), the cosine is zero, giving I = 0.
sin θ ≈ tan θ = y/D = (1.2 × 10⁻³) / 1.5 = 8.0 × 10⁻⁴
πs sin θ / λ = π × (2.0 × 10⁻⁴) × (8.0 × 10⁻⁴) / (6.0 × 10⁻⁷) = π × 0.267 = 0.838 rad
I/I₀ = 4 cos²(0.838) = 4 × (0.669)² = 4 × 0.448 = 1.79
This point lies between a bright fringe (at I/I₀ = 4) and a dark fringe (at I/I₀ = 0), and the cos² intensity profile gives a non-zero but reduced value — consistent with a position roughly 40% of the way from a bright to a dark fringe.
The pure-interference profile I(θ) = 4I₀ cos²(πs sin θ/λ) assumes the slits are infinitely narrow. Real slits have finite width a, producing a single-slit diffraction envelope of the form sinc²(πa sin θ/λ). The observed intensity is the product:
I_observed(θ) = 4I₀ cos²(πs sin θ/λ) × sinc²(πa sin θ/λ)
The cosine-squared term encodes the interference between the two slits; the sinc-squared term encodes the diffraction from each individual slit. The next lesson (single-slit diffraction) develops the second factor in detail.
For a stable interference pattern, the path-difference uncertainty between the two slit paths must be small compared with the coherence length L_c of the source — the longitudinal distance over which the source's wave-trains remain in phase. Coherence length is approximately L_c ≈ λ²/Δλ where Δλ is the spectral linewidth of the source.
| Source | Wavelength | Linewidth Δλ | Coherence length L_c |
|---|---|---|---|
| Sodium discharge lamp | 589 nm | ~0.1 nm | ~3.5 mm |
| He-Ne laser (red) | 633 nm | ~10⁻⁵ nm | ~40 m |
| Stabilised single-mode laser | 633 nm | ~10⁻⁹ nm | ~400 km |
| White light bulb | ~550 nm | ~300 nm | ~1 μm |
This explains why white light produces only a few visible fringes before they wash out: the coherence length of ~1 μm is barely longer than the wavelength itself, and beyond about 3–5 fringe spacings the wavetrains from one slit no longer overlap coherently with the wavetrains from the other. A laser, by contrast, supports clean fringes over thousands of orders.
L_c ≈ λ²/Δλ = (589 × 10⁻⁹)² / (0.10 × 10⁻⁹) = 3.47 × 10⁻³ m = 3.47 mm
For the nth-order fringe, the path difference between the slits is nλ. The path difference is observable while nλ < L_c, so n_max = L_c/λ = (3.47 × 10⁻³) / (5.89 × 10⁻⁷) = 5890 orders.
In practice this is far more than the visible field of view: with a 1 mm fringe spacing on a 1 m screen, only ~500 fringes are even geometrically visible. The coherence length is not the limiting factor here; the source is sufficiently coherent.
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