---
title: "Plane Electromagnetic Waves and Polarization"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 13
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/13-plane-electromagnetic-waves-and-polarization
---

# Chapter 13 — Plane Electromagnetic Waves and Polarization

Tilt your head while wearing polarizing sunglasses and the glare on the wet road comes and goes; turn the glasses in front of a phone screen and the screen goes black. The light that reaches your eye is not only a wave with a frequency and an intensity: it has a *direction of vibration*, and every polarizer, screen, 3D-cinema lens and optical fibre connector plays with it. This chapter solves [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) in empty space — the plane wave, with its transverse electric and magnetic fields locked at right angles — surveys the spectrum those waves span, and studies the polarization: how to describe it, how to select it with a polarizer ([Malus’s law](#prop-b2-plane-waves-polarization-malus)), and how to transform it with a birefringent plate.

## 13.1 Plane waves in vacuum

**Theorem 13.1 (Wave equation; plane progressive harmonic waves).**

In vacuum, free of charges and currents, [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) give

$$
\Delta\vect E = \frac1{c^2}\frac{\partial^2\vect E}{\partial t^2} , \qquad
\Delta\vect B = \frac1{c^2}\frac{\partial^2\vect B}{\partial t^2} , \qquad
c = \frac1{\sqrt{\varepsilon_0\mu_0}} = 2.998 \times 10^{8}\,\mathrm{m}/\mathrm{s} .
$$

The *plane progressive harmonic wave* (PPH wave) $\underline{\vect E} =
\underline{\vect E}_0\,\eu^{\iu(\omega t - \vect k\cdot\vect r)}$ is a solution provided $\omega = ck$ (no dispersion), and [Maxwell’s equations](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#thm-b2-maxwell-equations-maxwell) then require

$$
\vect k\cdot\vect E = 0 , \qquad
\vect B = \frac{\vect k\wedge\vect E}\omega = \frac{\vect u\wedge\vect E}c ,
$$

$\vect u = \vect k/k$ being the direction of propagation: the wave is *transverse*, $(\vect E, \vect B, \vect u)$ is a right-handed orthogonal triad, $\vect E$ and $\vect B$ are in phase and $B = E/c$. Its intensity is $I = \varepsilon_0cE_0^2/2$ and it carries the [momentum flux](https://one-course.com/books/physics/4/en/chapter/5-momentum-and-energy-balances-in-flows#thm-b2-flow-balances-momentum) $I/c$ ([Chapter 12](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#ch-b2-poynting-vector)).

**Proof.** The [wave equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation) was derived in [Exercise 11.11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#exo-b2-maxwell-equations-11). For $\eu^{\iu(\omega t - \vect k\cdot\vect r)}$, $\partial_t \to \iu\omega$ and $\vect\nabla \to
-\iu\vect k$: $\Delta \to -k^2$, so $k^2 = \omega^2/c^2$; Maxwell–Gauss gives $-\iu
\vect k\cdot\underline{\vect E} = 0$, Maxwell–flux $\vect k\cdot\underline{\vect B} = 0$, Maxwell–Faraday $-\iu\vect k\wedge\underline{\vect E} = -\iu\omega\underline{\vect B}$, whence $\underline{\vect B} = \vect k\wedge\underline{\vect E}/\omega$, of modulus $kE/\omega = E/c$ and perpendicular to both; Maxwell–Ampère is then satisfied identically. ∎

![A plane progressive harmonic wave at one instant: E and B perpendicular to each other and to the direction of travel, in phase, B = E/c; the whole pattern slides along u at c.](https://one-course.com/images/onecourse/chapters/physics-4/b2-plane-waves-polarization/fig-402199216cda.svg)

*A plane progressive harmonic wave at one instant: $\vect E$ and $\vect B$ perpendicular to each other and to the direction of travel, in phase, $B = E/c$; the whole pattern slides along $\vect u$ at $c$.*

**Remark 13.2 (The spectrum).**

One equation, one speed, and twenty orders of magnitude of frequency. Radio: $30\,\mathrm{kHz}$–$300\,\mathrm{MHz}$ ($\lambda$ from $10\,\mathrm{km}$ to $1\,\mathrm{m}$), antennas and circuits ([Chapter 17](https://one-course.com/books/physics/4/en/chapter/17-dipole-radiation-and-scattering#ch-b2-dipole-radiation)). Microwaves: $300\,\mathrm{MHz}$–$300\,\mathrm{GHz}$, radar, ovens, mobile phones, the cosmic background. Infrared: to $400\,\mathrm{THz}$ ($750\,\mathrm{nm}$), the thermal radiation of everything around us ([Chapter 26](https://one-course.com/books/physics/4/en/chapter/26-thermal-radiation#ch-b2-thermal-radiation)). Visible: $400\,\mathrm{THz}$–$750\,\mathrm{THz}$, $750\,\mathrm{nm}$ (red) to $400\,\mathrm{nm}$ (violet) — one octave, photons of $1.6$ to $3.1\,\mathrm{eV}$. Ultraviolet to $10\,\mathrm{nm}$; X-rays to $10\,\mathrm{pm}$ ($100\,\mathrm{keV}$), from inner atomic shells and braking electrons; gamma rays beyond, from nuclei. Only the sources and the detectors differ: the wave is the same.

