---
title: "Guided Waves and Cavities"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/16-guided-waves-and-cavities
---

# Chapter 16 — Guided Waves and Cavities

A radar’s pulse travels from the transmitter to the dish inside a rectangular copper pipe; the light of an internet link travels ten thousand kilometres inside a glass thread thinner than a hair; the microwaves of an oven bounce between its walls and settle into a pattern of hot and cold spots. A wave confined by conducting or refracting walls is no longer a plane wave free to go anywhere: it must satisfy the boundary conditions on the walls, and that selects the shapes it may take — the *modes* — and forbids it below a *cut-off* frequency. This chapter builds the simplest guide, two parallel conducting plates, from the reflections of [Chapter 15](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#ch-b2-wave-interfaces); then the [rectangular waveguide](#prop-b2-guided-waves-rectangular), the closed cavity, and, in the ray picture, the [optical fibre](#prop-b2-guided-waves-fibre).

## 16.1 Waves between two conducting plates

**Proposition 16.1 (Modes of the parallel-plate guide).**

Two perfectly conducting planes $y = 0$ and $y = a$ bound a vacuum. Seek a wave travelling along $z$ with $\vect E = E(y)\,\eu^{\iu(\omega t - k_gz)}
\vect e_x$ (the field parallel to the plates, perpendicular to the propagation). [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) in vacuum and the condition $E = 0$ on the plates impose

$$
E(y) = E_0\sin\frac{n\pi y}a , \qquad
k_g^2 = \frac{\omega^2}{c^2} - \Bigl(\frac{n\pi}a\Bigr)^2 , \qquad n = 1, 2, \dots
$$

Mode $n$ propagates only above its *cut-off frequency* $f_n = nc/2a$; below, $k_g$ is imaginary and the field decays along the guide. Its phase and group velocities are

$$
v_\varphi = \frac c{\sqrt{1 - f_n^2/f^2}} > c , \qquad
v_g = c\sqrt{1 - f_n^2/f^2} < c , \qquad v_\varphi v_g = c^2 ,
$$

and its magnetic field has components along $y$ and along $z$: a [guided wave](#prop-b2-guided-waves-plates) is not transverse in the direction of propagation.

**Proof.** Insert in $\Delta\vect E = \partial_t^2\vect E/c^2$: $E'' - k_g^2E = -(\omega^2/c^2)E$, i.e. $E'' + (\omega^2/c^2 - k_g^2)E = 0$, whose solutions vanishing at $y = 0$ and $y = a$ are the sines with $\omega^2/c^2 - k_g^2 = (n\pi/a)^2$; $\operatorname{div}
\vect E = 0$ holds since $E_x$ does not depend on $x$. Faraday gives $\underline B_y = (k_g/\omega)\underline E$ and $\underline B_z = (\iu/\omega)\partial_y\underline E$, the latter in quadrature. The velocities follow from the Klein–Gordon form of the [dispersion relation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) ([Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets)). ∎

**Remark 16.2 (The mode as two plane waves).**

$\sin(n\pi y/a)\eu^{-\iu k_gz}$ is the sum of two plane waves with wavevectors $(0, \pm n\pi/a, k_g)$, of modulus $\omega/c$, bouncing between the plates at the angle $\theta$ from the axis with $\cos\theta = k_gc/\omega$: a guided mode is a plane wave zigzagging between the walls, reflected with $r = -1$ at each one, the transverse [standing wave](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes) being the condition that it interferes constructively with itself. The energy travels along the zigzag at $c$, hence along the axis at $c\cos\theta = v_g$; the crests’ intersections with the axis run at $c/\cos\theta = v_\varphi$. At cut-off the wave bounces back and forth across the guide and goes nowhere.

![Left: a guided mode is a plane wave bouncing between the plates; the angle opens as the frequency approaches cut-off. Right: the transverse profiles E(y) of the first two modes, vanishing on the conducting walls.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/fig-8c498f244107.svg)

*Left: a guided mode is a plane wave bouncing between the plates; the angle $\theta$ opens as the frequency approaches cut-off. Right: the transverse profiles $E(y)$ of the first two modes, vanishing on the conducting walls.*

![Left: the dispersion curves of the first three modes of a guide, each a hyperbola starting at its cut-off. Right: phase and group velocities of a mode against frequency; the energy crawls near cut-off and approaches c far above.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/fig-d963aa069bd3.svg)

![Left: the dispersion curves of the first three modes of a guide, each a hyperbola starting at its cut-off. Right: phase and group velocities of a mode against frequency; the energy crawls near cut-off and approaches c far above.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/fig-6e42fc7b9cb7.svg)

*Left: the dispersion curves of the first three modes of a guide, each a hyperbola starting at its cut-off. Right: phase and group velocities of a mode against frequency; the energy crawls near cut-off and approaches $c$ far above.*

**Example 16.3 (Numbers).**

Plates $3\,\mathrm{cm}$ apart: $f_1 = 5\,\mathrm{GHz}$, $f_2 = 10\,\mathrm{GHz}$; between those frequencies only one mode propagates (*single-mode* operation, which keeps a pulse from splitting into several with different $v_g$). At $8\,\mathrm{GHz}$ the mode $n = 1$ has $v_g = 0.78c$, $v_\varphi =
1.28c$, and the guide wavelength $\lambda_g = 2\pi/k_g = 4.8\,\mathrm{cm}$ instead of $3.75\,\mathrm{cm}$ in free space. At $4\,\mathrm{GHz}$ nothing passes: the field decays as $\eu^{-z/\ell}$ with $\ell = c/2\pi\sqrt{f_1^2 - f^2} = 1.6\,\mathrm{cm}$ — the principle of the mesh in an oven door, of the waveguide below cut-off used as a calibrated attenuator, and of the metal ducts of ventilation that let air but no radio through.

