What is interference in the context of wave physics?

Interference is a phenomenon that occurs when two or more waves overlap and combine to form a new wave pattern. This can result in regions of constructive interference, where the waves add together to produce a larger amplitude, or destructive interference, where the waves cancel each other out. Understanding interference is crucial in the study of wave behavior, particularly in optics and acoustics, and is a key concept in the Cambridge International A Levels Physics curriculum.

How does the principle of superposition relate to interference?

The principle of superposition states that when two or more waves meet at a point, the resultant displacement is the sum of the displacements of the individual waves. This principle is fundamental to understanding interference, as it explains how waves can constructively or destructively interfere with each other. In the context of Cambridge International A Levels Physics, superposition is essential for analyzing wave interactions in various scenarios, such as light waves in optics.

What are the conditions necessary for constructive and destructive interference?

For constructive interference to occur, the waves must be in phase, meaning their peaks and troughs align, resulting in a wave with greater amplitude. Conversely, destructive interference happens when the waves are out of phase by half a wavelength, causing the peaks of one wave to align with the troughs of another, effectively canceling each other out. These conditions are crucial for experiments and applications involving wave interference, such as in the design of noise-canceling headphones.

Can you explain the difference between coherent and incoherent sources in interference?

Coherent sources are wave sources that maintain a constant phase relationship, which is essential for producing stable interference patterns. Incoherent sources, on the other hand, have random phase differences, making it difficult to observe clear interference patterns. In the Cambridge International A Levels Physics syllabus, understanding the distinction between these sources is important for experiments involving light and sound waves, such as in the double-slit experiment.

What is the significance of the double-slit experiment in understanding interference?

The double-slit experiment is a classic demonstration of wave interference, where light passing through two closely spaced slits creates an interference pattern of bright and dark fringes on a screen. This experiment provides strong evidence for the wave nature of light and illustrates the principles of superposition and interference. It is a pivotal experiment in the Cambridge International A Levels Physics curriculum, highlighting the dual nature of light and the fundamental concepts of wave behavior.

Why is "Mixing up $a$, $D$ in fringe-spacing formula" a common mistake in Interference?

Why it happens: Similar variables. How to avoid it: a = SLIT SEPARATION (small, mm). D = SCREEN DISTANCE (large, m). D / a — wavelength × screen / slit. Source: 9702 Examiner Reports 2022-2024

Cambridge International A Levels Physics (9702)

Interference

0% Complete

3 tasks remaining

Short Notes - Interference

Detailed Study Notes

Detailed notes on Superposition for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Interference Study Notes — Cambridge International A Level Physics 9702 Superposition (2025-2027 syllabus)

Constructive and destructive interference. Young's slits. Fringe spacing.

At a glance

Constructive: path diff = $n\lambda$ .
Destructive: path diff = $(n + 1/2)\lambda$ .
Fringe spacing $y = \lambda D / a$ .
Coherent sources required.
Same polarisation for EM.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

8.3.1 — State conditions for interference.
8.3.2 — Apply path-difference rules.
8.3.3 — Use Young's slits formula.

Interference basics

Superposition produces fringes.

Constructive. Crests align with crests → bigger amplitude. Path difference = $n\lambda$ .

Destructive. Crest meets trough → cancellation. Path difference = $(n + 1/2)\lambda$ .

Coherent sources required.

Same frequency.
Constant phase difference.
Approx same amplitude (for visible contrast).
Same polarisation (for EM).

Cambridge tip. Mark scheme rewards COHERENCE condition explicitly.

Figure 1 — Constructive and destructive superposition

Superposition simply adds the displacements at every point. The energy is not destroyed at a dark fringe — it is redistributed into the bright ones.

Figure 2 — Path difference decides constructive or destructive interference

What matters is the extra distance one wave travels, measured in wavelengths. A whole number puts them back in step; a half-integer puts them exactly out of step.

Constructive: $n\lambda$ .
Destructive: $(n + 1/2)\lambda$ .
Coherence required.