![The electromagnetic spectrum on a logarithmic frequency scale; the visible band is a sliver — one octave out of twenty decades.](https://one-course.com/images/onecourse/chapters/physics-4/b2-plane-waves-polarization/fig-98deb39a1748.svg)

*The [electromagnetic spectrum](#rem-b2-plane-waves-polarization-spectrum) on a logarithmic frequency scale; the visible band is a sliver — one octave out of twenty decades.*

## 13.2 Polarization

**Definition 13.3 (Polarization states).**

For a PPH wave along $z$ the field in a plane $z =$ const is

$$
\vect E = E_{0x}\cos(\omega t - kz)\,\vect e_x + E_{0y}\cos(\omega t - kz - \varphi)\,\vect e_y ,
$$

and the tip of $\vect E$ describes, in general, an ellipse: *elliptical polarization*. Two special cases: $\varphi = 0$ or $\pi$, *linear* polarization along a fixed direction at angle $\arctan(\pm E_{0y}/E_{0x})$ from $x$; $E_{0x} = E_{0y}$ and $\varphi = \pm\pi/2$, *circular* polarization, $\vect E$ of constant modulus turning at $\omega$ (right or left according to its sense of rotation, seen facing the oncoming wave). *Natural* (unpolarized) light, from a lamp or the Sun, is a succession of short wave trains with random, rapidly changing polarizations: no direction is preferred on average. Any polarization decomposes into two linear (or two circular) components.

**Proof.** With $X = E_x/E_{0x}$, $Y = E_y/E_{0y}$: $X = \cos\theta$, $Y = \cos(\theta - \varphi)$, so $X^2 + Y^2 - 2XY\cos\varphi = \sin^2\varphi$, an ellipse, degenerate into the lines $Y = \pm X$ for $\varphi = 0, \pi$ and into a circle for $E_{0x} = E_{0y}$, $\varphi = \pm\pi/2$. ∎

**Proposition 13.4 (Polarizers and Malus’s law).**

A *polarizer* transmits the component of $\vect E$ along its transmission axis $\vect p$ and absorbs the other (a sheet of aligned long molecules, which conduct along their length — the transmitted polarization is perpendicular to the molecules). Linearly polarized light of intensity $I_0$ whose direction makes the angle $\theta$ with $\vect p$ emerges linearly polarized along $\vect p$ with

$$
I = I_0\cos^2\theta \qquad\text{(Malus)} ;
$$

[natural light](#def-b2-plane-waves-polarization-states) emerges with $I_0/2$, polarized along $\vect p$. Two crossed polarizers transmit nothing; a third one inserted at $45{}^{\circ}$ between them transmits $I_0/8$.

**Proof.** The transmitted amplitude is $E_0\cos\theta$ and $I \propto E_0^2$. [Natural light](#def-b2-plane-waves-polarization-states): average of $\cos^2\theta$ over all $\theta$, $\tfrac12$. Three polarizers: $\tfrac12\cdot\cos^2 45^\circ\cdot\cos^2 45^\circ = \tfrac18$ — the middle polarizer does not merely attenuate, it *turns* the polarization. ∎

![Left: natural light through three polarizers at 0, 45 and 90 — the middle one lets an eighth through a pair that alone would block everything. Right: Malus’s law.](https://one-course.com/images/onecourse/chapters/physics-4/b2-plane-waves-polarization/fig-f0a30ed341a7.svg)

*Left: [natural light](#def-b2-plane-waves-polarization-states) through three polarizers at $0^\circ$, $45^\circ$ and $90^\circ$ — the middle one lets an eighth through a pair that alone would block everything. Right: [Malus’s law](#prop-b2-plane-waves-polarization-malus).*

**Example 13.5 (Sunglasses, screens, photographs).**

Light reflected at a glancing angle from water or asphalt is largely polarized horizontally (Brewster’s angle, [Chapter 15](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#ch-b2-wave-interfaces)): sunglasses with a vertical transmission axis cut the glare and keep the rest. A liquid-crystal screen emits polarized light: through a polarizer at $90{}^{\circ}$ it goes black. A photographer’s polarizing filter darkens the sky (scattered light is partly polarized, [Chapter 17](https://one-course.com/books/physics/4/en/chapter/17-dipole-radiation-and-scattering#ch-b2-dipole-radiation)) and removes reflections from glass.

