Reflection and Refraction

Beginner

Step 1, the normal. Wherever a ray of light lands on a surface, imagine a line standing straight out of the surface at that point. It is called the normal, and every angle in optics is measured from it, not from the surface.

Step 2, reflection. A ray bouncing off a smooth surface leaves at the same angle from the normal as it arrived, like a snooker ball rebounding from the cushion. A mirror is smooth, so all the rays bounce together and you see a clear picture. Paper is rough on a tiny scale: light leaves it in every direction, so you can see the page from anywhere in the room, but never your face in it.

Step 3, refraction. When light passes into water or glass at a slant, it changes direction at the surface. The reason is speed: light is slower in glass. Think of a shopping trolley rolling at an angle from smooth floor onto carpet: the wheel that reaches the carpet first is slowed first, and the trolley swings round. The rule: on the slower side, the ray is closer to the normal.

In the animation the yellow dot leaves a white mark every quarter of a second. Inside the glass the marks crowd together: slower light, shorter steps. Notice the faint ray that bounces back too.

Step 4, trapped light. Leaving glass, a ray bends away from the normal. Tilt it far enough and it cannot get out at all: all of it reflects back inside. This total internal reflection guides light along optical fibres.

Check yourself: a ray goes from water into air. Does it end up closer to the normal or farther from it? (Farther: light is faster in air.)

Intermediate

Rays, the normal and the refractive index

A lake mirrors the far shore, yet you see the stones at your feet: how much light a surface reflects depends on the angle at which the light arrives. A narrow beam of light can be drawn as a ray. A ray is a line along the direction in which the light's energy travels. Where a ray meets a surface, draw the normal, the line perpendicular to the surface there. Every angle in optics is measured from the normal, never from the surface: the angle of incidence ${\theta }_1$, the angle of reflection ${\theta }_r$ and the angle of refraction ${\theta }_2$. The reflected and refracted rays stay in the plane of incidence, which holds the incident ray and the normal.

The refractive index compares the speed of light in vacuum, $c\approx 3.00\times 10^{8}m/s$, with its speed $v$ in the material:
$n=\frac{c}{v}$

Air has $n=1.0003$, rounded to 1.00; water has 1.33, ordinary glass about 1.5 (the animation uses 1.50) and diamond 2.42. Of two media, the one with the larger $n$ is called optically denser.

The laws of reflection and refraction

At a smooth boundary, part of the light is reflected and part enters the second medium. Their directions obey two laws:

• ${\theta }_r={\theta }_1\text{(reflection)}$
• $n_1\sin {\theta }_1=n_2\sin {\theta }_2$

The reflected ray leaves on the other side of the normal at the angle at which the light arrived. The second law is Snell's law of refraction, described by Ibn Sahl in Baghdad around 984, found again by Willebrord Snell in 1621 and published by Descartes in 1637.

In Snell's law the product $n\sin \theta$ is the same on both sides. Entering a medium of larger $n$, $\sin \theta$ must shrink, so the ray bends towards the normal; entering one of smaller $n$, it bends away. A ray arriving along the normal goes straight on, though it still slows down, and every light path can be travelled backwards.

The frequency $f$, set by the source, never changes at the boundary. Since $v=f\lambda$, the wavelength shrinks in proportion to the speed:
$\lambda =\frac{{\lambda }_0}{n}$
where ${\lambda }_0$ is the wavelength in vacuum. Green light of 550 nm has a wavelength of about 367 nm in the animation's glass, and it is still green: colour goes with the frequency.

Why slower light bends

Picture a straight wavefront arriving at the surface; it makes the same angle ${\theta }_1$ with the surface as the ray makes with the normal. One end touches the surface at $A$. While the other end covers its last stretch, $v_1\Delta t$ in air, to reach the surface at $B$, the light that entered at $A$ advances only $v_2\Delta t$ into the glass. The new front starts at $B$ and just touches the circle of radius $v_2\Delta t$ around $A$: the wave has turned.

