What diffraction is, and Huygens' construction for a gap and for an obstacle

How the amount of spreading depends on the gap width compared with the wavelength

Why a diffraction grating gives sharp maxima, and deriving nλ= dsin θ

Slit spacing from lines per mm, angles of the orders and the highest order

Measuring the wavelength of laser light with a grating (Core Practical 6), with uncertainties

Electron diffraction as evidence that electrons behave as waves

The de Broglie wavelength λ= h/p, including electrons accelerated through a p.d.

1. Diffraction is the spreading of a wave as it passes through a gap or around the edge of an obstacle, and it is greatest when the gap is about the same size as the wavelength.

2. Huygens: every point on a wavefront acts as a source of secondary wavelets; the new wavefront is the surface that touches them all.

3. A diffraction grating gives sharp maxima where the path difference between adjacent slits is a whole number of wavelengths: nλ= dsin θ.

4. sin θ cannot exceed 1, so the highest order is the largest whole number not greater than d/λ.

5. Electrons diffract through thin crystals, so they have a wavelength: λ= h/p. Faster electrons have shorter wavelengths.

You can hear someone talking in the next room through an open door even when you cannot see them. Sound bends round the edges of the doorway; light, apparently, does not. Both are waves, and the difference comes down to the size of the wavelength compared with the gap.

Diffraction happens to every kind of wave (water waves, sound, light, microwaves) and is strong evidence that something is a wave.

It changes the direction of travel and the shape of the wavefronts. It does not change the wavelength, the frequency or the speed, because the medium is the same on both sides of the gap.

Energy is spread over a wider region, so the amplitude in any one direction beyond the gap is smaller.

Using the construction at a gap

Plane wavefronts reach the barrier, and only the part of each wavefront inside the gap carries on. Each point in the gap sends out a semicircular wavelet. In the middle of a wide gap the wavelets overlap side by side, so their envelope is still straight. At the edges there are no neighbouring wavelets on the outside, so the envelope curves round into the region behind the barrier: this is the spreading.

With a gap about one wavelength wide there is effectively only one source of wavelets, so the wave leaves as semicircular wavefronts. At an obstacle the same thing happens at each edge: wavelets from the edge of the unblocked wavefront curve into the "shadow" behind it.

What matters is the ratio of wavelength to gap width, λ/b, not either one on its own.

Gap width ≈ λ: maximum spreading; the wave fills almost the whole region beyond the gap.

Gap width much larger than λ: little spreading; the wave carries on almost as a straight-edged beam.

Narrowing the gap (closer to λ) or increasing the wavelength both increase the spreading.

If the gap is much smaller than λ, very little of the wave gets through at all.

Visible light has λ≈4 ×10⁻⁷ m to 7 ×10⁻⁷ m, far smaller than doorways, so light diffraction is noticeable only with very narrow slits or very fine edges. Sound has wavelengths from centimetres to metres, so it diffracts round everyday openings.

WORKED EXAMPLE · QUESTION

A courtyard is enclosed by a wall with a single opening 1.2 m wide. Outside, a bird sings at 2.8 kHz and a lorry engine produces a low note of 95 Hz. The speed of sound is 340 m s⁻¹. Calculate the wavelength of each sound and explain which one is heard better by a person inside the courtyard, to one side of the opening and out of the direct line.

SOLUTION

Wavelengths: bird: λ= 340 ÷2800 = 0.12 m (0.121 m). Engine: λ= 340 ÷95 = 3.6 m (3.58 m).

Compare with the gap: bird: λ/b ≈0.1; engine: λ/b ≈3.0.

The engine note is heard better. Its wavelength is larger than the gap, so it spreads strongly round behind the wall. The bird's wavelength is about a tenth of the gap, so it diffracts only a little and mostly continues straight on.

A diffraction grating is a plate with a very large number of equally spaced, parallel, very narrow slits, often specified as a number of lines per mm. Each slit is so narrow that light diffracts from it over a wide range of angles, so light from all the slits overlaps beyond the grating. In most directions the waves from the many slits arrive with a spread of phases and cancel. Only in certain directions do they all arrive in phase and superpose constructively. The result is a few sharp, bright maxima with darkness between them, much sharper and brighter than the fringes from two slits.

