Rainbow and Glory Scattering

A rainbow is a mathematical singularity you can see with your eyes. Sunlight enters a spherical raindrop, refracts, reflects once off the back surface, refracts out — and the total deviation angle, as a function of where the ray strikes the drop, has a minimum: 137.5° for water. Rays pile up at that extremal angle the way cars pile up where a highway's exit ramps merge, and the sky lights up in an arc 42.5° from the point opposite the Sun. Descartes traced the rays in 1637 and found the angle; Newton explained the colours; Airy supplied the wave theory that tames the infinity. Classical scattering theory absorbs all of it into one statement: the differential cross-section is a Jacobian, and Jacobians blow up where maps fold. This article is about the two generic ways the classical scattering map $b \to \Theta(b)$ can fold. When the deflection function has a stationary point, $\mathrm{d}\Theta/\mathrm{d}b = 0$ , a whole band of impact parameters lands at nearly the same angle: that is a rainbow, with its inverse-square-root intensity spike and its abrupt dark side. When a nonzero impact parameter scatters exactly forward or backward, an entire ring of trajectories focuses onto the axis and the $\sin\Theta$ in the cross-section's denominator kills the Jacobian: that is a glory, the ghostly ringed halo around an aircraft's shadow on a cloud. Neither infinity is physical — wave mechanics replaces the rainbow spike with an Airy fringe pattern and the glory with Bessel rings — but the classical deflection function correctly predicts where they sit, which is why beams of atoms, nuclei, and light are still analysed with the machinery this article builds. We recompute every number, from Descartes' 42° to the glory rings around your aeroplane's shadow.

Key takeaways

Classical scattering maps an incoming impact parameter $b$ to a deflection angle $\Theta(b)$ ; the differential cross-section diverges wherever this map becomes singular.
A rainbow occurs at a stationary point $\mathrm{d}\Theta/\mathrm{d}b = 0$ : rays pile up at an extremal deflection angle $\Theta_r$ and the classical cross-section blows up as $|\Theta-\Theta_r|^{-1/2}$ , with two branches on the bright side and none on the dark side.
A glory occurs at $\Theta = 0$ or $\pi$ with $b \neq 0$ : the $\sin\Theta$ in the cross-section denominator vanishes while $b$ stays finite — an entire ring of trajectories focuses onto the axis.
The water droplet is the anchor case: Descartes' ray (one internal reflection) has minimum deviation $137.5°$ for $n = 1.33$ , putting the primary bow $42.5°$ from the antisolar point; the secondary bow (two reflections) sits near $50°$ , with Alexander's dark band between them.
Both divergences are artefacts of the geometric limit; wave mechanics replaces them with finite Airy (rainbow) and Bessel-function $J_0$ (glory) patterns — the supernumerary arcs and glory rings.
Rutherford scattering has neither feature: its deflection function is strictly monotonic and reaches $\pi$ only at $b = 0$ .
Key idea: The classical cross-section $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\frac{b}{\sin\Theta}\left|\frac{\mathrm{d}b}{\mathrm{d}\Theta}\right|$ diverges at two kinds of points: rainbows where $\mathrm{d}\Theta/\mathrm{d}b=0$ (extremal angle), and glories where $\sin\Theta=0$ at finite $b$ (forward/backward focusing). Both mark the failure of the geometric limit, not true infinities.

Where this lives today

Rainbow scattering is a precision tool wherever the interaction potential is the unknown. In crossed-molecular-beam machines, the angular position $|\Theta_r|$ of the rainbow peak in atom–atom scattering measures the depth of the interatomic van der Waals well (deeper well, larger rainbow angle), while the spacing of the supernumerary oscillations — the atomic Airy fringes — measures its range: invert the fringe pattern and you have read the potential off the sky, exactly as Descartes read $n = 1.33$ off the bow. Nuclear physics runs the same playbook at MeV scales: "nuclear rainbows" in the elastic scattering of alpha particles and light heavy ions off nuclei fix the depth of the real optical potential, and through it the compressibility of nuclear matter — a number that propagates all the way into neutron-star modelling. The glory, meanwhile, has become a climate instrument. Earth-observing satellites routinely image glories in sunlight backscattered from liquid-water cloud decks, and since the $J_0$ ring radius scales as $1/(ka)$ — the physics of Worked example 2 — the ring pattern inverts directly into droplet size and its spatial variability, feeding the cloud-microphysics terms that remain among the largest uncertainties in 2026-generation climate models. The same backscatter enhancement matters closer to home: laser safety analyses and lidar returns from fog both carry glory-type corrections. Not bad for a halo once explained as the observer's personal sanctity.