## 16.2 The rectangular waveguide and the cavity

**Proposition 16.4 (The rectangular guide; the TE10_{10}10​ mode).**

A hollow metal pipe of rectangular section $a \times b$ ($a > b$) carries modes whose cut-off frequencies are

$$
f_{mn} = \frac c2\sqrt{\Bigl(\frac ma\Bigr)^2 + \Bigl(\frac nb\Bigr)^2} ;
$$

the lowest, the *TE$_{10}$ mode* ($f_{10} = c/2a$), has $\vect E = E_0\sin
(\pi x/a)\,\eu^{\iu(\omega t - k_gz)}\,\vect e_y$: one half-sine across the wide side, uniform across the narrow side, vanishing on the two walls $x = 0$, $x = a$ — exactly the parallel-plate mode, the side walls $y = 0$, $b$ being perpendicular to $\vect E$ and imposing nothing on it. The guide is used between $f_{10}$ and the next cut-off; its magnetic field lines close in the $xz$ plane and the currents they induce in the walls are what a slot must not cut. The power carried is $\tfrac14\varepsilon_0cE_0^2\,ab
\sqrt{1 - f_{10}^2/f^2}$.

**Proof.** The general mode is $\cos$/$\sin$ products in $x$ and $y$ with the boundary conditions on the four walls, giving $k_g^2 = \omega^2/c^2 - (m\pi/a)^2
- (n\pi/b)^2$ (admitted in general; the TE$_{10}$ case is the proposition above). Power: the mean [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) along $z$ is $E_0^2\sin^2(\pi x/a)
\,k_g/2\mu_0\omega$ integrated over the section. ∎

![The TE_10 mode of a rectangular guide: the electric field spans the narrow dimension with a half-sine profile across the wide one; the magnetic field closes in loops in the plane of the wide faces, a half guide wavelength long.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/fig-a4737721b108.svg)

*The TE$_{10}$ mode of a rectangular guide: the electric field spans the narrow dimension with a half-sine profile across the wide one; the magnetic field closes in loops in the plane of the wide faces, a half guide wavelength long.*

**Proposition 16.5 (Cavity resonator).**

Closing a guide with two conducting walls a length $d$ apart turns it into a *cavity*, in which the [guided wave](#prop-b2-guided-waves-plates) must form a [standing wave](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes) along $z$ too: $k_g = p\pi/d$, and the resonance frequencies of a rectangular box $a \times b \times d$ are

$$
f_{mnp} = \frac c2\sqrt{\Bigl(\frac ma\Bigr)^2 + \Bigl(\frac nb\Bigr)^2 + \Bigl(\frac pd\Bigr)^2} .
$$

The number of modes with frequency below $f$ grows as $8\pi V f^3/3c^3$ for a box of volume $V$ (admitted: count the points of the lattice $(m/2a, n/2b, p/2d)$ in an eighth of a sphere, with two polarizations) — the density of modes that the theory of thermal radiation will need ([Chapter 26](https://one-course.com/books/physics/4/en/chapter/26-thermal-radiation#ch-b2-thermal-radiation)). Each mode has a quality factor $Q$ set by the wall losses (the surface resistance of [Chapter 14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media)), typically $10^3$–$10^4$ for copper at microwave frequencies, $10^{10}$ for a superconducting cavity.

**Proof.** [Standing waves](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes) along the three directions with nodes on the six walls; the wavevector $(m\pi/a, n\pi/b, p\pi/d)$ has modulus $\omega/c$. ∎

**Example 16.6 (Ovens, radars, clocks).**

A microwave oven of $30 \times 30 \times 20$ cm$^3$ near $2.45\,\mathrm{GHz}$ has dozens of modes within a few percent of the magnetron’s frequency: the field is a superposition of [standing waves](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes), with maxima $6\,\mathrm{cm}$ apart, which the turntable (and a "mode stirrer") smooth out. A radar’s magnetron is itself a [cavity resonator](#prop-b2-guided-waves-cavity); a particle accelerator is a chain of superconducting cavities whose TM mode pushes the bunches; the atoms of a caesium clock cross a microwave cavity tuned to $9.19\,\mathrm{GHz}$.

## 16.3 The optical fibre in the ray picture

**Proposition 16.7 (Step-index fibre).**

A glass core of index $n_1$ surrounded by a cladding of slightly lower index $n_2$ guides light by [total internal reflection](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#thm-b2-wave-interfaces-snell) at the core boundary. A ray entering the end face from air at the angle $\alpha$ from the axis is trapped if $\sin\alpha < \sqrt{n_1^2 - n_2^2}$, the *[numerical aperture](#prop-b2-guided-waves-fibre)*; rays of different angles travel different lengths, and a pulse entering a fibre of length $L$ is spread by

$$
\Delta t = \frac{n_1L}c\Bigl(\frac{n_1}{n_2} - 1\Bigr) ,
$$

the *[modal dispersion](#prop-b2-guided-waves-fibre)* — some $50\,\mathrm{ns}$ per kilometre for $n_1
- n_2 = 0.01$, which limits a multimode fibre to a few megabits per second over ten kilometres. A *single-mode* fibre has a core so thin ($9\,\text{µ}\mathrm{m}$) that only one mode propagates, like the guide between its first two cut-offs; what remains is the chromatic dispersion of the glass ([Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets)).