The diffraction grating equation

Here n is the order of the maximum (0, 1, 2, …), λ the wavelength (m), d the slit spacing, centre to centre (m), and θ the angle between the maximum and the straight-through (zero-order) direction. The equation is in the formulae list supplied in the exam, but you must be able to derive it:

Plane waves arrive normally, so adjacent slits are in phase.

The path difference from adjacent slits to a distant point at angle θ is dsin θ.

Constructive superposition needs a path difference of a whole number of wavelengths, nλ. Then every pair of neighbouring slits is in step, so all the slits are.

The slit spacing comes from the lines per mm: d = 1 ÷ (number of lines per metre), so for N lines per mm, d = 1/(N ×10³) m. The zero order is straight through; the other orders come in pairs at ±θ. Larger λ or smaller d gives larger angles. With white light each order except zero becomes a spectrum, violet nearest the centre and red furthest out, and higher-order spectra can overlap.

WORKED EXAMPLE · QUESTION

A diode laser of wavelength 635 nm shines normally onto a grating with 400 lines per mm. Calculate (a) the slit spacing, (b) the angles of the first-order and second-order maxima, and (c) the angle between the two first-order maxima.

SOLUTION

(a) d = 1 ÷(400 ×10³) = 2.50 × 10⁻⁶ m.

(b) n = 1: sin θ₁ = 635 ×10⁻⁹ ÷2.50 ×10⁻⁶ = 0.254, so θ₁ = 14.7°. n = 2: sin θ₂ = 2 ×0.254 = 0.508, so θ₂ = 30.5°. 14.7° and 30.5°

(c) The first orders lie either side of the centre: 2 ×14.7° = 29.4°.

θ₂ is not 2θ₁: it is sin θ that doubles, not θ.

Since sin θ can be at most 1, nλ≤d, so n ≤d/λ. The highest order is the largest whole number not greater than d/λ: always round down. The total number of maxima is 2nₘₐₓ + 1, both sides plus the central one. Used in reverse, a measured angle and a known d give the wavelength: λ= dsin θ/n.

WORKED EXAMPLE · QUESTION

For the 400 lines per mm grating and 635 nm laser of worked example 2: (a) determine the highest order and the total number of maxima; (b) the grating is replaced by one with 800 lines per mm. How many maxima are seen now?

SOLUTION

(a) d/λ= 2.50 ×10⁻⁶ ÷635 ×10⁻⁹ = 3.94, so the highest order is n = 3 (at 49.6°). Counting both sides and the centre: 2 ×3 + 1 = 7 maxima.

(b) d = 1.25 ×10⁻⁶ m; d/λ= 1.97, so the highest order is 1: 3 maxima.

Rounding 3.94 up to 4 would need sin θ= 1.02, which is impossible.

WORKED EXAMPLE · QUESTION

Light from a gas discharge lamp passes normally through a grating with 450 lines per mm. For one bright line, the angle between the two second-order maxima, on opposite sides of the centre, is 58.0°. (a) Calculate the wavelength of the line. (b) Determine the highest order in which this line can be seen.

SOLUTION

θ is measured from the centre, so halve the angle between the pair: θ₂ = 58.0° ÷2 = 29.0°. d = 1 ÷(450 ×10³) = 2.222 ×10⁻⁶ m.

(a) λ= dsin θ/n = 2.222 ×10⁻⁶ ×sin 29.0° ÷2 = 5.39 × 10⁻⁷ m (539 nm).

(b) Test sin θ= nλ/d: n = 3: 0.727; n = 4: 0.970; n = 5: 1.21 > 1. Highest order: 4th.

The fourth order is at about 76° and would be faint and hard to see in practice.

This is Core Practical 6, described here in our own words.

Set-up: mount a grating of known lines per mm in a holder, with a laser shining normally onto it. Place a white screen (or a wall) parallel to the grating, about 1–2 m away.

Check it is normal: the maxima either side of the centre should be the same distance from the central spot; if not, rotate the grating until they are.

Measure D: the perpendicular distance from grating to screen, with a tape measure or metre rules (mm resolution).

Measure 2x: the distance between the two maxima of the same order on opposite sides of the centre, with a metre rule, after recording the spot centres on paper fixed to the screen. Repeat and take the mean.

