Rainbow Optics & Airy Caustic Diffraction: How Internal Reflection and Wave Interference Forge Spectral Arcs and Supernumerary Bands
The violent convective fury of the late-summer afternoon begins to yield just as the barometric pressure bottoms out and commences its sharp, post-frontal surge. Overhead, the churning underside of a retreating cumulonimbus cloud wall—bruised in charcoal and slate—lumbers eastward across the horizon. The ambient air cools precipitously by five degrees Celsius in minutes, charged with the unmistakable earthen perfume of geosmin and ozonated petrichor rising from rain-soaked topsoil. Behind you, breaking cleanly beneath the ragged shelf of the lifting cloud base in the western sky, the late afternoon sun descends toward the horizon, casting long, amber-gold shafts of low-angle illumination across the landscape.
When you turn your back squarely to the low sun and cast your gaze toward the retreating curtain of precipitation, the atmosphere transforms. Spanning across the darkened eastern sky, a luminous, concentric ribbon of spectral colour ignites against the rain. Its outer rim burns in deep crimson, stepping gracefully down through amber, emerald, and azure to an incandescent violet rim. Nestled immediately above it lies a desolate, uncanny corridor of darkened sky—a charcoal moat separating the blazing inner arc from a fainter, colour-inverted twin higher above. Beneath the primary violet edge, delicate pastel fringes of green and rose repeat like ripples upon water. This is not merely a poetic spectacle; it is a masterclass in macroscopic wave mechanics, geometrical caustics, and microphysics written across the canvas of the sky.
Sunlight Rays (Parallel)
===========================>
.---.
/ \
Incoming Ray -------------->( Droplet) (Refraction + Internal Reflection)
\ /
'---'
/ \
/ \ Descartes Ray (~42°)
v v
[ Observer Eye at Anti-Solar Apex ]
1. WHAT IS ACTUALLY HAPPENING: PLAIN ENGLISH FIRST
To understand why a rainbow appears as a crisp circular arc rather than a formless haze of scattered light, imagine millions of airborne raindrops acting as miniature, liquid glass spheres suspended in the air.
When a beam of white sunlight strikes a falling raindrop, it does not simply bounce off the droplet’s exterior like a ping-pong ball. Instead, the ray enters the water, bending as it slows down across the air-water boundary. Once inside, this light travels to the rear inner wall of the droplet. A portion of it breaks through and is lost, but a crucial fraction reflects off the interior curved surface, travelling back toward the front face, where it bends once more as it re-emerges into the ambient air.
Yet this three-step journey—refraction in, reflection at the back, refraction out—does not scatter light evenly in every direction. Think of how a car’s headlights sweep across a brick wall as it executes a sharp U-turn: while the headlights swing through intermediate angles rapidly, they appear to linger longest at the exact turning point where the turn reaches its outermost apex. In optical physics, this concentration of rays at a mathematical turning point is known as a caustic. For a single internal bounce inside a water drop, that turning point forces an immense concentration of exiting light into a conical sheath tilted at roughly 42 degrees relative to the incoming sunlight.
Because white sunlight comprises a spectrum of wavelengths—from long, sluggish red waves to short, energetic violet waves—the water bends each wavelength by a subtly different amount. Violet light bends more sharply than red light, causing the exiting caustic cones to separate spatially: violet emerges at an angle of roughly 40.4 degrees, while red emerges at roughly 42.3 degrees. When millions of raindrops simultaneously cast these 42-degree cones of concentrated light toward your eyes, your perspective aggregates them into a magnificent, luminous ring centered on the exact shadow of your own head: the anti-solar point.
2. THE GEOMETRICAL OPTICS OF DESCARTES: DERIVING THE 42° CONE
While natural philosophers since Aristotle contemplated the phenomenon, it was René Descartes in 1637 who provided the first rigorous mathematical foundation for the primary rainbow, later enriched by Isaac Newton’s discovery of chromatic dispersion. To derive the exact angular boundaries of the bow, we turn to the foundational law of classical refraction: Snell's Law.
