University Physics — Year 2 · Bachelor Year 2

16Guided 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; then the rectangular waveguide, the closed cavity, and, in the ray picture, the optical 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 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 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 (Chapter 8). ∎

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 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. — 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. — 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 TE $_{10}$ 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 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.

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 must form a standing wave 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). Each mode has a quality factor $Q$ set by the wall losses (the surface resistance of Chapter 14), typically $10^3$ – $10^4$ for copper at microwave frequencies, $10^{10}$ for a superconducting cavity.

Proof. Standing waves 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, with maxima $6\,\mathrm{cm}$ apart, which the turntable (and a "mode stirrer") smooth out. A radar’s magnetron is itself a cavity resonator; 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 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; 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 — 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).

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.

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 times a progressive factor along the guide; the boundary conditions quantize the transverse wavenumber. (3) The dispersion 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

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

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

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 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

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 on each plate (boundary relation). (c) Mean Poynting vector 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

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 across the sheet in dB (compare Chapter 14). (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 per centimetre (use the plate formula with $a = 1\,\mathrm{cm}$ as an estimate). (d) Why is the attenuation independent of the wall conductivity?

Solution

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 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 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 in dB/m. (d) Compare with a coaxial cable’s $0.5\,\mathrm{dB}/\mathrm{m}$ : why are guides used at these frequencies?

Solution

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

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

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 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

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

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 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

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.

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 ( $a = 8.6\,\mathrm{cm}$ , $b = 4.3\,\mathrm{cm}$ ) that opens into the oven cavity ( $30 \times 30 \times 20$ cm $^3$ ); walls steel, $\gamma = 1 \times 10^{6}\,\mathrm{S}/\mathrm{m}$ .

Cut-off frequencies of the guide’s TE $_{10}$ , TE $_{20}$ , TE $_{01}$ modes; check that only TE $_{10}$ propagates at $2.45\,\mathrm{GHz}$ .
Guide wavelength, phase and group velocities at $2.45\,\mathrm{GHz}$ .
Field amplitude $E_0$ in the guide for $800\,\mathrm{W}$ (TE $_{10}$ power formula); compare with the breakdown field of air.
Surface current amplitude on the broad wall (of order $E_0/\mu_0c$ ); skin depth and surface resistance of the steel; loss per metre of guide.
Modes of the cavity: list those with frequencies within $3\%$ of $2.45\,\mathrm{GHz}$ (try $(m, n, p)$ up to $4$ ); how many?
Distance between the hot spots of one such mode along each axis; why does the plate turn, and what is a "mode stirrer"?
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.
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?
Time for the energy to travel the $20\,\mathrm{cm}$ of guide from the magnetron to the cavity.
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.

The door mesh has $1.5\,\mathrm{mm}$ holes in a $1\,\mathrm{mm}$ sheet: decay length in a hole and attenuation in dB across the sheet.
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): depth of the slot at $2.45\,\mathrm{GHz}$ .
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?
A $30\,\mathrm{cm}$ duct of $4\,\mathrm{cm}$ diameter vents the cavity: is it below cut-off at $2.45\,\mathrm{GHz}$ ? Attenuation over its length.
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 $0.2\,\mathrm{dB}/\mathrm{km}$ .

Numerical aperture and acceptance angle; is it easy to couple light in?
Modal spread per kilometre for the multimode fibre; maximum bit rate over $10\,\mathrm{km}$ (one bit per spread time).
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 $1$ ).
In the single-mode fibre the remaining spreading is chromatic: with $\omega'' \approx -0.2\,\mathrm{m}^{2}/\mathrm{s}$ (Chapter 8) and $50\,\mathrm{ps}$ pulses, distance over which a pulse doubles; bit rate over $100\,\mathrm{km}$ .
Attenuation 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?
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 $100\,\mathrm{km}$ .
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 $1.6\,\text{µ}\mathrm{m}$ )?
Fresnel reflection at a cleaved fibre end facing air: loss in dB; why connectors use index-matching gel or polished physical contact.
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)?
Compare the two guides of this problem: what confines the wave, what limits the bandwidth, what the losses.

Solution

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 at a glass–glass boundary; bandwidth set by modal and chromatic dispersion; losses by scattering and absorption, $0.2\,\mathrm{dB}/\mathrm{km}$ .