OCR A Level Physics A: Superposition, interference and the double slit (H556)

This guide covers section 4.4.3 (Superposition), outcomes (a) to (g), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. It sits in Module 4, Electrons, waves and photons. The specification says that d sin θ = nλ and diffraction gratings are assessed at A level only, so the grating section is marked that way. Stationary waves, polarisation, refraction and the electromagnetic spectrum belong to other units and are not repeated here.

Pair this guide with the revision notes and the practice questions. The hub and checklist show where this unit sits, and the free 10-minute diagnostics help you find weak spots. For uncertainty methods used in the experiments, see units, measurement, uncertainty and vectors and practical endorsement skills.

What this unit covers

Outcome	What you must be able to do
4.4.3(a)(i)	State and apply the principle of superposition
4.4.3(a)(ii)	Describe techniques and procedures for superposition experiments with sound, light and microwaves (PAG5)
4.4.3(b)	Use graphical methods to show superposition
4.4.3(c)	Explain interference, coherence, path difference and phase difference
4.4.3(d)	Give the conditions for constructive and destructive interference in terms of path difference and phase difference
4.4.3(e)	Explain two-source interference with sound and microwaves
4.4.3(f)	Describe Young’s double-slit experiment with visible light, and why it confirmed the wave nature of light
4.4.3(g)(i)	Use λ = ax/D for all waves where a ≪ D
4.4.3(g)(ii)	Describe how to find the wavelength of light with a double slit and with a diffraction grating, using d sin θ = nλ (the d sin θ = nλ and grating parts are A level only)

4.4.3(a)(i) and (b): the principle of superposition

Principle of superposition: at any point where waves overlap, the net displacement equals the signed sum of the displacements each wave would cause alone. Displacements have direction, so a displacement of +3 cm and one of −2 cm add to +1 cm. The waves then carry on as if they had never met.

Graphical method

1. Draw both waves on the same axes, using the same horizontal scale.
2. Pick points along the axis, for example every quarter of a wavelength.
3. At each point read both displacements, with signs, and add them.
4. Plot the sums and join them with a smooth curve.

For pulses, slide one pulse across the other in steps and add at each step. Where a crest meets a trough the pulse heights subtract.

Worked example 1. Wave 1 has displacement y₁ = 3.0 sin θ (cm). Wave 2 has the same frequency, amplitude 2.0 cm and leads wave 1 by 90°, so y₂ = 2.0 sin(θ + 90°). Find the resultant.

θ	y₁ / cm	y₂ / cm	y₁ + y₂ / cm
0°	0.0	2.0	2.0
60°	2.6	1.0	3.6
120°	2.6	−1.0	1.6
180°	0.0	−2.0	−2.0
240°	−2.6	−1.0	−3.6
300°	−2.6	1.0	−1.6
360°	0.0	2.0	2.0

The sums form a sine curve of the same frequency. Its greatest value is about 3.6 cm, which is larger than 3.0 cm but smaller than 3.0 + 2.0 = 5.0 cm. Waves that are neither in phase nor in antiphase combine partially.

Two limiting cases matter. In phase, amplitudes add (3.0 + 2.0 = 5.0 cm). In antiphase they subtract (3.0 − 2.0 = 1.0 cm). If the amplitudes are equal in antiphase, the resultant is zero at every point.

4.4.3(c) and (d): interference, coherence, path and phase difference

Interference is the superposition of waves that gives a pattern of reinforcement and cancellation. Constructive interference gives a larger resultant amplitude. Destructive interference gives a smaller one, and complete cancellation if the amplitudes are equal.

Coherent sources have a constant phase difference, which requires the same frequency. Two separate filament lamps are not coherent: their light comes from atoms emitting in short, random bursts, so the phase difference changes too fast to see a steady pattern. A common way to get coherent sources is to split one source, as the double slit does. Equal amplitudes give the clearest cancellation.

Path difference is the extra distance one wave travels compared with the other to reach the same point. Phase difference is how far ahead one wave is in its cycle compared with the other. They are linked by:

phase difference (radians) = 2π × path difference / λ

phase difference (degrees) = 360° × path difference / λ

For two sources that emit in phase:

Interference	Path difference	Phase difference
Constructive	nλ (n = 0, 1, 2, …)	2πn rad (360°n)
Destructive	(n + ½)λ	(2n + 1)π rad (180°, 540°, …)

If the sources are in antiphase, the conditions swap: a path difference of nλ now gives destructive interference.

4.4.3(e): two-source interference with sound and microwaves

Sound. Connect two loudspeakers to one signal generator so they are coherent and in phase. Walk along a line parallel to the speakers, or move a microphone connected to an oscilloscope. The signal alternates between loud (maxima) and quiet (minima).

Worked example 2. The speakers emit 850 Hz and the speed of sound is 340 m s⁻¹. Point P is 2.60 m from one speaker and 3.40 m from the other. Point Q is 2.50 m and 3.10 m away. Describe the sound at each.

λ = v/f = 340 / 850 = 0.40 m

P: path difference = 3.40 − 2.60 = 0.80 m = 2.0λ, a whole number, so loud (constructive).

Q: path difference = 3.10 − 2.50 = 0.60 m = 1.5λ, a half-integer, so quiet (destructive). The phase difference is 2π × 1.5 = 3π rad (540°).

Microwaves. A transmitter faces a metal plate containing two narrow gaps. The gaps act as coherent sources. A probe receiver, connected to a meter, is moved along an arc or line beyond the plate. The reading rises and falls, with maxima where the path difference is nλ.