The physical picture

Follow one ray of sunlight into one raindrop. It refracts at the entry surface, crosses the drop, reflects once off the back, crosses again, and refracts out. Each step's geometry is fixed by Snell's law and the impact parameter $b$ — the ray's offset from the drop's central axis. Add up the direction changes and you get the total deviation $D(b)$ . The crucial fact is that $D(b)$ is not monotonic: near-axial rays deviate by nearly 180° (almost straight back), grazing rays also deviate strongly, and in between the deviation dips to a minimum — 137.48° for $n = 1.33$ , reached by the ray striking at $b = 0.862\,R$ :

Figure

Descartes' ray — sunlight refracts, reflects once inside the drop, and exits deviated by 137.5 degrees, the minimum over all impact parameters

Why does a minimum of the deviation make a bright arc? Because intensity is a Jacobian. Rays with impact parameters just below and just above $0.862\,R$ exit at almost exactly the same angle — the map from $b$ to angle is momentarily flat — so a finite band of incoming sunlight is compressed into an infinitesimal band of directions. On one side of the extremal angle two rays arrive for every direction; on the other side, none: the geometric intensity jumps from a divergence to zero. That dark side is real and visible — Alexander's dark band, the noticeably gloomy strip between the primary and secondary bows. Strip away the optics and the same structure lives in any central-force problem with competing tendencies. An attractive well pulls trajectories forward and around (negative deflection); a repulsive core bounces them back (positive deflection). The resulting deflection function rises to 180° at $b = 0$ , plunges through zero, bottoms out at some most-negative angle $\Theta_r$ , and relaxes to zero from below:

Figure

The deflection function of a well-plus-core potential — one glory crossing and one rainbow extremum, the two generic singularities

Two marked points on that curve are the whole subject. Where the curve crosses zero at finite $b$ , a full ring of impact parameters — area $2\pi b_g\,\mathrm{d}b$ — exits into the vanishingly thin forward cone, and by symmetry the same can happen at 180°: the glory, axial focusing pure and simple. Where the curve bottoms out, the mapping folds back on itself: two branches above the extremum, none below, with the fold point itself infinitely bright in the ray picture. Rainbow and glory are not exotic; they are the two generic catastrophes any smooth non-monotonic deflection function must exhibit.

The machinery of scattering singularities

Everything runs through the classical cross-section formula, inherited from the flux-conservation argument of the previous articles. It is worth restating with fresh eyes, because both singularities are visible in it by inspection:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b}{\sin\Theta}\left|\frac{\mathrm{d}b}{\mathrm{d}\Theta}\right| \qquad (1)

where: $b$ — impact parameter; $\Theta$ — scattering angle; when several impact parameters produce the same $\Theta$ , their contributions add. Two distinct denominators can vanish: $\mathrm{d}\Theta/\mathrm{d}b = 0$ makes $|\mathrm{d}b/\mathrm{d}\Theta|$ blow up (rainbow), and $\sin\Theta = 0$ at finite $b$ makes the prefactor blow up (glory). Every singular feature of classical scattering is one of these two, and the formula tells you so before any dynamics is specified. The dynamics enters through the deflection function. For a particle of energy $E$ in a central potential $V(r)$ , the orbit quadrature gives it exactly:

\Theta(b) = \pi - 2b\!\int_{r_{\min}}^{\infty}\frac{\mathrm{d}r}{r^{2}\sqrt{1 - \dfrac{b^{2}}{r^{2}} - \dfrac{V(r)}{E}}} \qquad (2)

where: $r_{\min}$ — the distance of closest approach, the outermost zero of the radicand. Note the sign convention: for attractive potentials the integral can exceed $\pi$ , making $\Theta$ negative — the trajectory is pulled around the far side. Angles $\Theta$ and $-\Theta$ arrive at the same detector (only $|\Theta|$ mod $2\pi$ is observable), which is precisely how a negative-angle extremum shows up as a bright ring at $|\Theta_r|$ . A rainbow exists wherever the deflection function goes stationary:

\left.\frac{\mathrm{d}\Theta}{\mathrm{d}b}\right|_{b_r} = 0 \quad\Longrightarrow\quad \Theta_r = \Theta(b_r) \qquad (3)

where: $b_r$ , $\Theta_r$ — the rainbow impact parameter and rainbow angle. What happens to the intensity there is a short Taylor expansion away:

Derivation ARainbow from an extremum of the deflection function

We derive the rainbow's signature intensity law (4): near a stationary point of $\Theta(b)$ the classical cross-section diverges as the inverse square root of the angular distance to the rainbow, illuminated from one side only. The strategy is a second-order Taylor expansion of the deflection function about $b_r$ — the first-order term vanishes by definition — followed by inversion of the quadratic and substitution into equation (1). Watch the two-branch structure emerge from the square root: it is the geometric origin of the supernumerary interference fringes that the wave theory will paint on the bright side.