See the full worked example for interference →

Young's double-slit experiment

Bright/dark fringe pattern.

Setup. Monochromatic coherent light through two narrow slits, projected onto screen.

Pattern. Equally-spaced bright fringes (constructive) and dark fringes (destructive).

Fringe spacing. $y = \frac{\lambda D}{a}$ where $a$ = slit separation, $D$ = slit-to-screen distance.

Example. $a = 0.5$ mm, $D = 2$ m, $\lambda = 600$ nm: $y = (600 \times 10^{-9})(2)/(0.5 \times 10^{-3}) = 2.4 \times 10^{-3}$ m = 2.4 mm.

Wavelength measurement. Rearrange to find $\lambda = ya/D$ .

Cambridge tip. Mark scheme rewards correct variable identification.

Figure 3 — Young's double-slit pattern

Light from both slits reaches every fringe — it is the difference in the two path lengths that decides whether that point is bright or dark.

Figure 4 — Worked example of fringe spacing

Fringes are only millimetres apart, which is why the slit separation must be small and the screen distance large.

Figure 5 — What changes the fringe spacing

Blue light gives narrower fringes than red because its wavelength is shorter — a quick check that you have the formula the right way up.

$y = \lambda D / a$ .
$a$ small, $D$ large.
Standard $\lambda$ measurement.

See the full worked example for interference →

Quick recap

Constructive vs destructive.
Coherence required.
$y = \lambda D / a$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Path-difference rules.
Coherence conditions.
$y = \lambda D / a$ .

How it’s examined

Interference on every Paper 1 or 2. Most-tested: Young's slits calculation (7 marks), conditions (5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Interference

Step-by-step solutions to past-paper-style questions on interference, written exactly the way a tutor would explain them at the board.

1Young's double-slit (7 marks)
Extended• Adapted from 9702/22 May/Jun 2024• Young's slits
Question
In Young's experiment, slit separation 0.50 mm, screen 2.0 m away. Light wavelength 600 nm. Find fringe spacing. (7 marks)
Step-by-step solution
1. Step 1
  Fringe spacing $y = \lambda D / a$ .
  $y = \dfrac{(600 \times 10^{-9})(2.0)}{0.50 \times 10^{-3}} = 2.4 \times 10^{-3} \text{ m}$
Answer
$y = 2.4$ mm.
2Coherence (5 marks)
Extended• coherence
Question
State the conditions required for observable interference fringes. (5 marks)
Step-by-step solution
1. Step 1
  Same frequency (constant relative phase).
2. Step 2
  Constant phase difference (coherent).
3. Step 3
  Similar amplitudes (for clear contrast).
4. Step 4
  Same plane of polarisation (for EM waves).
Answer
Coherent (same freq, constant phase diff), similar amplitudes, same polarisation.
3Path difference (6 marks)
Extended• path difference
Question
State the path differences for (a) constructive, (b) destructive interference with wavelength $\lambda$ . (6 marks)
Step-by-step solution
1. Step 1
  (a) Constructive (bright fringe).
  $\text{Path difference} = n\lambda, \quad n = 0, 1, 2, \ldots$
2. Step 2
  (b) Destructive (dark fringe).
  $\text{Path difference} = (n + \tfrac{1}{2})\lambda$
Answer
Constructive: $n\lambda$ . Destructive: $(n + 1/2)\lambda$ .

Key Formulae — Interference

The formulae you need to memorise for interference on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Fringe spacing (Young's slits)
$y = \dfrac{\lambda D}{a}$
$y$
fringe spacing
$\lambda$
wavelength
$D$
slit-to-screen distance
$a$
slit separation
When to use
Young's double-slit experiment.

Key Definitions and Keywords — Interference

Definitions to memorise and the exact keywords mark schemes credit for interference answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Interference
Examiner keyword
Superposition of two coherent waves producing alternating regions of high and low amplitude.
Constructive interference
Examiner keyword
Waves in phase reinforce → high amplitude. Path difference = $n\lambda$ .
Destructive interference
Examiner keyword
Waves out of phase by $\pi$ cancel → zero (or low) amplitude. Path difference = $(n + 1/2)\lambda$ .
Coherent sources
Examiner keyword
Sources with constant phase difference (same frequency, fixed timing).

