---
title: "Multiple-Wave Interference and Gratings"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 21
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/21-multiple-wave-interference-and-gratings
---

# Chapter 21 — Multiple-Wave Interference and Gratings

Tilt a compact disc under a lamp and it throws a rainbow: its spiral of pits, a micrometre and a half apart, is a grating — thousands of tiny reflecting strips, each sending back a copy of the wave, all interfering. Two waves give the soft $\cos^2$ fringes of the last chapters; a thousand give fringes a thousand times narrower, sharp enough to separate two wavelengths that differ by a part in a hundred thousand. That is the instrument with which the composition of the Sun and the speed of galaxies were read, and which sits in every spectrometer. This chapter adds up $N$ waves, derives the [grating equation](#prop-b2-gratings-equation), its dispersion and its resolving power, and meets the other multiple-wave device, the Fabry–Pérot cavity that will become the laser’s resonator.

![A compact disc in sunlight: its spiral of pits, 1.6\, µ m apart, is a reflection grating, and each order spreads the white light into a spectrum.](https://one-course.com/images/onecourse/chapters/physics-4/b2-gratings/img-7c0468b08e65.jpg)

*A compact disc in sunlight: its spiral of pits, $1.6\,\text{µ}\mathrm{m}$ apart, is a reflection grating, and each order spreads the white light into a spectrum.*

## 21.1 Interference of $N$ waves

**Theorem 21.1 (NNN equal waves in arithmetic phase progression).**

$N$ coherent waves of equal amplitude $a$, each lagging the previous by the phase $\varphi$, superpose into a wave of amplitude $A(\varphi)$ and intensity

$$
A = a\,\frac{\sin(N\varphi/2)}{\sin(\varphi/2)} , \qquad
I(\varphi) = I_0\,\frac{\sin^2(N\varphi/2)}{\sin^2(\varphi/2)} :
$$

*principal maxima* of intensity $N^2I_0$ where $\varphi = 2\pi p$ (all waves in phase), of angular half-width $2\pi/N$ in $\varphi$ (the first zeros are at $\varphi = 2\pi p \pm 2\pi/N$), separated by $N - 2$ *secondary maxima* no higher than about $4.5\%$ of the principal ones for large $N$. The more waves, the sharper and brighter the peaks: the width shrinks as $1/N$, the height grows as $N^2$, the energy (their product) as $N$.

**Proof.** Complex amplitudes $a\eu^{\iu k\varphi}$, $k = 0, \dots, N - 1$: a geometric series, $a(1 - \eu^{\iu N\varphi})/(1 - \eu^{\iu\varphi}) = a\eu^{\iu(N - 1)\varphi/2}\sin(N\varphi/2)
/\sin(\varphi/2)$. Zeros where $\sin(N\varphi/2) = 0$ but $\sin(\varphi/2) \ne 0$: $\varphi = 2\pi
m/N$, $m$ not a multiple of $N$. In the phasor picture the $N$ arrows close into a regular polygon at the zeros and align at the principal maxima. ∎

![Left: the phasor sum of five equal waves — a regular fan whose chord is the resultant; it closes into a pentagon, and vanishes, at = 2π/5. Right: the intensity for 2, 5 and 12 waves, normalized to its peak: the principal maxima sharpen as 1/N.](https://one-course.com/images/onecourse/chapters/physics-4/b2-gratings/fig-c62cb7fcf338.svg)

*Left: the phasor sum of five equal waves — a regular fan whose chord is the resultant; it closes into a pentagon, and vanishes, at $\varphi = 2\pi/5$. Right: the intensity for $2$, $5$ and $12$ waves, normalized to its peak: the principal maxima sharpen as $1/N$.*

## 21.2 The diffraction grating

**Proposition 21.2 (Grating equation).**

A *grating* is a set of $N$ identical, equidistant, parallel slits (or grooves) of period $a$. A plane wave incident at the angle $\theta_0$ from the normal is re-emitted by each slit; in the direction $\theta$ the waves from neighbouring slits have the [path difference](https://one-course.com/books/physics/4/en/chapter/19-two-wave-interference#thm-b2-two-wave-interference-formula) $\delta = a(\sin\theta
- \sin\theta_0)$, and they reinforce in the directions

$$
a\,(\sin\theta_p - \sin\theta_0) = p\lambda , \qquad p = 0, \pm1, \pm2, \dots
$$

— the *orders* of the grating. The zero order is the undeviated beam for every wavelength; each other order spreads the wavelengths, with the *[angular dispersion](#prop-b2-gratings-equation)*

$$
\frac{\dd\theta}{\dd\lambda} = \frac{p}{a\cos\theta} ,
$$

larger for a finer grating and a higher order; the maximum order is $|p| \le a(1 + |\sin\theta_0|)/\lambda$. A *reflection grating* (grooves ruled on a mirror) obeys the same equation with the reflected angles.

