Prisms, Dispersion and Colour

Beginner

Step 1, white is a mixture. Sunlight looks colourless, yet it carries every colour of the rainbow at once. In 1666 Isaac Newton sent a sunbeam through a glass prism and got a band of colours, always in the same order: red, orange, yellow, green, blue, violet. That band is a spectrum.

Step 2, each colour has its own speed in glass. In empty space all colours travel equally fast, but glass holds violet back the most and red the least. Since bending comes from slowing down, violet bends the most and red the least. This sorting of colours by refraction is called dispersion. A flat window sorts them a tiny bit too, but its second face, parallel to the first, undoes the effect; a prism's tilted faces add up instead, so the fan opens wide.

In the animation a torch shines white light into a prism: watch the beam cross the glass and fan out, painting a spectrum on the table.

Step 3, rainbows. Every raindrop works like a tiny prism with a mirror at the back: sunlight goes in, reflects off the back of the drop and comes out sorted, each colour at its own angle close to 42°. That is why a rainbow appears only with the Sun behind you and the rain in front, red on the outside of the bow.

Step 4, colours of things. A ripe tomato looks red because it swallows most colours and sends back mainly red. Lights, on the other hand, add up: red, green and blue together look white, which is how a phone screen builds every colour from three kinds of tiny dots.

Check yourself: under pure blue light, what colour does a ripe tomato look? Answer: almost black, because there is no red light for it to send back.

Intermediate

Index, wavelength and dispersion

In glass, violet light travels more slowly than red light. Light crosses empty space at $c=3.00\times 10^{8}m/s$, whatever its colour. In a transparent material it is slower, at a speed $v$, and the refractive index says how many times slower:
$n=\frac{c}{v}$

Crossing a surface, light keeps its frequency $f$, because the far side is driven at the rate of the arriving wave; since $v=f\lambda$, it is the wavelength that shrinks, by the factor $n$. Colour goes with the frequency, so colours are labelled by their wavelength $\lambda$ in vacuum: about 700 nm for red ($f=4.3\times 10^{14}\mathrm{Hz}$) and 400 nm for violet.

The index is slightly different for each colour, and that difference is dispersion. In glass and water it rises towards short wavelengths, following an empirical rule:
$n(\lambda )\approx A+\frac{B}{{\lambda }^2}\text{(Cauchy)}$
The prism in the animation is a dense flint glass with $A=1.700$ and $B=16000\mathrm{n}{\mathrm{m}}^2$, so $n_{\mathrm{red}}=1.700+16000/700^2=1.733$ and $n_{\mathrm{violet}}=1.700+16000/400^2=1.800$. Water disperses less: 1.3305 for red, 1.3435 for violet. These small differences, 0.067 and 0.013, are all that separates the colours.

Two refractions in a prism

At each face, Snell's law holds with every angle measured from the normal to that face:
$n_1\sin {\theta }_1=n_2\sin {\theta }_2$
For a prism of apex angle $\alpha$ in air, a ray is then followed in three steps:

• $\sin {\theta }_1=n\sin {\theta }_2$
• ${\theta }_3=\alpha -{\theta }_2$
• $\sin {\theta }_4=n\sin {\theta }_3$

The middle step is geometry: the ray inside and the two face normals form a triangle whose third angle is $180^{\circ }-\alpha$, so ${\theta }_2+{\theta }_3=\alpha$. The ray turns by ${\theta }_1-{\theta }_2$ at the first face and by ${\theta }_4-{\theta }_3$ at the second, both times towards the base, so its total deviation is
$\delta ={\theta }_1+{\theta }_4-\alpha \text{(prism of apex angle }\alpha \text{)}$

For a thin prism, with small angles, the sines can be replaced by the angles: ${\theta }_1+{\theta }_4\approx n({\theta }_2+{\theta }_3)=n\alpha$, and
$\delta \approx (n-1)\alpha \text{(thin prism)}$
Two colours then separate by about $(n_{\mathrm{violet}}-n_{\mathrm{red}})\alpha$. A window has $\alpha =0$: every colour leaves parallel to the incoming beam, and the colours overlap again.

Worked example: the prism in the animation

The white beam meets the first face at ${\theta }_1=55^{\circ }$, and $\sin 55^{\circ }=0.8192$. For red, $\sin {\theta }_2=0.8192/1.733=0.4727$, so ${\theta }_2=28.2^{\circ }$; for violet, $\sin {\theta }_2=0.8192/1.800=0.4551$ and ${\theta }_2=27.1^{\circ }$. Inside the glass the paths are only $1.1^{\circ }$ apart, so the animation draws one white beam there.