Common Mistakes and Misconceptions — Interference

The traps other students keep falling into on interference questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Mixing up $a$ , $D$ in fringe-spacing formula
9702 Examiner Reports 2022-2024
Why it happens
Similar variables.
How to avoid it
$a$ = SLIT SEPARATION (small, mm). $D$ = SCREEN DISTANCE (large, m). $\lambda D / a$ — wavelength × screen / slit.

Interference — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge International A Levels Physics (9702)

Interference

0% Complete

3 tasks remaining

Short Notes - Interference

Detailed Study Notes

Detailed notes on Superposition for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Interference Study Notes — Cambridge International A Level Physics 9702 Superposition (2025-2027 syllabus)

Constructive and destructive interference. Young's slits. Fringe spacing.

At a glance

Constructive: path diff = $n\lambda$ .
Destructive: path diff = $(n + 1/2)\lambda$ .
Fringe spacing $y = \lambda D / a$ .
Coherent sources required.
Same polarisation for EM.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

8.3.1 — State conditions for interference.
8.3.2 — Apply path-difference rules.
8.3.3 — Use Young's slits formula.

Interference basics

Superposition produces fringes.

Constructive. Crests align with crests → bigger amplitude. Path difference = $n\lambda$ .

Destructive. Crest meets trough → cancellation. Path difference = $(n + 1/2)\lambda$ .

Coherent sources required.

Same frequency.
Constant phase difference.
Approx same amplitude (for visible contrast).
Same polarisation (for EM).

Cambridge tip. Mark scheme rewards COHERENCE condition explicitly.

Figure 1 — Constructive and destructive superposition

Superposition simply adds the displacements at every point. The energy is not destroyed at a dark fringe — it is redistributed into the bright ones.

Figure 2 — Path difference decides constructive or destructive interference

What matters is the extra distance one wave travels, measured in wavelengths. A whole number puts them back in step; a half-integer puts them exactly out of step.

Constructive: $n\lambda$ .
Destructive: $(n + 1/2)\lambda$ .
Coherence required.

See the full worked example for interference →

Young's double-slit experiment

Bright/dark fringe pattern.

Setup. Monochromatic coherent light through two narrow slits, projected onto screen.

Pattern. Equally-spaced bright fringes (constructive) and dark fringes (destructive).

Fringe spacing. $y = \frac{\lambda D}{a}$ where $a$ = slit separation, $D$ = slit-to-screen distance.

Example. $a = 0.5$ mm, $D = 2$ m, $\lambda = 600$ nm: $y = (600 \times 10^{-9})(2)/(0.5 \times 10^{-3}) = 2.4 \times 10^{-3}$ m = 2.4 mm.

Wavelength measurement. Rearrange to find $\lambda = ya/D$ .

Cambridge tip. Mark scheme rewards correct variable identification.

Figure 3 — Young's double-slit pattern

Light from both slits reaches every fringe — it is the difference in the two path lengths that decides whether that point is bright or dark.

Figure 4 — Worked example of fringe spacing

Fringes are only millimetres apart, which is why the slit separation must be small and the screen distance large.

Figure 5 — What changes the fringe spacing

Blue light gives narrower fringes than red because its wavelength is shorter — a quick check that you have the formula the right way up.

$y = \lambda D / a$ .
$a$ small, $D$ large.
Standard $\lambda$ measurement.

See the full worked example for interference →

Quick recap

Constructive vs destructive.
Coherence required.
$y = \lambda D / a$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Path-difference rules.
Coherence conditions.
$y = \lambda D / a$ .

How it’s examined

Interference on every Paper 1 or 2. Most-tested: Young's slits calculation (7 marks), conditions (5 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Interference

Step-by-step solutions to past-paper-style questions on interference, written exactly the way a tutor would explain them at the board.

