Part 2 · Light

2.5 Young's slits, gratings and Bragg's law

Source: Postorino, week 3 slides, pp. 12–16

Objectives

Recipe. Young: phase difference $ϕ = \frac{2 π}{λ} d sin θ$ ; bright fringes at $d sin θ = mλ$ , spacing $λ D / d$ .
Picture. $N$ sources: principal maxima in the same places, $N^{2}$ times as bright and $1/ N$ as wide, with $N - 2$ small maxima between.
Theorem. Bragg’s law $2 d sin θ = nλ$ for reflection from crystal planes.
Key. A grating spreads light by wavelength: with a detector it measures the power spectrum.

Young’s experiment

A plane wave hits a screen with two slits, A and B, a distance $d$ apart. Light from both reaches a point P on a viewing screen at distance $D$ , at height $y$ from the centre, at angle $θ$ .

Same frequency: $ω_{A} = ω_{B} = ω$ . Nearly the same wave vector, $k_{A} ≅ k_{B} = k$ : the two rays to a distant P are almost parallel.
Same amplitude $E_{0}$ : the slits have the same size.
Different distances travelled: $r_{A} - r_{B} = Δ r$ .
The waves start in phase at the slits, because the same plane wave reaches both. At P their phase difference is $ϕ = k Δ r$ .

Fraunhofer geometry: point-like slits, screen and source very far away, so $D ≫ d$ and the incoming wave is plane. This is case 1 (Proposition 2.4.3): $Re {E} = 2 E_{0} cos [k x - ω t + \frac{ϕ}{2}] cos [\frac{ϕ}{2}]$ .

Theorem 2.5.1Young’s fringesslides w3, p. 12

$ϕ = k Δ r = \frac{2 π}{λ} d sin θ \approx \frac{2 π}{λ} d \frac{y}{D}, I (y) \propto 4 E_{0}^{2} cos^{2} (\frac{π}{λ} \frac{d}{D} y) .$ The amplitude of the field depends on the position on the screen.

Proof

In the small triangle at the slits, $d$ is the hypotenuse and $Δ r$ a leg, so $Δ r = d sin θ$ (the slide’s “ $d sin θ = ? r$ ” lost the $Δ$ ). In the large triangle to the screen, $tan θ = y / D$ . For small angles $sin θ \approx tan θ = y / D$ . The intensity goes as the squared amplitude, $4 E_{0}^{2} cos^{2} (ϕ /2)$ .

WarningSlide slip

The slide gives the intensity as $cos^{2} (\frac{2 π}{D} \frac{d}{λ} y)$ , which is $cos^{2} ϕ$ . The amplitude factor is $cos (ϕ /2)$ , so the argument has $π$ , not $2 π$ . With the slide’s version the fringes would be half as far apart.

Recipe 2.5.2Fringesbeyond the lectures

Bright fringes where $ϕ = 2 πm$ : $d sin θ = mλ$ , i.e. $y_{m} = m \frac{λ D}{d}$ , $m = 0, \pm 1, \pm 2, \dots$
Dark fringes where $ϕ = (2 m + 1) π$ : $d sin θ = (m + \frac{1}{2}) λ$ .
Fringe spacing $Δ y = λ D / d$ .

Example 2.5.3Measuring a wavelength with a rulerbeyond the lectures

$λ = 500$ nm, slits $d = 0.1$ mm apart, screen at $D = 1$ m: fringes $Δ y = 5 \times 1 0^{- 7} \cdot 1/1 0^{- 4} = 5$ mm apart. A wavelength far too small to see becomes a spacing you can measure with a ruler; this is how Young measured $λ$ in 1801.

Real, extended slits. With slits of finite width $a$ , the fringes are no longer equally bright: they fade away from the centre under a broad envelope.

RemarkWhy the envelopebeyond the lectures

Each slit is itself a row of point sources, and its far-field amplitude is the Fourier transform of its opening: a boxcar of width $a$ . So the envelope is a sinc², $[\frac{s i n ( πa s i n θ / λ )}{πa s i n θ / λ}]^{2}$ (Theorem 1.7.3). The narrower the slit, the wider the envelope.

