Diffraction and Single Slit Patterns

Diffraction is the spreading of waves when they pass through a gap or around an obstacle. It is a defining characteristic of wave behaviour and cannot be explained by a simple particle model.

Spec mapping: This lesson sits under AQA 7408 section 3.3.2 and covers single-slit diffraction: the qualitative explanation via Huygens' principle, the condition sin θ = nλ/a for minima, the characteristic broad central maximum with secondary maxima of diminishing intensity, the inverse relationship between slit width and central-maximum angular width, and the modulation of the double-slit pattern by the single-slit envelope ("missing orders"). (Refer to the official AQA specification document for exact wording.)

Synoptic links:

• Section 3.3.2 (Young's double slits): the observed Young's pattern is the two-slit interference factor multiplied by the single-slit envelope from each individual slit — both must be understood together.
• Section 3.3.2 (diffraction gratings): the grating pattern is the limit of many slits each producing its own single-slit diffraction envelope, with constructive interference between them producing sharp grating maxima.
• Section 3.2.2 (electron diffraction): the Davisson–Germer experiment and the more recent single-electron-at-a-time diffraction experiments show that the diffraction equation sin θ = nλ/a applies to matter waves with λ = h/p — a Paper 2 synoptic favourite.

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What is Diffraction?

When a wave passes through a gap (aperture) or encounters an obstacle, it spreads out into the region beyond. This spreading is called diffraction.

Key observations:

• Diffraction is most significant when the gap width is comparable to the wavelength (gap ≈ λ).
• If the gap is much larger than the wavelength, very little diffraction occurs — the wave passes through almost as a straight beam.
• If the gap is much smaller than the wavelength, the wave spreads out almost equally in all directions (semicircular wavefronts in 2D).

Huygens' Principle

Diffraction can be explained using Huygens' principle:

Every point on a wavefront can be considered as a source of secondary spherical wavelets. The new wavefront is the envelope (tangent surface) of all these secondary wavelets.

When a wave passes through a narrow slit, only the wavelets originating from within the slit opening contribute to the wave on the other side. These wavelets spread out, causing the wave to fan out beyond the slit.

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Single Slit Diffraction Pattern

When monochromatic light passes through a single narrow slit of width a and falls on a distant screen, the following pattern is observed:

Features of the Pattern

1. A broad central maximum — the brightest and widest part of the pattern.
2. Subsidiary (secondary) maxima on either side — much dimmer, decreasing in intensity with distance from the centre.
3. Dark minima between the maxima — positions of zero intensity (destructive interference).
4. The central maximum is twice as wide as each subsidiary maximum.

Positions of the Minima

The first minimum on either side of the central maximum occurs at an angle θ given by:

sin θ = λ/a

More generally, the minima occur at angles where:

sin θ = nλ/a (n = ±1, ±2, ±3, ...)

where:

• θ = angle from the central axis to the minimum
• λ = wavelength of light
• a = slit width
• n = order number (n ≠ 0; n = 0 corresponds to the central maximum)

Why Do the Minima Occur?

At the first minimum (n = 1), the path difference between wavelets from the top and bottom of the slit is exactly one wavelength (λ). The slit can be divided into two halves: every wavelet from the top half has a corresponding wavelet from the bottom half that is exactly λ/2 ahead, so they cancel in pairs. The result is complete destructive interference.

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Worked Examples

Worked Example 1 — Light of wavelength 589 nm passes through a single slit of width 0.10 mm. Calculate the angle of the first minimum.

λ = 589 nm = 5.89 × 10⁻⁷ m a = 0.10 mm = 1.0 × 10⁻⁴ m

sin θ = λ/a = (5.89 × 10⁻⁷)/(1.0 × 10⁻⁴) = 5.89 × 10⁻³

θ = arcsin(5.89 × 10⁻³) = 0.34°

The angle is very small because the slit width is much larger than the wavelength.

Worked Example 2 — In the same experiment, the screen is 3.0 m from the slit. Calculate the width of the central maximum on the screen.