Ray Trajectory Inside a Primary Droplet:
Incident Ray (Angle i)
\
---------*----------------- Normal at Entry
/ \ (Angle r) \
| \ |
| * Rear Reflection Point
| / |
\ / (Angle r) /
---------*----------------- Normal at Exit
/
v Emerging Descartes Ray (Angle i)
Snell's Law and the Angular Deviation Function
When a ray of sunlight strikes a spherical raindrop of refractive index $n$ at an angle of incidence $i$ relative to the droplet surface normal, it enters at an angle of refraction $r$. In ambient air where the refractive index is taken as $n_{\text{air}} \approx 1.000$, Snell’s law dictates:
$$\sin(i) = n \sin(r)$$
As the ray enters the droplet, its path is deflected by $(i - r)$. At the back wall, it undergoes specular internal reflection, changing its heading by $(\pi - 2r)$. Finally, as it exits through the droplet's front-facing lower boundary, it experiences an additional refraction deflection of $(i - r)$. The total mechanical deflection $\Delta(i)$ across the entire three-segment path is the sum of these angular deviations:
$$\Delta(i) = (i - r) + (\pi - 2r) + (i - r) = \pi + 2i - 4r$$
For an observer gazing away from the sun toward the anti-solar point, the visible angle of emergence $\theta(i)$ measured relative to the incoming sunlight axis is the supplementary angle of this total deviation:
$$\theta(i) = \pi - \Delta(i) = 4r - 2i$$
+-------------------------------------------------------------------------+
| PRIMARY RAINBOW EMERGENCE ANGLE |
| |
| θ(i) = 4r - 2i |
| |
| Where: |
| i = Angle of incidence at the droplet boundary |
| r = Angle of refraction, governed by sin(i) = n·sin(r) |
+-------------------------------------------------------------------------+
Finding the Descartes Minimum Deviation Ray
Descartes realized that the brilliant intensity of the rainbow occurs where rays of varying incident angles bunch together at a local extremum. To calculate the turning point where the emerging angular density diverges, we differentiate $\theta(i)$ with respect to $i$ and set the derivative to zero:
$$\frac{d\theta}{di} = 4\frac{dr}{di} - 2 = 0 \implies \frac{dr}{di} = \frac{1}{2}$$
Differentiating both sides of Snell's Law ($\sin i = n \sin r$) with respect to $i$ yields:
$$\cos(i) = n \cos(r) \frac{dr}{di}$$
Substituting our stationary condition $\frac{dr}{di} = \frac{1}{2}$ into this differential relation gives:
$$\cos(i) = \frac{n}{2}\cos(r)$$
To solve explicitly for the incident angle $i$, we square both sides and invoke the trigonometric identity $\cos^2(x) = 1 - \sin^2(x)$:
$$\cos^2(i) = \frac{n^2}{4}\cos^2(r) = \frac{n^2}{4}\left(1 - \sin^2(r)\right)$$
Recalling from Snell's law that $\sin^2(r) = \frac{\sin^2(i)}{n^2}$, we substitute this directly into our bracketed term:
$$\cos^2(i) = \frac{n^2}{4}\left(1 - \frac{\sin^2(i)}{n^2}\right) = \frac{n^2 - \sin^2(i)}{4}$$
Expressing $\cos^2(i)$ as $1 - \sin^2(i)$ yields:
$$1 - \sin^2(i) = \frac{n^2 - \sin^2(i)}{4}$$
$$4 - 4\sin^2(i) = n^2 - \sin^2(i)$$
$$3\sin^2(i) = 4 - n^2 \implies \sin(i) = \sqrt{\frac{4 - n^2}{3}}$$
Equivalently, in terms of the cosine of the incident angle:
$$\cos(i) = \sqrt{\frac{n^2 - 1}{3}}$$
+-------------------------------------------------------------------------+
| DESCARTES INCIDENT CAUSTIC CONDITION |
| |
| cos(i_D) = √[ (n² - 1) / 3 ] |
+-------------------------------------------------------------------------+
Worked Numerical Proof: The Primary Spectral Boundaries
Water exhibits chromatic dispersion: its index of refraction varies systematically with the optical wavelength ($dn/d\lambda < 0$). By applying the Descartes formula to the standard visible spectrum in liquid water at $20^\circ\text{C}$, we observe the exact physical origin of the rainbow's spectral separation:
-
