OxfordAQA International AS and A-level Physics: Interference and diffraction: double slits and gratings (9630)

This guide covers sections 3.5.6 (Interference) and 3.5.7 (Diffraction) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, written for exams from May/June 2020 onwards. Both sections are shared content, so they apply at International AS and at International A-level. Sections 3.6 to 3.13 are International A-level only; these two are not.

Work through the guide, then use the revision notes and the practice set. Course pages: physics hub and printable checklist. Not sure where you stand? Try the free 10-minute diagnostics.

What you must be able to do

Section	Outcome
3.5.6	Use path difference and coherence to explain interference
3.5.6	Describe interference and diffraction patterns made with a monochromatic laser beam
3.5.6	Describe Young’s double-slit experiment (two coherent sources, or one source and a double slit) and use fringe spacing w = λD/s
3.5.6	Describe the pattern produced with white light
3.5.6	Show awareness of laser safety (how a laser works is not required)
3.5.6	Describe and explain interference with sound and electromagnetic waves
3.5.6	Required practical 5: practical work on interference using Young’s slits and a grating
3.5.7	Describe the single-slit pattern for monochromatic and white light
3.5.7	Describe qualitatively how central maximum width varies with wavelength and slit width (the intensity graph is not required)
3.5.7	Plane transmission grating at normal incidence: derive and use d sin θ = nλ
3.5.7	Applications of diffraction gratings (use of the spectrometer is not tested)

Path difference and coherence

When two waves meet, their displacements add (superposition). Whether the sum is large or small depends on the path difference: how much further one wave has travelled to reach the point than the other.

• If the waves left their sources in phase and the path difference is a whole number of wavelengths, nλ, they arrive in phase. Constructive interference gives a maximum.
• If the path difference is half a wavelength plus any whole number of wavelengths, (n + ½)λ, they arrive in antiphase. Destructive interference gives a minimum.

For a steady pattern the sources must be coherent: they have the same frequency (and so wavelength) and a constant phase difference. Two separate lamps are not coherent, because their atoms emit at random, so the phase difference changes too quickly to see. Two loudspeakers driven by one signal generator are coherent. So are the two slits of a Young’s experiment, which are lit by one wavefront, so any phase change in the light reaches both slits together.

A laser gives a narrow beam of monochromatic light (a single wavelength). It is also coherent across the width of the beam, which is why it suits interference and diffraction work. You do not need to explain how a laser works.

Worked example 1: sound from two loudspeakers

Two loudspeakers A and B are connected to one signal generator set to 1700 Hz, with sound travelling at 340 m s⁻¹ in air. A microphone is at P, 3.10 m from A and 3.50 m from B. A second position Q is 2.30 m from A and 2.60 m from B. Describe the sound at P and Q.

1. λ = v ÷ f = 340 ÷ 1700 = 0.200 m.
2. At P, path difference = 3.50 − 3.10 = 0.40 m = 2.0λ. The waves arrive in phase, so the sound is loud.
3. At Q, path difference = 2.60 − 2.30 = 0.30 m = 1.5λ. The waves arrive in antiphase, so the sound is quiet.

Interference with sound and electromagnetic waves

You should be able to describe and explain interference with sound and with electromagnetic waves, not only with light.

• Sound. Two loudspeakers on one oscillator face a room. Walking parallel to the line joining them, you hear loud and quiet regions alternately.
• Microwaves. A microwave transmitter faces a metal plate with two slits. A detector moved along a line parallel to the plate gives a signal that rises and falls: maxima where the path difference is nλ, minima where it is (n + ½)λ.
• Radio waves. Two aerials fed from one transmitter produce regions of strong and weak signal in the same way.

In each case the explanation has the same three steps: coherent sources, path difference at the point, in phase or in antiphase.

Young’s double-slit experiment

A single narrow slit (or a laser beam) lights two parallel slits a distance s apart. Each slit acts as a source, and the two sources are coherent because they come from one wavefront. Light spreads out from each by diffraction and the waves overlap. On a screen a distance D away, you see equally spaced bright and dark fringes.

The bright fringes are at equal spacing w, where

w = λD ÷ s

Here w is the distance between the centres of adjacent bright fringes, λ the wavelength, s the slit separation (centre to centre) and D the distance from the slits to the screen. The relation applies when D is much larger than s, so the angles are small.

Why the formula works

For a point at distance x from the central fringe, the small-angle path difference from the two slits is about s·x ÷ D. A bright fringe needs path difference nλ, so x = nλD ÷ s. Adjacent fringes (n and n + 1) differ by λD ÷ s, which is w.

w increases with λ and D, and decreases when s increases.

Worked example 2: finding a wavelength

Slits 0.40 mm apart are 1.80 m from a screen. The centres of nine bright fringes are measured and the distance from the first to the ninth is 19.2 mm. Find the wavelength.

1. Nine fringes contain 8 spacings, so w = 19.2 mm ÷ 8 = 2.40 mm.
2. λ = ws ÷ D = (2.40 × 10⁻³ × 0.40 × 10⁻³) ÷ 1.80 = 5.33 × 10⁻⁷ m (533 nm).

White light

With white light, each wavelength makes its own fringe pattern with its own spacing. At the centre the path difference is zero for every colour, so all colours are bright together and the central fringe is white. On each side the fringes are coloured: violet and blue (shorter λ, smaller w) lie nearer the centre and red lies furthest out in each fringe. Further out the colours overlap.

