Part 2 · Light

2.3 Reflection and refraction

Source: Postorino, week 2 slides, p. 20, and week 3 slides, pp. 1–4; Bastieri’s note on Snell’s law

Objectives

Vocabulary. Refractive index, homogeneous, isotropic, TE (s) and TM (p) polarisation, critical angle, Brewster angle.
Theorem. Snell’s law $n_{1} sin ϑ_{1} = n_{2} sin ϑ_{2}$ , from Fermat’s principle of least time.
Recipe. Critical angle $ϑ_{C} = arcsin (n_{2} / n_{1})$ , Brewster angle $tan ϑ_{B} = n_{2} / n_{1}$ , normal reflectance $R = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2}$ .
Picture. From one medium to another the frequency stays, the speed and the wavelength change.

Waves in a medium

Replace $μ_{0} ε_{0} \to μ ε = μ_{r} ε_{r} μ_{0} ε_{0}$ : swap the properties of vacuum for those of a homogeneous isotropic medium, with $ε_{r}$ and $μ_{r}$ its relative permittivity and permeability. Everything stays the same except the speed, which becomes $v = \frac{1}{μ ε} = \frac{1}{μ _{r} ε _{r}} \frac{1}{μ _{0} ε _{0}} = \frac{c}{μ _{r} ε _{r}} .$

Definition 2.3.1Refractive indexslides w3, p. 1

$n = μ_{r} ε_{r}, v = \frac{c}{n} or n = \frac{c}{v} (n = 1 \to v = c) .$

For most transparent materials $μ_{r} \approx 1$ , so $n \approx ε_{r}$ beyond the lectures. Light is slower in matter, and $n \geq 1$ measures by how much. Why $ε_{r}$ , and so $n$ , depends on frequency is §3.1 Dielectrics: the Lorentz model.

Homogeneous: the property does not depend on position.
Isotropic: the property does not depend on direction.
Scale matters! (slides w3, p. 4). Glass is homogeneous for light, whose wavelength is about 500 nm, but not for X-rays, which see its atoms.

Fermat’s principle and Snell’s laws

Key observation 2.3.2Fermat’s principleslides w3, p. 1

Of all the paths an electromagnetic wave (a light ray) could take from one point to another, it follows the one that takes the shortest time.

Theorem 2.3.3Snell’s lawsslides w3, p. 1

Reflection: $ϑ_{inc} = ϑ_{refl}$ , angles measured from the normal.
Refraction: $n_{1} sin ϑ_{1} = n_{2} sin ϑ_{2}$ .
The incident, reflected and refracted rays and the normal are coplanar.

With $v_{1} = c / n_{1}$ and $v_{2} = c / n_{2}$ : if $n_{1} < n_{2}$ then $v_{1} > v_{2}$ and the ray bends towards the normal, $ϑ_{1} > ϑ_{2}$ .

Reflection from Fermatbeyond the lectures

The light goes from S to P by way of the mirror, in one medium, so the shortest time is the shortest path. Reflect P in the mirror to P′. Every path S → mirror → P has the same length as S → mirror → P′, and the shortest of those is the straight line SP′. It meets the mirror where the two angles are equal.

Refraction from Fermat: the lifeguard problem. This is Bastieri’s note, the “document on E-learning” the slides cite. A brachistochrone (Greek brachistos, shortest, and chronos, time) is the path between two points that takes the least time. A simple case:

the lifeguard is at A on the beach and must reach a swimmer at B in the water as fast as possible;
they run on sand at $v_{1}$ and swim at $v_{2} < v_{1}$ ;
A′ and B′ are the feet of the perpendiculars from A and B to the shoreline.

Two obvious choices are wrong. The straight line AB is the shortest path, but has too long a stretch in the water. The path A → B′ → B makes the swim as short as possible but stretches the run too much. In medio stat virtus: enter the water at some point O, and find where.