Calculate: x = (2x)/2, tan θ= x/D, then λ= dsin θ/n. Repeat for the second order and compare.

Safety: never look into the laser beam or its reflections; use a low-power laser, keep the beam below eye level, remove shiny surfaces such as watches, and switch the laser off when not in use.

Measuring across the centre doubles the length measured, so the same reading uncertainty is a smaller percentage, and you avoid locating the exact centre of the central spot. A larger D or a higher order gives larger distances and smaller percentage uncertainties.

Grating angles are often too large for the small-angle approximation, so always find θ from tan θ= x/D, then take the sine.

WORKED EXAMPLE · QUESTION

A student uses a grating labelled 300 lines per mm. The screen is 1.500 m from the grating. Three measurements of the distance between the two first-order maxima are 0.592 m, 0.586 m and 0.598 m. Determine the wavelength of the laser light, and the percentage uncertainty in the mean distance between the first-order maxima.

SOLUTION

Mean and uncertainty (half the range): 2x = 0.592 m; uncertainty = (0.598 - 0.586) ÷2 = ±0.006 m.

Angle: x = 0.296 m; tan θ= 0.296 ÷1.500 = 0.197, so θ= 11.2°.

Wavelength: d = 1 ÷(300 ×10³) = 3.33 ×10⁻⁶ m; λ= 3.33 ×10⁻⁶ ×sin 11.2° = 6.45 × 10⁻⁷ m.

Percentage uncertainty: 0.006 ÷0.592 ×100 = 1.0%. Halving 2x to get x leaves the percentage uncertainty unchanged.

In an electron diffraction tube, electrons released from a hot cathode are accelerated through a p.d. of a few kilovolts towards an anode and pass through a very thin film of graphite (carbon). On the fluorescent screen at the end you see a bright central spot surrounded by concentric bright rings.

The regularly spaced layers of atoms in the graphite act like a grating. The atomic spacing (about 10⁻¹⁰ m) is comparable with the electrons' wavelength, so the electrons diffract strongly.

Rings arise because graphite is made of many tiny crystals at random orientations: each set of atomic layers sends electrons out at the same angle from the beam, in every direction around it.

Why this is evidence for waves: diffraction and superposition are wave behaviours. Particles travelling in straight lines would give a single bright patch, not rings with dark regions between them.

The test: increasing the accelerating p.d. makes the rings smaller. Faster electrons have more momentum, so (by λ= h/p) a shorter wavelength, and they diffract through smaller angles, exactly as a wave would.

Here λ is the de Broglie wavelength (m), h the Planck constant, 6.63 ×10⁻³⁴ J s, and p the momentum of the particle (kg m s⁻¹), p = mv. Both λ= h/p and p = mv are in the formulae list. The relation applies to any moving particle, but large objects have such tiny wavelengths that their wave behaviour can never be observed.

Electrons accelerated through a potential difference

An electron accelerated from rest through a p.d. V gains kinetic energy equal to the work done on it:

This comes from V = W/Q and Eₖ = ½ mv²; derive it, as it is not given.

Find v, then p = mv, then λ= h/p, each on its own line. Combining the steps gives λ= h/√(2meV), so λ∝1/√V: four times the p.d. halves the wavelength.

WORKED EXAMPLE · QUESTION

Electrons are accelerated from rest through a potential difference of 2.5 kV. e = 1.60 ×10⁻¹⁹ C, mₑ = 9.11 ×10⁻³¹ kg, h = 6.63 ×10⁻³⁴ J s. (a) Calculate the speed, the momentum and the de Broglie wavelength of the electrons. (b) Explain why these electrons are diffracted by a crystal with atomic spacing about 1 ×10⁻¹⁰ m.

SOLUTION

(a) Eₖ = eV = 1.60 ×10⁻¹⁹ ×2500 = 4.0 ×10⁻¹⁶ J. Speed: v = √(2 ×4.0 ×10⁻¹⁶ ÷9.11 ×10⁻³¹) = 2.96 ×10⁷ m s⁻¹. Momentum: p = 9.11 ×10⁻³¹ ×2.96 ×10⁷ = 2.70 ×10⁻²³ kg m s⁻¹. Wavelength: λ= 6.63 ×10⁻³⁴ ÷2.70 ×10⁻²³ = 2.46 × 10⁻¹¹ m.