The two right-angled triangles share their longest side, the stretch $AB$ of length $L$ along the surface, so
$\sin {\theta }_1=\frac{v_1\Delta t}{L},\sin {\theta }_2=\frac{v_2\Delta t}{L}$
Dividing gives $\sin {\theta }_1/\sin {\theta }_2=v_1/v_2=n_2/n_1$, which is Snell's law: the beginner's shopping-trolley picture made exact.

Worked example: the ray in the animation

In the first part of the animation, light goes from air, $n_1=1.00$, into glass, $n_2=1.50$, at ${\theta }_1=50^{\circ }$. Snell's law gives
$\sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2}=\frac{1.00\cdot 0.766}{1.50}=0.511,{\theta }_2=30.7^{\circ }$
This is the line $1.00\sin 50^{\circ }=1.50\sin 30.7^{\circ }$ on the screen. The ray turns through $50^{\circ }-30.7^{\circ }=19.3^{\circ }$ towards the dashed normal, while the faint reflected ray leaves at $50^{\circ }$ on the other side of it.

The readout drops from $3.00\times 10^8$ to $v=c/n=2.00\times 10^{8}m/s$ as the dot enters the glass, so the white marks it leaves every quarter of a second are only two thirds as far apart there: the same time, less distance.

How bright is the faint ray? For light arriving along the normal, the fraction of the power reflected is
$R={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2\text{(normal incidence)}$
For air and glass, $R=(0.50/2.50{)}^2=0.04$, or 4 %. At $50^{\circ }$ it is about 6 %, and it climbs towards 100 % at grazing incidence: a lake mirrors the far shore, yet you see the stones at your feet.

Total internal reflection

In the second part of the animation the light travels inside the glass, now $n_1=1.50$, towards air, $n_2=1.00$. Snell's law reads $\sin {\theta }_2=1.50\sin {\theta }_1$: the ray leaves farther from the normal than it arrived, and it grazes the surface, ${\theta }_2=90^{\circ }$, at the critical angle
$\sin {\theta }_c=\frac{n_2}{n_1}(n_1>n_2)$
For glass and air, $\sin {\theta }_c=1/1.50=0.667$, so ${\theta }_c=41.8^{\circ }$.

Beyond ${\theta }_c$, Snell's law would need $\sin {\theta }_2>1$: there is no refracted ray, and all the light is reflected. The animation's ray meets the surface at $50^{\circ }$, where $1.50\sin 50^{\circ }=1.15$; the screen shows ${\theta }_1=50^{\circ }>{\theta }_c=41.8^{\circ }$, and the whole beam returns into the glass. This total internal reflection loses nothing, and it can happen only on the side with the higher index.

Worked example: a lamp in a swimming pool

A small lamp lies on the floor of a pool 2.0 m deep. How much of the surface does it light, what does a swimmer see looking up, and how deep does the pool look from straight above?

For water to air, $\sin {\theta }_c=1.00/1.33=0.752$, so ${\theta }_c=48.8^{\circ }$. Only rays inside a cone of that half-angle escape, so the lit patch is a circle of radius $r=d\tan {\theta }_c=2.0\cdot 1.14=2.28\mathrm{m}$, where $d$ is the depth; beyond it, the surface is a perfect mirror seen from below. Run the rays backwards: a swimmer looking up sees the whole sky squeezed into a window $2{\theta }_c=97.5^{\circ }$ across, ringed by a reflection of the pool floor.

Looking straight down, you receive rays close to the normal, where $\sin \theta \approx \theta$. A ray leaving the water is tilted $n$ times more than inside, so traced back it seems to come from a point only $1/n$ as deep. The pool looks $2.0/1.33=1.50\mathrm{m}$ deep, a quarter less than it is.

Common mistakes

• Angles are measured from the normal, not from the surface. A ray at $40^{\circ }$ to the glass surface has ${\theta }_1=50^{\circ }$; using 40° in Snell's law gives 25.4° instead of 30.7°.
• Snell's law relates sines, not angles. Dividing 30° by 1.5 gives 20° instead of 19.5°; at 50° the shortcut gives 33.3° instead of 30.7°.
• Light keeps its colour in water or glass. Orange light of 600 nm has a wavelength of 400 nm in glass of index 1.50, but its frequency, $5.0\times 10^{14}\mathrm{Hz}$, is unchanged.
• Total internal reflection needs light going from a higher index to a lower one. Entering glass from air, even at 89°, the ray gets in at 41.8°.
• Optically denser does not mean heavier. Olive oil floats on water, yet with $n\approx 1.47$ against 1.33 it bends light more.

