Much has been written about rainbows—in textbooks, in magazines, and on the Web. They are, after all, beautiful, fascinating, occasionally awe-inspiring, even a little mysterious. Any driver, coming around a bend in the road on a misty afternoon and encountering a rainbow arcing across the eastern sky, is likely to stop to admire it, perhaps to photograph it—all the more likely if there are children on board. And the physics is fascinating1—straightforward at the level of geometrical optics, more challenging when wave aspects are taken into account.
This essay is a supplement to an article entitled “Rainbows: A Graphical Approach” that appeared in The Physics Teacher, vol. 58, March 2020, page 152. It is available at https://aapt.scitation.org/doi/10.1119/1.5145401. That article (some parts of which are reproduced here) presents a graphical approach to answering the following questions about the primary rainbow:
| 1. | How can one use the physics of refraction to determine the angle at which the primary rainbow is seen? |
| 2. | Why is the primary rainbow apparently brighter than the sky beneath it, even though more light, after the refraction-reflection-refraction sequence, goes into the region below the rainbow than into the rainbow itself? |
| 3. | Since dispersion separates light into its component colors, why is color seen only in the rainbow, and not in the illuminated region below the rainbow? |
| 4. | About what fraction of the refracted-reflected-refracted light goes into the rainbow itself and what fraction into the region below it? |
In this essay, I consider
- rainbows of all orders (although third and higher orders are rarely glimpsed)
- primary rainbows caused by substances other than water.
A. “Bending angles”
Consider a narrow ray of light arriving at a water droplet, refracting into it, reflecting within it, and refracting back out of it (Figure 1). Since the droplet is so tiny, any slight variations in direction or intensity of the light approaching a single droplet can be ignored.2 For studying rainbows, it’s enough to consider parallel, equally intense light rays. The sample ray shown in Figure 1 arrives with an “impact parameter”3 b at an angle of incidenceθ, continues inside the drop at an angle of refraction φ, reflects off an inside surface with angles of both incidence and reflection being φ, and refracts out of the droplet (with, again, angles φ and θ). (There is also reflection at the refraction points and refraction at the reflection point, but these represent “lost” light that doesn’t contribute to the rainbow.)
At point A in the figure, the light ray undergoes a deflection (I’ll call it a bending) θ – φ. At B its bending is 180 – 2φ. At C its bending is againθ – φ. Since all three of these bendings are in the same direction (clockwise in the figure), the total bending (which I’ll call ψ) is the sum of the three:
ψ = 180 + 2θ – 4φ. (1)
This is for the light that will contribute to the primary rainbow. If one allows for k internal reflections instead of just one, and with the same number of refractions (2), one concludes that the total bending of the light beam is
ψ = 180k + 2θ – 2(k + 1)φ. (2)
This could, of course, be more than 360 degrees (which supports the use of the word “bending” instead of “deflection”).
B. Rainbow angles
As the impact parameter b varies from 0 to R or, equivalently, as sin θ varies from 0 to 1, the total bending angle will vary smoothly. Light arriving from a single direction and impacting all parts of the droplet will emerge in a wide range of directions. Will a rainbow be formed? Perhaps. It depends on how the total bending angle ψ varies as the impact parameter b varies. If the total bending angle ψ varies always in the same direction (that is, if ψ depends monotonically on b) , the answer is no. Then there will be a smooth spray of light over a range of directions (perhaps all directions). If ψ rises and then falls, or falls and then rises, there will be a rainbow. The amount of light per unit of bending angle will then be maximum where ψ is not changing. The rainbow will appear where dψ / db = 0—or, equivalently, where dψ / d(sin θ) = 0 (at which point also dψ / dθ = 0). The spray of light will have an edge, and the edge will be bright (or at least brighter than its immediate surroundings).
Now to use this reasoning to search for a formula for rainbow angles.
It might seem silly to write θ = arc sin (sin θ), but it’s actually helpful because it is the sines of angles that enter into Snell’s law of refraction (sin θ / sin φ = n, where n is the index of refraction4). As an independent variable, I define
u = sin θ. (3)
Since sin θ = b/R, this is equivalent to using the impact parameter b as the independent variable. As u varies from 0 to 1 (or b from 0 to R), the incident light ray moves from a head-on collision with the droplet’s “Equator” to a grazing collision with its “Pole.”
Eq. (2) for the total bending angle, written in what at first seems a cumbersome form, is
ψ = 180k + 2 arcsin (sin θ) – 2(k + 1) arcsin (sin φ). (4)
Sin θ can be replaced by u (as defined in Eq. (3)) and, with the help of Snell’s law, sin φ can be replaced by u/n. A formula for the total bending angle as a function of the single variable u (and the parameter k) is then
ψ = 180k + 2 arcsin (u) – 2(k + 1) arcsin (u/n). (5)
To find rainbow angles we need only set dψ/du = 0, solve this equation for ur (r for rainbow), and substitute this ur back into Eq. (5) to get ψr, a formula for rainbow bending angles that depends on n and k.
I omit the details here (a “problem for the student,” as professors like to say about problems they prefer not to work through). The solution for ur is
This can be substituted into Eq. (5) to provide a formula for rainbow angles (of any order for any substance, not necessarily just water). The result for the set of rainbow bending angles is
There is one small complication, how to get from the rainbow bending angle ψr to the rainbow viewing angle ωr. If there is a formula for it, it is probably laced with caveats. It makes sense just to look at ψr and decide how to peel ωr from it. This “ad hoc” procedure is illustrated in Table 1 for the first four rainbows of red light from water droplets (n = 1.331).
