Diffraction — waves bending around obstacles — Physics

Before this

This leads to

Standing waves
The diffraction grating · coming soon
Electron diffraction · coming soon

You can hear round a corner, but you cannot see round it

Stand in a corridor next to an open door. Someone in the room is talking, out of your sight. You hear every word. The sound has come out through the doorway and turned the corner to reach you. But the light from that person has not. If it had, you would see them.

Sound and light are both waves. So why does one turn the corner and the other go straight on?

The answer is not that light cannot bend round edges. It can, and it does. The answer is a comparison of two lengths: the size of the gap and the wavelength of the wave. This lecture is about that comparison. The Young's double slit lecture promised it twice: once for the spreading of light at each slit, and once for why the fringes fade towards the edges of the screen. Both promises are kept here.

What diffraction is

Diffraction is the spreading of a wave when it passes through a gap or past the edge of an obstacle. Every kind of wave does it: water waves, sound, light, microwaves and radio waves.

Watch water waves in a ripple tank meet a barrier with a gap in it. Straight wavefronts arrive. On the far side, the wavefronts curve round into the region behind the barrier. That region would be a sharp "shadow" if the waves simply went straight on.

Three things do not change. The wave is still in the same medium, so its speed is the same. Its frequency is set by the source, so that is the same too. Because $v = f λ$ , the wavelength is also the same. Only the direction of travel spreads out. This is how diffraction differs from refraction, where the wave enters a new medium and its speed and wavelength change.

The word is old. The Italian scientist Francesco Maria Grimaldi named it, from a Latin root meaning "breaking apart", in a book published in 1665, two years after his death. He had noticed that the edges of shadows were not perfectly sharp.

Why a wave spreads: every point is a new source

A useful picture comes from the Dutch scientist Christiaan Huygens. Treat every point on a wavefront as a tiny source that sends out its own small circular wave. The new wavefront is the line that touches all these small waves.

In open space, the small waves from a long straight wavefront add up to another long straight wavefront. Nothing spreads. Now put a barrier in the way, with a gap in it. Only the points in the gap send waves forward. At the two edges of the gap, there is nothing beside them to cancel their sideways spreading. So the wave curls round behind the barrier.

This picture makes a prediction. If the gap is very wide, the edges are a small part of the wavefront, and most of the wave goes straight on. If the gap is narrow, the "edges" are almost the whole wave. Then the gap behaves like a single point source, and the wave spreads out in circles. The size of the gap matters only compared with the wavelength. A doorway is "narrow" for sound and "wide" for light.

Gap size compared with wavelength

Write the width of the gap as $b$ and the wavelength as $λ$ . The ratio $b / λ$ decides what happens. Try it in the explorer below.

Explore — a wave through a gap

Straight wavefronts arrive from the left, and the dark bar is the barrier. Change the width of the gap, measured in wavelengths. The dashed grey lines show the direction of the first dark band on each side. Watch how far the wave spreads behind the barrier.

Gap width ÷ wavelength: 5.0

Try:

How far a wave spreads after a gap, for four gap widths. The angle is measured from the straight-on direction to the first dark band, using $sin θ_{1} = λ / b$ (explained below).

Gap 20 wavelengths wide. First dark band at about 2.9°. The wave goes almost straight on, with a fairly sharp shadow.
Gap 10 wavelengths wide (ripple tank: λ = 1.0 cm, gap 10 cm). About 5.7°. Clear spreading at the edges.
Gap 2 wavelengths wide (microwaves: λ = 3.0 cm, gap 6.0 cm). 30°. Strong spreading.
Gap 1 wavelength wide or less (sound through a doorway). No dark band in front of the barrier. The wave spreads into the whole region behind the barrier, almost as circles. Less energy gets through a gap this narrow.

What should you have seen? With a gap 20 wavelengths wide, the wave went almost straight through. As you narrowed the gap, the spreading grew. At about one wavelength, the dark bands disappeared and the wave filled the whole region behind the barrier.

Many textbooks say that diffraction is "greatest when the gap equals the wavelength". That is a useful guide, but read it carefully. The spreading keeps growing as the gap gets narrower. Below one wavelength the wave already spreads into every direction ahead, so it cannot spread further. But less and less of the wave gets through. So a gap about one wavelength wide is where diffraction is both strong and easy to notice.

Stop and think

A laser beam passes through a narrow slit. Then the slit is made narrower. Does the patch of light on the far screen get narrower, because less light gets through, or wider?