Where you meet it

Optical fibres carry internet traffic as light trapped by total internal reflection. The prisms in binoculars meet the light at $45^{\circ }$, beyond the $41.8^{\circ }$ critical angle, so they reflect better than silvered mirrors. Diamond, with $n=2.42$ and ${\theta }_c=24.4^{\circ }$, traps light and sends much of it back out through the top, which is why it sparkles. Mirages over hot roads come from light bending gradually in air of varying index.

Because $n$ depends slightly on colour, a prism splits white light, the subject of Prisms, Dispersion and Colour. Curved boundaries that form images follow in Lenses and Mirrors and The Eye, Cameras and Telescopes, and the polarisation of reflected light in Polarization of Light.

Advanced

Fermat's principle of stationary time

Reflection and refraction look like two separate rules, one for mirrors and one for glass. Both follow from a single statement about travel time, and seeing why is the most direct way into ray optics.

Start with the time itself. Light moves at $c/n$ in a medium of refractive index $n$, so covering a short length $ds$ takes $nds/c$. Adding these pieces along a whole path gives its travel time. The integral $\int nds$ is called the optical path length, and the time is that length divided by $c$:
$T=\frac{1}{c}\int nds$

Fermat's principle says that light follows a path for which $T$ is stationary. Stationary means that a small deformation of the path changes the time only to second order in the size of the deformation, never to first order, just as a smooth function is flat at the bottom of a valley or the top of a hill.

Stationary is not always least. Inside a mirror shaped as an ellipse, every path from one focus to the mirror and on to the other focus takes the same time, so all of them are stationary. In a mirror curved more tightly than that ellipse, the real path is the longest of its neighbours, not the shortest.

Why should light care about time at all? Because light is a wave. Take every conceivable path from the source to the detector and give each one a small arrow, a phasor, turned through the phase $\omega T$ that the wave accumulates along it, where $\omega$ is the angular frequency. The light that arrives is the sum of all these arrows. Label the paths by a parameter $x$, and let $x_0$ be the stationary one, with time $T_0$. Near it the time changes only quadratically, with $T^{\prime \prime }$ the second derivative of $T$ at $x_0$:
$T\approx T_0+\frac{1}{2}{T}^{\prime \prime }(x-x_0{)}^2$

Close to $x_0$ the times hardly differ, so neighbouring arrows point almost the same way and add up. Farther out the phase changes ever faster from one path to the next, and the arrows curl into tight spirals that cancel. What arrives therefore comes from a band of paths around the stationary one whose times differ by less than about half a period. That band is what we call a ray. In quantum theory the same arrows are a photon's probability amplitudes.

Snell's law from the travel time

The most useful consequence of Fermat's principle is the law of refraction. To derive it, write the travel time as a function of one number, the point where the light crosses the surface, and ask where that time is stationary.

Legend

$A$	Start of the route, a height $a$ above the surface.
$B$	End of the route, a depth $b$ below the surface.
$a$	Height of $A$ above the surface.	$\mathrm{m}$
$b$	Depth of $B$ below the surface.	$\mathrm{m}$
$D$	Horizontal distance from $A$ to $B$.	$\mathrm{m}$
$x$	Distance along the surface from the foot of $A$ to the crossing point.	$\mathrm{m}$
$D-x$	The rest of the horizontal distance: from the crossing point to the foot of $B$.	$\mathrm{m}$
${\theta }_1$	Angle between the first leg and the normal to the surface (the dashed vertical line).	$\mathrm{rad}$
${\theta }_2$	Angle between the second leg and the normal.	$\mathrm{rad}$
$n_1$	Refractive index of the upper medium.
$n_2$	Refractive index of the lower medium.

Figure 1 sets up the problem. The point $A$ is a height $a$ above a flat surface, in a medium of index $n_1$. The point $B$ is a depth $b$ below it, in a medium of index $n_2$, a horizontal distance $D$ further on.