Step 1. Expand the deflection function about the stationary point, where the linear term is absent by (3):

\Theta(b) = \Theta_r + \tfrac{1}{2}\Theta''_r\,(b-b_r)^2 + \mathcal{O}\big((b-b_r)^3\big), \qquad \Theta''_r \equiv \frac{\mathrm{d}^2\Theta}{\mathrm{d}b^2}\bigg|_{b_r}

Step 2. Invert. For a minimum ( $\Theta''_r > 0$ ; the mirrored argument covers a maximum):

b_\pm(\Theta) = b_r \pm \sqrt{\frac{2(\Theta - \Theta_r)}{\Theta''_r}}

Why one side is dark: the square root is real only for $\Theta > \Theta_r$ . On that side every angle is reached by two impact parameters — the two branches $b_+$ and $b_-$ ; below $\Theta_r$ no classical trajectory arrives at all. In the raindrop this is Alexander's dark band, and in an atomic beam it is the abrupt shadow boundary of the rainbow peak.

Step 3. Differentiate each branch:

\left|\frac{\mathrm{d}b_\pm}{\mathrm{d}\Theta}\right| = \frac{1}{\sqrt{2\,\Theta''_r\,(\Theta-\Theta_r)}}

Why the exponent is universal: it comes only from the quadratic shape of a generic extremum, not from the potential. Any smooth fold gives $-1/2$ — the same exponent as the fold caustics of optics, of which the rainbow is the oldest known example.

Step 4. Substitute both branches into (1) and add them:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \;\simeq\; \frac{b_r}{\sin\Theta_r}\,\frac{\sqrt{2}}{\sqrt{\Theta''_r\,|\Theta-\Theta_r|}} \;\xrightarrow{\;\Theta\to\Theta_r\;}\; \infty

Why the divergence is harmless in total: $\int|\Theta-\Theta_r|^{-1/2}\mathrm{d}\Theta$ converges — the antiderivative is $2|\Theta-\Theta_r|^{1/2}$ — so the rainbow concentrates a finite amount of scattered flux into an infinitesimally sharp (in ray optics) angular feature. The intensity is infinite; the energy is not.

For the record, the rainbow law just derived:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\bigg|_{\text{rainbow}} \simeq \frac{b_r}{\sin\Theta_r}\,\frac{\sqrt{2}}{\sqrt{\,|\Theta''_r|\,|\Theta-\Theta_r|\,}} \qquad (4)

where: $\Theta''_r$ — the curvature of the deflection function at the fold; a flatter extremum (small $\Theta''_r$ ) makes a brighter, angularly tighter rainbow. This inverse-square-root law, integrable but unbounded, is the classical fingerprint that experiments fit near any rainbow angle. The glory needs no expansion at all — only the observation that the spherical Jacobian has a zero of its own. Suppose some finite impact parameter $b_g$ scatters exactly backward:

\Theta(b_g) = \pi \quad (\text{or } 0), \qquad b_g \neq 0 \qquad (5)

where: the crucial requirement is $b_g \neq 0$ — the numerator of (1) must survive while $\sin\Theta$ dies.

Go deeper · Derivation B · 1 minGlory from axial focusing at finite impact parameter

Derivation BGlory from axial focusing at finite impact parameter

We derive the glory divergence (6) and locate its physical origin in rotational symmetry. The strategy is direct evaluation of equation (1) near $\Theta = \pi$ with a generic (non-stationary) deflection function — the glory needs no fold, only a crossing — and the geometric picture to hold onto is a ring collapsing onto a point: all azimuths of the ring $b = b_g$ exit along the single backward direction, and a one-dimensional ring of flux delivered into a zero-dimensional direction is infinite flux per solid angle. Watch why the head-on trajectory, which also scatters backward, produces no such singularity.