Theorem 2.3.4Snell’s law from least timeBastieri, p. 1–2

Let $d = \overline{A^{'} B^{'}}$ be fixed and $\overline{A^{'} O} = x$ , so $\overline{O B^{'}} = d - x$ . The total time is $T = \frac{A O}{v _{1}} + \frac{OB}{v _{2}} = \frac{A A ^{'} ^{2} + x ^{2}}{v _{1}} + \frac{B B ^{'} ^{2} + ( d - x ) ^{2}}{v _{2}}, (1)$ and it is least where $\frac{sin ϑ _{1}}{sin ϑ _{2}} = \frac{v _{1}}{v _{2}}, (3)$ with $ϑ_{1}$ , $ϑ_{2}$ the angles between AO, OB and the normal at O. For light, $v_{i} = c / n_{i}$ , which is $n_{1} sin ϑ_{1} = n_{2} sin ϑ_{2}$ .

Proof

The minimum is at a stationary point, $d T / d x = 0$ , with $d$ , $v_{1}$ , $v_{2}$ , $\overline{A A^{'}}$ , $\overline{B B^{'}}$ fixed: $0 = \frac{d T}{d x} = \frac{x}{v _{1} A A ^{'} ^{2} + x ^{2}} - \frac{d - x}{v _{2} B B ^{'} ^{2} + ( d - x ) ^{2}} = \frac{A ^{'} O}{v _{1} A O} - \frac{O B ^{'}}{v _{2} OB} . (2)$ The ratios are $\overline{A^{'} O} / \overline{A O} = sin ϑ_{1}$ and $\overline{O B^{'}} / \overline{OB} = sin ϑ_{2}$ , so $0 = \frac{s i n ϑ _{1}}{v _{1}} - \frac{s i n ϑ _{2}}{v _{2}}$ . It is a minimum: beyond the lectures $\frac{d ^{2} T}{d x ^{2}} = \frac{A A ^{'} ^{2}}{v _{1} A O ^{3}} + \frac{B B ^{'} ^{2}}{v _{2} OB ^{3}} > 0.$

Snell’s law is the condition for the brachistochrone between two points in two different media.

Loading the demo…

Figure 2.3.1 The lifeguard problem. Drag the entry point

O

along the shore and watch the total time

T

: the graph has a single

minimum

, and there, and only there,

sin ϑ_{1} / sin ϑ_{2} = v_{1} / v_{2}

Fresnel’s laws: how much is reflected

Snell’s laws give the directions. Fresnel’s laws give how much light is reflected and how much transmitted, which depends on the angle and on the polarisation.

Definition 2.3.5TE and TM polarisationslides w3, p. 2

TE (transverse electric, s polarisation): $E$ oscillates parallel to the interface, perpendicular to the plane of incidence.
TM (transverse magnetic, p polarisation): $B$ oscillates parallel to the interface, and $E$ lies in the plane of incidence.

Any polarisation is a combination of the two.

Proposition 2.3.6Fresnel formulasbeyond the lectures

For amplitudes, with $ϑ_{1}$ incident and $ϑ_{2}$ refracted, $r_{s} = \frac{n _{1} cos ϑ _{1} - n _{2} cos ϑ _{2}}{n _{1} cos ϑ _{1} + n _{2} cos ϑ _{2}}, r_{p} = \frac{n _{2} cos ϑ _{1} - n _{1} cos ϑ _{2}}{n _{2} cos ϑ _{1} + n _{1} cos ϑ _{2}}, R = ∣ r ∣^{2} .$ At $ϑ_{1} = 0$ both reduce to the normal-incidence formula (Proposition 2.3.9), up to sign.

The slide shows $R$ against the angle for both polarisations (Figure 2.3.2 draws the same curves):

$n_{1} = 1$ , $n_{2} = 2$ (into the denser medium): at normal incidence both reflect about 11 %. $R_{s}$ rises steadily to 100 % at grazing incidence; $R_{p}$ falls to zero at Brewster’s angle, about 63°, then rises to 100 %.
$n_{1} = 2$ , $n_{2} = 1$ (into the less dense medium): both reach 100 % at the critical angle, about 30°. Beyond it lies total internal reflection.

Proposition 2.3.7Critical angleslides w3, p. 2–3

If $n_{1} > n_{2}$ , light is totally reflected beyond $ϑ_{C} = arcsin (\frac{n _{2}}{n _{1}}) .$

Proof

From Snell, $sin ϑ_{2} = \frac{n _{1}}{n _{2}} sin ϑ_{1}$ . As $ϑ_{1}$ grows, so does $ϑ_{2}$ . If $n_{1} / n_{2} > 1$ there is a limiting angle where $ϑ_{2} = π /2$ : the refracted ray grazes the surface. Beyond it $sin ϑ_{2}$ would have to exceed 1, so no light is transmitted: everything is reflected. Setting $sin ϑ_{2} = 1$ gives $ϑ_{C}$ .