**Proof.** The phase between successive slits is $\varphi = 2\pi\delta/\lambda$; the principal maxima of [Theorem 21.1](#thm-b2-gratings-nwaves) are at $\delta = p\lambda$. Differentiate $a\sin\theta = p\lambda + a\sin\theta_0$ at fixed $p$, $\theta_0$. ∎

![Left: the grating — neighbouring slits re-emit with the path difference a at normal incidence. Right: on a screen, the white zero order and the spectra of the orders ±1, ±2, violet nearest the centre, wider (and eventually overlapping) as p grows.](https://one-course.com/images/onecourse/chapters/physics-4/b2-gratings/fig-9394d0051541.svg)

*Left: the grating — neighbouring slits re-emit with the [path difference](https://one-course.com/books/physics/4/en/chapter/19-two-wave-interference#thm-b2-two-wave-interference-formula) $a\sin\theta$ at normal incidence. Right: on a screen, the white zero order and the spectra of the orders $\pm1$, $\pm2$, violet nearest the centre, wider (and eventually overlapping) as $p$ grows.*

**Example 21.3 (A classroom grating and a CD).**

A grating of $600$ lines per millimetre ($a = 1.67\,\text{µ}\mathrm{m}$) at normal incidence sends $500\,\mathrm{nm}$ light to $\sin\theta = 0.3$, $\theta = 17.5{}^{\circ}$ in the first order, $36.9{}^{\circ}$ in the second, and has no third order ($3 \times 0.3 > 1$); its first-order spectrum runs from $13.9{}^{\circ}$ ($400\,\mathrm{nm}$) to $24.8{}^{\circ}$ ($700\,\mathrm{nm}$). A compact disc, $a =
1.6\,\text{µ}\mathrm{m}$, is such a grating in reflection — hence the rainbow, and the fact that a DVD ($a = 0.74\,\text{µ}\mathrm{m}$) spreads it wider. The second-order red ($700\,\mathrm{nm}$ at $\sin\theta = 0.84$) overlaps the third-order violet: orders overlap beyond the first, a nuisance cured with filters or a prism.

## 21.3 Resolving power

**Proposition 21.4 (Resolving power of a grating).**

Two close wavelengths $\lambda$ and $\lambda + \Delta\lambda$ are just resolved in order $p$ when the [principal maximum](#thm-b2-gratings-nwaves) of one falls on the first zero of the other (Rayleigh’s criterion), which happens for

$$
\frac\lambda{\Delta\lambda} = pN :
$$

the *resolving power* is the order times the number of lines lit. It can also be written $W(\sin\theta - \sin\theta_0)/\lambda$ with $W = Na$ the width of the grating: the largest [path difference](https://one-course.com/books/physics/4/en/chapter/19-two-wave-interference#thm-b2-two-wave-interference-formula) across the grating, in wavelengths — which is why large spectrographs use gratings tens of centimetres wide and work at grazing angles.

**Proof.** The maximum of order $p$ for $\lambda + \Delta\lambda$ is at $\varphi = 2\pi p$ for that wavelength, i.e. at $\delta = p(\lambda + \Delta\lambda)$; the first zero of the maximum for $\lambda$ is at $\delta = p\lambda + \lambda/N$. Equating: $p\Delta\lambda = \lambda/N$. Then $pN =
Na(\sin\theta - \sin\theta_0)/\lambda$. ∎

**Example 21.5 (Separating the sodium lines).**

The doublet, $\Delta\lambda = 0.6\,\mathrm{nm}$ at $589\,\mathrm{nm}$, needs $R = 1000$: a thousand lines in the first order, $2\,\mathrm{mm}$ of a $600\,\mathrm{lines}/\mathrm{mm}$ grating. Its own line width, $0.02\,\mathrm{nm}$, needs $R = 30\,000$: a $5\,\mathrm{cm}$ grating in the first order, or $2.5\,\mathrm{cm}$ in the second — and a $10\,\mathrm{cm}$ grating of $1200\,\mathrm{lines}/\mathrm{mm}$ in the second order reaches $240\,000$, resolving $0.0025\,\mathrm{nm}$: the width of a line broadened by the thermal motion of the atoms, and the Doppler shift of a star moving at $1\,\mathrm{km}/\mathrm{s}$.