Within one medium the quickest route is a straight line, so any candidate route is straight above the surface and straight below it. It is then fixed by one number: the distance $x$ along the surface from the foot of $A$ to the crossing point. Each leg is the hypotenuse of a right triangle, with sides $a$ and $x$ above the surface and $b$ and $D-x$ below it. The solid gold route is the quickest; the dashed routes cross the surface elsewhere, and each takes longer.

By Pythagoras the legs are $\sqrt{a^2+x^2}$ and $\sqrt{b^2+(D-x{)}^2}$ long. Light covers them at $c/n_1$ and $c/n_2$, so the route that crosses at $x$ takes
$T(x)=\frac{n_1\sqrt{a^2+x^2}+n_2\sqrt{b^2+(D-x{)}^2}}{c}$
The quickest route is where this function is stationary, so its derivative must vanish. Moving the crossing point to the right lengthens the upper leg and shortens the lower one, which is where the minus sign comes from:
$\frac{dT}{dx}=\frac{1}{c}\left(\frac{n_{1}x}{\sqrt{a^2+x^2}}-\frac{n_2(D-x)}{\sqrt{b^2+(D-x{)}^2}}\right)=0$

Now read the two fractions off the figure. At the crossing point the upper leg makes the angle ${\theta }_1$ with the normal, the dashed vertical line. The side $a$ is parallel to that normal, so the angle at $A$ between the leg and $a$ is also ${\theta }_1$. The side opposite this angle is $x$, and the first fraction, opposite over hypotenuse, is $x/\sqrt{a^2+x^2}=\sin {\theta }_1$. In the lower triangle the side $b$ is vertical too, the angle at $B$ is ${\theta }_2$, the opposite side is $D-x$, and the second fraction is $(D-x)/\sqrt{b^2+(D-x{)}^2}=\sin {\theta }_2$. The condition $dT/dx=0$ is therefore Snell's law:
$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Is this time a minimum? Differentiating once more gives a sum of two positive terms,
$\frac{d^{2}T}{d{x}^2}=\frac{1}{c}\left(\frac{n_1{a}^2}{(a^2+x^2{)}^{3/2}}+\frac{n_2{b}^2}{(b^2+(D-x{)}^2{)}^{3/2}}\right)>0$
so at a flat surface the stationary time is a true minimum.

Reflection follows from the same steps. Put $B$ above the surface, in the same medium as $A$, so that both legs have the index $n_1$. The condition then becomes $\sin {\theta }_r=\sin {\theta }_1$, where ${\theta }_r$ is the angle between the reflected leg and the normal: the angle of reflection equals the angle of incidence.

The principle also predicts something the angles alone cannot. Light must be slower, by the factor $n_2/n_1$, in the medium where it bends towards the normal. Newton's particle model explained the same bending by a pull of the glass on the light, which would make it faster inside. The two pictures predict opposite speeds, and in 1850 Foucault measured light to be slower in water than in air, as Fermat's principle requires.

Wavefronts and phase matching

Fermat's principle speaks of paths. The same law can be seen in the wave itself, as the condition that the waves on the two sides of the surface fit together. This view is the one that generalises: it explains mirages and links refraction to conservation laws.

Legend

${\lambda }_1$	Wavelength in the upper medium: the distance between neighbouring wavefronts there.	$\mathrm{m}$
${\lambda }_2$	Wavelength in the lower medium: ${\lambda }_2={\lambda }_1{n}_1/n_2$, shorter because light is slower there.	$\mathrm{m}$
$\Lambda$	Distance along the surface between the points where neighbouring fronts meet it, the same on both sides.	$\mathrm{m}$
${\theta }_1$	Angle between the incident ray and the normal; also the angle between a front and the surface.	$\mathrm{rad}$
${\theta }_2$	Angle between the refracted ray and the normal.	$\mathrm{rad}$
$n_1$	Refractive index of the upper medium.
$n_2$	Refractive index of the lower, optically denser medium (shaded).