**Proof.** Total reflection at the core–cladding interface needs $\sin\theta >
n_2/n_1$ for the angle $\theta$ from the normal to the boundary, i.e. $\cos
\theta' > n_2/n_1$ for the angle $\theta'$ from the axis inside; refraction at the entrance gives $\sin\alpha = n_1\sin\theta'$, whence $\sin\alpha < n_1\sqrt{1
- n_2^2/n_1^2}$. The steepest trapped ray travels $L/\cos\theta' = Ln_1/n_2$ against $L$ for the axial one, at $c/n_1$. ∎

![A step-index fibre in the ray picture: rays within the acceptance cone are trapped by total internal reflection; the zigzag ray arrives after the axial one — modal dispersion.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/fig-4daa9055592a.svg)

*A step-index fibre in the ray picture: rays within the acceptance cone are trapped by [total internal reflection](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#thm-b2-wave-interfaces-snell); the zigzag ray arrives after the axial one — [modal dispersion](#prop-b2-guided-waves-fibre).*

**Method 16.8 (Guided-wave bookkeeping).**

(1) Identify the walls and their condition (tangential $\vect E = 0$ on a conductor; total reflection for a dielectric). (2) Write the field as a transverse [standing wave](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes) times a progressive factor along the guide; the boundary conditions quantize the transverse wavenumber. (3) The [dispersion relation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) $k_g^2 = \omega^2/c^2 - k_\perp^2$ gives the cut-off, $v_\varphi$, $v_g$, $\lambda_g$. (4) Count the propagating modes at the working frequency; aim for one. (5) For a cavity add the longitudinal condition and find the resonance frequencies; for the losses use the surface resistance.

## 16.4 Exercises

**Exercise 16.1 ★.**

Plates $2.0\,\mathrm{cm}$ apart: cut-off frequencies of the first three modes; at $12\,\mathrm{GHz}$, which modes propagate, and for the first one $k_g$, $\lambda_g$, $v_\varphi$, $v_g$; at $6\,\mathrm{GHz}$, the decay length of the field.

**Solution of Exercise 16.1.**

$f_n = nc/2a = 7.5$, $15$, $22.5\,\mathrm{GHz}$. At $12\,\mathrm{GHz}$ only $n = 1$: $k_g =
(2\pi/c)\sqrt{f^2 - f_1^2} = 196\,\mathrm{rad}/\mathrm{m}$, $\lambda_g = 3.2\,\mathrm{cm}$, $v_\varphi = c/0.78
= 3.8 \times 10^{8}\,\mathrm{m}/\mathrm{s}$, $v_g = 2.3 \times 10^{8}\,\mathrm{m}/\mathrm{s}$. At $6\,\mathrm{GHz}$: $\ell = c/2\pi\sqrt{f_1^2 -
f^2} = 1.1\,\mathrm{cm}$.

**Exercise 16.2 ★.**

A standard X-band waveguide has $a = 22.9\,\mathrm{mm}$, $b = 10.2\,\mathrm{mm}$. Cut-off of TE$_{10}$, of TE$_{20}$ and TE$_{01}$; the single-mode band; at $10\,\mathrm{GHz}$: $\lambda_g$, $v_g$, the angle of the zigzag. Why is $b \approx a/2$ a good choice?

**Solution of Exercise 16.2.**

TE$_{10}$ $6.55\,\mathrm{GHz}$, TE$_{20}$ $13.1\,\mathrm{GHz}$, TE$_{01}$ $14.7\,\mathrm{GHz}$: single mode from $6.55$ to $13.1\,\mathrm{GHz}$. At $10\,\mathrm{GHz}$: $\sqrt{1 - (6.55/10)^2} =
0.756$, $\lambda_g = 4.0\,\mathrm{cm}$, $v_g = 0.76c$, $\cos\theta = 0.756$, $\theta = 41{}^{\circ}$. $b \approx a/2$ keeps TE$_{01}$ above TE$_{20}$ (the widest single-mode band) while leaving the largest gap for the field (power before breakdown).

**Exercise 16.3 ★.**

A cavity of $30 \times 30 \times 20$ cm$^3$: frequencies of the modes $(1,1,0)$, $(1,0,1)$, $(2,1,1)$, $(3,2,1)$; number of modes below $2.5\,\mathrm{GHz}$ from the counting formula; the lowest mode of a $9.19\,\mathrm{GHz}$ cubic cavity — its side.

**Solution of Exercise 16.3.**

$f = 1.5 \times 10^8\sqrt{(m/0.3)^2 + (n/0.3)^2 + (p/0.2)^2}$: $0.71\,\mathrm{GHz}$, $0.90\,\mathrm{GHz}$, $1.35\,\mathrm{GHz}$, $1.95\,\mathrm{GHz}$. $N = 8\pi Vf^3/3c^3 = 8\pi \times 0.018 \times 1.56 \times 10^{28}/
8.1 \times 10^{25} \approx 260$. Cube: $f = c\sqrt2/2L$, $L = 2.3\,\mathrm{cm}$.