![Rayleigh’s criterion: the peak of one wavelength on the first zero of the other; the sum (dashed) shows a dip of about 20\% between two resolved lines.](https://one-course.com/images/onecourse/chapters/physics-4/b2-gratings/fig-4f2ea9bc201f.svg)

*Rayleigh’s criterion: the peak of one wavelength on the first zero of the other; the sum (dashed) shows a dip of about $20\%$ between two resolved lines.*

## 21.4 The Fabry–Pérot cavity

**Proposition 21.6 (Multiple-wave interference by division of amplitude).**

Two parallel partially reflecting mirrors (intensity reflectance $R$) a distance $L$ apart transmit a wave at normal incidence only in so far as the waves that have bounced $0, 1, 2, \dots$ times between them add in phase: with $\varphi = 4\pi L/\lambda$ the round-trip phase, the transmitted intensity is (Airy’s function)

$$
\frac{I_t}{I_0} = \frac1{1 + F\sin^2(\varphi/2)} , \qquad F = \frac{4R}{(1 - R)^2} :
$$

sharp peaks of full transmission at $2L = p\lambda$ (the *resonances* of the cavity, spaced in frequency by the *[free spectral range](#prop-b2-gratings-fabryperot)* $c/2L$), of relative width $1/\mathcal F$ with the *finesse* $\mathcal F = \pi
\sqrt R/(1 - R)$ — $30$ for $R = 0.9$, $300$ for $R = 0.99$. It is the grating’s equal in resolving power ($p\mathcal F$ with $p = 2L/\lambda \sim 10^5$, hence $R \sim
10^7$) and, closed on itself, the resonator of the laser ([Chapter 23](https://one-course.com/books/physics/4/en/chapter/23-the-laser-stimulated-emission-and-gaussian-beams#ch-b2-laser)).

**Proof.** Each round trip multiplies the amplitude by $R\eu^{\iu\varphi}$ (up to a constant phase); the transmitted amplitude is $t^2a\sum(R\eu^{\iu\varphi})^k =
t^2a/(1 - R\eu^{\iu\varphi})$ with $t^2 = 1 - R$; its squared modulus is $(1 - R)^2/
(1 - 2R\cos\varphi + R^2) = (1 - R)^2/[(1 - R)^2 + 4R\sin^2(\varphi/2)]$. Peaks at $\varphi = 2\pi p$; half-maximum where $F\sin^2(\varphi/2) = 1$, i.e. $\Delta\varphi \approx
4/\sqrt F = 2(1 - R)/\sqrt R$, against the $2\pi$ between peaks. ∎

![The Airy transmission of a Fabry–Pérot cavity: full transmission at the resonances 2L = p, peaks narrowing as the mirrors’ reflectance grows.](https://one-course.com/images/onecourse/chapters/physics-4/b2-gratings/fig-cd76e0309f1b.svg)

*The Airy transmission of a Fabry–Pérot cavity: full transmission at the resonances $2L = p\lambda$, peaks narrowing as the mirrors’ reflectance grows.*

**Method 21.7 (Grating calculations).**

(1) Period $a = 1/(\text{lines per mm})$; [grating equation](#prop-b2-gratings-equation) with the signs of the angles. (2) List the orders that exist ($|\sin\theta| \le 1$) and check their overlap ($p\lambda_{\max}$ against $(p + 1)\lambda_{\min}$). (3) Dispersion $p/a\cos\theta$, converted to a position on the detector with the focal length: $\dd x = f\,\dd\theta$. (4) Resolving power $pN$ with $N$ the lines actually illuminated; compare $\lambda/R$ with the structure to be resolved. (5) For real instruments add the entrance slit’s width (it blurs the lines) and the detector’s pixels.

## 21.5 Exercises

**Exercise 21.1 ★.**

A grating of $500$ lines per millimetre at normal incidence, $\lambda =
550\,\mathrm{nm}$: angles of the orders; highest order; angular width of the first-order visible spectrum ($400$–$700\,\mathrm{nm}$); same grating at $30{}^{\circ}$ incidence: orders on each side.

**Solution of Exercise 21.1.**

$a = 2\,\text{µ}\mathrm{m}$, $\sin\theta_p = 0.275p$: $16.0{}^{\circ}$, $33.4{}^{\circ}$, $55.6{}^{\circ}$; no fourth order. First-order visible from $11.5{}^{\circ}$ to $20.5{}^{\circ}$: $9{}^{\circ}$. At $30{}^{\circ}$: $\sin\theta = 0.5 + 0.275p$: $p = +1$ ($51{}^{\circ}$) on one side, $p = -1$ to $-5$ ($13^\circ$, $-3^\circ$, $-19^\circ$, $-37^\circ$, $-61^\circ$) on the other.

**Exercise 21.2 ★.**

The sodium doublet with a $600$-lines/mm grating and a lens of $f =
50\,\mathrm{cm}$: angular and linear separation of the two lines on the detector in the first and second orders; pixel size needed to see them apart.