## 13.3 Birefringence and wave plates

**Proposition 13.6 (Wave plates).**

In a *birefringent* crystal (calcite, quartz, mica; also a stretched plastic sheet) a wave propagating along a given direction splits into two [linear polarizations](#def-b2-plane-waves-polarization-states), along the *fast* and *slow* axes of the plate, travelling with different indices $n_f <
n_s$ (admitted — the crystal responds differently along its axes). A plate of thickness $e$ delays the slow component by the phase

$$
\delta = \frac{2\pi}\lambda(n_s - n_f)\,e .
$$

A *[half-wave plate](#prop-b2-plane-waves-polarization-plates)* ($\delta = \pi$) turns a [linear polarization](#def-b2-plane-waves-polarization-states) at angle $\alpha$ from the [fast axis](#prop-b2-plane-waves-polarization-plates) into a [linear polarization](#def-b2-plane-waves-polarization-states) at $-\alpha$: it rotates it by $2\alpha$. A *[quarter-wave plate](#prop-b2-plane-waves-polarization-plates)* ($\delta = \pi/2$) turns a [linear polarization](#def-b2-plane-waves-polarization-states) at $45{}^{\circ}$ from its axes into circular light, and circular light back into linear; at other angles, into elliptical light with axes along the plate’s.

**Proof.** Write the incident field $E_0(\cos\alpha\,\vect e_f + \sin\alpha\,\vect e_s)\cos\omega t$; after the plate, $E_0[\cos\alpha\cos\omega t\,\vect e_f + \sin\alpha\cos(\omega t - \delta)
\vect e_s]$. For $\delta = \pi$ the $\vect e_s$ component changes sign: direction $(\cos\alpha, -\sin\alpha)$. For $\delta = \pi/2$ and $\alpha = 45^\circ$: $(E_0/\sqrt2)(\cos
\omega t, \sin\omega t)$, a circle. ∎

![The polarization states of a plane wave, seen facing the oncoming light, and a birefringent plate with its fast and slow axes.](https://one-course.com/images/onecourse/chapters/physics-4/b2-plane-waves-polarization/fig-ba0f35553928.svg)

*The polarization states of a plane wave, seen facing the oncoming light, and a birefringent plate with its fast and slow axes.*

**Example 13.7 (Where plates are used).**

A [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) between a polarizer and a mirror makes an *optical isolator*: the light goes out circular, comes back circular of the opposite handedness, is turned by the plate into linear light at $90{}^{\circ}$ from the polarizer and is absorbed — no reflection returns to the laser. The two eyes’ images of a 3D film are projected in opposite [circular polarizations](#def-b2-plane-waves-polarization-states) and sorted by the quarter-wave–polarizer sandwiches of the glasses, which still work when you tilt your head (linear polarizers would not). A transparent ruler between crossed polarizers shows coloured fringes: stress makes plastic birefringent, and engineers read the stresses of a model from the pattern.

**Method 13.8 (Following a polarization through optics).**

(1) Choose axes; write the incident field as two components with their phase difference. (2) A polarizer: keep the projection on its axis (amplitude $\times\cos$, intensity $\times\cos^2$; [natural light](#def-b2-plane-waves-polarization-states) $\times\tfrac12$). (3) A plate: project on its fast and slow axes, delay the slow one by $\delta$, recombine. (4) Read the result: phase $0$ or $\pi$ means linear, $\pm\pi/2$ with equal amplitudes means circular. (5) To test an unknown beam: rotate a polarizer (intensity varies: linear or elliptical part), then add a [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) (circular becomes linear and can be extinguished).

## 13.4 Exercises

**Exercise 13.1 ★.**

An FM station at $100\,\mathrm{MHz}$ gives, at a receiver, an intensity of $1.0\,\text{µ}\mathrm{W}/\mathrm{m}^{2}$. Wavelength, wavenumber, period; amplitudes $E_0$ and $B_0$; energy density; emf induced in a $1\,\mathrm{m}$ antenna aligned with $\vect E$; photon flux (per square metre and second).

**Solution of Exercise 13.1.**

$\lambda = 3.0\,\mathrm{m}$, $k = 2.1\,\mathrm{rad}/\mathrm{m}$, $T = 10\,\mathrm{ns}$; $E_0 = \sqrt{2I/\varepsilon_0c}
= 27\,\mathrm{mV}/\mathrm{m}$, $B_0 = 9 \times 10^{-11}\,\mathrm{T}$; $u = I/c = 3.3 \times 10^{-15}\,\mathrm{J}/\mathrm{m}^{3}$; emf $E_0 \times 1\,\mathrm{m} = 27\,\mathrm{mV}$; $I/hf = 10^{-6}/6.6 \times 10^{-26} = 1.5 \times 10^{19}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$.

**Exercise 13.2 ★.**

Frequency and photon energy (in eV) for $\lambda = 1\,\mathrm{km}$, $10\,\mathrm{cm}$, $10\,\text{µ}\mathrm{m}$, $550\,\mathrm{nm}$, $10\,\mathrm{nm}$, $0.1\,\mathrm{nm}$, $1\,\mathrm{pm}$; name the band of each; the wavelength of a $1\,\mathrm{MeV}$ gamma ray and of the $2.7\,\mathrm{K}$ cosmic background (peak near $160\,\mathrm{GHz}$).