For Red Light ($\lambda \approx 656\text{ nm}$, Fraunhofer C-line): The refractive index of water is $n_{\text{red}} \approx 1.3311$. $$\cos(i_D) = \sqrt{\frac{(1.3311)^2 - 1}{3}} = \sqrt{\frac{1.7718 - 1}{3}} = \sqrt{0.25728} \approx 0.50723$$ $$i_D = \arccos(0.50723) \approx 59.52^\circ$$ Using Snell's Law to calculate the refracted angle $r$: $$\sin(r) = \frac{\sin(59.52^\circ)}{1.3311} = \frac{0.8618}{1.3311} \approx 0.6474 \implies r \approx 40.35^\circ$$ Computing the maximum emergence angle $\theta_{\text{max}}$: $$\theta_{\text{red}} = 4(40.35^\circ) - 2(59.52^\circ) = 161.40^\circ - 119.04^\circ = 42.36^\circ \approx 42.4^\circ$$
-
For Violet Light ($\lambda \approx 404\text{ nm}$, Fraunhofer G'-line): The refractive index of water is $n_{\text{violet}} \approx 1.3445$. $$\cos(i_D) = \sqrt{\frac{(1.3445)^2 - 1}{3}} = \sqrt{\frac{1.8077 - 1}{3}} = \sqrt{0.26923} \approx 0.51887$$ $$i_D = \arccos(0.51887) \approx 58.74^\circ$$ Using Snell's Law: $$\sin(r) = \frac{\sin(58.74^\circ)}{1.3445} = \frac{0.8548}{1.3445} \approx 0.6358 \implies r \approx 39.48^\circ$$ Computing the maximum emergence angle $\theta_{\text{max}}$: $$\theta_{\text{violet}} = 4(39.48^\circ) - 2(58.74^\circ) = 157.92^\circ - 117.48^\circ = 40.44^\circ \approx 40.4^\circ$$
Thus, geometrical optics predicts that the primary rainbow manifests as an intense set of concentric coloured rings spanning from an inner violet boundary at $40.4^\circ$ to an outer red ceiling at $42.4^\circ$. Because $\theta(i)$ represents a maximum turning point, light can exit at any angle smaller than $42^\circ$, but geometrically zero light can emerge at angles greater than $42.4^\circ$. This asymmetry explains why the interior disk of a primary rainbow glows with a milky, bright diffuse illumination, while the sky immediately outside the red boundary drops off sharply in radiance.
3. THE SECONDARY BOW AND ALEXANDER'S DARK BAND
When ambient sunlight enters a droplet through its lower hemisphere, it can undergo two internal reflections before exiting back toward the observer. Each internal reflection incurs an optical energy loss (roughly 90–95% of the light refracts out of the drop at each bounce), making the secondary rainbow considerably fainter than the primary.
Ray Trajectory Inside a Secondary Droplet:
---------*----------------- Entry (Lower Hemisphere)
/ / \
| / * Reflection 1 (Upper Back)
| / |
| * Reflection 2 |
\ \ (Lower Back) /
-------*------------------- Exit (Upper Hemisphere)
\
v Emerging Secondary Descartes Ray (~51°)
Geometrical Derivation of the Secondary Rainbow
For two internal bounces, the total deflection angle $\Delta_2(i)$ is:
$$\Delta_2(i) = (i - r) + (\pi - 2r) + (\pi - 2r) + (i - r) = 2\pi + 2i - 6r$$
Measured relative to the anti-solar heading, the emergence angle $\theta_2(i)$ is:
$$\theta_2(i) = \Delta_2(i) - \pi = \pi + 2i - 6r$$
Differentiating with respect to $i$ to locate the stationary caustic gives:
$$\frac{d\theta_2}{di} = 2 - 6\frac{dr}{di} = 0 \implies \frac{dr}{di} = \frac{1}{3}$$
Invoking Snell's Law and solving the differential equation via identical trigonometric manipulation yields the secondary Descartes condition:
$$\cos(i_{D2}) = \sqrt{\frac{n^2 - 1}{8}}$$
+-------------------------------------------------------------------------+
| SECONDARY DESCARTES CAUSTIC CONDITION |
| |
| cos(i_D2) = √[ (n² - 1) / 8 ] |
+-------------------------------------------------------------------------+
Substituting $n \approx 1.333$ yields an incidence angle of $i_{D2} \approx 71.8^\circ$ and a refracted angle of $r \approx 45.5^\circ$. When substituted back into our secondary emergence angle formula:
$$\theta_2 = 180^\circ + 2(71.8^\circ) - 6(45.5^\circ) \approx 50.6^\circ \text{ (Red)} \quad \text{to} \quad 53.5^\circ \text{ (Violet)}$$
Unlike the primary bow, where the derivative yielded an angular maximum, the secondary rainbow’s stationary point is an angular minimum. Consequently: 1. Secondary rays can emerge only at angles greater than $\approx 50.6^\circ$. 2. Because higher refractive indices ($n_{\text{violet}}$) produce larger minimum deviation angles, the colour ordering of the secondary rainbow is completely inverted, displaying red along its inner arc ($50.6^\circ$) and violet at its outer perimeter ($53.5^\circ$).