**Solution of Exercise 21.2.**

$a = 1.67\,\text{µ}\mathrm{m}$. Order 1: $\theta = 20.7{}^{\circ}$, $\Delta\theta = \Delta\lambda/a\cos\theta =
3.9 \times 10^{-4}\,\mathrm{rad}$, $0.19\,\mathrm{mm}$; order 2: $\theta = 45{}^{\circ}$, $1.0 \times 10^{-3}\,\mathrm{rad}$, $0.51\,\mathrm{mm}$. Pixels of $50\,\text{µ}\mathrm{m}$ or less.

**Exercise 21.3 ★.**

Resolving power of a $2\,\mathrm{cm}$ grating of $600$ lines/mm in orders $1$, $2$, $3$; smallest $\Delta\lambda$ resolved at $550\,\mathrm{nm}$; does it separate the sodium doublet? The hydrogen H$\alpha$ line from its deuterium twin ($0.18\,\mathrm{nm}$ apart)?

**Solution of Exercise 21.3.**

$N = 12\,000$: $R = 12\,000$, $24\,000$, $36\,000$; $\Delta\lambda = 0.046$, $0.023$, $0.015\,\mathrm{nm}$; the doublet ($0.6\,\mathrm{nm}$) and the H/D pair ($0.18\,\mathrm{nm}$, needing $R = 3600$) are easily separated.

**Exercise 21.4 ★.**

A compact disc ($a = 1.6\,\text{µ}\mathrm{m}$) and a DVD ($a = 0.74\,\text{µ}\mathrm{m}$) lit at normal incidence: angles of first-order violet and red; which orders exist for each; why does the CD show two rainbows and the DVD one?

**Solution of Exercise 21.4.**

CD: violet $14.5{}^{\circ}$, red $25.9{}^{\circ}$; red exists to order 2, violet to order 4: two (partly overlapping) rainbows. DVD: violet $32.7{}^{\circ}$, red $71{}^{\circ}$; red has only the first order: one rainbow, wider.

**Exercise 21.5 ★★.**

*The $N$-slit pattern.* (a) Show that between two principal maxima there are $N - 1$ zeros and $N - 2$ secondary maxima. (b) For large $N$, intensity of the first [secondary maximum](#thm-b2-gratings-nwaves) relative to the principal one (at $N\varphi/2 \approx 3\pi/2$): about $1/22$. (c) Sketch the pattern for $N = 3$ and $N = 6$ and give the heights of the secondary maxima for $N = 3$. (d) Fraction of the total energy in the principal maxima for large $N$ (compare the integrals of the peak, width $\propto 1/N$, height $\propto N^2$, with the rest).

**Solution of Exercise 21.5.**

(a) Zeros at $\varphi = 2\pi m/N$, $m = 1, \dots, N - 1$: $N - 1$ zeros, $N - 2$ maxima between them. (b) At $N\varphi/2 = 3\pi/2$, $I \approx I_0/\sin^2(3\pi/2N) \approx I_0(2N/3\pi)^2$: $4/9\pi^2 = 0.045$ of $N^2I_0$. (c) $N = 3$: a single [secondary maximum](#thm-b2-gratings-nwaves) at $\varphi = \pi$ of height $I_0$, one ninth of the principal $9I_0$; $N = 6$: four secondary maxima of a few percent. (d) The central lobe of $\sin^2x/x^2$ holds $90\%$ of the energy.

**Exercise 21.6 ★★.**

*Overlapping orders.* (a) Show that order $p$ of $\lambda_{\max}$ overlaps order $p + 1$ of $\lambda_{\min}$ when $p\lambda_{\max} > (p + 1)\lambda_{\min}$; for $400$–$700\,\mathrm{nm}$, from which order? (b) [Free spectral range](#prop-b2-gratings-fabryperot) in order $p$: $\Delta\lambda_{\text{FSR}}
= \lambda/p$. (c) A spectrograph works in order $20$ near $500\,\mathrm{nm}$ (an echelle grating): [free spectral range](#prop-b2-gratings-fabryperot); how is the overlap removed (cross-disperser)? (d) The Fabry–Pérot’s [free spectral range](#prop-b2-gratings-fabryperot) is $c/2L$: write it in wavelength and compare with the grating’s.

**Solution of Exercise 21.6.**

(a) $p > \lambda_{\min}/(\lambda_{\max} - \lambda_{\min}) = 1.33$: from $p = 2$. (b) $\lambda/p$. (c) $25\,\mathrm{nm}$; a prism crossed with the grating stacks the orders one above the other. (d) $\Delta\lambda = \lambda^2/2L$, a fraction of a nanometre — hundreds of times smaller than a grating’s.