Figure 2 draws a beam crossing the surface into a medium where light is 1.5 times slower, $n_2/n_1=1.5$. The purple lines are wavefronts, lines of equal phase. They are also lines of equal travel time from the source, so they are Fermat's principle in wave form, and each front is perpendicular to its ray. Neighbouring fronts are one wavelength apart: ${\lambda }_1$ in the upper medium and ${\lambda }_2={\lambda }_1{n}_1/n_2$ in the lower one, because the frequency is the same on both sides while the speed is lower below.

The expert layer of the animation shows the law as this matching condition: its fronts are ${\lambda }_0$ apart in air, where ${\lambda }_0$ is the wavelength in vacuum, and ${\lambda }_0/n$, 1.5 times closer, in glass.

Now look along the surface. The wave is continuous across it, so the fronts above and below must join without a break: each front reaches the surface at the same point from both sides. Neighbouring fronts therefore meet the surface the same distance $\Lambda$ apart on both sides, the green bar.

Relate $\Lambda$ to each wavelength. A front is perpendicular to its ray and the surface is perpendicular to the normal, so the angle between a front and the surface equals the angle between the ray and the normal, ${\theta }_1$ above the surface. Two neighbouring fronts and the stretch $\Lambda$ of surface between their meeting points form a right triangle: $\Lambda$ is its hypotenuse, and the gap between the fronts, ${\lambda }_1$, is the side opposite ${\theta }_1$. Hence ${\lambda }_1=\Lambda \sin {\theta }_1$. Below the surface the same triangle gives ${\lambda }_2=\Lambda \sin {\theta }_2$. Eliminating $\Lambda$,
$\frac{{\lambda }_1}{\sin {\theta }_1}=\Lambda =\frac{{\lambda }_2}{\sin {\theta }_2}$
For the animation's angles, $50^{\circ }$ in air and $30.7^{\circ }$ in glass, both sides give $\Lambda =1.31{\lambda }_0$. Writing $\lambda ={\lambda }_0/n$ on each side turns the equation into $n_1\sin {\theta }_1=n_2\sin {\theta }_2$: Snell's law once more.

The tilt of the fronts also shows why the ray turns towards the normal in the slower medium. One end of an oblique front reaches the surface first and is slowed first, while the other end still moves fast in the upper medium. The front pivots about its slowed end, and the ray, perpendicular to the front, turns towards the normal.

The matching condition has a compact form. The wave vector points along the ray and has the size $k=n\omega /c=2\pi /\lambda$. Its component along the surface is $k\sin \theta =2\pi /\Lambda$, so phase matching says that this tangential component is the same on both sides. This mirrors two conservation laws of mechanics. A boundary that does not change in time conserves the frequency $\omega$, as time symmetry conserves energy. A surface that looks the same all along its length conserves the tangential wave vector, as translation symmetry conserves momentum.

The rule holds for any stack of parallel layers: crossing one boundary after another, $n(z)\sin \theta (z)$ stays the same at every height $z$. Where $n$ grows, $\sin \theta$ shrinks and the ray turns towards the normal. In air whose index varies smoothly with height, a ray therefore curves towards higher $n$. Over a hot road the air near the ground is warmer and its index lower, so light from the sky curves upwards into the eye and the road seems to shine like water: a mirage.

Refraction angle against incidence angle

Snell's law is not linear in the angles. A graph of one angle against the other shows at a glance where the small-angle rule of lens formulas holds, where it fails, and where total internal reflection begins.

Legend

${\theta }_1$	Angle of incidence, measured from the normal.	$\mathrm{rad}$
${\theta }_2$	Angle of refraction, measured from the normal.	$\mathrm{rad}$
${\theta }_c$	Critical angle for glass into air, $\sin {\theta }_c=1/1.5$; also the angle at which the air-into-glass curve ends.	$\mathrm{rad}$
$n_1$	Refractive index of the medium the light comes from.
$n_2$	Refractive index of the medium it enters.

Figure 3 plots the refraction angle ${\theta }_2$ against the incidence angle ${\theta }_1$, from $0$ to $90^{\circ }$, for glass of index 1.5 crossed in both directions. As in Snell's law, $n_1$ is the index of the medium the light comes from and $n_2$ that of the medium it enters.