Both rays reach the second face at ${\theta }_3=60^{\circ }-{\theta }_2$: $31.8^{\circ }$ for red, $32.9^{\circ }$ for violet. On the way out, $\sin {\theta }_4=1.733\cdot 0.5267=0.9128$ gives ${\theta }_4=65.9^{\circ }$ for red, and $1.800\cdot 0.5436=0.9785$ gives $78.1^{\circ }$ for violet. The deviations are $\delta =55^{\circ }+65.9^{\circ }-60^{\circ }=60.9^{\circ }$ and $55^{\circ }+78.1^{\circ }-60^{\circ }=73.1^{\circ }$, the values on the $\delta$ readouts. The $1.1^{\circ }$ gap inside has grown to $12.2^{\circ }$: violet leaves close to grazing, where a small change of $\sin {\theta }_4$ makes a large change of angle.

The speed readouts drop from $3.00\times 10^{8}m/s$ in air to $c/n=2.998\times 10^8/1.733=1.73\times 10^{8}m/s$ for red and $1.67\times 10^{8}m/s$ for violet inside the glass. The $\delta$ arc opens between the dashed straight-on direction and the red ray, and the colours land on the table in order, violet nearest the prism.

Worked example: a thin glass prism

A 10° prism of crown glass, the ordinary glass of lenses, has $n=1.513$ for red and $1.531$ for violet. A narrow white beam crosses it almost at right angles and lands on a wall 2.0 m away. How wide is the spectrum?

With the thin-prism rule,
${\delta }_{\mathrm{red}}\approx 0.513\cdot 10^{\circ }=5.13^{\circ },{\delta }_{\mathrm{violet}}\approx 0.531\cdot 10^{\circ }=5.31^{\circ }$
The spread is $0.18^{\circ }$, or $0.18\cdot \pi /180=3.1\times 10^{-3}\mathrm{rad}$, so on the wall the spectrum is about $2.0\mathrm{m}\cdot 3.1\times 10^{-3}=6.3\mathrm{mm}$ wide, while the whole spot moves about $2.0\tan 5.2^{\circ }\approx 0.18\mathrm{m}$. An exact trace with Snell's law at $5^{\circ }$ incidence gives $5.16^{\circ }$ and $5.35^{\circ }$: the rule is good to better than 1 %.

Cut from the animation's flint glass, the same prism gives $0.733\cdot 10^{\circ }=7.3^{\circ }$ for red and $8.0^{\circ }$ for violet: 1.4 times the deviation, but 3.7 times the spread, $0.67^{\circ }$, which makes a spectrum 23 mm wide.

Raindrops and the rainbow angle

Sunlight enters a spherical raindrop at an angle of incidence $\theta$, refracts to ${\theta }^{\prime }$ with $\sin \theta =n\sin {\theta }^{\prime }$, reflects once off the back and refracts out, meeting the surface inside at the same ${\theta }^{\prime }$ each time because the drop is round. It turns by $\theta -{\theta }^{\prime }$ on entering, by $180^{\circ }-2{\theta }^{\prime }$ at the back and by $\theta -{\theta }^{\prime }$ on leaving, $180^{\circ }+2\theta -4{\theta }^{\prime }$ in all, so it comes back at an angle from the direction opposite the Sun of
$\phi =4{\theta }^{\prime }-2\theta$

In the animation the white ray enters at $\theta =59.3^{\circ }$ ($\sin \theta =0.86$). For red, $\sin {\theta }^{\prime }=0.86/1.3305=0.6464$, so ${\theta }^{\prime }=40.27^{\circ }$ and $\phi =4\cdot 40.27^{\circ }-2\cdot 59.32^{\circ }=42.4^{\circ }$. Violet, with $n=1.3435$, has ${\theta }^{\prime }=39.80^{\circ }$ and $\phi =40.6^{\circ }$: the two $\phi$ readouts. Rays entering higher or lower return at smaller angles, so the light gathers near this largest angle.

Drops lying $42.4^{\circ }$ from the point opposite the Sun (the shadow of your head) therefore send you red, and those at $40.6^{\circ }$ send violet: red is the outer edge of the bow. The Sun must be behind you and less than 42° above the horizon, or the bow lies below the horizon.

Adding lights, subtracting paints

The eye has three kinds of colour receptors (cones), most sensitive to red, green and blue light. Lights falling on one spot add up: red plus green excites the cones much as spectral yellow at 580 nm does, so the spot looks yellow, and red, green and blue together look white. Screens build their colours from tiny red, green and blue dots in this way.

Paints, inks and filters subtract: each absorbs part of the light, and the eye receives what survives. Yellow paint absorbs blue and cyan paint absorbs red, so their mixture returns only green. An object's colour also depends on the light: under pure blue light a ripe tomato has no red to send back and looks almost black.

Common mistakes

• The prism does not create colours: it sorts those already in white light, and a reversed second prism merges them back into white.
• Light does not change colour in glass. Red light keeps its frequency of $4.3\times 10^{14}\mathrm{Hz}$; its wavelength shrinks from 700 nm to $700/1.733=404\mathrm{nm}$, yet it is still red.
• A larger index means slower light, not faster: violet, with $n=1.800$, crosses the glass at $1.67\times 10^{8}m/s$, red at $1.73\times 10^{8}m/s$.
• The thin-prism rule needs a thin prism. For the 60° prism of the animation, $(n-1)\alpha$ gives $44^{\circ }$ for red instead of $60.9^{\circ }$, and a spread of $4.0^{\circ }$ instead of $12.2^{\circ }$. And $n\alpha$ is wrong for any prism: only the excess of $n$ over 1 bends light.
• Mixing lights is not mixing paints: red and green light make yellow, but red and green paint make a dull brown, because each paint absorbs most of what the other returns.