**Exercise 21.7 ★★.**

*Blazed grating.* A reflection grating has its grooves tilted by the blaze angle $\beta$ so that each facet reflects specularly into the direction where the order $p$ is wanted. (a) For the Littrow mount ($\theta = \theta_0 = \beta$): show $2a\sin\beta = p\lambda$; blaze angle for first-order $500\,\mathrm{nm}$ with $1200$ lines/mm. (b) Why does the blaze send most of the light into one order (think of the envelope of a single facet’s diffraction, [Chapter 22](https://one-course.com/books/physics/4/en/chapter/22-fraunhofer-diffraction-and-spatial-filtering#ch-b2-diffraction))? (c) The same grating is blazed for $1\,\text{µ}\mathrm{m}$: in which order is it efficient at $500\,\mathrm{nm}$? (d) Resolving power of a $10\,\mathrm{cm}$ grating in Littrow at $\beta = 64{}^{\circ}$, $\lambda = 500\,\mathrm{nm}$ (use $R = 2W\sin\beta/\lambda$).

**Solution of Exercise 21.7.**

(a) $\sin\beta = \lambda/2a = 0.3$, $\beta = 17.5{}^{\circ}$. (b) The single facet diffracts into a broad lobe centred on its specular direction; the orders that fall inside it get the light. (c) Order 2 ($2 \times 500\,\mathrm{nm} = 1\,\text{µ}\mathrm{m}$ at the same angle). (d) $R = 2W\sin\beta/\lambda = 3.6 \times 10^5$.

**Exercise 21.8 ★★.**

*The slit.* A spectrometer’s entrance slit of width $s$ is imaged on the detector with magnification $1$; the grating ($600$ lines/mm, $5\,\mathrm{cm}$, order $1$, $f = 50\,\mathrm{cm}$) disperses. (a) Width of a line on the detector due to the slit alone for $s = 50\,\text{µ}\mathrm{m}$, in nm. (b) Width due to the grating’s resolving power. (c) Which dominates; slit width at which the two are equal. (d) Why not close the slit further?

**Solution of Exercise 21.8.**

(a) $\dd\lambda/\dd x = a\cos\theta/pf = 3.1\,\mathrm{nm}/\mathrm{mm}$: $50\,\text{µ}\mathrm{m}$ is $0.16\,\mathrm{nm}$. (b) $R = 30\,000$: $0.018\,\mathrm{nm}$. (c) The slit, by nine; equal at $s = 6\,\text{µ}\mathrm{m}$. (d) Diffraction at the slit and a starved detector.

**Exercise 21.9 ★★.**

*Fabry–Pérot.* Mirrors $R = 0.95$, $L = 5\,\mathrm{mm}$, $\lambda = 600\,\mathrm{nm}$. (a) Order $p$, [free spectral range](#prop-b2-gratings-fabryperot) in frequency and in wavelength. (b) Finesse; width of a transmission peak in frequency and wavelength. (c) Resolving power $p\mathcal F$; compare with the $10\,\mathrm{cm}$ grating. (d) Why is a Fabry–Pérot always used with a prefilter or a grating?

**Solution of Exercise 21.9.**

(a) $p = 2L/\lambda = 16\,700$; FSR $c/2L = 30\,\mathrm{GHz}$, $\lambda^2/2L = 0.036\,\mathrm{nm}$. (b) $\mathcal F
= \pi\sqrt{0.95}/0.05 = 61$; $0.5\,\mathrm{GHz}$, $0.6\,\mathrm{pm}$. (c) $p\mathcal F = 10^6$, four times the $10\,\mathrm{cm}$ grating’s. (d) Its [free spectral range](#prop-b2-gratings-fabryperot) is a fraction of a nanometre: the orders overlap unless the light is prefiltered to one of them.

**Exercise 21.10 ★★★.**

*The limit of a grating.* (a) Show that $R = pN = W(\sin\theta - \sin\theta_0)
/\lambda \le 2W/\lambda$: the resolving power cannot exceed the width of the grating in wavelengths, times two. (b) Interpret: the largest [path difference](https://one-course.com/books/physics/4/en/chapter/19-two-wave-interference#thm-b2-two-wave-interference-formula) between the extreme rays. (c) A $30\,\mathrm{cm}$ grating at $500\,\mathrm{nm}$: ultimate $R$; smallest resolvable velocity by Doppler shift ($\Delta\lambda/\lambda = v/c$). (d) Why do astronomers nevertheless reach $1\,\mathrm{m}/\mathrm{s}$ (think of measuring the position of a line to a fraction of its width, with many lines)?