Table 1. First four red-light rainbows from water droplets
| Order | Rainbow bending angle ψr | Viewing angle “algorithm” | Rainbow viewing angle ωr | Direction |
| 1 | 137.6 | 180 – ψr | 42.4 | away from Sun |
| 2 | 230.4 | ψr – 180 | 50.4 | away from Sun |
| 3 | 317.5 | 360 – ψr | 42.5 | toward Sun |
| 4 | 402.8 | ψr – 360 | 42.8 | toward Sun |
C. Alexander’s Dark Band
Not just at the rainbow angles, but at all angles reached by the sprays of light from the droplets, the same “algorithms” relate viewing angle to bending angle. Figures 2 and 3 show how the viewing angle ω varies as a function of θ for the first two orders (k = 1 and k = 2) for both red and violet light (and colors in between) in water. These graphs are based on Eq. (5), but with θ instead of sin θchosen as the x-axis variable. (Graphs of ωr vs. sin θ contain the same information but appear more “squashed.”)
From Figures 2 and 3 you can conclude that no visible light that has been once or twice reflected in water droplets shows up between the primary and secondary rainbows (between about 42 degrees and about 50 degrees). The once-reflected light is seen only at and below 42 degrees. The twice-reflected light is seen only at and above 50 degrees. When you are fortunate enough some time to view both a primary and a secondary rainbow, take note of the fact that the sky between them is indeed darker than elsewhere. If you would like to give this region a name, you are too late. It is called Alexander’s dark band,5 named after Alexander of Aphrodisias, who described it in about 200 AD.
D. Rainbows from other substances
Rainbows could, in principle, be formed by mists of liquids other than water, or even by a cloud of solid transparent microspheres. Table 2 shows the viewing angles for the primary rainbows of various materials.
Table 2. Primary rainbow viewing angles for various materials
| Substance | Index of refraction | Viewing angle of primary rainbow (in degrees) |
| Water | 1.33 | 42.5 |
| Kerosene | 1.39 | 34.5 |
| CCl4 | 1.46 | 26.7 |
| Benzene | 1.50 | 22.8 |
| Plate glass | 1.52 | 21.1 |
| Other glass | 1.47 to 1.61 | 25.7 to 14.2 |
| Hypotheticum | 2.00 | 0 |
Materials with an index of refraction of 2.00 or more do not produce primary rainbows. Diamond, for all its lovely optical effects, fails the primary rainbow test. Its index of refraction is 2.42. But, if rendered into microspheres, it could produce secondary and higher-order rainbows. Look at Eq. (6) to see that to be able to produce a rainbow of order k a material must have an index of refraction n < k + 1.
* * * * *
Given the artistic and emotional appeal of rainbows, it is not surprising that philosophers, scientists, and textbook authors have paid them a great deal of attention. If I attempted to do justice to past work, this essay would sink under the weight of its citations. So just a few remarks. Aristotle and other ancients (including Alexander of Aphrodisias) took an interest in rainbows. Quantitative treatment had to await the discovery of the law of refraction by the Dutch astronomer Willibrord Snell (1621), a law that may have been discovered earlier by Abu Sahl Wayjan Ibn Rustam al-Kuhl in Persia (984) and by Thomas Harriott in England (1602), as well as later by René Descartes in France (1637). What we now call Snell’s law was evidently known to Isaac Newton, who, despite his belief that light traveled faster in glass or water than in air, was able to use the law to calculate accurately the viewing angle of the rainbow (1704).
Authors of non-calculus textbooks understandably keep their distance from the formula dψ/du = 0. Some authors of calculus-based textbooks also sidestep a discussion of the real reason for rainbow angles, leaving many student to believe, I suspect (a suspicion I expressed near the beginning of this essay), that the refraction-reflection-refraction sequence itself concentrates the light into a band only a few degrees wide. One exemplary exception to the misperception waltz is provided by Harris Benson in his University Physics, in which he devotes a two-page spread to the derivation of the primary rainbow angle.6 (I also treat the problem with calculus in my Classical and Modern Physics textbook,7 although as an end-of chapter problem.)
I thank Paul Hewitt for stimulating my renewed work on rainbows, Alexander Haussmann for instructing me on certain rainbow subtleties, and Adam Ford for rendering the figures.
1 My one-time colleague Fred Reines, when he undertook for the first time to teach physics to students from the arts and humanities, titled his course “Rainbows and Things.” I know that he enjoyed the experience, and apparently so did the students, since word of mouth brought more to his classroom door.
2 Isaac Newton, in studying rainbow angles, actually took account of the fact that the Sun’s finite size causes its rays at Earth at any moment to be not quite parallel.
3 The impact parameter is the lateral distance of an “off-center” path from a path headed straight toward the center of a target. It is commonly used in describing scattering events in the nuclear and particle worlds.
4 Strictly, n is the ratio of the index of refraction of water to the index of refraction of air.
5 Alexander’s Ragtime Band came later.
6 Benson, University Physics (John Wiley, 1991), pp. 730-731.
7 Ford, Classical and Modern Physics (Xerox College Publishing, 1972), vol. 2, p. 964, P18.21