Where you meet it

Spectrometers spread light from stars, flames and lamps into spectra that reveal what the source is made of, and cut diamonds flash with colour because diamond has a large index, 2.42, and strong dispersion. In lenses dispersion is a nuisance: a simple lens focuses violet closer than red, and a crown lens cemented to a weaker diverging flint lens cancels most of the error, as Lenses and Mirrors and The Eye, Cameras and Telescopes show. In optical fibres it smears out short pulses. Gratings, in Diffraction and the Double Slit, sort colours by interference instead and bend red the most, the reverse of a prism.

Advanced

Why light is slower in glass

The refractive index $n$ says how many times more slowly light travels in a material than in vacuum, and the prism shows that $n$ depends on colour. To see why, look at what glass is made of: nuclei and bound electrons, with empty space between them in which every electric field travels at exactly $c$. The slower wave in glass must therefore be a combined effect. The arriving wave shakes the electrons, the shaking electrons radiate, and their radiation adds to the arriving wave. This section sets up the two pieces; the next one adds them.

In the Lorentz model each electron, of charge $q_e$ and mass $m$, is held in place by a spring-like force, so that left alone it would oscillate at a natural frequency ${\omega }_0$. The light's electric field at the electron is $E_0{e}^{i\omega t}$, where $\omega$ is the light's angular frequency and the physical field is the real part. Newton's second law and its steady solution are
$\ddot{x}+{\omega }_0^{2}x=\frac{q_e}{m}{E}_0{e}^{i\omega t}x=\frac{q_e{E}_0}{m({\omega }_0^2-{\omega }^2)}{e}^{i\omega t}$
To get the solution, try $x\propto e^{i\omega t}$, so that $\ddot{x}=-{\omega }^{2}x$, and solve for the amplitude. The electron oscillates at the driving frequency, not at its own. Below resonance, $\omega <{\omega }_0$, the factor ${\omega }_0^2-{\omega }^2$ is positive, so the dipole $q_{e}x$ moves in step with the field, and its amplitude grows as $\omega$ approaches ${\omega }_0$.

Now take a thin slab of the material, of thickness $\Delta z$, across the beam, with $N$ electrons per unit volume. Seen along the beam it is a flat sheet carrying a charge $N{q}_e\Delta z$ per unit area, all oscillating together in the plane of the sheet. Such a sheet radiates a plane wave on each side, with a field proportional to the velocity of its charges (a result from Electromagnetic Waves; ${\epsilon }_0$ is the permittivity of free space):
$E_a=-\frac{N{q}_e\Delta z}{2{\epsilon }_{0}c}\dot{x}$
Because the slab is thin, $E_a$ is small. What matters most is its timing: it follows the velocity $\dot{x}$, not the displacement $x$, and that shifts it by a quarter cycle.

One thin slab delays the wave

Two waves of the same frequency add like arrows. Figure 1 uses this to show how a small wave a quarter cycle behind turns into a delay of the whole wave, which is what a lower speed means.

Legend

$E_0$	Wave arriving at the slab, drawn as an arrow whose length is its amplitude and whose direction is its phase.	$V/m$
$E_a$	Wave radiated by the electrons of the slab: small, and a quarter cycle behind $E_0$, so drawn at a right angle, clockwise.	$V/m$
$E^{\prime }$	Wave beyond the slab: the sum of $E_0$ and $E_a$.	$V/m$
$\phi$	Angle from $E_0$ to $E^{\prime }$: the phase delay $\omega (n-1)\Delta z/c$ added by the slab, equal to $|E_a|/|E_0|$.	$\mathrm{rad}$

In Figure 1 each wave is an arrow in the complex plane: its length is the amplitude and its direction is the phase, with the common factor $e^{i\omega t}$ left out. Beyond the slab both waves travel on at $c$, so the factor they pick up on the way is also common to both and is left out too. The arriving wave $E_0$ is the horizontal arrow.

To find the direction of $E_a$, put the electron's motion into its formula. Since $\dot{x}=i\omega x$,
$E_a=-i\frac{\omega N{q}_e^2\Delta z}{2{\epsilon }_{0}mc({\omega }_0^2-{\omega }^2)}{E}_0$
Below resonance the fraction is positive (the sign of $q_e$ drops out, since only $q_e^2$ appears), so $E_a$ is $E_0$ times a small positive number times $-i$. Multiplying by $-i=e^{-i\pi /2}$ turns an arrow a right angle clockwise, and it is the same as replacing $t$ by $t-T/4$, where $T$ is the period: the wave is a quarter cycle late. That is the short arrow $E_a$, pointing straight down.