**Solution of Exercise 21.10.**

(a) $|\sin\theta - \sin\theta_0| \le 2$. (b) The extreme rays differ in path by $W(\sin\theta - \sin\theta_0)$ at most $2W$. (c) $2 \times 0.3/5 \times 10^{-7} = 1.2 \times 10^6$; $c/R
= 250\,\mathrm{m}/\mathrm{s}$. (d) A line’s centre is located to a hundredth of its width, and a thousand lines average down by $\sqrt{1000}$: metres per second, with heroic stability.

**Exercise 21.11 ★★★.**

*Bragg.* A crystal is a three-dimensional grating: X-rays reflected by successive atomic planes a distance $d$ apart, at the glancing angle $\theta$, interfere constructively when $2d\sin\theta = p\lambda$ (Bragg). (a) Derive it from the [path difference](https://one-course.com/books/physics/4/en/chapter/19-two-wave-interference#thm-b2-two-wave-interference-formula) between two planes. (b) Copper K$\alpha$ radiation ($0.154\,\mathrm{nm}$) on rock salt ($d = 0.282\,\mathrm{nm}$): the Bragg angles. (c) Why do visible wavelengths give no Bragg reflection from crystals, and what does the converse (no X-ray diffraction from glass) tell about glass? (d) Why must $\lambda < 2d$?

**Solution of Exercise 21.11.**

(a) The wave reflected by the lower plane travels $2d\sin\theta$ more. (b) $\sin\theta = 0.273p$: $15.8{}^{\circ}$, $33.1{}^{\circ}$, $55.0{}^{\circ}$. (c) $\lambda \gg 2d$: no angle satisfies the equation; glass has no regular planes, hence only diffuse rings — it is amorphous. (d) $\sin\theta \le 1$.

**Exercise 21.12 ★★★.**

*Phased array.* $N$ identical antennas in a line, spaced $a$, fed with the same amplitude and a phase step $\psi$ between neighbours. (a) Show that the radiated field in the direction $\theta$ is that of the $N$-wave sum with $\varphi = 2\pi a\sin\theta/\lambda - \psi$: the main beam points where $\sin\theta = \lambda\psi/2\pi a$. (b) Half-width of the main beam, $\Delta\theta
\approx \lambda/Na\cos\theta$. (c) Why must $a \le \lambda/2$ to avoid a second main beam (a grating lobe) when steering? (d) A radar with $64$ elements at $\lambda/2$, $\lambda = 3\,\mathrm{cm}$: beam width; time to scan $90^\circ$ in steps of one beam width if the phases switch in $1\,\text{µ}\mathrm{s}$.

**Solution of Exercise 21.12.**

(a) Neighbours differ in phase by $2\pi a\sin\theta/\lambda$ from the path and $-\psi$ from the feed; the [principal maximum](#thm-b2-gratings-nwaves) at $\varphi = 0$. (b) First zero at $\Delta\varphi = 2\pi/N$: $\Delta\theta = \lambda/Na\cos\theta$. (c) A second [principal maximum](#thm-b2-gratings-nwaves) at $\varphi = \pm2\pi$ needs $|\sin\theta \pm \lambda/a| \le 1$: impossible for every steering angle only if $a \le \lambda/2$. (d) $Na = 0.96\,\mathrm{m}$: $1.8{}^{\circ}$; $50$ steps, $50\,\text{µ}\mathrm{s}$.

## 21.6 Problem: A spectrograph for a star

**Problem 21.1.**

Weekend problem — designing the spectrograph that reads a star’s light: the grating, the detector, the lines of hydrogen, and the wobble of a planet

A spectrograph: entrance slit, collimator of focal length $f_1 =
1.0\,\mathrm{m}$, a reflection grating of $1200$ lines/mm and width $W =
12\,\mathrm{cm}$, a camera of $f_2 = 1.0\,\mathrm{m}$, a detector with $15\,\text{µ}\mathrm{m}$ pixels. It works in the first order, around $550\,\mathrm{nm}$, with the grating at $20{}^{\circ}$ incidence.

**Part I — The grating.**

1. Period of the grating; number of lines lit if the beam covers the whole width.
2. Diffraction angle of $550\,\mathrm{nm}$ in the first order; of $400\,\mathrm{nm}$ and $700\,\mathrm{nm}$ ; angular width of the visible spectrum.
3. [Angular dispersion](#prop-b2-gratings-equation) at $550\,\mathrm{nm}$ ; linear dispersion on the detector in nm per mm and nm per pixel.
4. Resolving power; smallest $\Delta\lambda$ resolved at $550\,\mathrm{nm}$ ; in pixels.
5. Does the second-order violet fall onto the first-order visible? Which filter removes it?
6. Length of detector needed for the whole visible spectrum in the first order.
7. The grating is blazed at $17{}^{\circ}$ : for which wavelength is it most efficient in the first order at this incidence (use the Littrow estimate $2a\sin\beta = \lambda$ )?
8. The grating returns $80\%$ of the light at the blaze wavelength and the telescope plus optics $40\%$ : fraction of the star’s photons that reach the detector.