**Solution of Exercise 13.2.**

$300\,\mathrm{kHz}$ ($1.2 \times 10^{-9}\,\mathrm{eV}$, radio); $3\,\mathrm{GHz}$ ($1.2 \times 10^{-5}\,\mathrm{eV}$, microwave); $30\,\mathrm{THz}$ ($0.12\,\mathrm{eV}$, infrared); $545\,\mathrm{THz}$ ($2.25\,\mathrm{eV}$, visible); $3 \times 10^{16}\,\mathrm{Hz}$ ($124\,\mathrm{eV}$, extreme UV); $3 \times 10^{18}\,\mathrm{Hz}$ ($12\,\mathrm{keV}$, X); $3 \times 10^{20}\,\mathrm{Hz}$ ($1.2\,\mathrm{MeV}$, gamma). $1\,\mathrm{MeV}$: $1.2\,\mathrm{pm}$; $160\,\mathrm{GHz}$: $1.9\,\mathrm{mm}$.

**Exercise 13.3 ★.**

Linearly polarized light of intensity $I_0$ falls on a polarizer at $30{}^{\circ}$, then a second at $60{}^{\circ}$ from the first, then a third at $90{}^{\circ}$ from the first: intensity after each. Same with [natural light](#def-b2-plane-waves-polarization-states). Remove the middle one: what changes?

**Solution of Exercise 13.3.**

$0.75I_0$, $0.56I_0$, $0.42I_0$; natural: $0.375$, $0.28$, $0.21$. Without the middle one: $0.75 \times \cos^260^\circ = 0.19I_0$ (natural $0.125I_0$): the intermediate polarizer lets *more* through by turning the polarization.

**Exercise 13.4 ★.**

Check that $\vect E = E_0\cos(\omega t - kz)\,\vect e_x$, $\vect B = (E_0/c)\cos(\omega t
- kz)\,\vect e_y$ satisfies the four Maxwell equations in vacuum when $\omega =
kc$. Which equation fixes the direction of $\vect B$, which its magnitude? What happens if one tries $\vect E$ along $\vect e_z$?

**Solution of Exercise 13.4.**

$\operatorname{div}\vect E = \partial_xE_x = 0$, $\operatorname{div}\vect B = 0$; $\operatorname{\vect{curl}}\vect E = E_0k\sin(\omega t - kz)\,\vect e_y = -\partial_t\vect B$ iff $\omega =
kc$ — Faraday fixes both the direction ($\vect e_y$) and the magnitude ($E_0/c$) of $\vect B$; $\operatorname{\vect{curl}}\vect B = -(E_0k/c)\sin(\omega t - kz)\,\vect e_x
= \mu_0\varepsilon_0\partial_t\vect E$. $\vect E$ along $\vect e_z$ would have $\operatorname{div}
\vect E \ne 0$ with no charge: impossible — the wave is transverse.

**Exercise 13.5 ★★.**

*Circular light.* (a) Write the field of a circularly polarized wave along $z$ and show that $|\vect E|$ is constant while its direction turns at $\omega$. (b) Show that the sum of a right- and a left-circular wave of equal amplitudes is linearly polarized, along a direction set by their relative phase. (c) Through a polarizer, what does circular light give, and does the intensity depend on the polarizer’s angle? (d) How then can one tell circular light from [natural light](#def-b2-plane-waves-polarization-states)?

**Solution of Exercise 13.5.**

(a) $\vect E = E_0(\cos(\omega t - kz), \pm\sin(\omega t - kz), 0)$: modulus $E_0$, angle $\pm(\omega t - kz)$. (b) $(\cos, \sin) + (\cos, -\sin) = (2\cos, 0)$: linear along $x$; a relative phase $\varphi$ turns the direction by $\varphi/2$. (c) $I_0/2$ at every angle. (d) A [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) turns circular light into linear light, which a polarizer then extinguishes; [natural light](#def-b2-plane-waves-polarization-states) stays unpolarized.

**Exercise 13.6 ★★.**

Quartz has $n_s - n_f = 0.0091$ at $589\,\mathrm{nm}$. (a) Thickness of a [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) of lowest order; of a [half-wave plate](#prop-b2-plane-waves-polarization-plates). (b) A [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) for $589\,\mathrm{nm}$ used at $450\,\mathrm{nm}$: phase delay (neglect the variation of the indices); what comes out for linear light at $45{}^{\circ}$? (c) Why are "multiple-order" plates (thickness $e +
m\lambda/(n_s - n_f)$) more sensitive to wavelength and temperature? (d) Why must a plate’s thickness be controlled to better than a micrometre?

**Solution of Exercise 13.6.**

(a) $e = \lambda/4(n_s - n_f) = 16\,\text{µ}\mathrm{m}$; half-wave $32\,\text{µ}\mathrm{m}$. (b) $\delta =
(\pi/2)(589/450) = 2.1\,\mathrm{rad}$: elliptical light. (c) $\delta \propto e/\lambda$: a plate of $m + \tfrac14$ waves has $\dd\delta/\dd\lambda$ and $\dd\delta/\dd T$ larger by $4m + 1$. (d) $\delta \propto e$: a micrometre on $16\,\text{µ}\mathrm{m}$ is $6\%$ of $\pi/2$.