Beyond the slab the two add, and their sum $E^{\prime }$ is the diagonal. Look at the right-angled triangle. Because $E_a$ is perpendicular to $E_0$ and much shorter, $E^{\prime }$ has the length of $E_0$ to first order (the change in length is of second order in $|E_a|/|E_0|$), but it is turned clockwise by a small angle $\phi =|E_a|/|E_0|$. Turned clockwise means behind in phase. The figure draws $\phi$ large so that the pieces can be seen; for a thin slab it is tiny.

Now compare with what a slower wave would do. Suppose the light really crossed the slab at speed $c/n$. It would take $n\Delta z/c$ instead of $\Delta z/c$, so it would arrive later by $(n-1)\Delta z/c$, and its arrow would be turned back by the phase $\phi =\omega (n-1)\Delta z/c$. For small $\phi$, the wave beyond the slab would be multiplied by
$e^{-i\phi }\approx 1-i\phi =1-\frac{i\omega (n-1)\Delta z}{c}$
that is, the original wave plus a small piece a quarter cycle behind, of relative size $\phi$: exactly the shape of $E_0+E_a$ in Figure 1. The two descriptions agree when the small pieces are equal. Setting $\omega (n-1)\Delta z/c$ equal to the fraction in $E_a$ above, the factors $\omega$, $\Delta z$ and $c$ cancel and leave
$n=1+\frac{N{q}_e^2}{2{\epsilon }_{0}m({\omega }_0^2-{\omega }^2)}$

This is the index of a dilute material. Each layer pulls the phase back a little, by an amount set by how strongly the electrons respond, and that depends on $\omega$: dispersion is built into the mechanism. As a check on the size, take air with one effective electron per molecule, $N=2.5\times 10^{25}{\mathrm{m}}^{-3}$, resonating near 100 nm. The formula gives $n-1\approx 1\times 10^{-4}$, the right order of magnitude for the measured $2.9\times 10^{-4}$.

Below resonance: Cauchy's law and the blue sky

Visible light sits at the low-frequency end of this formula, and expanding it there gives the simple law used for prisms. Glass and water have their strong resonances in the ultraviolet, so for visible light $\omega$ is well below ${\omega }_0$. Figure 3, in a later section, draws the whole curve.

First name the low-frequency limit. At $\omega =0$ the formula gives $n=A$, with $A-1=N{q}_e^2/(2{\epsilon }_{0}m{\omega }_0^2)$, so for any $\omega$
$n-1=\frac{A-1}{1-{\omega }^2/{\omega }_0^2}\approx (A-1)\left(1+\frac{{\omega }^2}{{\omega }_0^2}\right)$
using $1/(1-u)\approx 1+u$ for small $u$. With $\omega =2\pi c/\lambda$ and the resonance wavelength ${\lambda }_0=2\pi c/{\omega }_0$, the ratio ${\omega }^2/{\omega }_0^2$ equals ${\lambda }_0^2/{\lambda }^2$. The result is Cauchy's law, $n\approx A+B/{\lambda }^2$ with $B=(A-1){\lambda }_0^2$.

Read backwards, a measured $A$ and $B$ say where the resonance is. The animation's fits give ${\lambda }_0=\sqrt{16000/0.700}\approx 150\mathrm{nm}$ for the flint glass ($A=1.700$, $B=16000\mathrm{n}{\mathrm{m}}^2$) and $\sqrt{3090/0.3242}\approx 100\mathrm{nm}$ for water ($A=1.3242$, $B=3090\mathrm{n}{\mathrm{m}}^2$), both in the ultraviolet, as expected.

Cauchy's law also explains why a prism spreads the violet end of the spectrum more than the red. The animation's first expert graph plots $n(\lambda )$ from 400 to 700 nm, with a dot for each beam colour. Its slope, $dn/d\lambda =-2B/{\lambda }^3$, is $-9.3\times 10^{-5}\mathrm{n}{\mathrm{m}}^{-1}$ at 700 nm and $-5.0\times 10^{-4}\mathrm{n}{\mathrm{m}}^{-1}$ at 400 nm. Because of the cube, the curve is $(7/4{)}^3\approx 5.4$ times steeper at the violet end.

The prism then turns each difference in index into a difference in angle. Let its apex angle be $\alpha$ (60° in the animation). A ray meets the first face at the angle of incidence ${\theta }_1$ (55°) and refracts to ${\theta }_2$; it meets the second face at ${\theta }_3$ and leaves at ${\theta }_4$. Snell's law at the two faces gives $\sin {\theta }_1=n\sin {\theta }_2$ and $\sin {\theta }_4=n\sin {\theta }_3$, and the triangle gives ${\theta }_2+{\theta }_3=\alpha$. Now change $n$ slightly, keeping ${\theta }_1$ fixed. The first relation gives $n\cos {\theta }_{2}d{\theta }_2=-\sin {\theta }_{2}dn$, and the triangle gives $d{\theta }_3=-d{\theta }_2$. Put both into the differential of the second relation, $\cos {\theta }_{4}d{\theta }_4=\sin {\theta }_{3}dn+n\cos {\theta }_{3}d{\theta }_3$, and use $\sin {\theta }_3\cos {\theta }_2+\cos {\theta }_3\sin {\theta }_2=\sin \alpha$:
$\frac{d{\theta }_4}{dn}=\frac{\sin \alpha }{\cos {\theta }_2\cos {\theta }_4}$