**Part II — Lines.**

9. The hydrogen Balmer lines of the star: H $\alpha$ $656.3\,\mathrm{nm}$ , H $\beta$ $486.1\,\mathrm{nm}$ , H $\gamma$ $434.0\,\mathrm{nm}$ : their positions on the detector relative to $550\,\mathrm{nm}$ .
10. H $\alpha$ from a deuterium atom is $0.18\,\mathrm{nm}$ shorter: separated? By how many pixels?
11. The sodium doublet in the star’s spectrum: separation in pixels; and the width of each line if the star’s atmosphere broadens them to $0.05\,\mathrm{nm}$ .
12. The entrance slit is $100\,\text{µ}\mathrm{m}$ wide, imaged at magnification $f_2/f_1$ : its width on the detector in pixels and in nm; is the spectrograph slit-limited or grating-limited?
13. The light from the telescope comes as a disc of $1''$ on the sky, which the telescope ( $f = 20\,\mathrm{m}$ ) makes $100\,\text{µ}\mathrm{m}$ wide at the slit: fraction of light lost if the slit is narrowed to $50\,\text{µ}\mathrm{m}$ (take the disc as uniform).
14. Why does a narrower slit give sharper lines but noisier spectra?

**Part III — The star moves.**

15. A star recedes at $30\,\mathrm{km}/\mathrm{s}$ : Doppler shift of H $\alpha$ in nm and in pixels.
16. The Earth’s orbital motion adds $\pm30\,\mathrm{km}/\mathrm{s}$ over the year: why must it be subtracted, and to what precision if one wants $100\,\mathrm{m}/\mathrm{s}$ ?
17. A planet makes its star wobble at $10\,\mathrm{m}/\mathrm{s}$ : shift in nm and in pixels; is it below the resolution? Explain how measuring the centre of a line to a hundredth of its width, over a thousand lines, recovers it (statistical gain $\sqrt{N}$ ).
18. The temperature of the instrument changes by $1\,\mathrm{K}$ , and the grating (glass, expansion $1 \times 10^{-5}\,\mathrm{K}^{-1}$ ) dilates: shift of the lines in pixels; what stability is needed for $10\,\mathrm{m}/\mathrm{s}$ ?
19. A calibration lamp (thorium–argon) puts hundreds of known lines on the same detector: explain its role.

**Part IV — Alternatives.**

20. The same width of grating ruled at $300$ lines/mm and used in the fourth order (with a cross-disperser to sort the orders): diffraction angle, resolving power, dispersion and [free spectral range](#prop-b2-gratings-fabryperot) at $550\,\mathrm{nm}$ ; what has changed, what has not?
21. A Fabry–Pérot etalon ( $L = 1\,\mathrm{mm}$ , $R = 0.9$ ) placed before the slit: [free spectral range](#prop-b2-gratings-fabryperot) in nm, finesse, peak width; what does its comb of transmission peaks look like on the detector, and what is it used for?
22. A prism of $60{}^{\circ}$ in glass with $\dd n/\dd\lambda = -1 \times 10^{-4}\,\mathrm{nm}^{-1}$ has the [angular dispersion](#prop-b2-gratings-equation) $\dd D/\dd\lambda \approx 2\,\dd n/\dd\lambda$ near minimum deviation: compare with the grating’s; why did gratings replace prisms?
23. Why are the largest astronomical spectrographs built with gratings a metre wide and used at large angles (the limit $2W/\lambda$ )?
24. A magnitude-8 star delivers $1 \times 10^{4}\,$ photons per second and per nanometre at the telescope: photons per pixel per second on the detector, and the signal-to-noise ratio of a $100\,\mathrm{s}$ exposure (photon noise only).
25. Summarize: the three numbers that describe a spectrograph (dispersion, resolving power, [free spectral range](#prop-b2-gratings-fabryperot) ) and what sets each.

**Solution of Problem 21.1.**

**1.** $a = 0.833\,\text{µ}\mathrm{m}$; $N = 144\,000$.

**2.** $\sin\theta = \sin20^\circ - \lambda/a$ (the order lies on the other side of the normal): $550\,\mathrm{nm}$ at $18.5{}^{\circ}$, $400\,\mathrm{nm}$ at $7.9{}^{\circ}$, $700\,\mathrm{nm}$ at $29.9{}^{\circ}$: $22{}^{\circ}$ of spectrum.

