Why Is a Rainbow 42 Degrees? Descartes' Math Explained • The Science 360

Stand with your back to a low Sun and a rain shower ahead, and hold your arm straight out. A fist at arm's length covers about 10°. From the shadow of your head, the point directly opposite the Sun, count a little over four fists up and you will hit the red edge of the rainbow. Count five more and, if you are lucky, the secondary bow. Every observer, every shower, every century: 42° and 51°. Drop size does not change it, distance to the rain does not change it, and nor does how hard it is raining.

Most pages that come up for this question tell you the angle is "determined by refraction and reflection" and move on. That is true and explains nothing, because refraction and reflection send light out of a drop at every angle from 0° to 42°. The real reason is a turning point: a maximum in the exit angle, where a large bundle of rays all leave in nearly the same direction. René Descartes found it in 1637 by tracing rays by hand. We can do it in a few lines.

Quick answerSunlight entering a raindrop, reflecting once inside and leaving again comes out at an angle that depends on where it entered. That angle peaks at about 42° from the point opposite the Sun for water (n = 1.333). Near the peak many rays exit together, so light concentrates there and forms the bright bow.

Infographic: Why is a rainbow 42 degrees? A ray of sunlight enters a raindrop near its edge, reflects off the back and leaves, split into red and violet, at 42 degrees to the sunlight's direction. That angle, 42.0° for water, is a turning point: rays leaving after one reflection never exceed it, so light piles up there, about 20% of the drop's light within one degree. A chart of exit angle against where the ray hits shows one curve peaking at 42.0° and the two-reflection curve bottoming at 51.0°, the double rainbow. Tiles: violet at 40.6° and red at 42.4°, rainbow light 92% polarised. Myth: each drop does not make a whole rainbow. — Tap the image to view it full size. View the accessible, editable SVG.

What happens to light inside a raindrop?

A falling raindrop smaller than about 1 mm across is very nearly a sphere, and somewhat bigger drops are close enough. Parallel rays of sunlight hit it all over its face. Label each ray by its impact parameter $b$ : 0 for a ray aimed at the centre, 1 for one that just grazes the edge. Each ray refracts as it enters (Snell's law, $\sin i = n\sin r$ , where $i$ is the angle of incidence and $r$ the angle inside the water), crosses the drop, reflects off the back surface, crosses again and refracts out.

Ray diagram of parallel sunlight entering a spherical raindrop from the left, reflecting once off the back and exiting toward the lower left. The exiting rays fan out over a range of angles but crowd together near one direction. The highlighted Descartes ray, entering at impact parameter 0.86, exits at 42.0 degrees to the reversed sunlight direction. — Figure 1. Eleven rays traced exactly through a water drop (n = 1.3334 at 589 nm) with one internal reflection. The rays fan out but crowd together along the highlighted Descartes ray, which leaves at 42.0° to the direction back toward the Sun. Light that is transmitted straight through or reflected off the front surface is not drawn.

Add up the turning. Entering, the ray is deflected by (i − r). The internal reflection turns it by (180° − 2r). Leaving, it turns another (i − r). The total deviation from its original direction is:

\begin{aligned}D &= 2(i - r) + (180^\circ - 2r) \\ &= 180^\circ + 2i - 4r\end{aligned}

Angle from the antisolar point:

\theta = 180^\circ - D = 4r - 2i

For a ray through the centre, $i = r = 0$ and $\theta = 0$ : it comes straight back toward the Sun. As $b$ grows, $\theta$ grows, but not forever. Refraction bends steeply-hitting rays more, and eventually the $2i$ term wins and $\theta$ falls again. In between is a maximum.

Where does the 42 come from? The worked calculation

Set $d\theta/di = 0$ . From Snell's law, $dr/di = \cos i/(n\cos r)$ , so $d\theta/di = 4\cos i/(n\cos r) - 2$ , which is zero when $2\cos i = n\cos r$ . Squaring and using $\sin r = \sin i/n$ gives a clean result:

Condition for the turning point (one internal reflection)

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

Numbers for yellow light, 589 nm, water at 20 °C

$n = 1.33335$ (Daimon and Masumura, 2007), so $n^2 = 1.77782$ .