Check beyond the lectures: for $n_{1} = 2$ , $n_{2} = 1$ , $arcsin 0.5 = 30°$ , as in the figure; glass to air, $arcsin (1/1.5) = 41.8°$ . The slide’s picture: a source in glass ( $n_{1} = 1.5$ ) shining at air. Rays near the normal are partly refracted and partly reflected; at the critical angle the refracted ray grazes the surface; beyond it everything comes back into the glass.

Proposition 2.3.8Brewster’s angleslides w3, p. 2

At Brewster’s angle the reflected and refracted rays are at right angles, the TM reflection vanishes, and $tan ϑ_{B} = \frac{n _{2}}{n _{1}} .$ Unpolarised light comes off polarised: only the TE component, perpendicular to the plane of incidence, is reflected. The refracted ray is slightly polarised, having lost some TE.

The angle, from the right anglebeyond the lectures

$ϑ_{B} + ϑ_{r} = 90°$ , so $sin ϑ_{r} = cos ϑ_{B}$ , and Snell $n_{1} sin ϑ_{B} = n_{2} cos ϑ_{B}$ gives $tan ϑ_{B} = n_{2} / n_{1}$ . For $n_{2} / n_{1} = 2$ , $arctan 2 = 63.4°$ ; air to glass, $arctan 1.5 = 56.3°$ .

Why TM reflection vanishes there. The reflected light is radiated by the electrons of the second medium, which oscillate along the refracted ray’s $E$ . A dipole radiates nothing along its own axis. For TM light the refracted $E$ lies in the plane of incidence, across the refracted ray, so at Brewster’s angle it points exactly along the reflected direction: the dipoles cannot radiate that way.

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Figure 2.3.2 A ray hits the interface between medium

n_{1}

and medium

n_{2}

. Drag the

incident ray

. The reflected and refracted rays are drawn with thickness proportional to the power they carry, for TE or TM light; the graph on the right is

R_{s}

and

R_{p}

against the angle. It snaps at

ϑ_{B}

, where

R_{p} = 0

, and at

ϑ_{C}

, beyond which there is no refracted ray. Swap the media to see total internal reflection.

Polarisation is the direction of oscillation of $E$ .

$E$ , $B$ and the direction of propagation follow the right-hand rule.
Linear polarisation: $E$ oscillates in one fixed plane.
Circular polarisation: two perpendicular linear components of equal amplitude, $90°$ out of phase; the tip of $E$ turns as the wave advances (Figure 2.2.1). In the slide’s convention, if the wave approaches an observer and $E$ appears to turn anticlockwise, it is right-circular.

Consequences

Proposition 2.3.9Normal reflectivityslides w3, p. 3

At normal incidence $R = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2} .$

Air to glass, $n = 1.5$ : $R = (0.5/2.5)^{2} = 4%$ per surface. Air to $n = 2$ : $(1/3)^{2} = 11%$ , the left plot’s starting value.

Consequences of Snell’s and Fresnel’s laws (the slide shows pictures; the explanations are added):

Brilliance of diamonds. Diamond’s $n \approx 2.42$ gives a critical angle of only 24.4°. Light entering a well-cut stone is totally reflected inside it and comes back out of the top.
Anti-reflective coating. A thin layer on a lens makes the reflections from its two surfaces cancel by interference (§2.4 Adding waves: interference, standing waves, beats).
Optical fibres. Light is trapped in the core by total internal reflection. Only rays inside the acceptance cone are guided; its half-angle from air is $sin ϑ_{max} = n_{core}^{2} - n_{clad}^{2}$ , the numerical aperture.
Mirages. Air near hot ground is less dense, so $n$ decreases downward. Grazing rays from the sky bend up and are totally reflected, so the ground looks like water.
Polarised sunglasses. Glare off water or a road is reflected near Brewster’s angle, so it is largely TE, polarised parallel to the surface. A vertical polariser blocks it.