The Microphysics of Alexander's Dark Band
First documented by the Greek peripatetic philosopher Alexander of Aphrodisias around 200 AD, the dark corridor of sky nestled between the primary and secondary rainbows is a direct consequence of these directional constraints:
- Primary bow rays emerge exclusively at: $\theta \le 42.4^\circ$
- Secondary bow rays emerge exclusively at: $\theta \ge 50.6^\circ$
Between $42.4^\circ$ and $50.6^\circ$, single- and double-reflection ray paths are geometrically forbidden from redirecting light back to the observer. While the sky interior to the primary bow is flooded by overlapping sub-caustic single-reflection rays, and the sky exterior to the secondary bow is illuminated by diffuse double-reflection rays, the intermediate $8^\circ$ channel receives zero directly scattered internal light. The observer sees only the ambient background light of the cloud base, rendering Alexander’s dark band strikingly dim in comparison.
4. WAVE OPTICS: THE AIRY CAUSTIC AND SUPERNUMERARY INTERFERENCE
While classical geometrical optics successfully locates the primary and secondary rainbow angles, it suffers from two fatal physical failures: 1. The Singularity Problem: Geometrical ray density is proportional to $|d\theta/di|^{-1}$. Because $d\theta/di = 0$ at the Descartes ray, ray optics predicts an infinite intensity (an unphysical singularity) at the caustic boundary. 2. Missing Supernumeraries: Geometric optics fails to explain the faint, pastel green, pink, and violet arcs frequently observed immediately inside the primary violet bow.
Caustic Wavefront Profile and Two-Ray Path Interference:
Droplet Edge Cubic Wavefront W(u) ~ u³
|---| . . ' ' '
/ \ :
| *===+==============> Ray 1 (Impact Parameter b₁) \ Interference
| / | : ==> Phase ΔΦ
| / | ' . _ _ / (Supernumerary)
\ / ===> Ray 2 (Impact Parameter b₂)
|---|
The Fold Caustic and the Airy Diffraction Integral
In 1838, the English Astronomer Royal Sir George Biddell Airy resolved this breakdown by treating light as a continuous electromagnetic wave. As a planar wavefront traverses the droplet, the internal reflection and refraction distort the exiting wavefront into an inflected, cubic shape ($W(u) \propto -a u^3$) near the Descartes ray.
+-------------------------------------------------------------------------+
| AIRY CAUSTIC INTEGRAL |
| |
| Ai(z) = (1/π) ∫₀^∞ cos( t³/3 + z·t ) dt |
| |
| Where: |
| z = Dimensionless angular distance parameter from the caustic edge |
| t = Normalised integration variable across the exit pupil |
+-------------------------------------------------------------------------+
When this cubic wavefront propagates into the far field, the resulting electric field distribution $E(\theta)$ is given by the Fourier diffraction integral across the folded caustic, mapping directly to the canonical Airy function $\text{Ai}(z)$:
$$I(\theta) \propto \left| \text{Ai}\left( - \left[ \frac{12 \pi^2 a^2}{\lambda^2} \right]^{1/3} (\theta_{\text{max}} - \theta) \right) \right|^2$$
Where $\lambda$ is the optical wavelength and $a$ is the cubic curvature parameter of the emerging wavefront, which scales proportionally with the raindrop diameter $D$.