For the animation's prism this is 2.4 rad per unit of $n$ for red. For violet, whose exit ray leaves close to grazing so that $\cos {\theta }_4$ is small, it is 4.7. Both effects favour the violet end. Per nanometre of wavelength, the exit angle changes by $2.4\cdot 9.3\times 10^{-5}\approx 2.2\times 10^{-4}$ rad at 700 nm but by $4.7\cdot 5.0\times 10^{-4}\approx 2.4\times 10^{-3}$ rad at 400 nm, about ten times more: the violet end of the fan is stretched.

The same driven electrons also radiate sideways, and this gives the sky its colour. Far below resonance the amplitude $q_e{E}_0/(m({\omega }_0^2-{\omega }^2))$ hardly depends on $\omega$, because ${\omega }^2$ is small next to ${\omega }_0^2$. An oscillating dipole radiates power in proportion to the square of its acceleration, and the acceleration is ${\omega }^{2}x$, so the radiated power grows as ${\omega }^4{x}^2$, that is, as ${\omega }^4$. Air molecules therefore scatter sunlight in proportion to ${\omega }^4$ (Rayleigh scattering): $(650/450{)}^4\approx 4.4$ times more at 450 nm than at 650 nm. The blue light scattered out of the sunbeam fills the sky, and what is left after a long path through air, as at sunset, is red.

Phase velocity and group velocity

A prism separates colours in space. In a long piece of glass, dispersion also separates them in time, because a pulse does not travel at the speed of the crests inside it. Figure 2 shows the two speeds side by side.

Legend

$t_1$	First instant.	$\mathrm{s}$
$t_2$	A later instant; $\Delta t=t_2-t_1$.	$\mathrm{s}$
$E$	Electric field of the pulse: the blue curve, whose crests move at the phase velocity, under a shaded envelope.	$V/m$
$v\Delta t$	Distance a crest travels between $t_1$ and $t_2$: the phase velocity $v=c/n$ times $\Delta t$.	$\mathrm{m}$
$v_g\Delta t$	Distance the peak of the envelope travels in the same time: the group velocity $v_g=c/n_g$ times $\Delta t$.	$\mathrm{m}$

Figure 2 follows a short pulse of violet light through the flint glass, drawn at two instants $t_1$ and $t_2$, a time $\Delta t$ apart. The blue curve is the electric field $E$: a train of crests under a smooth envelope, the shaded outline. The crests move at the phase velocity $v=c/n$, the speed of the crests of an endless wave of one frequency, so the marked crest advances $v\Delta t$, the longer blue distance. The envelope moves at a different speed, the group velocity $v_g$, and advances only $v_g\Delta t$, the shorter purple distance. Watch the peak of the envelope: at $t_1$ the marked crest sits on it, and by $t_2$ the crest has moved ahead of it.

The figure uses the animation's glass at 400 nm, where $n=1.800$ and, as computed below, the group index $n_g=c/v_g$ is 2.000. In the same time the crests cross five wavelengths and the envelope only $5\cdot 1.800/2.000=4.5$. So the crest that sat on the peak at $t_1$ is half a wavelength ahead of it at $t_2$, and the peak now sits on a trough. Crests keep appearing at the tail of the pulse, running forward through it, and fading at its head.

Why does the envelope have its own speed? A pulse, like any signal, is a change in the wave, and a changing wave is a mixture of neighbouring frequencies. Take just two, at $\omega \pm \delta \omega$, with wavenumbers $k\pm \delta k$. Their sum is a wave at $\omega$ and $k$ whose amplitude varies as $\cos (\delta \omega t-\delta kz)$, and that pattern moves at $\delta \omega /\delta k$. For a narrow band of frequencies this becomes
$v_g=\frac{d\omega }{dk}$

In the glass $k=n\omega /c$, with $n$ depending on $\omega$, so $dk/d\omega =(n+\omega dn/d\omega )/c$. Since $\lambda =2\pi c/\omega$, a small change gives $d\lambda /\lambda =-d\omega /\omega$, hence $\omega dn/d\omega =-\lambda dn/d\lambda$. The group index is therefore
$n_g=\frac{c}{v_g}=n+\omega \frac{dn}{d\omega }=n-\lambda \frac{dn}{d\lambda }$

In normal dispersion $n$ falls as $\lambda$ grows, $dn/d\lambda <0$, so $n_g>n$: pulses move more slowly than the crests inside them. With Cauchy's law, $-\lambda dn/d\lambda =2B/{\lambda }^2$ and $n_g=A+3B/{\lambda }^2$. That is 1.798 for red and 2.000 for violet in the animation's glass, against phase indices of 1.733 and 1.800.