The central maximum extends from the first minimum on one side to the first minimum on the other. Its half-width is y₁ = D tan θ ≈ Dθ (for small angles).

y₁ = D sin θ = 3.0 × 5.89 × 10⁻³ = 1.767 × 10⁻² m

Full width of central maximum = 2y₁ = 2 × 1.767 × 10⁻² = 3.53 × 10⁻² m

Width = 35 mm (to 2 s.f.)

Worked Example 3 — Red light (λ = 650 nm) and blue light (λ = 470 nm) both pass through a single slit of width 0.080 mm. Compare the widths of their central maxima on a screen 2.0 m away.

For red light: sin θ_R = λ_R/a = (650 × 10⁻⁹)/(8.0 × 10⁻⁵) = 8.125 × 10⁻³ Width_R = 2Dsin θ_R = 2 × 2.0 × 8.125 × 10⁻³ = 3.25 × 10⁻² m = 32.5 mm

For blue light: sin θ_B = λ_B/a = (470 × 10⁻⁹)/(8.0 × 10⁻⁵) = 5.875 × 10⁻³ Width_B = 2Dsin θ_B = 2 × 2.0 × 5.875 × 10⁻³ = 2.35 × 10⁻² m = 23.5 mm

The red central maximum is wider because red light has a longer wavelength.

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Effect of Slit Width and Wavelength

Change	Effect on Central Maximum	Explanation
Decrease slit width a	Central maximum becomes wider	sin θ = λ/a increases as a decreases
Increase wavelength λ	Central maximum becomes wider	sin θ = λ/a increases as λ increases
Increase slit width a	Central maximum becomes narrower	Less diffraction when a >> λ
Decrease wavelength λ	Central maximum becomes narrower	Less diffraction when λ << a

Exam Tip: The narrower the slit, the more the light spreads out. This is counterintuitive but follows directly from sin θ = λ/a. Remember: significant diffraction requires the gap width to be comparable to the wavelength.

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Single Slit Diffraction with a Double Slit

In Young's double slit experiment, each slit also acts as a single slit, producing its own diffraction pattern. The observed pattern is actually the double slit interference pattern modulated by the single slit diffraction envelope.

This means:

• The overall intensity envelope follows the single slit diffraction pattern (determined by individual slit width a).
• Within this envelope, the equally spaced double slit fringes appear (determined by slit separation s).
• If a double slit fringe maximum coincides with a single slit minimum, that fringe is missing from the pattern.

Worked Example 4 — A double slit has slit width a = 0.10 mm and slit separation s = 0.50 mm. Monochromatic light of wavelength 600 nm is used. Which orders of the double slit interference pattern are missing?

The double slit maxima occur at angles where: sin θ = nλ/s The single slit minima occur at angles where: sin θ = mλ/a

A missing order occurs when both conditions are satisfied simultaneously:

nλ/s = mλ/a → n/s = m/a → n = ms/a

n = m × (0.50/0.10) = 5m

So orders n = 5, 10, 15, ... are missing.

The 5th, 10th, 15th (etc.) order double slit fringes coincide with single slit diffraction minima and are absent from the pattern.

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Diffraction of Other Waves

Diffraction is not limited to light:

• Sound waves diffract readily through doorways because their wavelengths (cm to m) are comparable to door widths (~1 m).
• Radio waves diffract around buildings and hills (long-wave radio has λ ~ 1 km).
• Water waves spread out when passing through harbour entrances.
• Microwaves (λ ~ cm) can be diffracted through gaps of similar size in laboratory demonstrations.

Worked Example 5 — A harbour entrance is 15 m wide. Ocean waves with wavelength 12 m approach the entrance. Describe the diffraction.

Ratio: a/λ = 15/12 = 1.25

Since the gap width is comparable to the wavelength (ratio ≈ 1), significant diffraction occurs. The waves spread out considerably after passing through the harbour entrance, forming roughly semicircular wavefronts inside the harbour.

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Intensity Profile of the Single-Slit Pattern

The full intensity distribution from a single slit of width a under monochromatic illumination of wavelength λ is:

I(θ) = I₀ × sinc²(πa sin θ / λ)

where sinc(x) ≡ sin(x)/x. The central maximum at θ = 0 has intensity I₀; the first zero on either side is at sin θ = ±λ/a; the subsidiary maxima decrease rapidly in intensity.