The Origin of Supernumerary Arcs
The wave-optical formulation reveals that for any viewing angle $\theta < \theta_{\text{max}}$ within the illuminated disk of the rainbow, two distinct geometric rays—one entering the droplet with an impact parameter $b_1 > b_{\text{Descartes}}$ and one with $b_2 < b_{\text{Descartes}}$—emerge in the exact same direction.
Because these two rays traverse paths of unequal physical distance inside the water drop, they acquire an optical path difference $\Delta L(D, \theta)$. As they exit the droplet, they interfere constructively and destructively: - Constructive Interference ($\Delta L = m\lambda$): Produces intense bright fringes—the supernumerary bows. - Destructive Interference ($\Delta L = (m + \frac{1}{2})\lambda$): Produces zero-intensity troughs between the fringes.
The angular spacing $\Delta \theta_{\text{super}}$ between successive supernumerary fringes scales inversely with droplet size according to the asymptotic zeros of the Airy function:
$$\Delta \theta_{\text{super}} \propto \left( \frac{\lambda}{D} \right)^{2/3}$$
When raindrops are exceptionally uniform and moderately small ($D \approx 0.5 - 1.0\text{ mm}$), the angular fringe spacing is broad and coherent, rendering multiple supernumerary arcs sharply visible beneath the violet rim. Conversely, if rain consists of large, polydisperse drops ($D > 2.0\text{ mm}$), the fringe spacing shrinks to sub-milliradian scales; the overlapping fringes of different droplet sizes blend together, washing out the supernumeraries and leaving only a smooth, featureless white glare interior to the primary bow.
5. OBSERVATIONAL MICROPHYSICS AND POLARIZATION
The visual appearance of a rainbow is governed by microphysical properties: the droplet size distribution, droplet oblateness, and the polarization physics of dielectric interfaces.
+-------------------------------------------------------------------------------+
| RAINBOW MORPHOLOGY VS. DROPLET DIAMETER |
+-------------------+-----------------------------------------------------------+
| Drop Diameter (D) | Observable Optical Manifestation |
+-------------------+-----------------------------------------------------------+
| D > 2.0 mm | Vivid, hyper-saturated red and orange; no supernumeraries;|
| (Heavy Downpour) | Bow flattened at crest due to droplet oblateness. |
+-------------------+-----------------------------------------------------------+
| D = 0.5 - 1.0 mm | Classic textbook rainbow: brilliant spectrum, narrow band;|
| (Moderate Rain) | 2 to 4 distinct supernumerary fringes below violet. |
+-------------------+-----------------------------------------------------------+
| D = 0.2 - 0.5 mm | Faded red band; green and blue dominate; prominent, wide |
| (Light Drizzle) | supernumerary arcs extending deep into the interior disk. |
+-------------------+-----------------------------------------------------------+
| D < 0.05 mm | Pure white "Fogbow" / Mistbow; diffraction completely |
| (Cloud / Mist) | broadens bands; faint orange outer rim, pale blue core. |
+-------------------+-----------------------------------------------------------+
Aerodynamic Oblateness
Falling raindrops do not maintain perfect spherical shapes. As documented by the World Meteorological Organization and Atmospheric Optics, hydrodynamic drag forces flatten the bottom of falling drops into oblate spheroids with cross-sectional eccentricity increasing with drop size.
Because horizontal cross-sections retain a circular radius of curvature while vertical cross-sections are flattened, large falling raindrops maintain the classical $42.4^\circ$ caustic along the vertical sides of the rainbow arc, while the apex (crest) of the bow is compressed and distorted. This asymmetry weakens and broadens the top of large-drop rainbows while keeping the vertical pillars near the ground razor-sharp and saturated.