Near a resonance: absorption and anomalous dispersion

Cauchy's law is only the low-frequency tail of the oscillator's index, and near ${\omega }_0$ the index formula above blows up, which no real material does. Real electrons also lose energy, to collisions and to their own radiation. Model that loss by a friction term $\gamma \dot{x}$ in the equation of motion, where $\gamma$ is the damping rate. The factor ${\omega }_0^2-{\omega }^2$ then becomes ${\omega }_0^2-{\omega }^2+i\gamma \omega$, and the index becomes a complex number, written $n=n^{\prime }-i{n}^{\prime \prime }$ with $n^{\prime }$ and $n^{\prime \prime }$ real:
$n=1+\frac{N{q}_e^2}{2{\epsilon }_{0}m({\omega }_0^2-{\omega }^2+i\gamma \omega )}$

Legend

$n$	Refractive index, here a complex number $n=n^{\prime }-i{n}^{\prime \prime }$; the two curves are its parts.
$n^{\prime }$	Real part of the index: the crests move at $c/n^{\prime }$.
$n^{\prime \prime }$	Imaginary part of the index: it sets how fast the wave is absorbed.
$\omega$	Angular frequency of the light.	$rad/s$
${\omega }_0$	Natural frequency of the electrons: the resonance.	$rad/s$
$\gamma$	Damping rate; it is also the width of the absorption peak at half its height.	$rad/s$
$A$	Index at very low frequency, where $n^{\prime }$ starts: the constant term of Cauchy's form.
$A+\frac{B}{{\lambda }^2}$	Cauchy's form (dashed): it follows $n^{\prime }$ well below the resonance and fails near it.

Figure 3 plots $n^{\prime }$ and $n^{\prime \prime }$ against $\omega$. Start at the left. At $\omega =0$ the real part $n^{\prime }$ starts at $A$ and rises slowly, and the dashed Cauchy form $A+B/{\lambda }^2$ follows it closely: this is the stretch where the prism and the rainbow live. Approaching ${\omega }_0$, the dashed form fails. The real part $n^{\prime }$ climbs to a peak, falls steeply through 1 exactly at $\omega ={\omega }_0$, dips to a minimum below 1, and then creeps back up to 1 from below. The imaginary part $n^{\prime \prime }$ is almost zero away from the resonance and forms a peak centred on ${\omega }_0$, whose width at half its height is $\gamma$.

What $n^{\prime \prime }$ means: a wave that has travelled a distance $z$ through the material carries the factor $e^{i(\omega t-n\omega z/c)}$. With $n=n^{\prime }-i{n}^{\prime \prime }$ this splits into $e^{i(\omega t-n^{\prime }\omega z/c)}$, a wave whose crests move at $c/n^{\prime }$, times $e^{-\omega n^{\prime \prime }z/c}$, an amplitude that shrinks with distance. So $n^{\prime \prime }$ is absorption. Figure 3 shows it strong only within about $\gamma$ of ${\omega }_0$: the material is opaque near its resonance and transparent away from it.

Across that absorption band $n^{\prime }$ falls as the frequency rises, the steep stretch of Figure 3. This is anomalous dispersion: inside the band the order of colours a prism would give is reversed. The link between absorption and dispersion is not a quirk of the model. The Kramers–Kronig relations, which follow from causality alone (no response before its cause), make it exact: a material whose index differs from 1 must absorb somewhere in the spectrum.

At the right of Figure 3, above all resonances, the correction to $n$ changes sign and $n^{\prime }$ is below 1. For X-rays every electron is nearly free, and glass has $n\approx 1-3\times 10^{-6}$ at 0.1 nm, so the crests travel faster than $c$. No signal does. Here $1-n$ is proportional to ${\lambda }^2$, so $-\lambda dn/d\lambda =2(1-n)$, and the group index formula gives $n_g=1+3\times 10^{-6}$. An index below 1 also means that X-rays striking a polished surface at grazing angles under about 0.15° are totally reflected, the principle of X-ray telescope mirrors.

Real materials have several resonances. Summing their contributions gives the Sellmeier form $n^2=1+{\sum }_k{B}_k{\lambda }^2/({\lambda }^2-{\lambda }_k^2)$, where each ${\lambda }_k$ is a resonance wavelength and $B_k$ its strength. The electronic terms lie in the ultraviolet and the terms from vibrating ions in the infrared. Between them, the group index of silica has a minimum near 1.3 µm, where pulses of slightly different wavelengths travel at the same speed: the zero-dispersion wavelength of telecom fibres.

Free electrons, as in a plasma, feel no restoring force, so ${\omega }_0=0$. The full treatment then gives $n^2=1-{\omega }_p^2/{\omega }^2$, with the plasma frequency ${\omega }_p$ defined by ${\omega }_p^2=N{q}_e^2/({\epsilon }_{0}m)$; the dilute formula with ${\omega }_0=0$ gives its first-order form, $n\approx 1-{\omega }_p^2/(2{\omega }^2)$. Below ${\omega }_p$, $n^2$ is negative and the wave cannot travel through: it is reflected. For the ionosphere's $10^{12}$ electrons per cubic metre, ${\omega }_p/2\pi \approx 9\mathrm{MHz}$, so radio waves of lower frequency sent straight up come back down.