Step 1. Parametrize the neighbourhood of the backward direction: $\Theta = \pi - \epsilon$ with $\epsilon \to 0^+$ . The spherical solid-angle factor vanishes linearly:

\sin\Theta = \sin(\pi - \epsilon) = \sin\epsilon \approx \epsilon

Step 2. Check the numerator. With $\Theta'(b_g) \neq 0$ , the deflection function is locally linear, $b(\Theta) \approx b_g + (\Theta-\pi)/\Theta'(b_g)$ , so as $\Theta \to \pi$ :

b \to b_g \neq 0, \qquad \left|\frac{\mathrm{d}b}{\mathrm{d}\Theta}\right| \to \frac{1}{|\Theta'(b_g)|} \quad\text{(finite)}

Why this is the whole point: neither factor from the deflection function misbehaves. The divergence will come purely from the coordinate Jacobian of the sphere — the glory is focusing geometry, not dynamical pile-up.

Step 3. Assemble (1):

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \simeq \frac{b_g}{|\Theta'(b_g)|}\cdot\frac{1}{\epsilon} \;\xrightarrow{\;\epsilon\to0\;}\; \infty

Why a ring focuses: by rotational symmetry every azimuthal position on the ring of impact parameters $b = b_g$ scatters into the same backward direction. A one-parameter family of trajectories converges on the axis — precisely the axial caustic that makes the glory a bright ring system around the antisolar point rather than a point of light.

Step 4. Contrast the head-on trajectory. At $b = 0$ , backscattering also occurs, but the numerator $b$ of equation (1) vanishes along with $\sin\Theta$ ; the ratio $b/\sin\Theta$ stays finite (Rutherford's finite backward floor is exactly this case). No ring, no focusing, no glory. The forward case $\Theta = 0$ is identical in structure, with $\epsilon = \Theta$ replacing $\epsilon = \pi - \Theta$ ; the $1/\sin\Theta$ pole resolves in the wave treatment into the $J_0^2(k b_g \sin\Theta)$ ring pattern of equation (7).

Stated as a formula:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\bigg|_{\text{glory}} \simeq \frac{b_g}{|\Theta'(b_g)|}\,\frac{1}{\sin\Theta} \;\xrightarrow{\;\Theta\to0,\pi\;} \infty \qquad (6)

where: the divergence is a simple pole in $\sin\Theta$ — stronger than the rainbow's inverse square root, but again integrable against the vanishing solid angle $\sin\Theta\,\mathrm{d}\Theta$ , so the total flux is finite. Now the anchor calculation — the oldest quantitative scattering theory in physics, and still the cleanest rainbow in nature.

Go deeper · Derivation C · 1 minDescartes and the 42 degree rainbow

Derivation CDescartes and the 42 degree rainbow

We compute the primary rainbow angle for a water droplet, $n = 1.33$ , reproducing Descartes' 1637 ray-tracing result with calculus he did not yet have. The system: a ray enters the spherical drop at incidence angle $i$ (set by the impact parameter through $\sin i = b/R$ ), refracts to the internal angle $r$ with $\sin i = n\sin r$ , reflects once off the back surface, and refracts out. The strategy is to write the total deviation $D(i)$ , minimize it over $i$ , and evaluate. The step to appreciate is why an extremum must exist: the deviation is $180°$ at central incidence and $166°$ at grazing, but dips in between — two competing geometric effects guarantee a fold.

Step 1. Add up the deviation. Entry refraction bends the ray by $i - r$ ; the internal reflection turns it by $\pi - 2r$ ; exit refraction bends it by $i - r$ again:

D(i) = 2(i - r) + (\pi - 2r) = \pi + 2i - 4r

Step 2. Impose the extremum condition $\mathrm{d}D/\mathrm{d}i = 2 - 4\,\mathrm{d}r/\mathrm{d}i = 0$ , i.e. $\mathrm{d}r/\mathrm{d}i = \tfrac12$ . Differentiating Snell's law $\sin i = n \sin r$ gives $\cos i = n\cos r\,(\mathrm{d}r/\mathrm{d}i)$ , so the rainbow ray satisfies:

2\cos i = n\cos r

Why a minimum must exist: near $i = 0$ , $r \approx i/n$ and $D$ decreases from $\pi$ at slope $2 - 4/n < 0$ ; near grazing ( $i \to 90°$ ), $\mathrm{d}r/\mathrm{d}i \to 0$ and $D$ rises at slope $+2$ . A function that starts falling and ends rising has a minimum between — the rainbow is guaranteed by the geometry, not fine-tuned by the material.

Step 3. Solve the pair (Snell plus extremum). Squaring and eliminating $r$ :

\cos^2 i = \frac{n^2 - 1}{3}

Step 4. Put in water, $n = 1.33$ : $\cos^2 i = (1.7689 - 1)/3 = 0.2563$ , so $\cos i = 0.5063$ , $i = 59.58°$ ; then $\sin r = \sin i/n = 0.8624/1.33 = 0.6484$ , $r = 40.42°$ . The impact parameter of the rainbow ray is $b_r = R\sin i = 0.862\,R$ .