Follow the blue curve first: air into glass, $n_1<n_2$. Near the origin it is a straight line of slope $n_1/n_2=0.667$, because for small angles $\sin \theta \approx \theta$ and Snell's law becomes ${\theta }_2\approx (n_1/n_2){\theta }_1$, the rule behind lens formulas. The curve then bends over and ends, at grazing incidence ${\theta }_1=90^{\circ }$, at ${\theta }_2=41.8^{\circ }$: the height of the horizontal dashed line marked ${\theta }_c$. Light from air never travels inside the glass at more than this angle from the normal.

Now the red curve: glass into air, $n_1>n_2$. It starts with slope 1.5, climbs ever faster, and reaches ${\theta }_2=90^{\circ }$ at ${\theta }_1={\theta }_c=41.8^{\circ }$, where it stops at the vertical dashed line. This is the critical angle, $\sin {\theta }_c=n_2/n_1$. Beyond it there is no refracted ray at all, and all the light is reflected: total internal reflection.

The two curves are mirror images of each other in the dashed diagonal ${\theta }_2={\theta }_1$. The reason is that every light path can be travelled backwards: if light from air at ${\theta }_1$ enters the glass at ${\theta }_2$, light inside the glass at ${\theta }_2$ leaves into air at ${\theta }_1$, which swaps the axes. That is why the same angle ${\theta }_c$ ends both curves.

Why does the red curve shoot up so steeply near its end? Differentiating Snell's law gives $n_1\cos {\theta }_{1}d{\theta }_1=n_2\cos {\theta }_{2}d{\theta }_2$, so
$\frac{d{\theta }_2}{d{\theta }_1}=\frac{n_1\cos {\theta }_1}{n_2\cos {\theta }_2}$
As ${\theta }_2\to 90^{\circ }$ the cosine in the denominator goes to zero and the slope diverges. Near ${\theta }_c$ a tiny tilt of the incoming ray swings the emerging ray through a wide angle.

Why light is slower in matter

So far the index $n$ has simply been given. This section explains where it comes from, and why it depends on colour.

The slowing is not light pausing at each atom. The wave drives the electrons of the material, which oscillate at the wave's frequency and radiate waves of their own. A thin layer of these electrons adds a weak wave that lags a quarter of a cycle behind the incoming one. Adding a small wave a quarter of a cycle behind shifts the phase of the total slightly back, as if the wave had arrived a little late. Layer after layer, these small delays add up, and the total wave travels at $c/n$.

The electrons respond more strongly as the frequency approaches their natural frequencies, which for glass lie in the ultraviolet. The index therefore rises with frequency: about 1.514 for red light and 1.522 for blue in a common crown glass. This dependence of $n$ on frequency is called dispersion, and it is what lets a prism split white light into colours. For X-rays, whose frequency lies above the resonances, $n$ is slightly below 1. The crests then move faster than $c$, which breaks no law: the crests of a steady wave carry no signal.

Amplitudes, Brewster's angle and the evanescent wave

Snell's law fixes the directions of the reflected and refracted rays, but not how much light goes each way. That depends on light being an electromagnetic wave, and it explains why a window reflects a little, why reflected glare is polarised, and what happens just beyond a totally reflecting surface.

Maxwell's equations require the components of the electric and magnetic fields along the surface to be continuous across it. Applying this to the incident, reflected and refracted waves gives the Fresnel equations, which give the fractions reflected and transmitted. They depend on polarisation, the direction in which the electric field oscillates. At normal incidence both polarisations behave alike, and the reflected fraction of the power is
$R={\left(\frac{n_1-n_2}{n_1+n_2}\right)}^2\text{(normal incidence)}$
This is 4.0 % for glass, 2.0 % for water and 17 % for diamond, the same from either side, because swapping $n_1$ and $n_2$ only changes the sign inside the square.

At oblique incidence the two polarisations part company. The plane of incidence is the plane that contains the incident ray and the normal. At the animation's $50^{\circ }$ from air into glass, light polarised perpendicular to that plane (called s) is reflected at 11.2 %, light polarised in that plane (called p) at only 0.3 %.

