Diffraction of Waves
In 2019, CSIRO engineers operating the Parkes radio telescope 'Murriyang' measured the diffraction limit for their 64 m dish at a working frequency of 2.4 GHz (λ ≈ 0.125 m). The ratio λ/d = 0.125/64 ≈ 1.95×10⁻³ rad gives a minimum resolvable angle of θ ≈ 2.4×10⁻³ rad = 0.14° — meaning two radio sources closer than that angle appear as one blurred source to the telescope.
Water waves pass through a narrow gap in a sea wall. Predict what the wavefronts look like on the other side. Does the width of the gap make a difference? Write your prediction.
Warm-up — diffraction is most noticeable when:
Know
- Diffraction: waves bend around obstacles and spread through gaps
- Maximum diffraction when $\lambda \approx$ gap width $d$
- Longer wavelengths diffract more than shorter wavelengths
Understand
- Why $\lambda/d$ ratio determines the extent of diffraction
- Why sound diffracts easily around corners but light does not
- The Huygens model of diffraction
Can Do
- Describe how changing gap size or wavelength affects diffraction
- Apply diffraction to real-world contexts
- Draw diffraction patterns for various gap/wavelength ratios
Core Content
Stand next to a doorway and clap your hands — someone around the corner hears the sound clearly, even though they have no line of sight to you. Now shine a torch through the same doorway: the beam travels in a straight line and no light curves around the corner. Both scenarios involve waves passing through a gap roughly 1 m wide, yet the outcomes are opposite. The difference is the wavelength-to-gap ratio.
The extent of diffraction is governed by the ratio of wavelength $\lambda$ to gap/obstacle size $d$:
- $\lambda \gg d$: Wave spreads almost as a semicircle — maximum diffraction.
- $\lambda \approx d$: Significant diffraction — waves spread noticeably around edges.
- $\lambda \ll d$: Waves mostly pass through straight with little spreading — minimal diffraction.
This explains why AM radio (wavelength ~200–600 m) diffracts around hills, while FM (~3 m) and visible light (~500 nm) do not bend around macroscopic obstacles.
Diffraction is the spreading of waves around obstacles or through gaps; it is greatest when wavelength $\lambda$ is approximately equal to the gap width $d$ ($\lambda \approx d$). All waves diffract; longer wavelengths diffract more than shorter wavelengths for the same gap.
Pause — copy the highlighted rule into your book before moving on.
Which gap produces the most diffraction for the same wavelength wave?
We just saw that diffraction is greatest when $\lambda \approx d$. That raises a question: what physical mechanism causes wavefronts to curve after passing through a gap? This card answers it → Huygens' principle: every point on a wavefront acts as a new circular wavelet source.
Huygens (1678) explained diffraction by treating every point on a wavefront as a new source of circular wavelets. The next wavefront is the envelope of all these secondary wavelets. When a wavefront reaches a gap, only the portion in the gap generates secondary wavelets — these spread in all forward directions, creating the characteristic curved pattern behind the gap.
After passing through a small gap ($\lambda \approx d$), wavefronts become nearly circular arcs centred on the gap. After a wide gap, the central portion remains flat with curved edges.
Huygens' principle: every point on a wavefront acts as a secondary source of circular wavelets; the new wavefront is their envelope. A gap of width $\approx \lambda$ produces near-circular wavefronts behind it; a gap much wider than $\lambda$ produces wavefronts that are flat in the centre with curved edges.
Add the highlighted Huygens principle to your notes before the check below.
Visible light does not diffract because it is not a mechanical wave.
Narrowing the gap for the same wave increases the amount of diffraction observed.
Activities
AM radio at 600 kHz has wavelength approximately 500 m; FM at 100 MHz has wavelength approximately 3 m. Explain why AM reception is reliably received behind a mountain while FM is often blocked.
For a wave of wavelength 2 cm, describe the diffraction pattern for each gap width: (a) 0.5 cm, (b) 2 cm, (c) 20 cm, (d) 200 cm.
A sonar system uses sound pulses to image underwater objects. Explain why using a higher-frequency (shorter wavelength) sound improves resolution of small objects. Your answer should refer to diffraction.
Which wave type does NOT diffract around everyday obstacles due to its very short wavelength?
A wave of wavelength 10 m passes through a gap of width 0.1 m. The $\lambda/d$ ratio is 100. The diffraction will be:
Huygens' principle explains diffraction by treating each point on a wavefront as:
UnderstandBand 3(3 marks) 1. Explain what diffraction is and describe the conditions under which it is most noticeable.
ApplyBand 4(3 marks) 2. A ripple tank has water waves of wavelength 3 cm. Describe and compare the diffraction patterns produced when the gap is (a) 2 cm and (b) 20 cm wide.
AnalyseBand 5(4 marks) 3. Explain why sound diffracts around a corner but visible light does not, using wavelength values to support your answer.
Show all answers
Short Answer — Model Answers
Q1 (3 marks): Diffraction is the spreading of waves around obstacles and through gaps. It is most noticeable when the wavelength is similar in size to the gap width or obstacle ($\lambda \approx d$). When $\lambda \ll d$ diffraction is negligible; when $\lambda \gg d$ the wave spreads broadly.
Q2 (3 marks): (a) 2 cm gap ≈ 3 cm wavelength — diffraction is significant; wavefronts spread in broad curves behind the gap, nearly semicircular. (b) 20 cm gap >> 3 cm wavelength — the central section passes relatively straight through, with only slight spreading at the edges of the gap.
Q3 (4 marks): Sound has wavelengths of roughly 0.02–20 m (at 17 m/s–20 000 Hz, 340 m/s base). Typical doorway ~1 m ≈ sound wavelengths → significant diffraction around corners. Visible light has wavelengths ~400–700 nm = $4 \times 10^{-7}$ to $7 \times 10^{-7}$ m; a doorway is 10⁶ times larger than the wavelength ($\lambda \ll d$) → negligible diffraction; light travels in straight lines past the doorway.
CSIRO's Parkes telescope 'Murriyang' working at 2.4 GHz (λ ≈ 0.125 m) with a 64 m dish cannot resolve two radio sources closer than θ ≈ λ/d ≈ 2.4×10⁻³ rad = 0.14°. This is a direct consequence of diffraction: the wave bends around the dish edge, smearing the wavefront. No signal processing can recover information that diffraction has already blurred.
Your Think First prediction was about water waves through a sea-wall gap. The wavefronts do spread in curved arcs behind the gap — nearly semicircular when the gap ≈ wavelength, nearly straight when the gap ≫ wavelength. The Parkes result makes the same λ/d rule concrete at radio-astronomy scale.