**Exercise 13.7 ★★.**

[Half-wave plate](#prop-b2-plane-waves-polarization-plates). (a) Show that linear light at angle $\alpha$ from the [fast axis](#prop-b2-plane-waves-polarization-plates) comes out linear at $-\alpha$. (b) Hence a plate rotated by $\beta$ rotates the polarization by $2\beta$: what rotation of the plate turns a vertical polarization horizontal? (c) What does a [half-wave plate](#prop-b2-plane-waves-polarization-plates) do to circular light? (d) Between crossed polarizers, a [half-wave plate](#prop-b2-plane-waves-polarization-plates) is rotated slowly: sketch the transmitted intensity against its angle.

**Solution of Exercise 13.7.**

(a) The slow component changes sign: $(\cos\alpha, \sin\alpha) \to (\cos\alpha,
-\sin\alpha)$. (b) A polarization at angle $\alpha$ from a plate at $\beta$ comes out at $2\beta - \alpha$: rotation $2\beta$; $45{}^{\circ}$. (c) Reverses the handedness. (d) $I = I_0\sin^22\beta$: four maxima and four extinctions per turn.

**Exercise 13.8 ★★.**

$N$ polarizers are placed one after the other, each rotated by $\pi/2N$ from the previous one, the last being at $90{}^{\circ}$ from the first. Transmission of linearly polarized light aligned with the first for $N = 1, 2, 5, 10, 100$; limit $N \to \infty$ (use $\cos^{2N}(\pi/2N) \approx
1 - \pi^2/4N$). Comment: a polarization can be turned by $90{}^{\circ}$ with no loss — by what?

**Solution of Exercise 13.8.**

$\cos^{2N}(\pi/2N)$: $0$, $0.25$, $0.61$, $0.78$, $0.976$, $\to 1$. A continuously twisting medium (optical activity, a liquid-crystal cell) rotates the polarization without loss.

**Exercise 13.9 ★★.**

*Partial polarization.* A rotating polarizer in front of a beam gives a maximum intensity $I_{\max}$ and a minimum $I_{\min}$. (a) Model the beam as [natural light](#def-b2-plane-waves-polarization-states) $I_n$ plus linearly polarized light $I_p$: express $I_{\max}$ and $I_{\min}$; the degree of polarization $p = I_p/(I_n + I_p) =
(I_{\max} - I_{\min})/(I_{\max} + I_{\min})$. (b) Blue sky at $90{}^{\circ}$ from the Sun: $I_{\max}/I_{\min} = 5$: $p$. (c) Light reflected from a lake near Brewster’s angle, $p = 0.9$: how much can polarizing sunglasses remove? (d) Can a polarizer alone distinguish partially linearly polarized light from partially circular light? What is needed?

**Solution of Exercise 13.9.**

(a) $I_{\max} = I_n/2 + I_p$, $I_{\min} = I_n/2$; $p = I_p/(I_n + I_p) = (I_{\max} -
I_{\min})/(I_{\max} + I_{\min})$. (b) $p = 4/6 = 0.67$. (c) The crossed polarizer passes $I_n/2 = 0.05I$: $95\%$ removed. (d) No — both give a constant intensity; a [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) is needed.

**Exercise 13.10 ★★★.**

*Two crossed beams.* Two PPH waves of equal amplitude and frequency, both polarized along $\vect e_y$, propagate in the $xz$ plane at angles $\pm\theta$ from $z$. (a) Write the total field and show it is a wave travelling along $z$ whose amplitude is modulated along $x$ as $\cos(kx\sin\theta)$. (b) Spacing of the dark planes (interference fringes); numbers for $\lambda = 633\,\mathrm{nm}$, $\theta = 5{}^{\circ}$, and for $\theta = 90{}^{\circ}$ (counter-propagating). (c) [Phase velocity](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#def-b2-dispersion-wave-packets-relation) of the pattern along $z$; compare with $c$; is anything travelling faster than light? (d) Compute the time-averaged [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) and show that its $x$ component vanishes: the energy flows along $z$ in the bright planes.

**Solution of Exercise 13.10.**

(a) Phases $\omega t - k(z\cos\theta \pm x\sin\theta)$: sum $2E_0\cos(kx\sin\theta)\cos(\omega t
- kz\cos\theta)\,\vect e_y$. (b) $\Delta x = \lambda/2\sin\theta$: $3.6\,\text{µ}\mathrm{m}$ at $5{}^{\circ}$, $\lambda/2 = 316\,\mathrm{nm}$ at $90{}^{\circ}$. (c) $v_\varphi = c/\cos\theta > c$: a pattern’s phase; the energy goes along $z$ at $c\cos\theta$ (and each wave at $c$). (d) $B_z = (\sin\theta/c)(E_1 - E_2)$, $E_y = E_1 + E_2$: $\langle\Pi_x\rangle \propto
\langle E_1^2 - E_2^2\rangle = 0$; $\Pi_z \propto \cos\theta(E_1 + E_2)^2$, maximal in the bright planes.