Brewster's Angle and Linear Polarization
One of the most dramatic yet overlooked characteristics of the rainbow is its extreme degree of linear polarization. When light reflects internally off the back wall of a droplet, the angle of internal incidence is:
$$r \approx 40.2^\circ$$
The Brewster's angle $\theta_B$ for an internal water-to-air boundary is:
$$\theta_B = \arctan\left(\frac{n_{\text{air}}}{n_{\text{water}}}\right) = \arctan\left(\frac{1.000}{1.333}\right) \approx 36.87^\circ$$
Because the internal reflection angle $r \approx 40.2^\circ$ is extraordinarily close to Brewster's angle $\theta_B \approx 36.9^\circ$, the parallel component ($p$-polarization, oscillating parallel to the plane of incidence) is almost completely transmitted through the back of the droplet and lost. The perpendicularly polarized light ($s$-polarization, oscillating perpendicular to the plane of incidence) undergoes strong Fresnel reflection.
Consequently, the light emerging from both the primary and secondary rainbows is over 94% linearly polarized tangential to the circular arc. If an observer views a primary rainbow through a linear polarizing filter or polarized sunglasses and rotates the lens by 90 degrees, the entire multi-coloured bow can be extinguished completely from the sky.
6. PRACTICAL OUTDOOR GUIDANCE
Measuring the Arc in the Field:
[ Top of Primary Arc: 42° Elevation ]
/\
/ \ 4 Fists Stacked
/ \ (≈ 40° - 42°)
/ \
Anti-Solar Point --------+------+-------- (Shadow of Observer's Head)
/ \
/ \
/____________\
1. Locating the Anti-Solar Point and Bow Geometry
- The Shadow Baseline: Always orient yourself so that your shadow points directly toward the center of your field of view. The anti-solar point lies precisely at the center of the shadow of your head.
- The $42^\circ$ Rule of Solar Altitude: Because the primary rainbow forms a cone with a semi-angle of $42.4^\circ$, you can only see a rainbow from level ground if the sun’s elevation is less than $42^\circ$ above the horizon.
- If the sun is higher than $42^\circ$, the rainbow cone is projected entirely into the ground below your feet.
- If the sun is at the horizon ($0^\circ$ elevation), exactly half of the complete circle (a semicircle) rises above the landscape.
- In an aircraft or atop an isolated mountain peak with rain below you, you can observe the elusive, unbroken $360^\circ$ full circle rainbow.
2. Angular Estimations with Hand Gestures
To measure rainbow angles without scientific instrumentation, use standard outdoor celestial angle rules at arm’s length: - One Closed Fist ($\approx 10^\circ$): Stacking four vertical fists from your head's shadow upward reaches the fiery red outer crest of the primary bow ($\approx 42^\circ$). - One Extended Palm Width (Thumb to Pinky $\approx 20^\circ$): Spanning two palm widths upward from the anti-solar point locates the primary arc. - Alexander’s Band Check: Place one fist between the $42^\circ$ primary red rim and the $51^\circ$ secondary red rim to confirm the $8^\circ$ span of Alexander's dark band.
3. Meteorological Diagnostic Indicators
When tracking rainbow conditions, monitor these real-time atmospheric shifts via local National Weather Service and Met Office radar feeds or handheld barometers: - Barometric Pressure: Watch for the sharp pressure jump signaling the gust front and cessation of heavy convective rain. - Wind Vector Shift: A rapid veering of surface winds (e.g., southwesterly to northwesterly) confirms the cold-frontal passage, heralding the dry slot required for clear sunlight to illuminate rain curtains downwind. - Sky Luminance Contrast: If the rain shaft to your east appears nearly black against a bright western horizon, the high contrast will optimize the visibility of subtle wave phenomena like supernumeraries and tertiary bows.
7. TODAY'S METEOROLOGICAL RULE OF THUMB
The Rainbow Observer's Law:
To see a rainbow, stand with the sun at your back and below 42 degrees in the sky; the lower the sun, the higher the arc. If the sky between the two bows appears noticeably dark and the violet rim ripples with pastel fringes, you are witnessing both Descartes' forbidden geometric zone and Airy’s quantum-like wave interference etched across the atmosphere.
Authoritative Scientific Resources & Further Reading
- Met Office Optical Guide to Atmospheric Rainbows
- NOAA JetStream: The Physics of Atmospheric Rainbows
- WMO International Cloud Atlas: Optical Meteors & Rainbow Classifications
- Atmospheric Optics: Primary Ray Tracing & Caustic Formations
- Airy Diffraction Integral and Caustic Wave Singularity Formations