The rainbow as a stationary angle

Water's index changes by only about 1% across the visible spectrum, from 1.3305 for red at 700 nm to 1.3435 for violet at 400 nm, yet a rainbow separates the colours cleanly. The reason is geometric: a drop sends the light it reflects once inside mostly in one direction, and that direction depends on $n$. Figure 4 shows where the preferred direction comes from.

Legend

$b$	Impact parameter of a ray: its distance from the axis of the drop in units of the radius, $b=\sin \theta$, with $\theta$ the angle of incidence.
$\phi$	Angle between the exit ray and the antisolar direction (straight away from the Sun), $\phi =4{\theta }^{\prime }-2\theta$, with ${\theta }^{\prime }$ the angle of refraction.	$\mathrm{rad}$
${\phi }_{\max }$	The largest exit angle, at the flat top of the curve: the rainbow angle.	$\mathrm{rad}$
$\Delta \phi$	Height of each shaded band: the same small range of exit angles in both.	$\mathrm{rad}$
$\Delta b_1$	Range of $b$ whose rays leave in the upper band: wide, because the curve is flat there.
$\Delta b_2$	Range of $b$ on the rising slope whose rays leave in the lower band: narrow, because the curve is steep there.

Figure 4 is for red light and for the rays that enter a spherical drop, are reflected once inside and leave. Sunlight falls evenly on the drop's face. Label each ray by its impact parameter $b$: its distance from the axis through the centre of the drop parallel to the sunlight, in units of the drop's radius. A ray that meets the surface at the angle of incidence $\theta$ has $b=\sin \theta$, and it refracts into the drop at ${\theta }^{\prime }$ with $\sin \theta =n\sin {\theta }^{\prime }$. Each of the two refractions turns the ray by $\theta -{\theta }^{\prime }$ and the reflection turns it by $180^{\circ }-2{\theta }^{\prime }$, so altogether it turns by $180^{\circ }+2\theta -4{\theta }^{\prime }$. The vertical axis is the angle $\phi$ between the exit ray and the antisolar direction, the direction straight away from the Sun; it is $180^{\circ }$ minus the total turn:
$\phi (b)=4{\theta }^{\prime }-2\theta$

Now follow the curve. It rises from zero at the centre, $b=0$, where the ray goes straight in and straight back. It levels off at a maximum ${\phi }_{\max }$, then falls to about 15° at the edge, $b=1$, where its tangent becomes vertical. The animation's second expert graph shows the top of this curve, for $b$ from 0.5 to 1.

The two shaded bands have the same height $\Delta \phi$: one on the rising slope, one at the top. Follow each band down to the $b$ axis to see which rays end up in it. On the steep slope only a narrow range $\Delta b_2$ of rays falls into the band, plus a thin sliver of edge rays near $b=1$, where the curve plunges back down through it. At the flat top a range $\Delta b_1$ about three times wider does.

The rays must also be counted by area: those with impact parameter between $b$ and $b+db$ hit a ring of the drop's face whose area is proportional to $bdb$, so rays far from the centre count more. With this weighting the upper band collects about 40% of all the once-reflected red rays and the lower band only about 9%. That is the whole effect: where $\phi$ is stationary, many rays leave in almost the same direction, and that direction looks bright.

To find the top, set $d\phi /d\theta =0$. From $\phi =4{\theta }^{\prime }-2\theta$ this needs $d{\theta }^{\prime }/d\theta =\frac{1}{2}$. Differentiating Snell's law gives $\cos \theta d\theta =n\cos {\theta }^{\prime }d{\theta }^{\prime }$, so at the top $2\cos \theta =n\cos {\theta }^{\prime }$. Square this and replace $n^2{\cos }^2{\theta }^{\prime }=n^2-n^2{\sin }^2{\theta }^{\prime }=n^2-{\sin }^2\theta$. Then $4{\cos }^2\theta =n^2-{\sin }^2\theta =n^2-1+{\cos }^2\theta$, which gives
$\phi =4{\theta }^{\prime }-2\theta ,{\cos }^2\theta =\frac{n^2-1}{3}\text{(rainbow)}$

For red, ${\cos }^2\theta =(1.3305^2-1)/3=0.2567$, so $\theta =59.6^{\circ }$ ($b=0.862$), ${\theta }^{\prime }=40.4^{\circ }$ and ${\phi }_{\max }=42.4^{\circ }$. For violet, $n=1.3435$ gives $b=0.855$ and ${\phi }_{\max }=40.6^{\circ }$. Red light thus comes back at up to 42.4° from the antisolar point and violet only up to 40.6°, so red is on the outside of the bow. The animation's ray, at $b=0.86$, sits almost on both maxima, which its dashed lines mark.