Show the answer

Wider. Less light gets through, so the patch is dimmer, but it spreads further. A narrower gap means a larger ratio $λ / b$ , and the angle of spreading grows with that ratio. If light were a stream of tiny particles moving in straight lines, a narrower slit would give a narrower strip. The fact that it gives a wider one is strong evidence that light behaves as a wave.

Why there are dark bands: the slit cancels itself

This is the hard part of the lecture. Spend time here. It uses only the path-difference rule from the superposition lecture.

Shine a laser through a single slit of width $b$ onto a screen far away. Look at the light that leaves the slit at an angle $θ$ to the straight-on direction. Huygens says every point across the slit sends out a wave. Split the slit into a top half and a bottom half. Now pair each point in the top half with the point exactly $b /2$ below it in the bottom half.

Light from the two points of a pair travels in the same direction, $θ$ . The lower point's light travels further by $\frac{b}{2} sin θ$ . Suppose that extra distance is exactly half a wavelength:

\frac{b}{2} sin θ = \frac{λ}{2} ⟹ b sin θ = λ

Then the two waves of every pair arrive in antiphase and cancel. Every point in the top half has a partner in the bottom half, so the whole slit cancels itself. In that direction, the screen is dark.

The first dark band of a single slit

For a slit of width $b$ and light of wavelength $λ$ , the first dark band on each side is at an angle $θ_{1}$ . For a screen far from the slit, $sin θ_{1} = \frac{λ}{b}$ .

If $b$ is less than or equal to $λ$ , there is no dark band in front of the slit, and the wave spreads into the whole region ahead.

Look at what just happened. The formula explains everything the explorer showed. A narrower slit makes $λ / b$ bigger, so $θ_{1}$ is bigger and the central bright band is wider. When $b$ is many wavelengths, $λ / b$ is tiny, and the light goes almost straight on. And when $b \leq λ$ , the sine would have to be 1 or more. At exactly 1 the angle is 90°, along the barrier itself, so there is no dark band in front of the slit.

Further out, other directions also cancel, and between them are dimmer bright bands. The graph shows the pattern a red laser makes through a slit 0.10 mm wide, on a screen 2.0 m away.

Brightness across the screen for a single slit (λ = 633 nm, slit width 0.10 mm, screen 2.0 m away). The central maximum is twice as wide as each secondary maximum. The first secondary maximum is less than 5% as bright as the centre.

Two features are worth remembering, with their reasons. The central maximum is twice as wide as the others, because it runs from the first dark band on one side to the first dark band on the other. And the secondary maxima are much dimmer, because in those directions most of the slit's waves still cancel each other, and only a small part is left over.

Predict — halving the slit width

The red laser above makes a central bright band about 25 mm wide on the screen. The slit is changed from 0.10 mm to 0.05 mm wide. What happens to the central band? Choose before you see the answer.

The central band becomes about twice as wide, about 51 mm, and dimmer. Halving $b$ doubles $λ / b$ . For small angles that doubles $θ_{1}$ , so it doubles the width of the band on the screen.

The promise from Young's double slit

The double-slit lecture said that each slit is so narrow that the light spreads out from it. Now you can see why. A typical slit is a fraction of a millimetre wide, only a few hundred wavelengths. So each slit makes its own single-slit pattern, and the two patterns overlap on the screen.

That lecture also said that the fringes fade towards the edges of the screen. Here is the reason. The two-slit fringes can only be as bright as the light that each slit sends in that direction. So the evenly spaced fringes sit inside the single-slit pattern, like fruit inside a bowl shaped like the central maximum.

Double-slit fringes (λ = 630 nm, slit separation 0.50 mm, screen 2.0 m away, fringe spacing 2.52 mm) inside the single-slit pattern of each slit (slit width 0.10 mm, dashed). The fringes fade towards the edges, and the fifth bright fringe on each side, at 12.6 mm, is missing.

Notice something unexpected in the figure. The fifth bright fringe on each side is missing. It should be at $5 \times 2.52 = 12.6$ mm. But the first dark band of each slit is also at $λ D / b = 12.6$ mm, so neither slit sends any light there. Two separate cancellations happen to fall at the same place.

Using the ratio

Worked Example 1 — Will it spread?

For each case, find $b / λ$ and decide how much the wave spreads.

(a) Microwaves, $λ = 3.0$ cm, through a gap 4.0 cm wide. $b / λ \approx 1.3$ . Strong spreading: $sin θ_{1} = 0.75$ , so $θ_{1} \approx 4 9^{\circ}$ .