Order	Approximate angle of maximum	Approximate fraction of I₀
Central	sin θ = 0	1.000
1st subsidiary	sin θ ≈ 1.430λ/a	0.0472
2nd subsidiary	sin θ ≈ 2.459λ/a	0.0165
3rd subsidiary	sin θ ≈ 3.471λ/a	0.00834

The first subsidiary maximum has only ~4.7% of the central intensity; the second is ~1.6%. This explains why photographs of single-slit diffraction show an overwhelmingly bright central peak with progressively dimmer secondary peaks.

Worked Example 6 — Light of wavelength 633 nm passes through a slit of width 0.040 mm onto a screen 2.0 m away. Calculate (a) the angular width of the central maximum, (b) the linear width on the screen, (c) the position of the first subsidiary maximum.

(a) sin θ₁ = λ/a = (633 × 10⁻⁹) / (4.0 × 10⁻⁵) = 1.58 × 10⁻²

Angular width = 2θ₁ ≈ 2 × 1.58 × 10⁻² = 3.16 × 10⁻² rad = 1.81°

(b) Linear width = 2D tan θ₁ ≈ 2 × 2.0 × 1.58 × 10⁻² = 6.33 × 10⁻² m = 63.3 mm

(c) The first subsidiary maximum is at sin θ ≈ 1.430 × λ/a = 1.430 × 1.58 × 10⁻² = 2.26 × 10⁻²

Linear position on screen: y = D × sin θ = 2.0 × 2.26 × 10⁻² = 4.53 × 10⁻² m = 45.3 mm from the central maximum on each side.

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Diffraction Around Obstacles

So far we have treated diffraction at apertures (gaps); diffraction also occurs around the edges of obstacles. According to Babinet's principle, the diffraction pattern from an opaque obstacle and from a complementary aperture of the same shape are identical (apart from the central beam), because they produce wavefronts that sum to the unobstructed wave.

The most famous historical demonstration is the Poisson bright spot (1818): when light is diffracted around a small opaque disk, a bright spot appears at the centre of the geometric shadow — predicted by Fresnel's wave theory and considered absurd by Poisson, who proposed it as a reductio ad absurdum against the wave theory. Arago performed the experiment and found the spot, confirming the wave theory triumphantly.

Worked Example 7 — A 1.0 mm-diameter coin is held in a parallel laser beam of wavelength 550 nm, with a screen 5.0 m beyond. Estimate the angular position of the first Fresnel-zone minimum around the shadow edge.

For a circular obstacle of radius r, the first off-axis minimum lies at sin θ ≈ 1.22 λ/(2r) (the Airy-disc analogue):

sin θ ≈ 1.22 × (5.5 × 10⁻⁷) / (1.0 × 10⁻³) = 6.71 × 10⁻⁴

Linear position on screen: y = D × sin θ = 5.0 × 6.71 × 10⁻⁴ = 3.35 × 10⁻³ m = 3.35 mm from the shadow centre.

(This is an A-Level-adjacent calculation; the Airy formula is not on the spec but is conceptually within reach.)

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Resolution and the Rayleigh Criterion

Diffraction sets a fundamental limit on how well two closely spaced sources can be distinguished by an optical system — telescope, microscope, or eye. According to the Rayleigh criterion, two point sources are just resolved when the central maximum of one source's diffraction pattern falls on the first minimum of the other. For a circular aperture of diameter D illuminated by light of wavelength λ, the minimum resolvable angular separation is:

θ_min = 1.22 λ/D

This is the diffraction-limited resolution; it applies whether the aperture is a telescope objective, a microscope condenser, or the pupil of the human eye.

Worked Example 8 — The human eye has a pupil diameter of approximately 4.0 mm in normal lighting. Estimate (a) the angular resolution at λ = 550 nm, (b) the smallest detail visible at a reading distance of 30 cm.

(a) θ_min = 1.22 × (5.5 × 10⁻⁷) / (4.0 × 10⁻³) = 1.68 × 10⁻⁴ rad ≈ 35 arcseconds