\begin{aligned}\cos i &= \sqrt{0.77782/3} = \sqrt{0.25927} \\ &= 0.50919 \Rightarrow i = \mathbf{59.39^\circ} \\ \sin r &= \sin 59.39^\circ / 1.33335 \\ &= 0.86065/1.33335 = 0.64548 \\ r &= \mathbf{40.20^\circ} \\ D &= 180^\circ + 2(59.39^\circ) - 4(40.20^\circ) \\ &= 137.97^\circ \\ \theta &= 180^\circ - 137.97^\circ = \mathbf{42.03^\circ}\end{aligned}

The ray that does this enters at $b = \sin i = 0.861$ , about 86% of the way from the centre to the edge. That is the Descartes ray in Figure 1.

Now the reason it matters. Move the entry angle 5° either side of the Descartes ray and the exit angle changes by only about 0.55°. Move it 10° and the exit changes by about 2.2°. Compare that with rays near the centre of the drop, where every extra degree of entry angle moves the exit by about a degree.

How much light piles up near the maximum

Rays with exit angle between 41.03° and 42.03° enter between $b = 0.79$ and $b = 0.91$ . Share of the drop's cross-section in that ring:

0.913^2 - 0.792^2 = \mathbf{20\%}

Share that exits between 30° and 31°: 2.6%. Light per degree near the bow is about 8 times higher, before any other effect.

Plot of exit angle measured from the antisolar point versus impact parameter from 0 to 1. The primary curve rises almost linearly to a flat maximum of 42.0 degrees at b equals 0.86, then drops. The secondary curve comes down from above to a flat minimum of 51.0 degrees at b equals 0.95. The band between 42 and 51 degrees is shaded and labelled Alexander's dark band. — Figure 2. Exit angle versus where the ray hits the drop, computed exactly for n = 1.3334. The flat tops are the rainbows: a maximum at 42.0° for one internal reflection and a minimum at 51.0° for two. No ray of either kind ends up between them.

In optics this kind of pile-up is called a caustic, the same thing as the bright curve on the bottom of a coffee mug in sunlight. Geometrical optics actually predicts infinite brightness exactly at the maximum; the wave nature of light (and the Sun's 0.53° width) smooths it out. Descartes called the rainbow angle the angle of minimum deviation, because 42° from the antisolar point is the same as a minimum of 138° in total deviation. Both descriptions are the same turning point.

The misconception: “each drop makes a rainbow”

A popular picture shows a single drop splitting white light into a neat fan of colours, as if each drop were a tiny prism projecting a rainbow. That is not what you see. Each drop sends you exactly one colour, or none, depending on its position. The drops that look red to you are the ones sitting on a cone of half-angle about 42.3° around the line from the Sun through your head. The drops that look violet sit on a slightly narrower cone, about 40.6°. Your rainbow is assembled from millions of different drops, and the person next to you sees a rainbow made of different drops.

Two consequences follow. A rainbow is not at any particular distance: it is a direction, like the Sun. And you can never reach its end, because the cone moves with you. The circle shape comes from the symmetry of the cone: every point 42° from the antisolar point is equivalent, so the bow is a circle centred on the shadow of your head.

Why is red on the outside?

Water's refractive index rises toward the blue, so blue light bends more on entering and leaving. Put $n = 1.3431$ (405 nm) into the same formula and the maximum drops to 40.6°. Put $n = 1.3305$ (700 nm) and it rises to 42.4°.

Plot of rainbow radius versus wavelength from 400 to 700 nm, drawn in the colour of each wavelength. The primary bow rises from 40.6 degrees at 400 nm to 42.4 degrees at 700 nm. The secondary bow falls from 53.6 degrees at 400 nm to 50.2 degrees at 700 nm. — Figure 4. Rainbow radius versus wavelength, computed from the Daimon and Masumura (2007) dispersion formula for water at 20 °C. The primary bow is about 1.8° wide; the secondary is about 3.3° wide with the colours reversed. Add the Sun's 0.53° disk to each and the colours overlap, which is why rainbow colours are never as pure as a spectrometer's.

Since each colour's bow is a maximum, the sky just outside 42.4° gets no once-reflected light of any colour, and the sky just inside 40.6° gets some of every colour. So the outer edge is a sharp red, while the inner edge fades into a whitish glow. That brightening inside the bow is real and easy to see once you look for it: the sky inside a rainbow is noticeably lighter than outside.