Technique and procedure (PAG5)

• Use one signal generator or one transmitter so the sources stay coherent.
• Measure positions with a ruler or metre rule and interpolate between scale markings.
• Move the detector slowly and locate each maximum or minimum, then repeat to check the position.
• Measure across several maxima and divide by the number of spacings.
• Keep reflecting objects away, since reflections add extra waves.

4.4.3(f): Young’s double-slit experiment

Light from a single narrow slit (or a laser) falls on two slits very close together. Diffraction at each slit spreads the light so the two beams overlap, and the two slits act as coherent sources because they are lit by the same wavefront. Alternating bright and dark fringes form on a screen. Bright fringes occur where the path difference from the slits is nλ, and dark ones where it is (n + ½)λ.

This was a classical confirmation of the wave nature of light. Newton favoured a particle (corpuscular) picture, while Huygens developed a wave picture. Interference, with cancellation of light by light, can be explained with waves, and particle models of the time could not account for it.

With white light the central fringe is white, because every wavelength has zero path difference there. The other fringes are coloured because each colour has a different spacing.

4.4.3(g)(i): λ = ax/D

For slit separation a, slit-to-screen distance D, and fringe spacing x (centre of one bright fringe to the centre of the next):

λ = ax/D, valid when a ≪ D.

Derivation. A point on the screen at angle θ to the central line has path difference a sin θ between the slits. For small θ, sin θ ≈ tan θ ≈ θ (in radians). A bright fringe at distance y from the centre has θ ≈ y/D, and the first bright fringe needs a θ = λ. So y = λD/a, and since fringes are equally spaced, x = λD/a.

The spacing grows with λ and D and shrinks as a increases.

Worked example 3. In a laser double-slit experiment a = 0.25 mm and D = 3.10 m. Twelve fringe spacings cover 79.2 mm. Find λ.

x = 79.2 / 12 = 6.60 mm

λ = ax/D = (0.25 × 10⁻³ × 6.60 × 10⁻³) / 3.10 = 5.3 × 10⁻⁷ m (530 nm, to 2 s.f.)

Measuring across 12 spacings matters. If the ruler reading uncertainty is ±0.5 mm, a single spacing of 6.6 mm has a percentage uncertainty of about 7.6%, whereas 79.2 mm has about 0.63%.

Worked example 4: when does a ≪ D fail? Microwaves of frequency 10.0 GHz (λ = c/f = 3.00 × 10⁸ / 10.0 × 10⁹ = 3.00 cm) pass through two gaps 12.0 cm apart, and the detector line is 1.50 m beyond.

Formula: x = λD/a = (0.0300 × 1.50) / 0.120 = 0.375 m.

Exact first maximum: sin θ = λ/a = 0.250, so θ = 14.5° and y = D tan θ = 1.50 × 0.2582 = 0.387 m.

The formula is about 3% out, because a is not much smaller than D here. For light, a ~ 10⁻⁴ m and D ~ 1 m, so the formula is very accurate.

4.4.3(g)(ii): finding the wavelength of light

Double slit. Place the laser, slits and screen on a bench. Measure D with a metre rule or tape, and the slit separation a from the slit slide (or a microscope). Measure the distance across several fringes and divide to find x. Use λ = ax/D. A graph of x against D has gradient λ/a, which reduces random error. Take care with lasers: never look along the beam, and avoid reflections into eyes.

Diffraction grating (A level only). A grating has many equally spaced slits. For light at normal incidence, adjacent slits give a path difference d sin θ, and maxima occur when this is a whole number of wavelengths:

d sin θ = nλ

Here d is the slit spacing (1/N, where N is lines per metre) and n is the order. Because sin θ cannot exceed 1, the highest order is the largest whole number below d/λ. Maxima from a grating are sharper than double-slit fringes, since many beams must be in phase, so angles can be measured more precisely.

Worked example 5. A grating has 750 lines per mm. The second-order maximum is at 39.0°. Find λ.

d = 1 / (750 × 10³) = 1.333 × 10⁻⁶ m

λ = d sin θ / n = (1.333 × 10⁻⁶ × sin 39.0°) / 2 = 4.2 × 10⁻⁷ m (420 nm)

Highest order: d/λ = 1.333 × 10⁻⁶ / 4.2 × 10⁻⁷ ≈ 3.2, so 3.

Procedure. Shine a laser through the grating at normal incidence onto a screen a distance L away. Measure the distance y from the central maximum to the order n spot. Then tan θ = y/L gives θ, and λ = d sin θ / n. For a grating, θ is large, so do not replace sin θ with tan θ. Averaging the angles on the left and right of the centre reduces error.

With white light, each order other than the zeroth is a spectrum with violet nearest the centre. For two slits (d = a) and small angles, d sin θ = nλ reduces to the fringe spacing in λ = ax/D.

Common errors

• Adding amplitudes instead of signed displacements.
• Mixing up path difference and phase difference. Path difference is in metres, phase difference in radians or degrees.
• Using nλ for destructive interference when the sources are in phase.
• Using the distance from the centre to the nth fringe as x. Spacing x is between neighbouring fringes.
• Leaving a in mm or λ in nm without converting to metres.
• Using lines per millimetre as d. Convert to metres, then take the reciprocal.
• Rounding the highest grating order up. It must be a whole number below d/λ.

Official syllabus

OCR, Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, section 4.4.3 Superposition.