Why can you hear someone around a corner before you can see them? Because sound waves bend around obstacles far more readily than light does. This spreading of waves as they pass through gaps or around edges is diffraction, and how much they spread depends on a single comparison: the size of the gap against the wavelength of the wave.
The same waves pass through a wide gap and then through a gap about the size of one wavelength.
Before reading: through which gap do the waves spread out (diffract) more, the wide one or the narrow one? Write your prediction and one sentence explaining why.
Waves spread into the space behind a gap or obstacle
Diffraction is what lets a wave reach places it could not get to by travelling in a straight line, it bends around edges and fans out after passing through an opening.
When straight wavefronts meet a gap or the edge of an obstacle, the wavefronts curve at the edges and spread into the "shadow" region behind. This is diffraction. It happens with all waves, water, sound and light, and does not change the wave's wavelength, frequency or speed; only its direction of travel spreads out.
Diffraction is the spreading of waves as they pass through a gap or around an obstacle. The wave bends into the region behind the edge, but its wavelength, frequency and speed are unchanged, only its direction spreads out.
Pause, copy the highlighted definition into your book before moving on.
Did you get this? True or false: when a wave diffracts through a gap, its wavelength gets longer.
The gap-to-wavelength ratio controls the spreading
We just saw that waves spread behind a gap. That raises a question: what decides whether they barely spread or fan out widely? This card answers it → compare the gap width $w$ with the wavelength $\lambda$.
The closer the gap size is to one wavelength, the more dramatically the wave spreads, the controlling factor is the ratio of the gap width to the wavelength.
The amount of diffraction depends on the ratio $\dfrac{\lambda}{w}$ of wavelength to gap width:
Maximum spreading when $w \approx \lambda$ · little spreading when $w \gg \lambda$
For a fixed gap, a longer wavelength diffracts more; for a fixed wavelength, a narrower gap diffracts more.
Water waves of wavelength $2.0\ \text{cm}$ pass first through a $2.0\ \text{cm}$ gap and then through a $20\ \text{cm}$ gap. Through which gap do the waves diffract more, and by what factor do the gap-to-wavelength ratios differ?
The amount of diffraction depends on the gap-to-wavelength ratio. Spreading is greatest when the gap width is about one wavelength ($w \approx \lambda$); for a fixed gap, longer wavelengths diffract more, and for a fixed wavelength, narrower gaps diffract more.
Add the highlighted rule to your notes before the check below.
Quick check: Sound (long wavelength) and light (very short wavelength) pass through the same doorway. Which diffracts more, and why?
Pattern A, Reason then sketch
In a ripple tank, waves of wavelength $1.5\ \text{cm}$ are sent at two gaps: gap A is $1.5\ \text{cm}$ wide and gap B is $9\ \text{cm}$ wide. In your book: (1) calculate $w/\lambda$ for each gap; (2) state which gap produces more diffraction; (3) sketch the wavefront pattern emerging from each gap and label which is which.
Fill the gaps: Complete the account of diffraction, then press Check.
Diffraction is the of a wave as it passes through a gap or around an obstacle. The effect is most pronounced when the size of the gap is comparable to the of the wave. Making the gap narrower makes the spreading .
Dot point: Conduct a practical investigation to model how waves diffract through openings and around obstacles of different sizes
A practical investigation of diffraction through openings and around obstacles
We just learned that the gap-to-wavelength ratio controls spreading. That raises a question: how can you observe this for yourself and test different sized openings and obstacles? This card answers it → a ripple-tank investigation.
A ripple tank lets you watch diffraction happen in real time by sending straight water waves at barriers with gaps and obstacles of different sizes.
A practical to model diffraction through openings and around obstacles of different sizes:
The tank shows that diffraction is strongest when the gap (or obstacle) is comparable to the wavelength, and that waves bend noticeably around small obstacles but pass an obstacle much larger than $\lambda$ leaving a clear shadow behind it.
Model diffraction in a ripple tank: make straight wavefronts, then observe spreading through wide vs one-wavelength gaps and around small vs large obstacles, keeping wavelength constant as the control. Diffraction is strongest when the gap or obstacle is comparable in size to the wavelength.
Pause, copy the highlighted practical outline into your book before the check below.
Last check: Decide if the statement is true or false.
Waves bend noticeably around an obstacle that is small compared with their wavelength, but a much larger obstacle leaves a clear shadow.
Match them up: Match each ripple-tank observation to the condition that produces it.