The p reflection vanishes altogether at Brewster's angle, $\tan {\theta }_B=n_2/n_1$: $56.3^{\circ }$ from air into glass. There the refracted ray, at $33.7^{\circ }$, is exactly perpendicular to the direction of the reflected ray. The electrons of the previous section explain why. For p light they oscillate perpendicular to the refracted ray, within the plane of incidence, and at Brewster's angle that line of oscillation points straight along the reflected direction. An oscillating dipole radiates nothing along its own axis, so no p light is reflected. Glare reflected near this angle is therefore mostly s-polarised, which is why polarising sunglasses cut it.

Total internal reflection hides a subtlety. Take light inside glass of index $n_1$, meeting a surface with air, of index $n_2$, beyond ${\theta }_c$, and write $k_0=2\pi /{\lambda }_0$ for the wave number in vacuum. Phase matching still demands the tangential wave number $n_1{k}_0\sin {\theta }_1$ on the air side. Beyond ${\theta }_c$ this exceeds $n_2{k}_0$, the whole wave number available in air. The component normal to the surface, whose square is $n_2^2{k}_0^2-n_1^2{k}_0^2{\sin }^2{\theta }_1$, is then the square root of a negative number: it is imaginary. An imaginary wave number describes decay, not oscillation. The field in the air falls off with the distance $z$ from the surface as $e^{-z/\delta }$, with
$\delta =\frac{{\lambda }_0}{2\pi \sqrt{n_1^2{\sin }^2{\theta }_1-n_2^2}}$

At an incidence of $50^{\circ }$ inside the animation's glass, $\delta =0.28{\lambda }_0$, about 155 nm for green light of 550 nm. This is the evanescent wave. On average it carries no energy away from the surface, so the reflection really is total. But bring a second block of glass within a few $\delta$ and the evanescent field reaches into it: light leaks across the gap. This frustrated total internal reflection follows the same mathematics as quantum tunnelling through a barrier.

Worked problem: a glass rod as a light pipe

Total internal reflection can trap light inside a rod or a fibre over kilometres. This problem finds when trapping works, how wide a cone of light a fibre accepts, and what that cone costs in signalling speed.

Legend

$n_1$	Refractive index of the core.
$n_2$	Refractive index of the cladding, lower than $n_1$.
${\alpha }_{\max }$	Largest angle from the axis at which light entering the end face is still trapped.	$\mathrm{rad}$
${\beta }_{\max }$	Angle from the axis, inside the core, of the ray that entered at ${\alpha }_{\max }$.	$\mathrm{rad}$
${\theta }_c$	Critical angle at the core wall, $\sin {\theta }_c=n_2/n_1$: the extreme ray meets the wall at exactly this angle from the normal.	$\mathrm{rad}$
$L$	Length of the fibre.	$\mathrm{m}$

Figure 4 shows a step-index fibre of length $L$: a core of index $n_1$ inside a cladding of index $n_2<n_1$. Light enters through the flat end face, which is perpendicular to the axis, the dashed horizontal line. The ray drawn is the extreme one that is still trapped. It enters at the angle ${\alpha }_{\max }$ from the axis, at the edge of the shaded cone, and refracts at the end face to the angle ${\beta }_{\max }$ from the axis inside the core.

The core wall is parallel to the axis, so its normal is perpendicular to the axis, and the ray meets the wall at $90^{\circ }-{\beta }_{\max }$ from that normal. For the extreme ray this is exactly the critical angle: ${\theta }_c=90^{\circ }-{\beta }_{\max }$. A ray inside the cone travels at a smaller angle to the axis, meets the wall at more than ${\theta }_c$, and is trapped as well, bouncing from wall to wall as drawn.

Start with a bare rod. Take a long rod of the animation's glass, $n=1.50$, in air, with a flat end face perpendicular to its axis: the air itself plays the cladding. Can light that enters the end at any angle escape through the side?