Why is a double rainbow at 51 degrees, and why the dark band?

Rays that reflect twice inside the drop turn by another $(180^\circ - 2r)$ . The deviation becomes $D = 360^\circ + 2i - 6r$ , and its turning point is now a minimum in angle from the antisolar point.

Secondary rainbow, two internal reflections, n = 1.33335

\begin{aligned}\cos^2 i &= \frac{n^2 - 1}{8} = \frac{0.77782}{8} = 0.097228 \\ \cos i &= 0.31181 \Rightarrow i = \mathbf{71.83^\circ} \\ \sin r &= 0.95014/1.33335 = 0.71260 \\ r &= \mathbf{45.45^\circ} \\ D &= 2(71.83^\circ - 45.45^\circ) \\ &\quad + 2(180^\circ - 2\times45.45^\circ) \\ &= 230.98^\circ\end{aligned}

Angle from the antisolar point: $230.98^\circ - 180^\circ = \mathbf{50.98^\circ}$

Because the secondary bow is a minimum, the light piles up on its inner edge and spreads outward, the opposite of the primary. Colours reverse, with red on the inside. Each internal reflection keeps only around 5% of the light at these angles, so the secondary is much fainter.

Between 42° and 51° neither family of rays arrives. That strip is Alexander's dark band, named after Alexander of Aphrodisias, who described it around 200 AD. It is not totally black, because light reflected off the outside of drops goes in every direction. Figure 3 shows how big the contrast is.

Log-scale plot of relative brightness versus angle from the antisolar point between 30 and 60 degrees, for red and blue light. Brightness rises steadily to a peak at 41 degrees for blue and 42 degrees for red, drops by about a factor of 100 into a flat dark band, then rises to a smaller peak at 50.5 degrees for red and 52.8 degrees for blue before falling again. — Figure 3. Computed brightness of light scattered by a sky full of identical spherical drops: zero, one and two internal reflections with Fresnel reflection coefficients for unpolarised sunlight, smoothed by the Sun's disk. Geometric optics only, so supernumerary fringes and diffraction are not shown.

In the model, the sky between 30° and 38° is about 17 times brighter than the dark band, and the sky beyond the secondary bow is about twice as bright as the band. You can check the first part with your own eyes on the next good rainbow.

Why is rainbow light polarised?

Here is a detail almost nobody mentions. At the Descartes ray, the light hits the back of the drop at 40.2° from the normal. The internal Brewster angle for water, at which light polarised in the plane of incidence is not reflected at all, is $\arctan(1/1.333) = 36.9^\circ$ . The rainbow ray reflects close to that angle, so most of the light polarised in that plane leaks out of the back of the drop and only the other polarisation is reflected toward you.

Fresnel calculation for the Descartes ray

Fraction surviving two refractions and one reflection: 8.8% (s-polarised) versus 0.34% (p-polarised).

Degree of polarisation:

\begin{aligned}P &= \frac{8.8 - 0.34}{8.8 + 0.34} \\ &= \mathbf{92\%}\end{aligned}

That polarisation is tangential to the bow. Secondary bow (two reflections): 81%.

Published values are a little higher, around 96% for the primary and 90% for the secondary, because the full calculation averages over the whole caustic and includes wave effects. Either way, a polarising filter or polarised sunglasses rotated the right way will make a rainbow almost vanish. Try it.

Why can’t you see a rainbow when the Sun is high?

The centre of the rainbow circle is the antisolar point, which is below the horizon by exactly the Sun's elevation. The top of the primary bow is therefore 42° minus the Sun's elevation above the horizon. With the Sun at 42°, the top of the bow touches the horizon. Above that, on flat ground, there is no primary rainbow at all.

Plot of the height of the rainbow top above the horizon versus the Sun's elevation. The primary top falls linearly from 42 degrees at sunrise or sunset to zero when the Sun is 42 degrees high; the secondary falls from 51 degrees to zero at 51 degrees. Labels note that 50 percent of the primary circle is above a flat horizon with the Sun at 0 degrees, 37 percent at 20 degrees and 22 percent at 35 degrees. — Figure 5. Rainbow height above a flat horizon versus Sun elevation. The visible fraction of the primary circle is arccos(tan e / tan 42°) / 180°, where e is the Sun's elevation. That is why rainbows are mostly a morning and late-afternoon sight, and a summer-midday impossibility at lower latitudes.