(b) The spacing is only about 4 times λ: similar orders of magnitude, so significant diffraction.

The speed is about 10% of the speed of light, low enough for ½ mv² to be a good approximation at this level.

WORKED EXAMPLE · QUESTION

A researcher wants electrons with a de Broglie wavelength of 2.1 ×10⁻¹⁰ m, equal to an atomic spacing in a crystal under study. mₚ = 1.67 ×10⁻²⁷ kg. (a) Calculate the accelerating p.d. needed. (b) Calculate the speed a proton would need to have the same de Broglie wavelength.

SOLUTION

Momentum: p = h/λ= 6.63 ×10⁻³⁴ ÷2.1 ×10⁻¹⁰ = 3.16 ×10⁻²⁴ kg m s⁻¹.

(a) v = p/m = 3.47 ×10⁶ m s⁻¹; Eₖ = ½ mv² = 5.47 ×10⁻¹⁸ J. Then V = Eₖ/e = 5.47 ×10⁻¹⁸ ÷1.60 ×10⁻¹⁹ = 34 V (34.2 V).

(b) Same wavelength means the same momentum: vₚ = 3.16 ×10⁻²⁴ ÷1.67 ×10⁻²⁷ = 1.9 × 10³ m s⁻¹ (1891 m s⁻¹).

The proton is about 1830 times more massive, so it needs a speed about 1830 times smaller for the same momentum.

2 Superposition and stationary waves

The grating equation is the condition for constructive superposition, path difference = nλ, using the same path-difference and phase ideas as two-source interference.

5 Photons, the photoelectric effect and line spectra

Light diffracts like a wave but also arrives as photons; electrons are particles that also diffract. Together these make up wave–particle duality, and h links both sides. A grating is also how line spectra are measured.

Unit 1 mechanics, Unit 4 particle physics

The electron calculation joins V = W/Q, Eₖ = ½ mv² and p = mv. To probe smaller structures such as nuclei, physicists use particles with much larger momentum, because a shorter de Broglie wavelength resolves finer detail.

1. Explain why sound from a room can be heard through an open window from round the side, but the room cannot be seen from there.

2. A grating with 120 lines per mm gives a first-order maximum at 4.2°. Calculate the wavelength.

3. Why are the maxima from a diffraction grating much sharper than the fringes from two slits?

4. An electron has a de Broglie wavelength of 5.0 × 10⁻¹¹ m. Calculate its momentum.

5. In an electron diffraction tube, the accelerating p.d. is increased. Explain what happens to the rings.

6. Electrons are accelerated from rest through 10 kV. Calculate their de Broglie wavelength.

1. Sound wavelengths (roughly 0.1 m to a few metres) are similar to the window width, so the sound diffracts strongly round the edges. Light wavelengths (about 10⁻⁷ m) are millions of times smaller than the window, so light diffracts negligibly and travels in almost straight lines.

2. d = 1 ÷1.20 ×10⁵ = 8.33 ×10⁻⁶ m; λ= 8.33 ×10⁻⁶ ×sin 4.2° = 6.10 × 10⁻⁷ m (610 nm).

3. With many slits, the waves only all arrive in phase when the path difference between neighbours is exactly nλ. Slightly away from that angle the many waves have a spread of phases and cancel, so each maximum is narrow; more slits also let through more light, so the maxima are bright.

4. p = h/λ= 6.63 ×10⁻³⁴ ÷5.0 ×10⁻¹¹ = 1.3 × 10⁻²³ kg m s⁻¹.

5. The electrons gain more kinetic energy, so they have greater speed and momentum. Their de Broglie wavelength h/p is shorter, so they are diffracted through smaller angles: the rings become smaller (closer to the centre).

6. λ= h/√(2meV) = 6.63 ×10⁻³⁴ ÷√(2 ×9.11 ×10⁻³¹ ×1.60 ×10⁻¹⁹ ×10 000) = 1.23 × 10⁻¹¹ m, half the 2.5 kV value, as λ∝1/√V predicts.