A ray entering at an angle $\alpha$ from the axis refracts to $\beta$, with $\sin \beta =\sin \alpha /1.50$ by Snell's law at the end face. Even at grazing entry, $\alpha =90^{\circ }$, this gives only $\sin \beta =0.667$, so $\beta$ never exceeds $41.8^{\circ }$. The ray then meets the side wall at $90^{\circ }-\beta \ge 48.2^{\circ }$ from the wall's normal, beyond ${\theta }_c=41.8^{\circ }$: every ray that gets in is trapped. In general the steepest ray has $\sin {\beta }_{\max }=1/n$, and it is trapped if it meets the wall beyond ${\theta }_c$, that is if $\cos {\beta }_{\max }\ge \sin {\theta }_c$. With $\sin {\theta }_c=1/n$ this reads $\sqrt{1-1/n^2}\ge 1/n$, or $n\ge \sqrt{2}=1.414$.

A bare rod leaks wherever dust or a finger touches it, because there the outside index is no longer 1. A real fibre therefore wraps its core in a cladding of slightly lower index, say $n_2=1.46$ around $n_1=1.50$. Now $\sin {\theta }_c=1.46/1.50=0.973$ and ${\theta }_c=76.7^{\circ }$, so inside the core the ray may make at most ${\beta }_{\max }=13.3^{\circ }$ with the axis. Snell's law at the end face, with air outside, gives $\sin {\alpha }_{\max }=n_1\sin {\beta }_{\max }$, and $\sin {\beta }_{\max }=\cos {\theta }_c=\sqrt{1-n_2^2/n_1^2}$. So the end accepts only rays with
$\sin {\alpha }_{\max }=n_1\cos {\theta }_c=\sqrt{n_1^2-n_2^2}=\sqrt{1.50^2-1.46^2}=0.344$
a cone of half-angle $20.1^{\circ }$. In Figure 4 the cone is far wider, because the figure exaggerates the difference between the indices. The sine, 0.344, is called the fibre's numerical aperture.

The price of accepting a cone is that rays at different angles travel different lengths. Over $L=1.00\mathrm{km}$ the axial ray takes $n_{1}L/c=5.00\mu \mathrm{s}$. The steepest trapped ray zigzags along a path longer by the factor $1/\cos {\beta }_{\max }$; since $\cos {\beta }_{\max }=\sin {\theta }_c=n_2/n_1$, that factor is $n_1/n_2$, and the ray takes 5.14 μs. A short pulse sent into the fibre therefore comes out smeared over 137 ns, so pulses must be at least that far apart to stay distinct: roughly 7 million per second over 1 km. Telecom fibres avoid this with cores only about 9 μm wide, which carry a single mode of the wave, and there the ray picture fails.

Limits of validity

Ray optics is a model. The paragraphs below show where it stops working.

It is the short-wavelength limit of wave optics. The band of contributing paths around a ray, met in the first section, is about $\sqrt{\lambda L}$ wide over a length $L$: 0.7 mm for light of 500 nm over 1 m. An opening narrower than this band cuts into it, and the ray picture breaks down. A slit of width $a=0.1\mathrm{mm}$, for example, spreads such a beam by about $\lambda /a=0.005\mathrm{rad}$, or 0.3°: this is diffraction.

A single index $n$ also assumes that the medium is linear (its response is proportional to the field), isotropic (the same in every direction) and non-absorbing. Each assumption can fail. Calcite is not isotropic: it has two indices, 1.486 and 1.658 at the extremes, so one ray entering it splits into two. In metals $n$ is complex, and light dies out within a few tens of nanometres. Finally, a moving boundary breaks the time symmetry that conserved the frequency: a mirror receding at speed $u$ lowers the frequency of the reflected light by a fraction of about $2u/c$.

Where this leads

Fermat's principle leads into Lenses and Mirrors, where a focus is a point that many paths reach in exactly the same time. Dispersion is the subject of Prisms, Dispersion and Colour, and Brewster's angle returns in Polarization of Light. The light wave itself is described in Electromagnetic Waves, the failure of rays in Diffraction and the Double Slit, and the quantum twin of frustrated reflection in Schrödinger's Equation and Tunnelling.

In Gravity as Curved Spacetime the Sun bends starlight that grazes it by 1.75 arcseconds, as if the space around it had a refractive index $n\approx 1+2GM/(r{c}^2)$, where $M$ is the Sun's mass and $r$ the distance from its centre. And the stationary-phase argument of the first section, by which rays emerge from waves, is the same argument by which classical mechanics, with its principle of least action, emerges from quantum mechanics.