Interference with Sound and EM Waves and Practical Activity Revision Notes | A Level AQA Physics

1. Why Waves Interfere — The Core Idea

Drop two stones into a still pond at the same time. Where the ripples meet, some regions become calmer and others choppier. That is interference — and it is the clearest proof that waves can overlap and combine. The same physics governs sound in a concert hall, light through a double slit, and radio signals bouncing between buildings.

Interference arises directly from the principle of superposition: when two waves occupy the same point in space, the resultant displacement equals the vector sum of the individual displacements. This is not a special property of light or sound — it applies to all wave types.

The opposite situation — destructive interference — occurs when a crest from one wave meets a trough from the other. The waves cancel, producing a point of minimum (or zero) amplitude. The path difference here is an odd number of half-wavelengths: $\Delta = (n + \frac{1}{2})\lambda$.

This is why you cannot produce a stable interference pattern by shining two separate torches at a screen — their phase relationship fluctuates billions of times per second. A laser, or a single lamp behind a double slit, provides the coherence needed for a steady pattern.

2. Interference with Sound Waves

Sound is a longitudinal wave, but it obeys the same superposition rules as any other wave. Connect two loudspeakers to the same signal generator so they emit sound of identical frequency and in phase — you now have two coherent sources. Walk slowly in a line parallel to the speakers and you will hear the volume rise and fall repeatedly.

The loud regions are where path differences equal whole-number multiples of the wavelength (constructive interference). The quiet regions — sometimes near-silence — are where path differences equal half-integer multiples of the wavelength (destructive interference). The pattern is real and measurable with a microphone and oscilloscope.

A common classroom demonstration uses two speakers separated by about 1 m, driven at roughly 1 kHz. At that frequency, the wavelength in air is approximately 0.34 m, so the interference fringes are spaced widely enough to walk between them. Reducing the frequency increases the wavelength (since v = fλ and the speed of sound in air is approximately constant), which spreads the fringes further apart — a direct, audible demonstration of the wave equation in action.

3. Young's Double-Slit Experiment — Light as a Wave

In 1801, Thomas Young directed light through two narrow, closely-spaced slits and observed alternating bright and dark bands — fringes — on a screen beyond. This was decisive evidence that light behaves as a wave, because only waves can interfere. At the time, Newton's corpuscular (particle) model of light was dominant, and Young's result was deeply controversial.

Picture the setup: a monochromatic light source illuminates a single slit, which then illuminates the double slit. Each slit acts as a coherent secondary source (by Huygens' principle), and the two sets of circular wavefronts overlap on the screen. Where crests coincide, a bright fringe forms; where a crest meets a trough, a dark fringe forms.

Notice the relationships encoded in this formula. Increasing $D$ (moving the screen further away) spreads the fringes out — each fringe is wider. Decreasing $s$ (moving the slits closer together) also spreads the fringes further apart — fringe spacing is inversely proportional to slit separation. Using a longer wavelength (e.g. red light instead of blue) increases $w$. These predictions are all testable and all confirmed experimentally.

Worked example: Calculating fringe spacing in Young's experiment

In a Young's double-slit experiment, monochromatic light of wavelength 590 nm passes through two slits separated by 0.50 mm. The screen is placed 1.8 m from the slits. Calculate the fringe spacing.

1. $\lambda = 590 \text{ nm} = 590 \times 10^{-9} \text{ m}$ $s = 0.50 \text{ mm} = 0.50 \times 10^{-3} \text{ m} = 5.0 \times 10^{-4} \text{ m}$ $D = 1.8 \text{ m}$ — Unit conversion is a common source of error. Always convert nm to m and mm to m before substituting. [M1]
2. $w = \frac{\lambda D}{s}$ — State the formula explicitly — this earns the method mark even if arithmetic goes wrong later. [M1]
3. $w = \frac{590 \times 10^{-9} \times 1.8}{5.0 \times 10^{-4}}$ — Substitute with units attached to each value. [M1]
4. $590 \times 10^{-9} \times 1.8 = 1062 \times 10^{-9} = 1.062 \times 10^{-6}$ — 590 multiplied by 1.8: $590 \times 1.8 = 590 \times 1 + 590 \times 0.8 = 590 + 472 = 1062$. Then carry the $10^{-9}$ factor.
5. $w = \frac{1.062 \times 10^{-6}}{5.0 \times 10^{-4}} = \frac{1.062}{5.0} \times 10^{-6-(-4)} = 0.2124 \times 10^{-2} = 2.124 \times 10^{-3} \text{ m}$ — Dividing powers of ten: $10^{-6} \div 10^{-4} = 10^{-6+4} = 10^{-2}$. Then $1.062 \div 5.0 = 0.2124$.
6. $w = 2.1 \times 10^{-3} \text{ m} = 2.1 \text{ mm}$ — Round to 2 significant figures to match the precision of the given data. Include units. [A1]

Answer: $w = 2.1 \times 10^{-3}$ m (2.1 mm)

Young's experiment also allows the wavelength of light to be measured: rearrange the formula to give $\lambda = \frac{ws}{D}$. Measure $w$ from the fringe pattern, $s$ with a travelling microscope, and $D$ with a ruler. This is the basis of Required Practical 2.

4. The Diffraction Grating

A diffraction grating takes the principle of Young's experiment and scales it up dramatically. Instead of two slits, a grating has hundreds or thousands of equally-spaced slits per millimetre. The result is that constructive interference only occurs at very specific, sharply-defined angles — producing bright, narrow maxima that are far easier to measure precisely than the broad fringes from a double slit.

The condition for a bright maximum (order $n$) from a diffraction grating is derived from the same path-difference logic used for the double slit. For adjacent slits separated by distance $d$, the path difference to a distant screen at angle $\theta$ is $d\sin\theta$. For constructive interference, this must equal a whole number of wavelengths.