From one medium to another

Theorem 2.3.10What changes at an interfaceslides w3, p. 4

In two homogeneous isotropic media, $v = c / n$ , $v = λ ν$ , $ω = k v$ . Going from one to the other:

the frequency $ν$ (and so $T$ and $ω$ ) does not change;
the speed $v$ changes;
the wavelength $λ$ changes: if $n_{2} > n_{1}$ , $v_{2} < v_{1}$ and $λ_{2} < λ_{1}$ ; precisely, $λ_{2} = λ_{1} n_{1} / n_{2}$ .

Why the frequency staysbeyond the lectures

The field on the two sides must match at the interface at every instant. The material oscillates at the frequency it is driven at (Theorem 1.4.1). So the crests reaching the interface per second equal those leaving it. The wavefronts then bunch up in the slower medium, and their kink at the interface is refraction.

Exercises

Exercise 2.3.1beyond the lectures

Water has $n = 1.33$ . Find the speed of light in it, the critical angle from water to air, and Brewster’s angle from air to water.

Solution

$v = c / n = 2.26 \times 1 0^{8}$ m/s; $ϑ_{C} = arcsin (1/1.33) = 48.8°$ ; $ϑ_{B} = arctan 1.33 = 53.1°$ .

Exercise 2.3.2beyond the lectures

Red light of $λ = 633$ nm in air enters glass with $n = 1.5$ . Give its frequency and wavelength in the glass.

Solution

$ν = c / λ = 4.74 \times 1 0^{14}$ Hz, the same in both media. In glass $λ = 633/1.5 = 422$ nm, but it is still red light: colour goes with frequency.

Exercise 2.3.3beyond the lectures

A lifeguard runs at 7 m/s and swims at 2 m/s. They are 20 m from the shoreline; the swimmer is 10 m out and 30 m further along. Write $T (x)$ and say how Snell’s law locates the entry point.

Solution

$T (x) = 400 + x^{2} /7 + 100 + (30 - x)^{2} /2$ . The optimum has $sin ϑ_{1} / sin ϑ_{2} = 7/2$ : $\frac{x}{400 + x ^{2}} = 3.5 \frac{30 - x}{100 + ( 30 - x ) ^{2}}$ . Numerically $x \approx 27.6$ m, entering the water about 2.4 m short of the point opposite the swimmer, for $T \approx 10.0$ s.

Exercise 2.3.4beyond the lectures

How much light gets through a glass window ( $n = 1.5$ ) at normal incidence, ignoring absorption and multiple reflections?

Solution

Each surface reflects 4 %, so $0.9 6^{2} = 92%$ gets through.

Exercise 2.3.5beyond the lectures

An optical fibre has $n_{core} = 1.48$ and $n_{clad} = 1.46$ . Find the critical angle at the core–cladding interface and the acceptance half-angle from air.

Solution

$ϑ_{C} = arcsin (1.46/1.48) = 80.6°$ from the normal: rays must travel within $9.4°$ of the axis. $sin ϑ_{max} = 1.4 8^{2} - 1.4 6^{2} = 0.242$ , so $ϑ_{max} = 14°$ . (Refraction at the entry face, $sin 14° = 1.48 sin 9.4°$ , connects the two.)

Exercise 2.3.6beyond the lectures

Check that $r_{p} = 0$ at $tan ϑ_{1} = n_{2} / n_{1}$ .

Solution

$r_{p} = 0$ needs $n_{2} cos ϑ_{1} = n_{1} cos ϑ_{2}$ , and Snell says $n_{1} sin ϑ_{1} = n_{2} sin ϑ_{2}$ . Multiply the two: $n_{1} n_{2} sin ϑ_{1} cos ϑ_{1} = n_{1} n_{2} sin ϑ_{2} cos ϑ_{2}$ , i.e. $sin 2 ϑ_{1} = sin 2 ϑ_{2}$ . With $ϑ_{1} \neq = ϑ_{2}$ this means $ϑ_{1} + ϑ_{2} = 90°$ : the rays are at right angles. Then $sin ϑ_{2} = cos ϑ_{1}$ , and Snell gives $tan ϑ_{1} = n_{2} / n_{1}$ .