**Exercise 16.4 ★.**

A fibre has $n_1 = 1.48$, $n_2 = 1.46$. [Numerical aperture](#prop-b2-guided-waves-fibre) and acceptance half-angle; modal spread per kilometre; maximum bit rate over $2\,\mathrm{km}$ if one bit must last longer than the spread; the same with $n_1 - n_2 = 0.003$.

**Solution of Exercise 16.4.**

$\mathrm{NA} = \sqrt{1.48^2 - 1.46^2} = 0.24$, $14{}^{\circ}$; $\Delta t = (n_1L/c)(n_1/n_2
- 1) = 68\,\mathrm{ns}/\mathrm{km}$; $135\,\mathrm{ns}$ over $2\,\mathrm{km}$: $7\,\mathrm{Mbit}/\mathrm{s}$; with $0.003$: $10\,\mathrm{ns}/\mathrm{km}$, $50\,\mathrm{Mbit}/\mathrm{s}$.

**Exercise 16.5 ★★.**

For the parallel-plate mode $n = 1$: (a) find $\vect B$ from Faraday’s law and show it has a $z$ component in quadrature with $E$. (b) [Surface current](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#prop-b2-maxwell-equations-boundary) on each plate (boundary relation). (c) Mean [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) along $z$ and the power per unit width; check that its transverse component averages to zero. (d) Show that the energy velocity, power over mean energy per unit length, equals $v_g$.

**Solution of Exercise 16.5.**

(a) $\underline B_y = (k_g/\omega)\underline E$, $\underline B_z = (\iu/\omega)\partial_y\underline E = (\iu\pi/a\omega)
E_0\cos(\pi y/a)\eu^{\iu(\omega t - k_gz)}$: in quadrature. (b) At $y = 0$, $B_y = 0$ and $\vect j_s = \vect e_y\wedge\vect B/\mu_0 = (B_z/\mu_0)\vect e_x$: along the field. (c) $\langle\Pi_z\rangle = E_0^2k_g\sin^2(\pi y/a)/2\mu_0\omega$, power per unit width $E_0^2k_ga/4\mu_0\omega$; $\langle\Pi_y\rangle \propto \operatorname{Re}(\underline E\,\iu\underline B_z^*)
= 0$. (d) Mean energy per unit length and width $\tfrac14\varepsilon_0E_0^2a$ (electric and magnetic halves each $\tfrac18\varepsilon_0E_0^2a$); the ratio is $c^2k_g/\omega = v_g$.

**Exercise 16.6 ★★.**

*Below cut-off.* (a) A $1\,\mathrm{mm}$ hole in the $0.5\,\mathrm{mm}$ mesh of an oven door at $2.45\,\mathrm{GHz}$: treat it as a guide of width $a = 1\,\mathrm{mm}$; decay length and [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) across the sheet in dB (compare [Chapter 14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media)). (b) A ventilation duct of $20\,\mathrm{cm}$ section in a shielded room: up to what frequency does it block radio, and by how much does it attenuate $100\,\mathrm{MHz}$ over $1\,\mathrm{m}$? (c) A "waveguide-beyond-cutoff attenuator" uses a tube of $1\,\mathrm{cm}$ diameter at $1\,\mathrm{GHz}$: [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) per centimetre (use the plate formula with $a =
1\,\mathrm{cm}$ as an estimate). (d) Why is the [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) independent of the wall conductivity?

**Solution of Exercise 16.6.**

(a) $f_1 = 150\,\mathrm{GHz} \gg f$: $\ell \approx a/\pi = 0.32\,\mathrm{mm}$; $\eu^{-0.5/0.32}$: $-14\,\mathrm{dB}$ in amplitude, $-27$ in power. (b) $f_1 = 750\,\mathrm{MHz}$; at $100\,\mathrm{MHz}$ $\ell = 6.4\,\mathrm{cm}$: $-136\,\mathrm{dB}$ over a metre. (c) $f_1 = 15\,\mathrm{GHz}$, $\ell =
3.2\,\mathrm{mm}$: $27\,\mathrm{dB}/\mathrm{cm}$. (d) The wave does not fit; evanescence is geometry, not dissipation.

**Exercise 16.7 ★★.**

*Power and breakdown.* The X-band guide of [Exercise 16.2](#exo-b2-guided-waves-2) carries TE$_{10}$ at $10\,\mathrm{GHz}$. (a) Power for $E_0 = 1\,\mathrm{MV}/\mathrm{m}$. (b) Air breaks down at $3\,\mathrm{MV}/\mathrm{m}$: maximum power; how do radars carry megawatts (pressurization, gases)? (c) Wall losses: the [surface current](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#prop-b2-maxwell-equations-boundary) at the broad wall is of order $E_0/\mu_0c$; with $R_s = 0.026\,\Omega$ for copper at $10\,\mathrm{GHz}$, estimate the loss per metre and the [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) in dB/m. (d) Compare with a [coaxial cable](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#prop-b2-dispersion-wave-packets-coax)’s $0.5\,\mathrm{dB}/\mathrm{m}$: why are guides used at these frequencies?