**Exercise 13.11 ★★★.**

*The liquid-crystal pixel.* A twisted-nematic cell between crossed polarizers rotates the polarization of light by $90{}^{\circ}$ when no voltage is applied (the molecules’ alignment twists through the cell and the polarization follows it — admitted), and not at all when a few volts untwist them. (a) Transmission in both states: which one is bright? (b) Real cells rotate by $90^\circ \pm 5^\circ$ at the design wavelength: contrast ratio $I_{\text{bright}}/I_{\text{dark}}$ if the dark state leaks through a residual $5{}^{\circ}$ error. (c) Why does a cell designed for $550\,\mathrm{nm}$ leak some blue and red in the dark state (think of the rotation as a stack of many thin half-wave-like retarders)? (d) A colour pixel uses three such cells with filters; why does the screen go black through a polarizer at some angle, and what does that angle tell you?

**Solution of Exercise 13.11.**

(a) No voltage: rotated by $90{}^{\circ}$, passes the crossed polarizer: bright; with voltage: dark. (b) Leak $\sin^25^\circ = 7.6 \times 10^{-3}$: contrast $130$. (c) The twist works as a stack of retarders whose $\delta \propto 1/\lambda$: exact at $550\,\mathrm{nm}$, the rotation is off at the ends of the spectrum, which leak — the dark state looks purplish. (d) When the external polarizer is crossed with the screen’s output polarizer; the angle is the screen’s polarization axis.

**Exercise 13.12 ★★★.**

*Optical activity.* In a sugar solution the two [circular polarizations](#def-b2-plane-waves-polarization-states) travel with slightly different indices $n_L$ and $n_R$. (a) Decompose a [linear polarization](#def-b2-plane-waves-polarization-states) into circular components and show that after a length $\ell$ the polarization is still linear but rotated by $\theta = \pi(n_L - n_R)\ell/\lambda$. (b) A $1\,\mathrm{dm}$ tube of a $0.10\,\mathrm{g}/\mathrm{mL}$ sucrose solution rotates sodium light by $6.65{}^{\circ}$: $n_L - n_R$. (c) A saccharimeter measures $\theta$ to $0.01{}^{\circ}$: precision on the concentration. (d) The same effect is produced in glass by a magnetic field along the beam (Faraday effect, $\theta = VB\ell$ with $V \approx
4\,\mathrm{rad}\,\mathrm{T}^{-1}\,\mathrm{m}^{-1}$): why does a Faraday rotator, unlike a sugar tube, not cancel its rotation when the light is sent back, and how does that make an isolator?

**Solution of Exercise 13.12.**

(a) $\vect e_x = \tfrac12[(\vect e_x + \iu\vect e_y) + (\vect e_x - \iu\vect e_y)]$; after $\ell$ the two components have phases $k_L\ell$ and $k_R\ell$, and recombine into linear light at the angle $(k_L - k_R)\ell/2 = \pi(n_L - n_R)\ell/\lambda$. (b) $n_L -
n_R = \theta\lambda/\pi\ell = 0.116 \times 589 \times 10^{-9}/0.314 = 2.2 \times 10^{-7}\,$. (c) $0.01/6.65 \times 0.1 = 1.5 \times 10^{-4}\,\mathrm{g}/\mathrm{mL}$. (d) The Faraday rotation is fixed by $\vect B$, not by the direction of travel: on the return trip it adds ($45{}^{\circ}$ $+$ $45{}^{\circ}$ $=$ $90{}^{\circ}$), and the input polarizer blocks the returning light — an isolator.

![Polarizing sunglasses held before a lake: the glare, reflected near Brewster’s angle and polarized horizontally, is cut by the vertical transmission axis of the lenses, and the lake bed shows through.](https://one-course.com/images/onecourse/chapters/physics-4/b2-plane-waves-polarization/img-41876249925c.jpg)

*Polarizing sunglasses held before a lake: the glare, reflected near Brewster’s angle and polarized horizontally, is cut by the vertical transmission axis of the lenses, and the lake bed shows through.*

## 13.5 Problem: The wave, the screen and the sail

**Problem 13.1.**

Weekend problem — a laser beam taken apart into its fields, a screen that paints with polarization, and a spacecraft that sails on light

**Part I — The beam.** A helium–neon laser emits $10\,\mathrm{mW}$ at $632.8\,\mathrm{nm}$ in a beam of $1.0\,\mathrm{mm}$ diameter, linearly polarized along $x$, travelling along $z$.