The zeroth order ($n = 0$) is always straight ahead regardless of wavelength — it is the undeviated beam. The first order ($n = 1$) appears at angle $\theta_1 = \arcsin(\lambda/d)$ on either side. Higher orders appear at larger angles. Because $\sin\theta$ cannot exceed 1, there is a maximum possible order: $n_{\text{max}} = \lfloor d/\lambda \rfloor$.

The zeroth order ($n = 0$) is always straight ahead regardless of wavelength — it is the undeviated beam. The first order ($n = 1$) appears at angle $\theta_1 = \arcsin(\lambda/d)$ on either side. Higher orders appear at larger angles. Because $\sin\theta$ cannot exceed 1, there is a maximum possible order: $n_{\text{max}} = \lfloor d/\lambda \rfloor$. Note that if $d/\lambda$ is exactly a whole number, that order would require $\theta = 90°$ and is not practically observable; in that edge case the highest usable order is $\lfloor d/\lambda \rfloor - 1$.

Worked example: Using the diffraction grating equation to find wavelength

A diffraction grating has 300 lines per mm. When monochromatic light is directed normally at the grating, the first-order maximum is observed at an angle of 10.2° from the central maximum. Calculate the wavelength of the light.

1. $N = 300 \text{ lines per mm} = 300 \times 10^3 \text{ lines per m} = 3.0 \times 10^5 \text{ lines m}^{-1}$ $d = \frac{1}{N} = \frac{1}{3.0 \times 10^5} = 3.33 \times 10^{-6} \text{ m}$ — Convert lines per mm to lines per metre first, then take the reciprocal. This step is where many students lose marks by forgetting the unit conversion. [M1]
2. $d\sin\theta = n\lambda$ $d = 3.33 \times 10^{-6} \text{ m},\quad \theta = 10.2°,\quad n = 1$ — State the formula and list known quantities. [M1]
3. $\lambda = \frac{d\sin\theta}{n} = \frac{3.33 \times 10^{-6} \times \sin(10.2°)}{1}$ — For $n = 1$, dividing by $n$ does not change the value, but write it explicitly for clarity.
4. $\sin(10.2°) = 0.1771$ — Use a calculator in degree mode. A common error is leaving the calculator in radian mode.
5. $\lambda = 3.33 \times 10^{-6} \times 0.1771 = 5.898 \times 10^{-7} \text{ m}$ $\lambda \approx 5.9 \times 10^{-7} \text{ m} = 590 \text{ nm}$ — $3.33 \times 0.1771 = 0.5898$, so $\lambda = 0.5898 \times 10^{-6} = 5.898 \times 10^{-7}$ m. This is in the yellow-orange region of the visible spectrum — a sanity check that the answer is physically reasonable. [A1]

Answer: $\lambda \approx 5.9 \times 10^{-7}$ m (590 nm) — yellow-orange light

5. Required Practical 2 — Investigating Interference

Required Practical 2 asks you to investigate interference effects using both Young's double-slit setup and a diffraction grating. The two methods complement each other: the double-slit gives a direct measurement of fringe spacing that can be used to calculate wavelength, while the grating gives a more precise angle measurement that yields a more accurate wavelength value.

When evaluating your results, consider the main sources of uncertainty. In the Young's experiment, the fringe spacing is small (typically 1–3 mm), so measuring across 10 fringes and dividing reduces the percentage uncertainty significantly. In the grating experiment, the main uncertainty comes from locating the exact centre of each maximum and measuring the angle accurately.

6. The Changing Understanding of Electromagnetic Radiation

The history of how physicists have understood light is one of the most instructive stories in science — a reminder that models are provisional and evidence drives revision. Newton's corpuscular model (light as a stream of particles) dominated for over a century, supported by his authority and by the fact that it explained reflection and refraction adequately.

Young's 1801 interference experiment was the first strong evidence that light must be a wave — particles cannot cancel each other out. Fresnel's mathematical wave theory in the 1810s–1820s consolidated this view, and Maxwell's electromagnetic theory (1865) provided a physical mechanism: light is an oscillating electric and magnetic field propagating through space. This wave model predicted a finite speed of light and the existence of the full electromagnetic spectrum, both subsequently confirmed.

The wave model appeared complete — until the photoelectric effect (observed in the late 19th century and explained by Einstein in 1905) showed that light also behaves as discrete packets of energy called photons. Neither the pure wave model nor the pure particle model is sufficient on its own. Modern physics describes light through quantum electrodynamics, which incorporates both wave-like and particle-like behaviour — a concept known as wave-particle duality.

For AQA, you need to appreciate this historical progression: corpuscular model $\rightarrow$ wave model (supported by interference and diffraction evidence) $\rightarrow$ photon model (supported by photoelectric effect) $\rightarrow$ wave-particle duality. Each transition was driven by experimental evidence that the existing model could not explain.

7. Exam Technique and Common Mistakes

Interference questions reward students who are precise with language and careful with unit conversions. The most common mark losses come from a small set of recurring errors — knowing them in advance is a significant advantage.

Across this lesson, the central thread has been the same: interference is the superposition of coherent waves, and the pattern of maxima and minima is determined entirely by path difference. Whether you are analysing sound from two loudspeakers, light through a double slit, or light through a diffraction grating, the conditions $\Delta = n\lambda$ (constructive) and $\Delta = (n + \frac{1}{2})\lambda$ (destructive) always apply. The historical development of the wave model of light — from Newton through Young and Maxwell to Einstein — shows how experimental evidence from interference and diffraction forced a fundamental revision of how physicists understand electromagnetic radiation.