Step 5. Evaluate the deviation:

D_{\min} = 180° + 2(59.58°) - 4(40.42°) = 137.5°

The bow therefore appears at $180° - 137.5° = 42.5°$ from the antisolar point. Dispersion spreads it into colours: $n_{\mathrm{red}} = 1.331$ gives $42.4°$ and $n_{\mathrm{violet}} = 1.343$ gives $40.7°$ — red on the outside, violet inside, a width of about $1.8°$ before adding the Sun's own half-degree disc. Minimizing $D(i)$ numerically for $n = 1.33$ reproduces $i = 59.58°$ , $r = 40.42°$ , $D = 137.48°$ , and the dispersion ordering places red on the outer rim — consistent with observation.

Wave mechanics rounds off both infinities. Near a rainbow the two branches $b_\pm$ are two coherent paths to the same angle; their interference, worked out by Airy in 1838, replaces the spike with a smooth peak plus fringes. Near a glory the interfering ring of azimuths gives a Bessel pattern:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\bigg|_{\text{rainbow}}^{\text{wave}} \propto \mathrm{Ai}^2\big[c\,(\Theta - \Theta_r)\big], \qquad \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\bigg|_{\text{glory}}^{\text{wave}} \propto J_0^2\big(k\,b_g \sin\Theta\big) \qquad (7)

where: $\mathrm{Ai}$ — the Airy function; $J_0$ — the zeroth Bessel function; $k = 2\pi/\lambda$ — the wavenumber; $c$ — a scale set by $k$ and the fold curvature $\Theta''_r$ (large for large drops, which is why big raindrops give sharp bows). The Airy oscillations on the bright side are the supernumerary bows — faint green-violet arcs hugging the inside of the primary — and the $J_0$ rings are the glory's coloured halos:

Figure

What the wave theory does to the rainbow spike — the Airy pattern is finite, shifted into the bright side, and ringed with supernumeraries

Which formula answers which question

you want	use	notes
where any singular feature sits	deflection function, eq. (2)	rainbows at $\Theta' = 0$ ; glories at $\Theta = 0,\pi$ with $b \neq 0$
intensity near a rainbow	eq. (4)	$\vert\Theta-\Theta_r\vert^{-1/2}$ , two branches, one-sided
intensity near a glory	eq. (6)	$1/\sin\Theta$ pole from axial focusing
the optical rainbow angle	$\cos^2 i = (n^2-1)/3$ , then $D = \pi + 2i - 4r$	$42.5°$ from antisolar for $n = 1.33$
finite wave patterns	eq. (7)	Airy for rainbows, $J_0$ rings for glories
whether a system has these features	monotonicity of $\Theta(b)$	pure Coulomb: monotonic, neither feature

Limiting cases

Case	Result	Meaning
pure Coulomb (Rutherford)	$\Theta(b) = 2\arctan(\kappa/2Eb)$ , strictly monotonic	no stationary point and $\Theta = \pi$ only at $b=0$ : no rainbow, no glory — the featureless benchmark
attractive well plus repulsive core	$\Theta(b)$ dips to $\Theta_r < 0$ at finite $b_r$	atomic and molecular rainbow scattering; $\vert\Theta_r\vert$ measures the well depth
orbiting (well deep enough that $V_{\mathrm{eff}}$ peak equals $E$ )	$\Theta(b) \to -\infty$ at $b_{\mathrm{orb}}$	the deflection winds through many turns, crossing $0, \pm\pi, \dots$ repeatedly — a ladder of forward and backward glories
water droplet, one internal reflection	$D_{\min} = 137.5°$ , bow at $42.5°$	Descartes ray at $b = 0.862R$ ; red outer rim, violet inner
water droplet, two internal reflections	$D = 230.1°$ , bow at $50.1°$	the secondary bow, colours reversed; Alexander's dark band between the two bows
droplet radius $\gg \lambda$	Airy fringes crowd toward $\Theta_r$	geometric optics recovered as the wave fringes become unresolvable

Common trapThe forward peak in Rutherford scattering is a forward glory

It is not — and the distinction is the cleanest test of whether the two singularity mechanisms are understood. The Rutherford forward divergence comes from the infinite range of the Coulomb tail: an unbounded reservoir of large- $b$ trajectories each deflecting a little, piling intensity into $\Theta \to 0$ with $b \to \infty$ . A forward glory requires the opposite: a finite impact parameter $b_g$ whose trajectory returns exactly to the forward direction, so that the $b/\sin\Theta$ factor in equation (1) has a genuinely nonzero numerator over a vanishing denominator. Same direction, different physics: one is a statement about the potential's reach, the other about a focusing accident of the deflection function. The companion trap sits at the backward pole: head-on trajectories ( $b = 0$ ) always bounce straight back, but they produce no glory — the numerator $b$ dies with $\sin\Theta$ and the ratio stays finite, which is exactly why Rutherford's 180° cross-section is a finite floor rather than a singular ring.

