Path Difference and Coherence Revision Notes | A Level AQA Physics

1. Why Two Waves Can Cancel Each Other Out

Two identical ripples spreading across a pond can meet and, at certain points, produce nothing at all — no displacement, no movement. That is not a coincidence; it is interference, and it is one of the most powerful pieces of evidence that light behaves as a wave.

To predict exactly where those calm spots and those amplified peaks appear, you need two ideas: path difference and coherence. Master these and the whole interference picture falls into place.

2. Path Difference — The Core Idea

Picture two loudspeakers side by side, both playing the same note. You walk across the room and notice the sound gets louder, then quieter, then louder again. The reason is that the distance from each speaker to your ear is different — one wave has travelled further than the other. That extra distance is the path difference.

The value of the path difference — measured in wavelengths — determines whether the two waves reinforce or cancel at that point. This is the rule that governs every interference pattern you will encounter at A-level.

2a. Constructive and Destructive Interference

When the path difference is a whole number of wavelengths (0, λ, 2λ, 3λ,...), the two waves arrive in phase: crest meets crest and trough meets trough. The amplitudes add together — this is constructive interference, producing a bright fringe or a loud region.

When the path difference is an odd number of half-wavelengths (λ/2, 3λ/2, 5λ/2,...), the waves arrive in antiphase: crest meets trough. The amplitudes cancel — this is destructive interference, producing a dark fringe or a quiet region.

Notice that destructive interference only produces complete cancellation when the two waves have equal amplitudes. In most A-level exam questions, this condition is assumed unless stated otherwise.

3. Coherence — Why the Source Matters

Here is the catch: you cannot produce a stable, observable interference pattern with just any two wave sources. Imagine two people clapping at slightly different rhythms — sometimes they clap together, sometimes they are out of step, and the pattern keeps shifting. The same thing happens with light sources that are not synchronised.

An ordinary light bulb emits photons in random bursts from millions of independent atoms. The phase relationship between any two points in the beam changes millions of times per second. Any interference fringes form and vanish so rapidly that the eye sees only a uniform glow — no pattern at all.

A laser produces light that is both monochromatic (single frequency) and highly coherent. Every photon is emitted in step with the others, so the phase difference between two points in the beam stays constant. This is why lasers are the standard source for interference and diffraction experiments at A-level.

4. Lasers as a Source of Monochromatic, Coherent Light

The word monochromatic means single-wavelength (or, more precisely, a very narrow band of wavelengths). White light contains a continuous spread of wavelengths, so each wavelength produces its own fringe pattern at a slightly different position. The patterns overlap and wash each other out, making fringes very difficult to observe clearly.

A laser solves both problems at once: it provides monochromatic light (so all fringes from a given order land at the same position) and coherent light (so the fringe pattern is stable over time). This makes laser light the ideal source for demonstrating interference and diffraction in the laboratory.

You are not required to describe how a laser works internally — the AQA specification explicitly excludes this. What you do need to know is what a laser produces and why those properties matter for interference experiments.

5. Laser Safety in Practical Work

The AQA specification requires you to show awareness of safety issues when using lasers. This is not a box-ticking exercise — laser light can cause permanent retinal damage in a fraction of a second, even from relatively low-power class 2 or class 3 lasers used in school laboratories.

• Never look directly into the laser beam or at its specular (mirror-like) reflections.
• Do not allow the beam to travel at eye level — set up the apparatus so the beam travels horizontally across a bench, not towards observers.
• Avoid placing reflective surfaces (jewellery, watches, metal rulers) in the path of the beam.
• Use the minimum power setting sufficient for the experiment.
• Ensure the beam is terminated by a non-reflective screen at the end of its path.
• Be aware of where the beam goes after diffraction or reflection — secondary beams can be just as hazardous as the primary beam.

In an exam question asking you to describe safety precautions, stating 'do not look into the beam' alone is unlikely to gain full marks. Examiners expect you to justify the precaution — for example, 'do not look directly into the beam because the focused spot on the retina can cause permanent damage before the blink reflex can respond.'

6. Interference and Diffraction with a Laser — Putting It Together

Now that you have path difference, coherence, and the properties of laser light in place, you can analyse what actually happens in a laser diffraction or double-slit experiment. The laser beam passes through a narrow slit or a pair of slits. Diffraction spreads the beam, and the two diffracted waves overlap in the region beyond the slits.

At any point on the screen, the path difference from the two slits determines whether that point is bright or dark. Because the laser is coherent, the phase difference between the two slits stays constant, so the bright and dark fringes remain fixed in position — you can observe and measure them.

The fringe spacing depends on the wavelength of the laser light, the slit separation, and the distance to the screen. Longer wavelengths produce wider fringe spacing; closer slits also produce wider fringe spacing. This relationship is captured in the double-slit equation $w = \frac{\lambda D}{s}$, where $w$ is the fringe spacing, $D$ is the distance to the screen, and $s$ is the slit separation.

Worked example: Calculating Fringe Spacing in a Double-Slit Experiment

A student shines a red laser (wavelength 650 nm) through a double slit with slit separation 0.40 mm. The screen is placed 1.8 m from the slits. Calculate the expected fringe spacing. [3 marks]

1. $w = \frac{\lambda D}{s}$ — [M1] Correct formula for double-slit fringe spacing. w = fringe spacing, λ = wavelength, D = distance to screen, s = slit separation.
2. $\lambda = 650 \times 10^{-9}$ m, $D = 1.8$ m, $s = 0.40 \times 10^{-3}$ m — Wavelength is given in nm: 1 nm = 10^{-9} m, so 650 nm = 650 × 10^{-9} m. Slit separation is given in mm: 1 mm = 10^{-3} m, so 0.40 mm = 0.40 × 10^{-3} m. Always convert to metres before substituting.
3. $w = \frac{650 \times 10^{-9} \times 1.8}{0.40 \times 10^{-3}}$ — [M1] Correct substitution with consistent SI units.
4. $650 \times 10^{-9} \times 1.8 = 1170 \times 10^{-9} = 1.170 \times 10^{-6}$ m — 650 × 1.8 = 1170, then carry the power of ten: × 10^{-9}.
5. $w = \frac{1.170 \times 10^{-6}}{0.40 \times 10^{-3}} = \frac{1.170 \times 10^{-6}}{4.0 \times 10^{-4}}$ — 0.40 × 10^{-3} = 4.0 × 10^{-4}. Dividing: 1.170 ÷ 4.0 = 0.2925, and 10^{-6} ÷ 10^{-4} = 10^{-2}.
6. $w = 0.2925 \times 10^{-2}$ m $= 2.9 \times 10^{-3}$ m $= 2.9$ mm — [A1] Answer to 2 significant figures, matching the precision of the data given. Always include units.

Answer: $w = 2.9 \times 10^{-3}$ m (2.9 mm)

Notice how the path difference idea underpins this result: the bright fringes occur where the path difference from the two slits is exactly 0, λ, 2λ, and so on. The geometry of the slits and screen translates those path difference conditions into physical positions on the screen.

7. Exam Technique and Common Mistakes

The path difference and coherence topic rewards precise language. Examiners award marks for specific terminology, and several common phrasings lose marks even when the underlying understanding is sound.

For extended-response questions on interference, structure your answer as: state the condition (path difference = nλ for constructive) → explain the phase relationship (waves arrive in phase) → link to the observable outcome (bright fringe / maximum amplitude). This three-part chain reliably picks up all available marks.