Many sources: gratings

Extending to many interfering sources gives diffraction. The slide shows the pattern of $N = 2$ , 4 and 8 coherent sources against $sin θ$ (from Mazzoldi, Nigro, Voci, Elementi di Fisica – Onde).

Theorem 2.5.4N sourcesslides w3, p. 13

The principal maxima stay at $sin θ = mλ / d$ , as for two slits.
Their peak intensity grows as $N^{2}$ : $4 I_{1}$ , $16 I_{1}$ , $64 I_{1}$ , with $I_{1}$ the intensity from one source.
They get narrower as $N$ grows, with $N - 2$ small secondary maxima between neighbouring principal ones.

Proofbeyond the lectures

Adjacent sources differ in phase by $ϕ = \frac{2 π}{λ} d sin θ$ . The sum of $N$ unit phasors is a geometric series: $j = 0 \sum N - 1 e^{ij ϕ} = \frac{e ^{i Nϕ} - 1}{e ^{i ϕ} - 1} \Rightarrow I = I_{1} \frac{sin ^{2} ( Nϕ /2 )}{sin ^{2} ( ϕ /2 )},$ using $∣ e^{i θ} - 1∣ = 2∣ sin (θ /2) ∣$ .

At $ϕ = 2 πm$ every phasor lines up: $I = N^{2} I_{1}$ .
The first zeros are at $ϕ = 2 πm \pm 2 π / N$ , so a principal maximum has width $\propto 1/ N$ .
Between principal maxima $sin (Nϕ /2)$ vanishes $N - 1$ times, leaving $N - 2$ secondary maxima.
For $N = 2$ , $sin^{2} ϕ / sin^{2} (ϕ /2) = 4 cos^{2} (ϕ /2)$ : Young again.

Loading the demo…

Figure 2.5.1 The intensity of

N

coherent slits against

sin θ

, in units of one slit’s

I_{1}

, with the

single-slit envelope

when the slits have a width. On the right, the

N

phasors added head to tail at the angle you drag to: they close into a loop at the zeros and line up into a straight arrow at the principal maxima. Raise

N

and watch the peaks narrow.

Gratings separate colours. The positions $sin θ = mλ / d$ depend on $λ$ , so wavelengths come out at different angles. White light leaves a grating as a white central maximum ( $m = 0$ , where all wavelengths agree) flanked by first- and second-order rainbows. With more lines the peaks are sharper and close wavelengths can be told apart.

Key observation 2.5.5A grating measures the power spectrumslides w3, p. 13

With a diffraction grating and a detector you can measure the power spectrum $∣ c (ω) ∣^{2}$ of the radiation falling on it: its intensity at each frequency (Key observation 1.7.2).

Interference in everyday life and in the lab

Everyday examples (pictures on the slide; the explanations are added):

Soap bubbles and oil films on wet asphalt: light reflected from the top and bottom of a thin film interferes. Which colour is constructive depends on the thickness, so a film of varying thickness shows bands of colour.
Dispersion prism, index $n (ν)$ : not interference. $n$ depends on frequency, so each colour refracts by a different angle (§3.2 The complex refractive index and absorption).
CDs: their closely spaced tracks are a reflection grating.

Instruments:

Fabry–Pérot interferometer. A slab of index $n$ and thickness $l$ with partly reflecting faces. A ray entering at angle $θ$ bounces back and forth, giving reflected rays $R_{0}, R_{1}, \dots$ and transmitted rays $T_{1}, T_{2}, \dots$ Many beams interfere, like a grating with very many sources: transmission peaks only when a round trip $2 n l cos θ$ fits a whole number of wavelengths. With high reflectivity (high finesse) the rings are very sharp.
Grating monochromator. Light enters through slit A, is collimated by a mirror and spread by a rotating grating; a second mirror focuses one wavelength on the exit slit. Rotating the grating scans the wavelength. A flame seen through a grating shows its spectrum in orders $m = - 1, 0, + 1$ .

X-ray diffraction and Bragg’s law

To see interference the spacing of the sources must be comparable with the wavelength, $λ \sim d$ . Atoms in a crystal are a few ångström apart, so they diffract X-rays, whose wavelengths are about 1 Å.