Dot point: Conduct a practical investigation to model how waves diffract through openings and around obstacles of different sizes
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
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UnderstandBand 4(4 marks) 1. Describe what diffraction is, and explain why a wave passing through a gap about the size of its wavelength diffracts much more than the same wave passing through a very wide gap. Refer to a ripple-tank investigation in your answer.
1 mark: diffraction = spreading of waves through a gap/around an obstacle · 1 mark: amount depends on gap-to-wavelength ratio · 1 mark: max diffraction when $w \approx \lambda$, little when $w \gg \lambda$ · 1 mark: links to a ripple-tank observation
ApplyBand 4(3 marks) 2. Describe a practical investigation using a ripple tank to model how waves diffract through openings of different sizes, stating the independent and dependent variables and the expected result.
1 mark: describes generating straight wavefronts and passing them through a gap between two barriers · 1 mark: IV is the gap width, DV is the amount of spreading beyond the gap, with wavelength controlled · 1 mark: narrower gaps produce greater spreading, greatest when the gap is comparable to the wavelength
Dot point: Conduct a practical investigation to model how waves diffract through openings and around obstacles of different sizes
AnalyseBand 5(3 marks) 3. Sound can be heard clearly around the corner of an open doorway, but a person standing there cannot be seen. Explain this difference in terms of diffraction and the wavelengths involved.
1 mark: audible sound wavelengths are comparable to the doorway width, so sound diffracts strongly · 1 mark: visible light wavelengths are hundreds of thousands of times smaller than the doorway, so light diffracts negligibly · 1 mark: concludes that the amount of diffraction depends on the ratio of wavelength to gap size
Dot point: Conduct a practical investigation to model how waves diffract through openings and around obstacles of different sizes
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Diffraction is the spreading out of waves as they pass through a gap or around the edge of an obstacle. The amount of diffraction depends on the ratio of the gap width to the wavelength: spreading is greatest when the gap is about one wavelength wide ($w \approx \lambda$) and very small when the gap is much wider than the wavelength ($w \gg \lambda$). A wide gap lets the wavefronts pass almost straight through with only slight curving at the edges, whereas a gap about one wavelength across forces the wavefronts to fan out strongly into curved waves. In a ripple tank, sending straight water waves of fixed wavelength at an adjustable barrier shows exactly this: narrowing the gap to roughly one wavelength produces strongly curved, spreading wavefronts, confirming the dependence on the gap-to-wavelength ratio.
Q2 (3 marks): Set up a ripple tank with a straight vibrating bar to generate straight, parallel wavefronts travelling towards two barriers placed to leave an adjustable gap between them. The independent variable is the width of the gap, changed in steps by sliding the barriers. The dependent variable is the amount of spreading of the wavefronts beyond the gap, judged from the projected pattern or from a photograph taken from directly above. The frequency of the vibrating bar and the depth of the water must be controlled, since both would change the wavelength and so confound the result. The expected result is that when the gap is much wider than the wavelength the waves pass through with only slight spreading at the edges and the wavefronts stay largely straight, whereas as the gap is narrowed the spreading increases, and when the gap is comparable to the wavelength the waves emerge as almost semicircular fronts centred on the gap.
Q3 (3 marks): The amount by which a wave spreads on passing through a gap depends on the ratio of its wavelength to the width of the gap. Audible sound has wavelengths of roughly a few centimetres to several metres, which is comparable to the width of a doorway of about a metre. The condition for strong diffraction is therefore met, the sound spreads out widely on emerging, and it reaches a listener who is well off to one side and out of the direct line. Visible light has wavelengths of around $5 \times 10^{-7}\ \text{m}$, which is smaller than the doorway by a factor of about a million. The gap is enormously wider than the wavelength, so the diffraction is utterly negligible and the light continues in effectively straight lines, casting a sharp-edged shadow. The person standing round the corner therefore cannot be seen. The same physical rule governs both cases; only the scale of the wavelength relative to the opening differs.
You can now describe diffraction, predict how much a wave spreads from the gap-to-wavelength ratio, and outline a ripple-tank investigation. Next lesson combines two waves that overlap, the principle of superposition and the interference patterns it creates.
↩ Back to focus-area hubAt the start you predicted which gap spreads the waves more. The waves diffract more through the narrow gap, specifically when the gap is about one wavelength wide. If you guessed the wide gap, revisit Card 2, a much wider gap lets the waves pass almost straight through.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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