**3.** $1/a\cos\theta = 1.27 \times 10^{-3}\,\mathrm{rad}/\mathrm{nm}$; $f_2\times$: $1.27\,\mathrm{mm}/\mathrm{nm}$, i.e. $0.79\,\mathrm{nm}/\mathrm{mm}$, $0.012\,\mathrm{nm}$ per pixel.

**4.** $R = N = 144\,000$: $0.0038\,\mathrm{nm}$, a third of a pixel.

**5.** Second-order $400\,\mathrm{nm}$ lands where first-order $800\,\mathrm{nm}$ would, beyond the red: no overlap in the visible; the overlap would start at $350\,\mathrm{nm}$, which a glass filter removes.

**6.** $22{}^{\circ}$ $\times$ $1\,\mathrm{m}$ $= 38\,\mathrm{cm}$: a detector covers a slice; the grating is turned to choose it.

**7.** $\lambda = 2a\sin17^\circ = 490\,\mathrm{nm}$.

**8.** $0.8 \times 0.4 = 32\%$.

**9.** Angles $26.5{}^{\circ}$, $14.0{}^{\circ}$, $10.3{}^{\circ}$: $+14$, $-8$ and $-14\,\mathrm{cm}$ from the $550\,\mathrm{nm}$ position.

**10.** $0.18\,\mathrm{nm}$ is $15$ pixels: separated.

**11.** $50$ pixels apart, each $4$ pixels wide.

**12.** $100\,\text{µ}\mathrm{m}$, $6.7$ pixels, $0.08\,\mathrm{nm}$ — twenty times the grating’s limit: slit-limited.

**13.** The image is $97\,\text{µ}\mathrm{m}$ wide: a $50\,\text{µ}\mathrm{m}$ slit passes half the light.

**14.** Less light for the same sharpness: the photon noise grows.

**15.** $\Delta\lambda = \lambda v/c = 0.066\,\mathrm{nm}$, $5.5$ pixels.

**16.** The Earth’s velocity shifts every line by up to $\pm5.5$ pixels through the year; it is known from the ephemerides to far better than the $0.3\%$ needed for $100\,\mathrm{m}/\mathrm{s}$.

**17.** $2.2 \times 10^{-5}\,\mathrm{nm}$, $0.002$ pixel — far below the line width. A centroid to a hundredth of a $7\,\mathrm{px}$ line is $0.07\,\mathrm{px}$; a thousand lines give $\sqrt{1000}$: $0.002$ pixel — the signal, just. Real instruments add resolving power and stability.

**18.** $\Delta\sin\theta = (\lambda/a) \times 10^{-5} = 6.6 \times 10^{-6}$, $\Delta\theta = 7 \times 10^{-6}\,\mathrm{rad}$, $7\,\text{µ}\mathrm{m}$: half a pixel per kelvin — millikelvins for $10\,\mathrm{m}/\mathrm{s}$.

**19.** Its lines, recorded at the same instant on the same pixels, give the wavelength scale and track every drift.

**20.** $a = 3.33\,\text{µ}\mathrm{m}$, $N = 36\,000$, same angle ($18.5{}^{\circ}$), $R = 4N
= 144\,000$, same dispersion, FSR $\lambda/4 = 137\,\mathrm{nm}$: resolving power and dispersion depend on $W$ and the angle only; the [free spectral range](#prop-b2-gratings-fabryperot) shrank.

**21.** FSR $\lambda^2/2L = 0.15\,\mathrm{nm}$ ($12$ pixels); $\mathcal F = 30$; peaks $0.005\,\mathrm{nm}$ wide: a comb of sharp lines across the detector, a wavelength ruler.

**22.** $2 \times 10^{-4}$ against $1.3 \times 10^{-3}\,\mathrm{rad}/\mathrm{nm}$: six times less, and a prism’s resolving power ($b\,\dd n/\dd\lambda \sim 10^4$) and non-linear scale lose too; gratings also work in the ultraviolet.

**23.** $R \le 2W/\lambda$: $4 \times 10^6$ for a metre at $500\,\mathrm{nm}$ — width and angle are the only ways up.

**24.** $10^4 \times 0.012 \times 0.32 = 38$ photons per pixel per second; $3800$ in $100\,\mathrm{s}$: SNR $\approx 60$.

**25.** Dispersion, set by $p/a\cos\theta$ and the camera’s focal length; resolving power, by the illuminated width and the angles ($pN$); [free spectral range](#prop-b2-gratings-fabryperot), by the order ($\lambda/p$).