Worked examples

1 · The primary rainbow, from scratch (confidence builder)

Compute the incidence angle, internal angle, minimum deviation, and sky position of the primary rainbow for water, $n = 1.33$ . Then find the width of the coloured band given $n_{\mathrm{red}} = 1.331$ , $n_{\mathrm{violet}} = 1.343$ .

Step 1. The rainbow ray satisfies $\cos^2 i = (n^2-1)/3 = (1.7689-1)/3 = 0.2563$ , so $\cos i = 0.5063$ and $\boxed{i = 59.6°}$ .

Step 2. Snell's law gives the internal angle: $\sin r = \sin(59.58°)/1.33 = 0.8624/1.33 = 0.6484$ , so $\boxed{r = 40.4°}$ .

Step 3. Deviation: $D = 180° + 2i - 4r = 180° + 119.2° - 161.7° = \boxed{137.5°}$ . The bow appears where deviated sunlight comes back toward the observer: $180° - 137.5° = \boxed{42.5°}$ from the antisolar point (the shadow of your own head).

Step 4. Dispersion. Repeating with $n = 1.331$ : $i = 59.53°$ , $r = 40.36°$ , $D = 137.6°$ , bow radius $42.4°$ . With $n = 1.343$ : $i = 58.83°$ , $r = 39.58°$ , $D = 139.3°$ , bow radius $40.7°$ . The band spans $\boxed{\approx 1.8°}$ , red outside, violet inside — plus half a degree of smearing from the Sun's angular size, which is why real bows are never razor sharp.

2 · The secondary bow, Alexander's band, and a glory's rings (exam level)

Find the deviation and sky position of the secondary rainbow (two internal reflections, $n = 1.33$ ). Locate Alexander's dark band. Then estimate the angular radius of the first dark ring of a glory produced by cloud droplets of radius $10\ \mathrm{μm}$ in green light ( $\lambda = 550$ nm).

Step 1. With two internal reflections the deviation is $D_2(i) = 2(i - r) + 2(\pi - 2r) = 2\pi + 2i - 6r$ . Setting $\mathrm{d}D_2/\mathrm{d}i = 0$ gives $\mathrm{d}r/\mathrm{d}i = 1/3$ , and differentiating Snell's law as before yields $\cos^2 i = (n^2-1)/8 = 0.0961$ , so $i = 71.9°$ , $\sin r = 0.9507/1.33 = 0.7148$ , $r = 45.6°$ .

Step 2. Evaluate: $D_2 = 360° + 143.9° - 273.8° = \boxed{230.1°}$ . A deviation beyond $180°$ means the ray crosses the axis; the effective scattering angle is $360° - 230.1° = 129.9°$ , so the secondary bow sits at $180° - 129.9° = \boxed{50.1°}$ from the antisolar point (about $51°$ if one uses $n = 4/3$ — the familiar textbook figure). This extremum is a maximum of the scattering angle from below, and the second refraction reverses the colour order: violet outside, red inside.

Step 3. Alexander's dark band: the primary illuminates only inside $42.5°$ , the secondary only outside $50.1°$ . The annulus between $\boxed{42.5°\ \text{and}\ 50.1°}$ receives no light from either one-reflection or two-reflection rays — the visibly dark strip between the bows, first described by Alexander of Aphrodisias around 200 AD.

Step 4. The glory. The wave pattern is $J_0^2(k b_g \sin\theta)$ with $b_g \approx a$ (the effective ring radius is close to the droplet radius) and $k = 2\pi/\lambda = 2\pi/5.5\times10^{-7} = 1.14\times10^{7}\ \mathrm{m^{-1}}$ . The first zero of $J_0$ is at argument $2.405$ , so $\sin\theta \approx 2.405/(k a) = 2.405/(1.14\times10^{7} \times 10^{-5}) = 0.0211$ , giving $\boxed{\theta \approx 1.2°}$ . A ring system a couple of degrees across, hugging the shadow of the aircraft — and inversely proportional to droplet size, which is exactly how glory images are inverted to measure cloud microphysics.