**Solution of Exercise 16.7.**

(a) $P = \tfrac14\varepsilon_0cE_0^2ab\sqrt{1 - f_{10}^2/f^2} = 120\,\mathrm{kW}$. (b) $\times9$: $1\,\mathrm{MW}$; pressurized nitrogen or SF$_6$ raise the breakdown field. (c) $j_s \approx E_0/\mu_0c = 2.7\,\mathrm{kA}/\mathrm{m}$; $\tfrac12R_sj_s^2 = 9 \times 10^{4}\,\mathrm{W}/\mathrm{m}^{2}$ over $2a =
0.046\,\mathrm{m}^{2}$ per metre: $4\,\mathrm{kW}/\mathrm{m}$, $3.5\%$ per metre, about $0.15\,\mathrm{dB}/\mathrm{m}$. (d) Three times less than the cable, and megawatts instead of kilowatts.

**Exercise 16.8 ★★.**

*Cavity quality.* A copper cubic cavity of side $10\,\mathrm{cm}$ in its lowest mode. (a) Frequency. (b) The stored energy is $\tfrac12\varepsilon_0E_0^2V/4$ (admitted) and the wall loss $\tfrac12R_sH_0^2\times$ area with $H_0 \approx E_0/
\mu_0c$ and $R_s = 1/\gamma\delta$: quality factor $Q = \omega W/P$ — compute it. (c) Bandwidth $f/Q$ of the resonance; ring-down time $Q/\omega$. (d) A superconducting cavity has $R_s \sim 1 \times 10^{-8}\,\Omega$: $Q$, and why accelerators use them.

**Solution of Exercise 16.8.**

(a) $f = c\sqrt2/2L = 2.12\,\mathrm{GHz}$. (b) $\delta = 1.4\,\text{µ}\mathrm{m}$, $R_s = 12\,\mathrm{m}\Omega$; $W = \varepsilon_0E_0^2V/8 = 1.1 \times 10^{-15}\,E_0^2$; $P = \tfrac12R_s(E_0/377)^2 \times 6L^2 =
2.5 \times 10^{-9}\,E_0^2$: $Q = \omega W/P \approx 6000$. (c) $360\,\mathrm{kHz}$; $0.45\,\text{µ}\mathrm{s}$. (d) $Q \sim 10^{10}$: the field of tens of megavolts per metre is sustained by kilowatts instead of gigawatts.

**Exercise 16.9 ★★.**

*Mode counting.* (a) Show that the number of modes of a box of volume $V$ with frequency below $f$ is $N(f) = 8\pi Vf^3/3c^3$ (points $(m, n, p)$ in the eighth of a sphere of radius $2fL/c$ for a cube, two polarizations). (b) The number of modes per unit volume and frequency, $\dd N/V\dd f = 8\pi f^2/c^3$. (c) A $1\,\mathrm{m}^{3}$ room: modes per hertz at $1\,\mathrm{GHz}$; at $500\,\mathrm{THz}$ (visible light). (d) Why does a small cavity have sparse modes while a big room has a continuum — and what is the oven’s case at $2.45\,\mathrm{GHz}$?

**Solution of Exercise 16.9.**

(a) Points $(m, n, p)$ with $m^2 + n^2 + p^2 \le (2fL/c)^2$ fill an eighth of a sphere: $\tfrac18\cdot\tfrac43\pi(2fL/c)^3$, times two polarizations: $8\pi L^3f^3/3c^3$. (b) $8\pi f^2/c^3$. (c) $9 \times 10^{-7}$ per hertz (one per megahertz) at $1\,\mathrm{GHz}$; $2 \times 10^5$ per hertz at $500\,\mathrm{THz}$. (d) Modes are discrete when their spacing exceeds their width; in the oven, one mode per $10\,\mathrm{MHz}$ or so around $2.45\,\mathrm{GHz}$ — dozens within a few percent.

**Exercise 16.10 ★★★.**

*TE$_{10}$ in full.* In the guide $0 < x < a$, $0 < y < b$: $\vect E =
E_0\sin(\pi x/a)\eu^{\iu(\omega t - k_gz)}\vect e_y$. (a) From Faraday, find $\underline B_x$ and $\underline B_z$; check $\operatorname{div}\vect B = 0$. (b) Check Maxwell–Ampère in vacuum, and recover $k_g^2 = \omega^2/c^2 - (\pi/a)^2$. (c) [Surface currents](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#prop-b2-maxwell-equations-boundary) on the four walls (the broad walls $y = 0$, $b$ and the narrow walls $x = 0$, $a$); which currents flow along $z$, which across? (d) A slot cut along $z$ in the centre of a broad wall does not radiate, a slot across it does: explain.

**Solution of Exercise 16.10.**

(a) From $\operatorname{\vect{curl}}\vect E = -\partial_t\vect B$:

$$
\underline B_x = -\frac{k_g}\omega\underline E , \qquad
\underline B_z = \frac{\iu\pi}{a\omega}E_0\cos\frac{\pi x}a\,\eu^{\iu(\omega t - k_gz)} ,
$$

and $\partial_xB_x + \partial_zB_z = -(k_g\pi/a\omega)E_0\cos(\pi x/a) + (k_g\pi/a\omega)E_0
\cos(\pi x/a) = 0$. (b) $(\operatorname{\vect{curl}}\vect B)_y = \partial_zB_x - \partial_xB_z =
(\iu/\omega)(k_g^2 + \pi^2/a^2)\underline E = (\iu\omega/c^2)\underline E$: $k_g^2 = \omega^2/c^2 - \pi^2/a^2$. (c) Broad walls: $\vect j_s = \pm(B_z\vect e_x - B_x\vect e_z)/\mu_0$ — along $z$ as $\sin(\pi x/a)$, across as $\cos(\pi x/a)$ (zero at the centre); narrow walls: along $y$ only. (d) A central longitudinal slot cuts only transverse currents, which vanish there; a transverse slot cuts the longitudinal ones and radiates: the slotted-waveguide antenna.