Worked example 3: white light fringes

A double slit has s = 0.50 mm and D = 1.5 m. Find the spacing for red light of 700 nm and violet light of 400 nm, and the distance between the third-order red and violet maxima.

1. w (red) = 700 × 10⁻⁹ × 1.5 ÷ 0.50 × 10⁻³ = 2.1 × 10⁻³ m = 2.1 mm.
2. w (violet) = 400 × 10⁻⁹ × 1.5 ÷ 0.50 × 10⁻³ = 1.2 mm.
3. Position of the third maximum = 3w, so the separation = 3 × (2.1 − 1.2) = 2.7 mm.

Required practical 5: interference effects

The practical covers Young’s slits and a diffraction grating. For Young’s slits:

1. Set up a laser, a double slit and a screen, with D of about 2 m.
2. Measure D with a metre rule or tape. Measure across several fringes (say ten spacings) with a ruler, then divide, so the ruler’s own uncertainty is shared out over ten spacings rather than one.
3. Calculate λ = ws ÷ D. Repeat for several values of D and plot w against D: the gradient is λ ÷ s.
4. For the grating, measure the angle to each order on both sides of the centre, halve the difference, and use d sin θ = nλ.

Quote the percentage uncertainty in each measurement and compare them.

Laser safety

Lasers can damage the retina. In a school setting:

• use a low-power laser only;
• never look into the beam, and never look at its reflection from a shiny surface;
• stop the beam with a matt screen or block at the end of its path;
• keep the beam below eye level.

Single-slit diffraction

When a wave passes through a gap, it spreads into the region behind. This is diffraction, and it is greatest when the gap width is close to the wavelength.

For monochromatic light through a single narrow slit, the screen shows a bright central maximum with dimmer, narrower maxima either side, separated by dark minima. The central maximum is much brighter than the others.

How the width of the central maximum depends on the conditions (qualitative only):

• Narrower slit: the central maximum becomes wider, because diffraction is stronger.
• Longer wavelength: the central maximum becomes wider.
• Wider slit: the maximum becomes narrower and brighter.

With white light the central maximum is white, because every colour has a maximum at the centre. The side maxima are coloured, with blue at the inner edge and red at the outer edge, because red is diffracted through a larger angle than blue.

You are not asked for the intensity against angle graph.

The plane transmission diffraction grating

A grating has a very large number of equally spaced parallel slits. Let d be the spacing between adjacent slits (the grating spacing). If a grating has N lines per metre, d = 1 ÷ N.

Derivation of d sin θ = nλ (normal incidence)

1. A plane wave strikes the grating at 90°, so all slits are in phase.
2. Take waves from two adjacent slits leaving at angle θ to the normal.
3. A perpendicular from the upper slit to the lower wave’s path forms a right-angled triangle with hypotenuse d and angle θ.
4. The path difference is d sin θ.
5. Waves from all the slits reinforce when this equals a whole number of wavelengths: d sin θ = nλ, with n = 0, 1, 2, … the order of the maximum.

The n = 0 maximum lies straight ahead. Maxima at ±1, ±2 and so on appear at either side. Because sin θ has a largest value of 1, n can be no bigger than d ÷ λ, rounded down to a whole number.

Worked example 4: angles and number of orders

A grating has 600 lines per mm. Light of wavelength 532 nm hits it at normal incidence. Calculate the first and second order angles and the total number of maxima.

1. d = 1 ÷ (600 × 10³) = 1.67 × 10⁻⁶ m.
2. n = 1: sin θ = 532 × 10⁻⁹ ÷ 1.67 × 10⁻⁶ = 0.319, so θ = 18.6°.
3. n = 2: sin θ = 0.638, so θ = 39.7°.
4. Maximum order: d ÷ λ = 3.13, so n ≤ 3 (n = 3 is at 73.3°).
5. Maxima are at n = 0, ±1, ±2, ±3: 7 maxima.

Worked example 5: finding a wavelength

A grating of 400 lines per mm gives a second order maximum at 33.0°. Find λ.

1. d = 1 ÷ (400 × 10³) = 2.50 × 10⁻⁶ m.
2. λ = d sin θ ÷ n = 2.50 × 10⁻⁶ × sin 33.0° ÷ 2 = 6.81 × 10⁻⁷ m (681 nm, red).

Grating versus double slit

A grating gives sharp, bright maxima that are widely spaced, so angles are easy to measure. A double slit gives broad, faint fringes. White light through a grating gives a central white maximum and, on each side, spectra with violet nearest the centre and red furthest. Spectra of different orders can overlap.

Applications of gratings

• Measuring the wavelengths of light emitted by lamps and stars, and so identifying elements from their line spectra.
• X-ray diffraction by regular layers of atoms in crystals works on the same principle, with a much smaller spacing.

Common errors

• Dividing by the number of fringes instead of the number of spacings.
• Using lines per millimetre directly as d. Convert to metres per line first.
• Using θ from the wrong place: the angle is measured from the straight-through direction.
• Calculating a non-whole order and rounding up. Only whole orders exist, so round down.
• Saying a wider slit gives a wider central maximum: it is the narrower slit that spreads more.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), sections 3.5.6 and 3.5.7, Version 4.4 (for exams May/June 2020 onwards), published by OxfordAQA.