Origins · 1 minRené Descartes and how this was discovered

René Descartes

1596–1650 · La Flèche, Paris, the Dutch Republic

Descartes gave the rainbow its first quantitative theory in the appendix Les Météores of his 1637 Discourse on Method — the same book that introduced Cartesian coordinates. Lacking calculus (Newton would not be born for another five years), he traced the paths of many parallel rays through a glass sphere of water numerically, computing the deviation ray by ray, and discovered that the exit angles accumulate near 138° for one internal reflection and near 130° (measured his way) for two: an extremum located by brute-force tabulation, a fold caustic found by hand. He correctly explained both bows' positions and the dark region between them; the colours defeated him — that final step took Newton's dispersion experiments of 1666–1704. The rainbow calculation is quintessential Descartes: a mystery of the heavens reduced to geometry, method over inspiration, and the swaggering confidence that nature's most beautiful apparition would yield to a man with a diagram. It did.

History

1637

— René Descartes publishes the ray-tracing theory of the rainbow in Les Météores, locating the primary bow near 42° by numerical minimization of the deviation.

1704

— Isaac Newton's Opticks explains the rainbow's colours through the dispersion of the refractive index, red deviated least.

1838

— George Biddell Airy derives the wave theory of the rainbow caustic; his integral (the Airy function) replaces the infinite ray-optics spike with a finite fringed peak — the supernumerary bows.

1959

— Ford and Wheeler's semiclassical scattering theory names and classifies rainbow and glory singularities of the deflection function for general potentials.

1992

— Nussenzveig's complex angular momentum theory gives the modern quantitative account of both effects, including the subtle surface-wave origin of the water-droplet glory.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 3 (scattering, deflection function) · Landau & Lifshitz, Mechanics, 3rd ed., §18-19 (disintegration and scattering of particles) · Taylor, Classical Mechanics, Ch. 14 (collision theory and cross-sections) · Morin, Introduction to Classical Mechanics, Ch. 7 (central forces) · H. M. Nussenzveig, Diffraction Effects in Semiclassical Scattering (Cambridge, 1992) · K. W. Ford & J. A. Wheeler, Ann. Phys. 7, 259 (1959) (semiclassical scattering: rainbows and glories). The optical primary-bow deviation $\approx138^\circ$ for $n=1.33$ follows from minimizing $D(\theta_i)=2\theta_i-4\theta_r+\pi$ with Snell's law. The worked-example numbers throughout this article ( $i = 59.58°$ , $r = 40.42°$ , $D_{\min} = 137.5°$ , bow at $42.5°$ , red/violet spread $42.4°/40.7°$ ; secondary $i = 71.9°$ , $D_2 = 230.1°$ , bow at $50.1°$ ; glory first dark ring $\approx 1.2°$ for $a = 10\ \mathrm{μm}$ , $\lambda = 550$ nm) are arithmetic consequences of the displayed formulas.

Builds on: Central Force Fundamentals · Angular Momentum Conservation in Central Fields · The Effective Potential · Reduction to the Radial Equation · The Two-Body Problem and Reduced Mass. Leads to: The Kepler Problem · Classification of Orbits by Energy.

Why is the sky visibly darker between the primary and secondary bows?

Because that annulus of angles — Alexander's dark band, 42.5° to 50.1° from the antisolar point — is unreachable by both of the dominant ray families. The one-reflection deviation has a minimum at 137.5°, so its light comes back only at radii inside 42.5°; the two-reflection deviation has its extremum placing light only outside 50.1°. Between the bows neither family delivers a single classical ray, and only the faint residue of higher-order paths and diffuse background fills it in. The band is the "dark side" of both rainbows simultaneously — the two-sided version of the one-sided shadow that every fold caustic casts, and honestly the easiest scattering singularity to verify with your own eyes on a wet afternoon.

What are the supernumerary bows, and why do they only appear sometimes?

They are the Airy oscillations of equation (7) — interference between the two ray branches $b_+$ and $b_-$ that reach each angle on the bright side of the rainbow. Their angular spacing scales like $(kR)^{-2/3}$ : big raindrops (millimetres) give fringes so fine they blur into the main bow, while small droplets (a few tenths of a millimetre) give fringes of a degree or so — visible as faint pink-green arcs inside the primary. The catch is uniformity: the fringe position depends on drop radius, so a shower with a broad size distribution smears the pattern away. Supernumeraries demand recent, fine, uniform drizzle — which is why they are a connoisseur's rainbow, and why their spacing was the first practical raindrop-size gauge.

Is the glory just a rainbow at 180 degrees?