(b) X-rays, $λ = 0.10$ nm, through a slit 1.0 mm wide. $b / λ = 10000000$ . $θ_{1}$ is far less than a thousandth of a degree. No noticeable spreading.

(c) Sound, $λ = 0.50$ m, past the corner of a building. An edge is not a gap, but the same idea applies: some sound curls round every edge, and longer wavelengths curl further.

(d) A long-wave radio signal, $λ \approx 1500$ m (a frequency of 200 kHz), and a valley 500 m wide between hills. The gap is a third of a wavelength. The signal spreads into the whole region behind the hills, which is why long-wave radio can be received far from a transmitter, without a straight line of sight.

Check (d): $λ = v / f = 3.0 \times 1 0^{8} \div 2.0 \times 1 0^{5} = 1500$ m. ✓

Worked Example 2 — The same doorway for sound and light

A doorway is 0.90 m wide. Compare sound of wavelength 1.0 m with green light of wavelength 500 nm.

Sound: $b / λ = 0.90 \div 1.0 = 0.9$ . This is less than 1, so there is no dark band. The sound spreads into the whole corridor. You hear the person round the corner.

Light: $b / λ = 0.90 \div (5.00 \times 1 0^{- 7}) = 1.8 \times 1 0^{6}$ . Then $sin θ_{1} \approx 5.6 \times 1 0^{- 7}$ , an angle of about 0.000 03°. The light goes straight on. You do not see the person.

Same doorway, same wall. The ratio is different by a factor of about two million, and that is the whole difference.

Worked Example 3 — Measuring the central band

A laser of wavelength 633 nm shines through a slit 0.10 mm wide onto a screen 2.0 m away. How wide is the central bright band?

Convert first: $λ = 633 \times 1 0^{- 9}$ m and $b = 0.10 \times 1 0^{- 3}$ m.

$sin θ_{1} = \frac{633 \times 1 0 ^{- 9}}{0.10 \times 1 0 ^{- 3}} = 0.006 33$ , so $θ_{1} \approx 0.36 3^{\circ}$ .

Distance from the centre to the first dark band: $2.0 \times tan 0.36 3^{\circ} \approx 0.0127$ m, which is 12.7 mm. The central band runs from $- 12.7$ mm to $+ 12.7$ mm, so it is about 25 mm wide.

Check with the small-angle shortcut: $λ D / b = 633 \times 1 0^{- 9} \times 2.0 \div 1 0^{- 4} = 0.012 66$ m. ✓

A prediction that looked absurd

In 1818 the French Academy of Sciences set a prize competition on diffraction. Augustin Fresnel entered a wave theory of light. One of the judges, Siméon Poisson, preferred the particle view. He worked out a strange consequence of Fresnel's theory: a small round disc should cast a shadow with a bright spot at its exact centre. To Poisson this seemed obviously impossible. Light from every point round the edge of the disc travels the same distance to the centre of the shadow, so on Fresnel's theory it all arrives in phase.

Another judge, François Arago, tried the experiment with a metal disc about 2 mm across. The bright spot was there, and Fresnel won the prize in 1819. The spot is now called the Arago spot or Poisson's spot. Waves had made a prediction that particles could not, and it was right. A century later, other experiments showed that light also arrives in small packets of energy, called photons. That story continues in the Pre-University lectures.

Three classic mistakes

Trap 1 — "A narrower gap makes a narrower beam"

For waves, a narrower gap makes the wave spread more. Less energy gets through, but it fans out over a wider angle. A narrower beam comes from a wider gap.

Trap 2 — "Diffraction changes the wavelength"

The wave stays in the same medium, so its speed, frequency and wavelength are all unchanged. Only its direction spreads. In a diagram, the wavefronts after the gap must be the same distance apart as before it.

Trap 3 — Mixing up diffraction and refraction

Refraction is a change of direction when a wave enters a new medium and changes speed. Diffraction is spreading at a gap or an edge in the same medium. Light bends at a glass surface by refraction. Sound turns a corner by diffraction.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Water waves of wavelength 2.0 cm pass through three gaps: 1.5 cm, 4.0 cm and 60 cm wide. Put them in order from most spreading to least, and say for which gap(s) there is no dark band.

Hint 1
Work out $b / λ$ for each gap.

Hint 2
A dark band needs $sin θ_{1} = λ / b$ to be less than 1, so $b$ must be bigger than $λ$ .