**Exercise 16.11 ★★★.**

*Dielectric guide.* A slab of index $n_1$ and thickness $a$ in a medium $n_2 < n_1$ guides light by total reflection. (a) For a ray at angle $\theta$ from the normal to the faces, the guided condition is that the round-trip phase across the slab, $2k_0n_1a\cos\theta - 2\varphi_r$ ($\varphi_r$ the phase shift of total reflection, taken $\approx 0$ here), is a multiple of $2\pi$: number of modes for $a = 50\,\text{µ}\mathrm{m}$, $\lambda = 1.5\,\text{µ}\mathrm{m}$, $n_1 = 1.48$, $n_2 = 1.46$ (admit the mode count $\approx 2a\sqrt{n_1^2 - n_2^2}/\lambda$ per polarization). (b) Thickness for a single mode. (c) Compare with a single-mode fibre core of $9\,\text{µ}\mathrm{m}$. (d) Why can the dielectric guide, unlike the metal one, not have a cut-off for its lowest mode?

**Solution of Exercise 16.11.**

(a) $2a\sqrt{n_1^2 - n_2^2}/\lambda = 2 \times 50 \times 0.242/1.5 \approx 16$ per polarization. (b) $a < \lambda/2\mathrm{NA} = 3.1\,\text{µ}\mathrm{m}$. (c) A round core with $\mathrm{NA} =
0.12$ is single-mode below $d = 2.405\lambda/\pi\mathrm{NA} = 10\,\text{µ}\mathrm{m}$: the $9\,\text{µ}\mathrm{m}$ standard. (d) A grazing ray is always totally reflected: the lowest mode exists at every frequency, its evanescent tails merely spreading into the cladding.

**Exercise 16.12 ★★★.**

*Graded-index fibre.* To reduce [modal dispersion](#prop-b2-guided-waves-fibre) the core index decreases from the axis outward, $n(r) = n_1\sqrt{1 - 2\Delta(r/a)^2}$ with $\Delta \approx 0.01$. (a) Explain qualitatively why an off-axis ray, though longer, can take the same time as the axial one. (b) Using the ray equation in a stratified medium, $n(r)\cos\theta(r) =$ const, show that a ray launched from the axis at a small angle oscillates sinusoidally about it (expand $n$ to second order). (c) Period of the oscillation for $a = 25\,\text{µ}\mathrm{m}$. (d) The residual modal spread is of order $n_1L\Delta^2/2c$: value per kilometre, compared with the step-index $n_1L\Delta/c$.

**Solution of Exercise 16.12.**

(a) The off-axis ray spends its time in lower index, where light is faster: the longer path is compensated. (b) $n(r)\cos\theta = n_1\cos\theta_0$ with $n \approx n_1(1 - \Delta r^2/a^2)$ and $\cos\theta \approx 1 - \theta^2/2$: $(\dd r/\dd z)^2 =
\theta_0^2 - 2\Delta r^2/a^2$, a harmonic oscillation $r = r_0\sin(\sqrt{2\Delta}\,z/a)$. (c) $2\pi a/\sqrt{2\Delta} = 1.1\,\mathrm{mm}$. (d) $n_1L\Delta^2/2c = 0.25\,\mathrm{ns}/\mathrm{km}$ against $49\,\mathrm{ns}/\mathrm{km}$: two hundred times better.

![Optical fibres lit from their far ends: light trapped by total internal reflection in a core thinner than a hair, and guided round every gentle bend.](https://one-course.com/images/onecourse/chapters/physics-4/b2-guided-waves/img-a4675afec71d.jpg)

*[Optical fibres](#prop-b2-guided-waves-fibre) lit from their far ends: light trapped by [total internal reflection](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#thm-b2-wave-interfaces-snell) in a core thinner than a hair, and guided round every gentle bend.*

## 16.5 Problem: The oven cavity and the optical fibre

**Problem 16.1.**

Weekend problem — two guided-wave systems in every kitchen and every city: the oven that cooks with standing waves and the fibre that carries the internet

**Part I — From the magnetron to the cavity.** A magnetron feeds $800\,\mathrm{W}$ at $2.45\,\mathrm{GHz}$ into a [rectangular waveguide](#prop-b2-guided-waves-rectangular) ($a = 8.6\,\mathrm{cm}$, $b = 4.3\,\mathrm{cm}$) that opens into the oven cavity ($30 \times 30 \times 20$ cm$^3$); walls steel, $\gamma = 1 \times 10^{6}\,\mathrm{S}/\mathrm{m}$.

