Interference and Diffraction of Light: Young's Slits, Gratings

Proof That Light Is a Wave

In 1801, Thomas Young shone light through two narrow slits and saw on a screen not two bright lines, but a pattern of many bright and dark bands. Only waves can do that: where light waves meet in step they reinforce, and where they meet out of step they cancel. Interference and diffraction are the clearest evidence that light behaves as a wave. They are also the basis of spectrometers, the colours of soap bubbles and the rainbow sheen of a CD. This tutorial covers interference and diffraction of light as taught in PHY 102 and met in WAEC physics: superposition, coherence, path difference, Young's double-slit experiment, single-slit diffraction, the diffraction grating, thin films and polarisation.

Superposition and Coherence

The principle of superposition states that when two waves meet, the resulting displacement at any point is the sum of their individual displacements.

Constructive interference: crest meets crest, so the waves reinforce, giving bright light or a loud sound.
Destructive interference: crest meets trough, so the waves cancel, giving darkness or quiet.

A stable interference pattern needs coherent sources: sources with the same frequency (and wavelength) and a constant phase difference. Two separate torches or bulbs are not coherent, because their light is emitted in random bursts. Young solved this by splitting light from one source with two slits. Lasers are naturally coherent.

Path Difference

Whether two waves arriving at a point reinforce or cancel depends on the difference in the distances they have travelled, the path difference:

Constructive (bright): path difference = nλ, where n = 0, 1, 2, …
Destructive (dark): path difference = (n + ½)λ, that is ½λ, 3λ/2, 5λ/2, …

Worked example 1

Two coherent sources emit waves of wavelength 2 cm. At a point P, the path difference is 5 cm. Is P a point of constructive or destructive interference?

5 cm ÷ 2 cm = 2.5 wavelengths, which is (n + ½)λ, so P is a point of destructive interference.

Young's Double-Slit Experiment

Monochromatic light (light of a single wavelength) passes through two narrow slits a distance d apart and falls on a screen a distance D away. Equally spaced bright and dark fringes appear. The fringe spacing (the distance between neighbouring bright fringes) is:

y = λD ÷ d

The fringes are wider for longer wavelengths (red more than blue), for a more distant screen, and for slits that are closer together.

Worked example 2

Light of wavelength 600 nm passes through two slits 0.30 mm apart onto a screen 1.5 m away. Find the fringe spacing.

y = (600 × 10⁻⁹ × 1.5) ÷ (0.30 × 10⁻³) = 3.0 × 10⁻³ m, or 3.0 mm.

Worked example 3: measuring wavelength

Fringes 2.4 mm apart are produced by slits 0.50 mm apart, with the screen 2.0 m away. Find the wavelength of the light.

λ = yd ÷ D = (2.4 × 10⁻³ × 0.50 × 10⁻³) ÷ 2.0 = 6.0 × 10⁻⁷ m (600 nm, orange light).

In practice, measure across several fringes and divide, to reduce the error.

With white light, the central fringe is white, but the others show coloured edges, because each wavelength produces a different fringe spacing.

Diffraction

Diffraction is the spreading of waves as they pass through a gap or around an obstacle. It is most noticeable when the gap is about the same size as the wavelength. Light has a tiny wavelength (400–700 nm), so it diffracts noticeably only through very narrow slits. That is why light usually seems to travel in sharp straight lines, while sound (wavelength around 1 m) bends easily around doorways.

Light through a single narrow slit produces a wide, bright central band with much fainter, narrower bands on either side. A narrower slit spreads the pattern wider.

The Diffraction Grating

A diffraction grating has thousands of equally spaced parallel slits (lines) per millimetre. Light from all the slits interferes to give very sharp, bright maxima at angles θ given by:

d sin θ = nλ

Here d is the spacing between neighbouring lines (d = 1 ÷ the number of lines per metre) and n = 0, 1, 2, … is the order of the maximum.