1. Frequency, wavenumber, photon energy; photons emitted per second.
2. Intensity; amplitudes $E_0$ and $B_0$ ; write $\vect E(z, t)$ and $\vect B(z, t)$ .
3. Energy density (mean) and the energy contained in $1\,\mathrm{m}$ of beam; how long is the beam emitted in $1\,\mathrm{ns}$ , and how many wavelengths does it contain?
4. [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) : mean value and its oscillation; momentum carried per second; force on a black target and on a mirror.
5. Compare $E_0$ with the field that binds the electron in a hydrogen atom ( $5 \times 10^{11}\,\mathrm{V}/\mathrm{m}$ ) and $B_0$ with the Earth’s field ( $50\,\text{µ}\mathrm{T}$ ).
6. [Radiation pressure](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#prop-b2-poynting-vector-wave) on a black beam stop, in pascals.
7. A free electron placed in the beam: amplitude of its velocity oscillation in the electric field (neglect the magnetic force first); ratio of the magnetic to the electric force on it — why is the magnetic force negligible for matter in ordinary light?
8. Write the fields of the same beam if it were circularly polarized; what changes in the intensity, in the [Poynting vector](https://one-course.com/books/physics/4/en/chapter/12-electromagnetic-energy-and-the-poynting-vector#thm-b2-poynting-vector-poynting) and in its time dependence?
9. The beam passes through a polarizer whose axis makes $30{}^{\circ}$ with $x$ : transmitted power; then through a second one crossed with the first: power; insert between them a third at $45{}^{\circ}$ from the first: power.

**Part II — The screen.** A liquid-crystal screen: backlight ([natural light](#def-b2-plane-waves-polarization-states)), a polarizer, the cell, a crossed polarizer. With no voltage the cell rotates the polarization by $90{}^{\circ}$; with voltage it does nothing.

10. Fraction of the backlight’s intensity transmitted in the bright state (perfect polarizers); in the dark state.
11. An intermediate voltage makes the cell behave as a retarder of phase $\delta$ between its axes, which are at $45{}^{\circ}$ to the polarizers (no rotation): show that the transmission is $\sin^2(\delta/2)$ — the grey scale.
12. Viewed through a polarizer rotated to $90{}^{\circ}$ from the screen’s output polarizer, what is seen? At $45{}^{\circ}$ ?
13. A [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) glued on the screen at $45{}^{\circ}$ to its polarizer turns the output circular: advantage for a viewer wearing polarizing sunglasses?
14. Perfect polarizers would pass half of the backlight in the bright state; real sheets transmit $85\%$ of the ideal each, and the colour filters of a pixel pass a third of the light: fraction of the backlight’s power reaching the viewer in white; name the two places where most of the light is lost.
15. Real crossed sheets leak $0.1\%$ of the light: contrast ratio of the screen (bright over dark) in the dark room.
16. Photographers’ "circular polarizers" are a linear polarizer followed by a [quarter-wave plate](#prop-b2-plane-waves-polarization-plates) at $45{}^{\circ}$ : what does the camera receive, and why is that better for the polarization-sensitive beam splitter of its autofocus than linear light?
17. Why do 3D-cinema glasses use circular rather than [linear polarization](#def-b2-plane-waves-polarization-states) ? What happens to the crosstalk between the eyes if the viewer tilts the head by $20{}^{\circ}$ with linear glasses (intensity leaking into the wrong eye)?

**Part III — The sail.** A solar sail of area $S = 1000\,\mathrm{m}^{2}$ and total mass $m = 50\,\mathrm{kg}$, perfectly reflecting, at $D = 1.0\,\mathrm{au} = 1.5 \times 10^{11}\,\mathrm{m}$ from the Sun ($I = 1.36\,\mathrm{kW}/\mathrm{m}^{2}$; Sun’s gravity $5.9 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ there).

18. Radiation force when the sail faces the Sun; acceleration; ratio to solar gravity.
19. The sail is tilted by $\theta$ from facing the Sun: show that the force is normal to the sail and proportional to $\cos^2\theta$ (intercepted power $\times\cos\theta$ , momentum change along the normal $\times\cos\theta$ ).
20. To spiral outward the sail keeps a force component along its orbital velocity: for $\theta = 35{}^{\circ}$ , tangential acceleration; speed gained per year; compare with the Earth’s orbital speed ( $30\,\mathrm{km}/\mathrm{s}$ ).
21. Why does the same sail, near Mercury ( $D = 0.39\,\mathrm{au}$ ), get $6.6$ times the force, and why is the ratio to gravity unchanged?
22. The sail’s film is $2\,\text{µ}\mathrm{m}$ of aluminized plastic: its temperature in sunlight if it absorbs $10\%$ and radiates from both faces as a black body ( $\sigma = 5.67 \times 10^{-8}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{K}^{-4}$ ).
23. Time needed to gain $1\,\mathrm{km}/\mathrm{s}$ facing the Sun; what would an absorbing (black) sail of the same size achieve?
24. Check the momentum of light with photons: number of photons hitting the sail per second ( $2\,\mathrm{eV}$ average) and the momentum $hf/c$ of each; recover the force.
25. Sum up: the three quantities carried by the wave (energy, momentum, polarization) and the device of this problem that exploits each.

**Solution of Problem 13.1.**

**1.** $f = 4.74 \times 10^{14}\,\mathrm{Hz}$, $k = 9.9 \times 10^{6}\,\mathrm{rad}/\mathrm{m}$, $hf = 3.1 \times 10^{-19}\,\mathrm{J} =
1.96\,\mathrm{eV}$; $3.2 \times 10^{16}$ photons per second.