1. Cut-off frequencies of the guide’s TE $_{10}$ , TE $_{20}$ , TE $_{01}$ modes; check that only TE $_{10}$ propagates at $2.45\,\mathrm{GHz}$ .
2. Guide wavelength, phase and group velocities at $2.45\,\mathrm{GHz}$ .
3. Field amplitude $E_0$ in the guide for $800\,\mathrm{W}$ (TE $_{10}$ power formula); compare with the breakdown field of air.
4. [Surface current](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#prop-b2-maxwell-equations-boundary) amplitude on the broad wall (of order $E_0/\mu_0c$ ); [skin depth](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) and surface resistance of the steel; loss per metre of guide.
5. Modes of the cavity: list those with frequencies within $3\%$ of $2.45\,\mathrm{GHz}$ (try $(m, n, p)$ up to $4$ ); how many?
6. Distance between the hot spots of one such mode along each axis; why does the plate turn, and what is a "mode stirrer"?
7. The cavity’s quality factor when loaded with food is about $200$ : bandwidth of its response; why a mismatch between the magnetron’s frequency and the modes does not matter much.
8. The empty cavity has $Q \approx 3000$ : stored energy at $800\,\mathrm{W}$ input, mean energy density, field amplitude; compare with question 3 and with breakdown. What protects the magnetron when the oven runs empty?
9. Time for the energy to travel the $20\,\mathrm{cm}$ of guide from the magnetron to the cavity.
10. The magnetron is fed by a half-wave rectified supply and emits only during half of each mains cycle: peak power during the bursts for a mean of $800\,\mathrm{W}$ ; what does the cavity do between bursts (use $Q \approx 200$ )?

**Part II — The door and the cut-off.**

11. The door mesh has $1.5\,\mathrm{mm}$ holes in a $1\,\mathrm{mm}$ sheet: decay length in a hole and [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) in dB across the sheet.
12. The door’s edge is sealed by a quarter-wave "choke": a slot $\lambda/4$ deep around the door frame, which presents an open circuit at the gap (recall [Chapter 15](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#ch-b2-wave-interfaces) ): depth of the slot at $2.45\,\mathrm{GHz}$ .
13. Legal leakage is $5\,\mathrm{mW}/\mathrm{cm}^{2}$ at $5\,\mathrm{cm}$ : to what fraction of $800\,\mathrm{W}$ through a $100\,\mathrm{cm}^{2}$ door does that correspond?
14. A $30\,\mathrm{cm}$ duct of $4\,\mathrm{cm}$ diameter vents the cavity: is it below cut-off at $2.45\,\mathrm{GHz}$ ? [Attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) over its length.
15. Why must the mesh be electrically bonded to the door frame all round (what would a gap in the bond behave like)?

**Part III — The fibre.** A step-index fibre: core $n_1 = 1.465$, cladding $n_2 = 1.460$, core diameter $50\,\text{µ}\mathrm{m}$ (multimode) or $9\,\text{µ}\mathrm{m}$ (single-mode); $\lambda = 1.55\,\text{µ}\mathrm{m}$; [attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) $0.2\,\mathrm{dB}/\mathrm{km}$.

16. [Numerical aperture](#prop-b2-guided-waves-fibre) and acceptance angle; is it easy to couple light in?
17. Modal spread per kilometre for the multimode fibre; maximum bit rate over $10\,\mathrm{km}$ (one bit per spread time).
18. Number of modes of the multimode fibre (admit $N \approx  \tfrac12(\pi d\,\mathrm{NA}/\lambda)^2$ ); of the single-mode one ( $d =  9\,\text{µ}\mathrm{m}$ : show the same formula gives about $1$ ).
19. In the single-mode fibre the remaining spreading is chromatic: with $\omega'' \approx -0.2\,\mathrm{m}^{2}/\mathrm{s}$ ( [Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets) ) and $50\,\mathrm{ps}$ pulses, distance over which a pulse doubles; bit rate over $100\,\mathrm{km}$ .
20. [Attenuation](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-complex) over $100\,\mathrm{km}$ in dB and as a power ratio; input $1\,\mathrm{mW}$ : output power; with amplifiers every $80\,\mathrm{km}$ , how many for a $6000\,\mathrm{km}$ transatlantic link?
21. Frequency and photon energy at $1.55\,\text{µ}\mathrm{m}$ ; photons per second in $1\,\mathrm{mW}$ , and per bit at $10\,\mathrm{Gbit}/\mathrm{s}$ , at the input and after $100\,\mathrm{km}$ .
22. Why $1.55\,\text{µ}\mathrm{m}$ (think of the two loss mechanisms of silica: Rayleigh scattering, falling as $1/\lambda^4$ , and infrared absorption rising beyond $1.6\,\text{µ}\mathrm{m}$ )?
23. Fresnel reflection at a cleaved fibre end facing air: loss in dB; why connectors use index-matching gel or polished physical contact.
24. Why does a fibre not radiate at a gentle bend, and why does it lose light at a sharp one (think of the angle of the ray on the core boundary)?
25. Compare the two guides of this problem: what confines the wave, what limits the bandwidth, what the losses.

**Solution of Problem 16.1.**

**1.** $c/2a = 1.74\,\mathrm{GHz}$; TE$_{20}$ and TE$_{01}$: $3.49\,\mathrm{GHz}$: only TE$_{10}$ at $2.45\,\mathrm{GHz}$.

**2.** $\sqrt{1 - (1.74/2.45)^2} = 0.70$: $\lambda_g = 17.4\,\mathrm{cm}$, $v_\varphi = 1.42c$, $v_g = 0.70c = 2.1 \times 10^{8}\,\mathrm{m}/\mathrm{s}$.

**3.** $E_0^2 = 4P/\varepsilon_0c\,ab\sqrt{\cdot} = 3200/6.9 \times 10^{-6}$: $E_0 = 21\,\mathrm{kV}/\mathrm{m}$, a hundredth of breakdown.