Worked example 4

Light of wavelength 500 nm falls on a grating with 500 lines per mm. Find the angle of the first-order maximum and the highest order visible.

d = 1 ÷ 500,000 m = 2.0 × 10⁻⁶ m. First order: sin θ = 500 × 10⁻⁹ ÷ 2.0 × 10⁻⁶ = 0.25, so θ ≈ 14.5°.

The highest order needs sin θ ≤ 1: n ≤ d ÷ λ = 4.0. So the 4th order would be at exactly 90°, which cannot actually be seen. The highest order observed in practice is 3.

Worked example 5

Find the angle of the second-order maximum for light of wavelength 600 nm and a grating with 300 lines per mm.

d = 1 ÷ 300,000 m ≈ 3.33 × 10⁻⁶ m, so sin θ = 2 × 600 × 10⁻⁹ ÷ 3.33 × 10⁻⁶ = 0.36, and θ ≈ 21.1°.

Because each wavelength diffracts at a different angle, a grating separates white light into spectra, more sharply than a prism. Spectrometers use gratings to identify elements from their spectral lines, including in distant stars. The colours seen on a CD or DVD are a natural grating effect from its closely spaced tracks.

Thin-Film Interference

The swirling colours of soap bubbles and of oil films on wet roads come from interference. Light reflected from the top and bottom surfaces of the thin film travels slightly different distances, and depending on the film's thickness, some colours reinforce while others cancel. The same principle gives anti-reflection coatings on spectacles and camera lenses, which are thin films designed to cancel reflected light.

Polarisation

Light is a transverse wave: its electric field vibrates at right angles to its direction of travel, in many directions. A Polaroid filter transmits only the vibrations in one plane, producing plane-polarised light. Two Polaroids at right angles block the light completely. Only transverse waves can be polarised, so this confirms that light is transverse (sound, which is longitudinal, cannot be polarised). Polarising sunglasses cut glare, because light reflected from water and roads is partly polarised.

Common Mistakes with Interference and Diffraction

Forgetting to convert units: nm (10⁻⁹ m), mm (10⁻³ m) and lines per mm into d in metres.
Writing y = λd/D. The correct formula is y = λD/d.
Using lines per mm directly as d. d is the reciprocal of the number of lines per metre.
Claiming an order with sin θ > 1 is visible. The maximum order is the whole-number part of d/λ, and an order exactly at 90° is not seen.
Expecting interference from two ordinary lamps. The sources must be coherent.

Interference and Diffraction: Practice Questions

Find the fringe spacing for light of wavelength 500 nm, slits 0.40 mm apart and a screen 2.0 m away.
What happens to the fringe spacing if the slit separation is doubled?
Find the first-order angle for light of wavelength 625 nm with a grating of 400 lines per mm.
A grating has d = 2.0 × 10⁻⁶ m. What is the highest order visible for light of wavelength 700 nm?
Two coherent waves meet with a path difference of 3λ. What kind of interference occurs?
Why do two separate torches not produce an interference pattern?

Answers: 1. 500 × 10⁻⁹ × 2 ÷ 0.4 × 10⁻³ = 2.5 mm. 2. It halves. 3. d = 2.5 × 10⁻⁶ m, so sin θ = 0.25 and θ ≈ 14.5°. 4. d/λ ≈ 2.86, so the 2nd order. 5. Constructive. 6. Their light is not coherent: the phase difference between them changes randomly and rapidly, so any pattern blurs out.

Conclusion

Interference and diffraction of light show that light is a wave. Coherent waves superpose: a path difference of nλ gives bright fringes, and (n + ½)λ gives dark ones. Young's double slit produces fringes of spacing y = λD/d, a diffraction grating gives sharp maxima at d sin θ = nλ, and thin films create the colours of soap bubbles. Polarisation shows that light is transverse. These effects power spectrometers, anti-reflection coatings and precise measurements of wavelength.