Wave optics: interference and diffraction

Light is an electromagnetic wave, so it shows typical wave phenomena: interference, diffraction, dispersion and polarisation. The rainbow colours of a soap bubble, the coloured bands on a CD, the rainbow itself and polarising sunglasses are all explained by wave optics. Geometric optics (rays) cannot explain them.

§1. Interference of light

Interference is the stable pattern of bright (maxima) and dark (minima) regions produced when coherent waves add up.
Coherent waves have the same frequency and a phase difference that does not change with time. Two separate lamps are not coherent, so coherent beams are obtained by splitting the light of one source.

The result depends on the path difference Δ:

k = 0, 1, 2, ...

Worked example

The path difference of two coherent waves is 1.5 μm. What is observed for λ = 500 nm and for λ = 600 nm?

1. Δ/λ = 1.5·10⁻⁶/5·10⁻⁷ = 3 — a whole number, a maximum.
2. Δ/λ = 1.5·10⁻⁶/6·10⁻⁷ = 2.5 — an odd number of half-waves (5), a minimum.

§2. Young’s experiment and thin films

In Young’s experiment two narrow slits lit by one source act as coherent sources. Bright and dark fringes appear on the screen with spacing

where L is the slit-to-screen distance and d is the slit separation.

Thin-film interference: light reflects from the top and bottom surfaces of a film and the two beams combine. This explains the colours of soap bubbles and oil patches on water, and Newton’s rings. Anti-reflection coating ("blooming"): a film of thickness λ/(4n) is put on a lens; the reflected rays cancel and more light is transmitted.

Worked example

d = 0.5 mm, L = 1 m, λ = 500 nm. Fringe spacing?

1. Δx = 5·10⁻⁷·1/(5·10⁻⁴).
2. Δx = 10⁻³ m = 1 mm.

§3. Diffraction of light

Diffraction is the bending of waves around obstacles into the region of geometric shadow. It is noticeable when the size of the obstacle or slit is comparable to the wavelength. Light’s wavelength is tiny (~0.5 μm), so everyday diffraction of light is hard to notice, whereas sound (~1 m) is heard around a wall.

The Huygens–Fresnel principle: every point of a wavefront is a source of secondary coherent waves, which interfere. Diffraction limits the resolving power of microscopes and telescopes.

§4. The diffraction grating

A diffraction grating is a large number of identical parallel slits. The grating spacing d is the distance between neighbouring slits; if there are N lines per mm, d = 1/N mm.

• k is the order of the spectrum; since sin φ ≤ 1, k ≤ d/λ.
• White light is split into spectra: the central maximum is white; in the others violet is closest to the centre and red farthest (larger λ — larger angle).

Worked example

d = 2·10⁻⁶ m, λ = 500 nm. Direction of the first-order maximum and the highest order?

1. sin φ₁ = λ/d = 5·10⁻⁷/2·10⁻⁶ = 0.25, φ₁ ≈ 14.5°.
2. k ≤ d/λ = 4. For k = 4, φ = 90° — that maximum never reaches the screen, so in practice orders up to 3 are seen (tests often accept k_max = d/λ = 4 — read the question carefully).

Worked example

A grating with 100 lines per mm shows a second-order maximum at sin φ = 0.1. Wavelength?

1. d = 1/100 mm = 10⁻⁵ m.
2. λ = d·sin φ/k = 10⁻⁵·0.1/2 = 5·10⁻⁷ m = 500 nm.

§5. Dispersion and polarisation

Dispersion is the dependence of the refractive index on wavelength. In glass n is larger for violet light and smaller for red. So a prism splits white light into a spectrum, with violet refracted most and red least (Newton’s experiment). A rainbow is dispersion in water drops.

Polarisation is the ordering of the E⃗ oscillations into one plane. Natural light is unpolarised; a polaroid passes oscillations in one direction only and halves the intensity. Turning a second polaroid (the analyser) changes the transmitted light: I = I₁·cos² φ (Malus’s law). Polarisation proves that light is a transverse wave.

§6. Wave optics in practice: the compact disc and measuring wavelength

The coloured bands on a compact disc (CD) mentioned in the introduction are an everyday diffraction grating. Data on the disc is written along a spiral track, and the spacing between neighbouring turns on a CD is about 1.6 μm. For reflected light these tracks act as a reflection grating, and the condition of §4, d·sin φ = kλ, still applies. Each colour in white light is reinforced at its own angle, so the disc shimmers like a rainbow. On a DVD the tracks are closer (≈ 0.74 μm), so the colours are deflected through larger angles and the disc holds more data.

Another important practical result of wave optics is that the wavelength of light can be measured with simple equipment. Measuring Δx, L and d in Young’s experiment gives λ = Δx·d/L; with a grating λ = d·sin φ/k. In 1801–1803 T. Young first estimated the wavelengths of different colours in exactly this way. Today spectrometers use the same principle to identify the composition of substances.

Worked example

The track spacing of a CD is 1.6 μm. Find the angle of the first-order maximum for a red laser λ = 650 nm and the highest order.

1. sin φ₁ = λ/d = 650/1600 ≈ 0.406 → φ₁ ≈ 24°.
2. d/λ = 1600/650 ≈ 2.46, so k_max = 2 — only first- and second-order maxima are seen.

Worked example

In Young’s experiment d = 0.2 mm, L = 2 m, and the spacing between neighbouring bright fringes is 6 mm. Wavelength?

1. λ = Δx·d/L = 6·10⁻³·2·10⁻⁴/2.
2. λ = 6·10⁻⁷ m = 600 nm — orange light.

Check yourself

1. Path difference 2 μm, λ = 400 nm. Maximum or minimum?

2. Path difference 0.9 μm, λ = 600 nm?

3. A grating with 200 lines per mm, λ = 500 nm. sin φ for the first maximum and the highest order?

4. In Young’s experiment the wavelength is doubled. Fringe spacing?

5. Thickness of an anti-reflection film (n = 1.25) for 550 nm?

6. Which colour is deviated more by a prism, and which by a grating?

7. The track spacing of a DVD is 0.74 μm and λ = 650 nm. Angle of the first-order maximum? Is there a second order?

8. In Young’s experiment d = 0.4 mm, L = 2 m, Δx = 3 mm. Wavelength?

Common mistakes

• Not converting nm to metres (1 nm = 10⁻⁹ m).
• Using the number of lines per mm instead of the grating spacing (d = 1/N).
• Writing the minimum condition as Δ = kλ/2: the number of half-waves must be odd.
• Mixing up the order of colours in a prism and a grating.
• Thinking incoherent sources (two lamps) also produce interference.
• Swapping d and L when finding λ from Young’s formula: λ = Δx·d/L, where d is the slit separation.