No — the mechanisms are genuinely different, which is the article's central taxonomy. A rainbow is a stationary point of the deflection function: $\mathrm{d}\Theta/\mathrm{d}b = 0$ , a fold in the map, divergence exponent $-1/2$ . A glory needs no fold at all: it is the coordinate singularity $\sin\Theta = 0$ struck at finite $b$ , a whole ring focusing on the axis, divergence exponent $-1$ . A system can have either without the other. The water-droplet glory has an extra subtlety worth knowing: for $n = 1.33$ no simple geometric ray actually exits at exactly 180° — the backward return involves surface waves creeping along the droplet boundary, unravelled by Nussenzveig's complex angular momentum theory. The classical condition (5) still locates the feature; the amplitude bookkeeping is where the wave theory earns its keep.

Why does the rainbow angle depend only on the refractive index and not on the size of the drop?

Ray optics in a sphere is scale-free: Snell's law and the reflection geometry involve only angles, and the impact parameter enters only through the ratio $b/R$ . Minimizing the deviation therefore gives a universal $i$ , $r$ , and $D_{\min}$ for a given $n$ — every spherical drop from drizzle to storm rain puts its fold at 137.5°. Size re-enters exactly where ray optics ends: through the wavelength ratio $kR$ . It sets the Airy fringe spacing (supernumeraries), the sharpness of the bow, and the glory ring radii $\propto 1/kR$ . So the position of the rainbow measures the material, while its fine structure measures the drops — a clean division of labour that atomic-beam experimenters copied wholesale: rainbow angle for well depth, fringes for range.

What plays the role of the rainbow's colours in atomic or nuclear rainbow scattering?

Energy takes over from wavelength. In the optical rainbow, colour separates because $n$ varies with $\lambda$ , moving the fold angle. In particle scattering the deflection function (2) depends on $E$ : raising the beam energy shallows the effective well, shrinking the rainbow angle $|\Theta_r|$ roughly as (well depth)/ $E$ for weak attraction. Scanning beam energy therefore sweeps the rainbow across the detector exactly as dispersion sweeps colours across the sky, and the map $\Theta_r(E)$ traces the potential's depth profile. Meanwhile the de Broglie wavelength controls the fringe structure, in perfect analogy with $kR$ . In nuclear rainbows there is even an absorption twist: the imaginary optical potential damps the inner branch $b_-$ , muting the fringes — a rainbow seen through smoked glass.

If the cross-section diverges at the rainbow angle, why is the total scattered flux finite?

Because the divergence is integrable. Near the rainbow, $\mathrm{d}\sigma/\mathrm{d}\Omega \sim |\Theta - \Theta_r|^{-1/2}$ , and $\int |\Theta-\Theta_r|^{-1/2}\,\mathrm{d}\Theta \propto |\Theta-\Theta_r|^{1/2}$ is finite over any interval containing $\Theta_r$ . The same holds for the glory: $\int (1/\sin\Theta)\,\sin\Theta\,\mathrm{d}\Theta$ — the pole is exactly cancelled by the solid-angle factor it came from. What diverges is the angular density of scattered particles, not their number; the caustic is a redistribution, not a creation. This is the sharp contrast with the Coulomb forward divergence of the total-cross-section article, which does make the integrated cross-section infinite because there it is fed by an infinite reservoir of impact parameters rather than a focusing accident of finitely many.

Can a purely repulsive potential produce a rainbow?

Not if the deflection decreases monotonically with impact parameter — which is the case for any repulsive potential whose strength falls off monotonically, Coulomb included: closer aim always means a harder kick, $\mathrm{d}\Theta/\mathrm{d}b < 0$ everywhere, no stationary point, no rainbow. The recipe for a rainbow is competition: two length scales pulling the deflection in opposite directions, so that $\Theta(b)$ must turn around somewhere. The classic realization is attraction outside, repulsion inside (the van der Waals well plus hard core), giving the negative-angle rainbow of atomic beams. More exotic routes exist — a repulsive potential with a sufficiently sharp shoulder can produce a stationary point too — but the theorem stands for simple monotone repulsion. Diagnose any potential the same way: sketch $\Theta(b)$ from equation (2) and count its turning points and zero crossings; the singularities follow by inspection.

Rainbow and Glory Scattering

Where this lives today

The physical picture

The machinery of scattering singularities

Which formula answers which question

Limiting cases

Worked examples

1 · The primary rainbow, from scratch (confidence builder)

2 · The secondary bow, Alexander's band, and a glory's rings (exam level)

Going further

Before this

Where this leads