**4.** $j_s \approx E_0/\mu_0c = 57\,\mathrm{A}/\mathrm{m}$; $\delta = 10\,\text{µ}\mathrm{m}$, $R_s = 0.1\,\Omega$; $\tfrac12R_sj_s^2 \times 2a = 28\,\mathrm{W}/\mathrm{m}$: $3.5\%$ per metre — steel is lossy, the guide is short.

**5.** $f = 1.5 \times 10^8\sqrt{11.1(m^2 + n^2) + 25p^2}$ within $\pm3\%$ of $2.45\,\mathrm{GHz}$: $(1, 2, 3)$, $(2, 1, 3)$ at $2.51\,\mathrm{GHz}$, $(4, 0, 2)$, $(0, 4, 2)$, $(4, 3, 0)$, $(3, 4, 0)$ at $2.50\,\mathrm{GHz}$ — about six.

**6.** Half-wavelengths $a/m$, $b/n$, $d/p$: $7$ to $10\,\mathrm{cm}$ apart. The turntable drags the food through maxima and minima; a stirrer (a rotating metal fan) reshuffles the modes.

**7.** $f/Q = 12\,\mathrm{MHz}$: the loaded resonances overlap into a continuum; the magnetron always finds a mode to feed.

**8.** $W = QP/\omega = 3000 \times 800/1.54 \times 10^{10} = 0.16\,\mathrm{J}$; $u = 8.7\,\mathrm{J}/\mathrm{m}^{3}$; $E_0 \approx \sqrt{4u/\varepsilon_0} = 2\,\mathrm{MV}/\mathrm{m}$ — near breakdown: arcs; the reflected power goes back to the magnetron, protected (a little) by a dummy load or circulator — hence "never run it empty".

**9.** $0.2/2.1 \times 10^8 = 1\,\mathrm{ns}$.

**10.** $1.6\,\mathrm{kW}$ peak; between bursts the field rings down in $Q/\omega = 13\,\mathrm{ns}$: the cavity is empty most of the time.

**11.** $\ell \approx a/\pi = 0.48\,\mathrm{mm}$; $\eu^{-1/0.48}$: $-18\,\mathrm{dB}$ in amplitude, $-36\,\mathrm{dB}$ in power.

**12.** $\lambda/4 = 3.1\,\mathrm{cm}$.

**13.** $5\,\mathrm{mW}/\mathrm{cm}^{2}$ $\times$ $100\,\mathrm{cm}^{2}$ = $0.5\,\mathrm{W}$: $0.06\%$.

**14.** $f_1 \approx c/2d = 3.75\,\mathrm{GHz} > 2.45\,\mathrm{GHz}$: below cut-off; $\ell =
1.7\,\mathrm{cm}$, $\eu^{-30/1.7}$: $-150\,\mathrm{dB}$.

**15.** The wall currents must flow continuously into the mesh; a gap in the bond is a slot that cuts them — a slot antenna radiating outward.

**16.** $\mathrm{NA} = \sqrt{1.465^2 - 1.460^2} = 0.12$, $7{}^{\circ}$: a narrow cone, needing a laser and a lens.

**17.** $(n_1L/c)(n_1/n_2 - 1) = 17\,\mathrm{ns}/\mathrm{km}$; $170\,\mathrm{ns}$ over $10\,\mathrm{km}$: $6\,\mathrm{Mbit}/\mathrm{s}$.

**18.** $\tfrac12(\pi d\,\mathrm{NA}/\lambda)^2 = 75$; for $9\,\text{µ}\mathrm{m}$: $2.4$ — the single-mode regime ($V = \pi d\,\mathrm{NA}/\lambda = 2.2 < 2.4$).

**19.** $\Delta x_0 = (c/n)\tau = 1\,\mathrm{cm}$: $t_{\text{sp}} = 10^{-4}/0.2 = 0.5\,\mathrm{ms}$, $100\,\mathrm{km}$: $10\,\mathrm{Gbit}/\mathrm{s}$ over that distance before compensation.

**20.** $20\,\mathrm{dB}$, a factor $100$: $10\,\text{µ}\mathrm{W}$; $6000/80 = 75$ amplifiers.

**21.** $f = 1.9 \times 10^{14}\,\mathrm{Hz}$, $hf = 0.8\,\mathrm{eV}$; $8 \times 10^{15}$ photons per second; $8 \times 10^5$ per bit at the input, $8 \times 10^3$ after $100\,\mathrm{km}$.

**22.** The sum of the two losses is minimal near $1.55\,\text{µ}\mathrm{m}$ ($0.2\,\mathrm{dB}/\mathrm{km}$).

**23.** $R = ((1.465 - 1)/2.465)^2 = 3.6\%$: $0.16\,\mathrm{dB}$ per face, twice at a connector plus the gap’s interference; gel or physical contact removes the air.

**24.** On a gentle bend the ray still meets the boundary beyond the critical angle; on a sharp one the incidence on the outer wall falls below it and light leaks out.

**25.** Oven: metal walls, reflection with $r = -1$; bandwidth set by the cavity’s modes; losses in the steel skin. Fibre: [total internal reflection](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#thm-b2-wave-interfaces-snell) at a glass–glass boundary; bandwidth set by modal and chromatic dispersion; losses by scattering and absorption, $0.2\,\mathrm{dB}/\mathrm{km}$.