How sharp is the edge of the bow? Near its top, ${\phi }_{\max }-\phi$ grows as the square of the distance from the top along the $b$ axis. Turned round, the range of $b$ that lands within an angle $\delta$ of the top grows only like $\sqrt{\delta }$, so the number of rays per unit angle grows like $1/\sqrt{{\phi }_{\max }-\phi }$ as $\phi$ approaches ${\phi }_{\max }$. In numbers, about a fifth of the once-reflected red rays leave within 1° of the maximum, some fifteen times more than in a 1° band around 20°. Beyond ${\phi }_{\max }$ none arrives at all, so the sky just inside the bow is brighter than just outside it.

Rays reflected twice inside the drop give a second bow. For them ${\phi }_2=180^{\circ }+2\theta -6{\theta }^{\prime }$, which is stationary, now at a minimum, where ${\cos }^2\theta =(n^2-1)/8$: at 50.2° for red and 53.6° for violet. This secondary bow is fainter, because light is lost at each reflection, and its colours are reversed, with red inside. Between 42.4° and 50.2° neither kind of ray arrives, so the band of sky between the two bows is darker.

Worked problem: a flash through the flint glass

Suppose the prism's flint glass ($A=1.700$, $B=16000\mathrm{n}{\mathrm{m}}^2$) is drawn into a fibre 1.00 km long, and a short flash of white light is sent in. How far apart in time do its red (700 nm) and violet (400 nm) parts arrive, and what would the ordinary phase indices predict?

Each colour travels as a pulse, so it moves at its group velocity, and the group index is the one to use. With Cauchy's law, $dn/d\lambda =-2B/{\lambda }^3$, so
$n_g=n-\lambda \frac{dn}{d\lambda }=A+\frac{3B}{{\lambda }^2}$
Here $B/{\lambda }^2$ is $16000/700^2=0.03265$ for red and $16000/400^2=0.1000$ for violet, giving $n_{g,\mathrm{red}}=1.700+3\cdot 0.03265=1.798$ and $n_{g,\mathrm{violet}}=1.700+3\cdot 0.1000=2.000$. Each part needs $t=L{n}_g/c$: $5.997\mu \mathrm{s}$ for red and $6.671\mu \mathrm{s}$ for violet. The delay between them is
$\Delta t=\frac{L(n_{g,\mathrm{violet}}-n_{g,\mathrm{red}})}{c}\left(n_g=n-\lambda \frac{dn}{d\lambda }\right)$
$\Delta t=\frac{1000\cdot 0.202}{2.998\times 10^8}=674\mathrm{ns}$

With the phase indices instead, $n_{\mathrm{violet}}-n_{\mathrm{red}}=0.1000-0.03265=0.06735$ and $1000\cdot 0.06735/(2.998\times 10^8)=225\mathrm{ns}$, three times too little. What stretches a pulse is how fast $n$ changes with colour, not the size of $n$ itself. A 1 ns white flash therefore leaves the fibre as a streak about 0.67 µs long, red first. Telecom links avoid this by sending laser light only 0.1 nm wide through silica near 1.55 µm, where the spread is about 17 ps per nanometre of bandwidth per kilometre: under 2 ps per kilometre.

Limits of validity

The oscillator model is classical. Quantum mechanics keeps its form but changes its meaning: each resonance is an atomic transition from the ground state, of energy $E_{\mathrm{g}}$, to an excited state of energy $E_k$, with $\hslash {\omega }_k=E_k-E_{\mathrm{g}}$, and each enters with a weight called its oscillator strength. The dilute result also ignores the field of neighbouring dipoles. In a dense solid that local field changes the result to $(n^2-1)/(n^2+2)=N{q}_e^2/(3{\epsilon }_{0}m({\omega }_0^2-{\omega }^2))$, which reduces to ours when $n$ is close to 1. Inside an absorption band the group velocity can exceed $c$ or even turn negative. The pulse is then strongly reshaped, its peak no longer marks a signal, and the front of any signal still never outruns $c$.

The ray picture of the rainbow needs drops much larger than the wavelength. In smaller drops, rays that leave at the same angle by different paths interfere and add faint supernumerary bands just inside the violet. In fog, droplets about $d=10$ µm across blur each colour by roughly $\lambda /d\approx 3^{\circ }$, more than the nearly 2° between red and violet, so a fog bow is almost white.

Finally, colour is not wavelength. The eye reduces any spectrum to three cone signals, so a mixture of red and green light can match spectral yellow exactly, and no screen with three primaries can show every spectral colour.

Where this leads

The driven oscillator is the one of Damping and Resonance, now at optical frequencies, and the radiating sheet of charge belongs to Electromagnetic Waves. The rainbow's stationary angle echoes Fermat's principle from Reflection and Refraction: there the travel time is stationary, here the exit angle, and in both cases neighbouring rays reinforce one another. Colour errors of lenses and their cure return in Lenses and Mirrors and in The Eye, Cameras and Telescopes; gratings, which sort colours by interference, in Diffraction and the Double Slit; and the partial polarisation of the blue sky in Polarization of Light. The resonance frequencies themselves come from atomic energy levels, which Atomic Models: the Bohr Atom begins to explain.