Full answer
The ratios are 0.75, 2 and 30. Most spreading: the 1.5 cm gap (ratio 0.75), then 4.0 cm ( $θ_{1} = 3 0^{\circ}$ ), then 60 cm ( $sin θ_{1} = 1/30$ , $θ_{1} \approx 1. 9^{\circ}$ ). Only the 1.5 cm gap has no dark band, because it is narrower than the wavelength. Check: $sin^{- 1} (0.5) = 3 0^{\circ}$ . ✓
Microwaves of wavelength 2.8 cm pass through a gap 6.0 cm wide. At what angle is the first dark band on each side?

Hint 1
Use $sin θ_{1} = λ / b$ . Both lengths are in cm, so they can be divided directly.

Hint 2
$sin θ_{1} = 2.8 \div 6.0$ .

Full answer
$sin θ_{1} = 2.8 \div 6.0 \approx 0.467$ , so $θ_{1} \approx 27. 8^{\circ}$ on each side of the straight-on direction. Check: $sin 27. 8^{\circ} \approx 0.466$ . ✓
A student shines a green laser (532 nm) through a single slit onto a wall 1.5 m away. The central bright band is 30 mm wide. Estimate the width of the slit.

Hint 1
The first dark band is half the band's width from the centre: 15 mm.

Hint 2
Find $θ_{1}$ from $tan θ_{1} = 0.015 \div 1.5$ , then use $b = λ / sin θ_{1}$ .

Full answer
$tan θ_{1} = 0.015 \div 1.5 = 0.010$ , so $θ_{1} \approx 0.57 3^{\circ}$ and $sin θ_{1} \approx 0.0100$ . Then $b = 532 \times 1 0^{- 9} \div 0.0100 \approx 5.3 \times 1 0^{- 5}$ m, about 0.053 mm. Check with the shortcut: $λ D / b = 532 \times 1 0^{- 9} \times 1.5 \div 5.32 \times 1 0^{- 5} = 0.015$ m. ✓
Many bats hunt insects using echoes of their own calls, at frequencies of about 25–100 kHz. Human speech is mostly a few hundred hertz. Take the speed of sound in air as 343 m/s. (a) Find the wavelength of a 50 kHz bat call and of a 300 Hz sound. (b) Use diffraction to explain why a bat cannot find a 5 mm insect with a 300 Hz call, but can with a 50 kHz call.

Hint 1
$λ = v / f$ .

Hint 2
Compare each wavelength with the insect's size. Think about what a wave does when it meets an obstacle much smaller than its wavelength.

Full answer
(a) 50 kHz: $λ = 343 \div 50000 \approx 0.0069$ m, about 6.9 mm. 300 Hz: $λ = 343 \div 300 \approx 1.14$ m. (b) A 1.14 m wave meets a 5 mm insect as an obstacle about 200 times smaller than its wavelength. The wave diffracts round it almost as if it were not there, so almost no echo comes back. A 6.9 mm wave is about the size of the insect, so the insect blocks and reflects a useful part of it. Short wavelengths are needed to "see" small things, which is the same reason a microscope's detail is limited by the wavelength of its light. Check: $6.9 mm \times 50000 Hz \approx 345$ m/s. ✓

The bigger idea

Diffraction sets a limit on how much detail any wave can show. In 1873 Ernst Abbe worked out that an ordinary light microscope cannot separate details much smaller than about half the wavelength of light, about 0.2 micrometres. In 2014 the Nobel Prize in Chemistry went to Eric Betzig, Stefan Hell and William Moerner, who found clever ways round this limit using glowing molecules. Electron microscopes go further, because fast electrons behave as waves with far shorter wavelengths.

The same physics also works in reverse, as a tool. When X-rays, with wavelengths similar to the spacing of atoms, pass through a crystal or a fibre, they diffract into a pattern that carries the structure inside. In May 1952, Rosalind Franklin and her student Raymond Gosling recorded an X-ray diffraction image of DNA, now called Photo 51. It was key evidence for the double helix described in 1953. To read a pattern like that, you need exactly the idea in this lecture: what is seen depends on the size of the structure compared with the wavelength.

Key concepts in this lesson

Diffraction is the spreading of a wave at a gap or an edge, in the same medium, so the speed, frequency and wavelength do not change.
How much a wave spreads depends on the gap width compared with the wavelength, so a doorway spreads sound strongly but light hardly at all.
A narrower gap makes the wave spread more, and when the gap is no wider than one wavelength, the wave spreads into the whole region ahead.
For a single slit of width $b$ , the first dark band is at $sin θ_{1} = λ / b$ , because light from the two halves of the slit cancels in pairs.
Each slit in Young's experiment makes its own single-slit pattern, so the double-slit fringes fade towards the edges and can go missing where a single-slit dark band falls.