1Young's double-slit (7 marks)
Extended• Adapted from 9702/22 May/Jun 2024• Young's slits
Question
In Young's experiment, slit separation 0.50 mm, screen 2.0 m away. Light wavelength 600 nm. Find fringe spacing. (7 marks)
Step-by-step solution
1. Step 1
  Fringe spacing $y = \lambda D / a$ .
  $y = \dfrac{(600 \times 10^{-9})(2.0)}{0.50 \times 10^{-3}} = 2.4 \times 10^{-3} \text{ m}$
Answer
$y = 2.4$ mm.
2Coherence (5 marks)
Extended• coherence
Question
State the conditions required for observable interference fringes. (5 marks)
Step-by-step solution
1. Step 1
  Same frequency (constant relative phase).
2. Step 2
  Constant phase difference (coherent).
3. Step 3
  Similar amplitudes (for clear contrast).
4. Step 4
  Same plane of polarisation (for EM waves).
Answer
Coherent (same freq, constant phase diff), similar amplitudes, same polarisation.
3Path difference (6 marks)
Extended• path difference
Question
State the path differences for (a) constructive, (b) destructive interference with wavelength $\lambda$ . (6 marks)
Step-by-step solution
1. Step 1
  (a) Constructive (bright fringe).
  $\text{Path difference} = n\lambda, \quad n = 0, 1, 2, \ldots$
2. Step 2
  (b) Destructive (dark fringe).
  $\text{Path difference} = (n + \tfrac{1}{2})\lambda$
Answer
Constructive: $n\lambda$ . Destructive: $(n + 1/2)\lambda$ .

Key Formulae — Interference

The formulae you need to memorise for interference on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Fringe spacing (Young's slits)
$y = \dfrac{\lambda D}{a}$
$y$
fringe spacing
$\lambda$
wavelength
$D$
slit-to-screen distance
$a$
slit separation
When to use
Young's double-slit experiment.

Key Definitions and Keywords — Interference

Definitions to memorise and the exact keywords mark schemes credit for interference answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Interference
Examiner keyword
Superposition of two coherent waves producing alternating regions of high and low amplitude.
Constructive interference
Examiner keyword
Waves in phase reinforce → high amplitude. Path difference = $n\lambda$ .
Destructive interference
Examiner keyword
Waves out of phase by $\pi$ cancel → zero (or low) amplitude. Path difference = $(n + 1/2)\lambda$ .
Coherent sources
Examiner keyword
Sources with constant phase difference (same frequency, fixed timing).

Common Mistakes and Misconceptions — Interference

The traps other students keep falling into on interference questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Mixing up $a$ , $D$ in fringe-spacing formula
9702 Examiner Reports 2022-2024
Why it happens
Similar variables.
How to avoid it
$a$ = SLIT SEPARATION (small, mm). $D$ = SCREEN DISTANCE (large, m). $\lambda D / a$ — wavelength × screen / slit.

Interference — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Interference

Short Notes - Interference

Detailed Study Notes

At a glance

What you’ll learn

Interference basics

Young's double-slit experiment

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Young's double-slit (7 marks)

2Coherence (5 marks)

3Path difference (6 marks)

Fringe spacing (Young's slits)

Interference

Constructive interference

Destructive interference

Coherent sources

✕Mixing up aaa, DDD in fringe-spacing formula

Interference — frequently asked questions

Interference

Short Notes - Interference

Detailed Study Notes

At a glance

What you’ll learn

Interference basics

Young's double-slit experiment

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Young's double-slit (7 marks)

2Coherence (5 marks)

3Path difference (6 marks)

Fringe spacing (Young's slits)

Interference

Constructive interference

Destructive interference

Coherent sources

✕Mixing up aaa, DDD in fringe-spacing formula

Interference — frequently asked questions

✕Mixing up $a$ , $D$ in fringe-spacing formula

✕Mixing up $a$ , $D$ in fringe-spacing formula