Theorem 2.5.6Bragg’s lawslides w3, p. 15

A wave hits a family of parallel crystal planes, spacing $d$ , at a glancing angle $θ$ measured from the planes. Each plane reflects a little; the reflections add up to a maximum of intensity for $2 d sin θ = nλ .$

Proofbeyond the lectures

The ray reflected by the second plane travels an extra $d sin θ$ on the way in and $d sin θ$ on the way out, $2 d sin θ$ in all: the $d sin θ$ marked twice in the slide’s figure. Constructive interference needs this extra path to be a whole number $n$ of wavelengths.

Loading the demo…

Figure 2.5.2 X-rays on a stack of crystal planes. Drag the glancing angle

θ

. The readout compares the extra path

2 d sin θ

with

λ

; the graph is the reflected intensity from 8 planes against

θ

, with sharp peaks at the

Bragg angles

. Change

d / λ

: fewer orders fit when

d

is small.

There are many planes in a crystal. A crystallographic plane passes through at least three lattice points, and through every lattice point passes one plane of each family. In a 2D lattice with basis vectors $a_{1}$ , $a_{2}$ , the families (10) and (21) have different spacings $d_{(hk l)}$ , and each family gives its own Bragg peaks.

Experiment. An X-ray beam hits a crystal, say a crystallised molecule, and the diffracted beams are recorded on film or a detector. In every variant one observes the peaks of maximum intensity of X-rays, or neutrons, and deduces from them the distances between the planes. This is how crystal and protein structures are solved. Neutrons diffract because they are waves too (§5.5 Franck–Hertz, Stern–Gerlach and wave–particle duality).

Wrap-up

With superposition and Fourier, the problem is studied one frequency at a time (monochromatic waves).
Diffractive devices are rows of interfering sources, and their response depends on frequency: the basis of instruments that measure the power spectrum, how a wave’s energy is shared over frequencies.
No wave is truly monochromatic (Key observation 2.4.8).

Exercises

Exercise 2.5.1beyond the lectures

In a Young experiment, $λ = 600$ nm, $d = 0.2$ mm and $D = 2$ m. Find the fringe spacing and the position of the third dark fringe from the centre.

Solution

$Δ y = λ D / d = 6 \times 1 0^{- 7} \cdot 2/2 \times 1 0^{- 4} = 6$ mm. Dark fringes are at $(m + \frac{1}{2}) λ D / d$ ; the third ( $m = 2$ ) is at $2.5 \times 6 = 15$ mm.

Exercise 2.5.2beyond the lectures

X-rays of $λ = 1.54$ Å reflect from planes with $d = 2.0$ Å. Find the Bragg angles and the largest possible order.

Solution

$sin θ = nλ /2 d = 0.385 n$ . $n = 1$ : $θ = 22.6°$ ; $n = 2$ : $θ = 50.4°$ ; $n = 3$ would need $sin θ = 1.155 > 1$ . The largest order is 2.

Exercise 2.5.3beyond the lectures

A grating has 600 lines/mm. At what angles does first-order light of 400 nm and 700 nm come out, and how wide is the first-order rainbow?

Solution

$d = 1/600$ mm $= 1667$ nm. $sin θ = λ / d$ : 0.240 for 400 nm ( $13.9°$ ), 0.420 for 700 nm ( $24.8°$ ). The rainbow spans about $11°$ .

Exercise 2.5.4beyond the lectures

For $N = 5$ slits, how many secondary maxima lie between two principal ones, and how bright is a principal maximum compared with one slit?

Solution

$N - 2 = 3$ secondary maxima; $N^{2} = 25$ times one slit’s intensity.

Exercise 2.5.5beyond the lectures

Electrons accelerated through 54 V have a de Broglie wavelength of 1.67 Å (§5.5 Franck–Hertz, Stern–Gerlach and wave–particle duality). Diffracted by planes 0.91 Å apart, at what glancing angle is the first Bragg peak?

Solution

$sin θ = λ /2 d = 1.67/1.82 = 0.918$ , $θ = 66.6°$ . (This is close to the Davisson–Germer geometry for nickel.)