You can hear someone talking through an open door before you can see them. Sound curls round the door frame; light doesn’t. Same doorway, same physics — but the two waves have wavelengths a million times apart, and that is what decides how much a wave bends around a corner. This page is about waves spreading where geometry says they shouldn’t.
📘 What you need to know
Diffraction is the spreading out of waves after passing through a narrow gap or around an obstruction
Diffraction is greatest when the wavelength is similar to the gap width
A gap much wider than λ gives very little spreading
Around a barrier, a longer wavelength means more diffraction and a smaller shadow
The diffracted wave has a smaller amplitude, but λ, f and v are all unchanged
All waves diffract — sound, water, light, radio
What is diffraction?
Definition — diffraction
The spreading out of waves after they pass through a narrow gap or around an obstruction
Notice what diffraction is not. The wave doesn’t hit a boundary and bounce (reflection), and it doesn’t cross into a new material and bend (refraction). It simply meets an edge, and afterwards it fills space that a straight-line ray would never have reached.
Diffraction happens in two situations:
When waves pass through an aperture — a gap or slit
When waves pass around a barrier — the edge of an obstacle
Diffraction through a gap
Send straight (plane) wavefronts at a barrier with a narrow gap in it. On the far side the wavefronts are no longer straight: they have curvature, bulging outwards as if a brand-new point source were sitting in the gap.
The gap behaves like a new point source. Crucially, the two orange markers are the same length — diffraction changes the shape of the wavefronts, never the wavelength.
Two things change on the far side of the gap:
The wavefronts acquire curvature — they fan out into the region behind the barrier
The amplitude falls, because the barrier either side of the gap absorbs wave energy
Never changes: the wavelength, the frequency and the speed of a diffracted wave are exactly what they were before the gap. If you sketch diffraction, keep the spacing between wavefronts constant — examiners look for it.
How much does it spread? Gap size vs wavelength
The amount of spreading depends on one comparison only: the size of the gap against the wavelength of the wave.
The condition for strong diffraction
gap width ≈ wavelength λ
Same wave, same wavelength, different gaps. When the gap matches λ the wave fans out into a semicircle; when the gap is several wavelengths wide the wave marches on almost straight, curling only at the edges.
Gap ≈ λ — maximum spreading; the wave fans out over a wide angle
Gap much bigger than λ — very little diffraction; the beam carries on almost straight
As the gap grows compared with λ, the curvature of the emerging wavefronts gets less pronounced
Here’s the trick: never ask “is the gap small?” — ask “is the gap small compared with the wavelength?” A 1 mm slit is enormous for light (λ ≈ 0.0005 mm) and hopelessly tiny for sound (λ ≈ 1 m). It is the ratio that matters, never the raw size.
Wavelength λ
compare with
Gap width
similar → lots gap ≫ λ → little
Amount of spreading
Diffraction around a barrier
Waves also curl around the edge of an obstacle, spreading into the region behind it. Directly behind the obstacle there is a shadow region that the wavefronts have not reached — and diffraction slowly eats into it.
The wavefronts bend around both edges and creep into the shadow. A longer wavelength curls in further, shrinking the shadow; a shorter one leaves a crisp, dark shadow.
How big the shadow is depends, once again, on the barrier size compared with λ:
Barrier compared with λ
Diffraction around it
Shadow region behind
Barrier larger than λ
Some — and much of the wave is reflected back
Large shadow, no wavefronts inside
Barrier about equal to λ
More diffraction around the edges
Smaller shadow
Barrier smaller than λ
The wave barely notices the obstacle
Very small shadow
The same rule stated the other way round: the greater the wavelength, the greater the diffraction. Which is why long-wave radio reaches you in a valley while FM cuts out.
🔎 Judging how much a wave will diffract
Find the wavelength of the wave, using v = fλ if you’re given a frequency.
Find the size of the gap or obstacle.