**2.** $I = 0.01/\pi(0.5 \times 10^{-3})^2 = 1.3 \times 10^{4}\,\mathrm{W}/\mathrm{m}^{2}$; $E_0 = 3.1\,\mathrm{kV}/\mathrm{m}$, $B_0 = 10\,\text{µ}\mathrm{T}$; $\vect E = E_0\cos(\omega t - kz)\vect e_x$, $\vect B = (E_0/c)\cos(\omega t
- kz)\vect e_y$.

**3.** $u = I/c = 4.2 \times 10^{-5}\,\mathrm{J}/\mathrm{m}^{3}$; $P/c = 3.3 \times 10^{-11}\,\mathrm{J}$ per metre of beam; $30\,\mathrm{cm}$, $4.7 \times 10^5$ wavelengths.

**4.** $\langle\Pi\rangle = I$ along $z$, oscillating as $\cos^2$ at $2f$; $P/c = 3.3 \times 10^{-11}\,\mathrm{N}$ on black, $6.7 \times 10^{-11}\,\mathrm{N}$ on a mirror.

**5.** $E_0/5 \times 10^{11} = 6 \times 10^{-9}$; $B_0$ is a fifth of the Earth’s field.

**6.** $I/c = 4.2 \times 10^{-5}\,\mathrm{Pa}$.

**7.** $v_0 = eE_0/m\omega = 0.18\,\mathrm{m}/\mathrm{s}$; $F_m/F_e = v_0B_0/E_0 = v_0/c = 6 \times
10^{-10}$: the magnetic force on slow charges is negligible.

**8.** $\vect E = E_0[\cos(\omega t - kz)\vect e_x \pm \sin(\omega t - kz)\vect e_y]$ with $E_0 = 2.2\,\mathrm{kV}/\mathrm{m}$ (same power), $\vect B = \vect e_z\wedge\vect E/c$: the intensity is unchanged but $\Pi = \varepsilon_0cE_0^2$ is now *constant* in time — the energy flow no longer pulses.

**9.** $10\cos^230^\circ = 7.5\,\mathrm{mW}$; $0$; $7.5 \times \cos^445^\circ = 1.9\,\mathrm{mW}$.

**10.** Bright: $\tfrac12$; dark: $0$.

**11.** After the first polarizer, equal components on the axes; after the retarder, phase $\delta$ between them; projection on the crossed direction $\propto(1 - \eu^{\iu\delta})/2$: $I \propto \sin^2(\delta/2)$.

**12.** Black; half intensity.

**13.** Circular output looks equally bright through polarizing sunglasses at any head angle.

**14.** $\tfrac12 \times 0.85 \times 0.85 \times \tfrac13 = 12\%$; the first polarizer (half absorbed) and the colour filters (two thirds).

**15.** Bright $0.36$, dark $\tfrac12 \times 10^{-3}$: contrast $\approx 700$.

**16.** Circular light, whatever the filter’s rotation: the polarization-sensitive splitter receives the same power however the filter is turned.

**17.** [Circular polarization](#def-b2-plane-waves-polarization-states) keeps its handedness when the head tilts; linear glasses tilted by $20{}^{\circ}$ leak $\sin^220^\circ = 12\%$ into the wrong eye.

**18.** $F = 2IS/c = 9.1\,\mathrm{mN}$; $a = 1.8 \times 10^{-4}\,\mathrm{m}/\mathrm{s}^{2}$; $3\%$ of solar gravity.

**19.** Intercepted power $\propto\cos\theta$, reflected momentum change along the normal $2p\cos\theta$: $F = (2IS/c)\cos^2\theta$ along the normal.

**20.** $F = 9.1 \times \cos^235^\circ = 6.1\,\mathrm{mN}$, tangential part $F\sin35^\circ
= 3.5\,\mathrm{mN}$: $a_t = 7 \times 10^{-5}\,\mathrm{m}/\mathrm{s}^{2}$, $2.2\,\mathrm{km}/\mathrm{s}$ per year — $7\%$ of the orbital speed.

**21.** $(1/0.39)^2 = 6.6$; gravity scales the same way.

**22.** $0.1 \times 1360 = 2\sigma T^4$: $T = 190\,\mathrm{K}$ — the mirror stays cold.

**23.** $1000/1.8 \times 10^{-4} = 5.6 \times 10^{6}\,\mathrm{s}$, two months; a black sail gets half the force.

**24.** $1.36 \times 10^6/3.2 \times 10^{-19} = 4.3 \times 10^{21}\,$ photons per second, each with $hf/c = 1.1 \times 10^{-27}\,\mathrm{kg}\,\mathrm{m}/\mathrm{s}$; reflected: $2Np = 9.1\,\mathrm{mN}$.

**25.** Energy ($u$, $I$): the beam stop and the photodiode. Momentum ($I/c$): the sail. Polarization (Malus, the plates): the screen and its polarizers.