Compare them. Similar sizes → strong diffraction. Gap or obstacle far bigger than λ → hardly any.
Sanity-check the physics: sound (λ ≈ metres) diffracts round doors; light (λ ≈ hundreds of nm) needs a slit that narrow.
WE 1
A door is left open by a gap of 0.90 m. Taking the speed of sound as 340 m s⁻¹, determine the frequencies of sound that will be best diffracted through the gap.
Step 1 — diffraction is strongest when λ is comparable to (or larger than) the gap
λ ≈ 0.90 m
Step 2 — rearrange the wave equation for f
f = v / λ
Step 3 — substitutef = 340 / 0.90 = 377.8 HzStep 4 — longer λ means smaller f, so those diffract at least as wellf ≤ 3.8 × 10² HzLow notes spill round the door; the high, hissy consonants don’t. That’s why muffled speech through a door sounds so bass-heavy.
WE 2
Explain, with a calculation, why you can hear someone speaking through that same 0.90 m doorway but cannot see them. Take a typical speech frequency as 500 Hz and green light as 5.0 × 10−7 m.
Step 1 — wavelength of the soundλ = 340 / 500 = 0.68 mStep 2 — compare each with the 0.90 m gap
sound: 0.68 m is about the same as 0.90 m
light: 0.90 / (5.0 × 10⁻⁷) = 1.8 × 10⁶Step 3 — interpret
The gap is 1.8 million wavelengths wide for light.
Sound diffracts, light does notLight does diffract through the doorway — by an utterly unmeasurable amount. Same physics, wildly different ratio.
WE 3
A hill about 300 m across stands between a radio and two transmitters: a long-wave station at 200 kHz and an FM station at 100 MHz. Determine which signal is received behind the hill. Take c = 3.00 × 108 m s⁻¹.
Step 1 — wavelength of each wave, λ = c / flong wave: λ = (3.00 × 10⁸) / (2.0 × 10⁵) = 1500 mFM: λ = (3.00 × 10⁸) / (1.00 × 10⁸) = 3.0 mStep 2 — compare each with the 300 m obstacle
1500 m is much bigger than the hill; 3.0 m is much smaller.
Step 3 — bigger λ means more diffraction, smaller shadowThe long-wave signal gets throughFM leaves a deep radio shadow behind the hill. Same hill, same rule — only the ratio λ to obstacle differs.
💡 Top tips
Always compare λ with the gap or obstacle. Absolute sizes mean nothing on their own.
When sketching, keep the wavefront spacing constant — diffraction never changes λ.
The diffracted wave has a smaller amplitude, because the barrier absorbs energy.
Longer λ → more diffraction → smaller shadow behind an obstacle.
Use v = fλ first if the question gives you a frequency instead of a wavelength.
⚠ Common mistakes
Thinking the wavelength changes when a wave diffracts — only the shape and amplitude change
Saying “the gap is small, so it diffracts a lot” without comparing it to λ
Confusing diffraction with refraction — diffraction needs no change of medium, just an edge
Believing light never diffracts — it does, but only through gaps a few hundred nanometres wide
Drawing diffracted wavefronts that bunch up or spread out as they curve
Forgetting that sound and water waves diffract just as readily as light
Quick recap:Diffraction is the spreading of waves through a gap or around an obstacle. It is most pronounced when λ is similar to the gap width, and negligible when the gap is far wider. Around a barrier, a longer λ means more curling and a smaller shadow region. The diffracted wave loses amplitude, but its λ, f and v are untouched.
One gap made the wave spread. Now imagine two gaps side by side: two spreading waves that overlap, reinforcing in some directions and cancelling in others. That’s where the beautiful stuff begins — superposition and interference. Before that, though, we put real numbers on the bending from the last page, with refractive index and Snell’s law, in Refraction of Waves.
Want diffraction to finally click?
Book a free meeting and we’ll work through gap-size reasoning, sketches and past-paper diffraction questions together.