In summer at mid-latitudes the Sun is above 42° for several hours around noon, so there is a rainbow-free window in the middle of the day. In winter at higher latitudes the Sun may never get that high, and a rainbow is possible at any hour. From a plane or a mountain top, where you can look down below the horizon, you can see much more of the circle, and occasionally all of it.

Does drop size change the rainbow angle?

The 42° does not depend on drop size at all, because the geometry above scales: a big sphere and a small sphere bend rays identically. What drop size changes is how much the wave nature of light blurs the result. With drops around 1 mm or smaller, faint extra pastel bands called supernumerary bows appear just inside the primary, which is interference between pairs of rays leaving at the same angle from either side of the Descartes ray. Thomas Young used them to argue for the wave theory of light, and George Airy worked out their theory in 1838. With the tiny droplets in fog, diffraction smears the colours together so thoroughly that you get a broad, nearly white fogbow, still centred near 42°.

Field check: with the Sun behind you and low, the primary bow's red edge is about 42° (just over four fists at arm's length) from the shadow of your head; the secondary is about 51°; the sky inside the primary is brighter than the band between the two; and polarised sunglasses tilted 90° will dim the bow.

FAQ

Why is a rainbow always 42 degrees?

Light that enters a spherical raindrop, reflects once inside and comes back out leaves at an angle that depends on where it hit the drop. That angle has a maximum of about 42 degrees from the antisolar point for water's refractive index of 1.333. Near the maximum, a wide range of rays leave at almost the same angle, so light piles up there and forms a bright ring.

Why is the secondary rainbow at 51 degrees?

The secondary bow is made by rays that reflect twice inside the drop. Their exit angle has a minimum of about 51 degrees from the antisolar point, so light piles up there instead. The second reflection loses more light, which is why the secondary bow is fainter, and it reverses the order of colours, putting red on the inside.

Why is the sky darker between a double rainbow?

Once-reflected light can only reach angles up to 42 degrees and twice-reflected light only down to 51 degrees, so no light from either path reaches the band between them. Only weak reflection from drop surfaces lights it. That darker strip is called Alexander's dark band.

Why can’t you see a rainbow at midday?

The rainbow is a circle 42 degrees around the point directly opposite the Sun, which is below the horizon by the Sun's elevation. When the Sun is higher than about 42 degrees, the whole primary bow is below the horizon. From an aircraft or a high cliff you can still see it, sometimes as a full circle.

Is the rainbow angle exactly 42 degrees?

No. It depends on wavelength because water's refractive index does. Using measured data for water at 20 °C, the primary rainbow is at 42.4 degrees for 700 nm red and 40.6 degrees for 400 nm violet. The Sun's half-degree disk and diffraction by the drops blur it further.

Sources

M. Daimon and A. Masumura, "Measurement of the refractive index of distilled water from the near-infrared region to the ultraviolet region", Applied Optics 46, 3811 (2007), via the refractiveindex.info database (20.0 °C dispersion formula): refractiveindex.info
Wikipedia, "Rainbow" (Alexander's band, Theodoric of Freiberg 1307, Descartes 1637, supernumeraries with drops about 1 mm or less, polarisation 96% primary and 90% secondary): en.wikipedia.org/wiki/Rainbow
Scientific American, "Why are rainbows curved?": scientificamerican.com
HyperPhysics (Georgia State University), "Rainbow" and "Brewster's angle": hyperphysics.phy-astr.gsu.edu
Atmospheric Optics (Les Cowley), primary and secondary rainbows, Alexander's band: atoptics.co.uk

Method note: all angles and curves are computed by exact ray tracing through a sphere with the Daimon and Masumura dispersion of water at 20 °C and Fresnel coefficients for unpolarised light. Brightness curves ignore diffraction and drop-size distributions, so they show the geometry, not the fine structure of a real bow.

Header image: NASA/Ben Smegelsky (image KSC-20220612-PH-JBS01_0008), regraded for this article.

What happens to light inside a raindrop?

Where does the 42 come from? The worked calculation

The misconception: “each drop makes a rainbow”

Why is red on the outside?

Why is a double rainbow at 51 degrees, and why the dark band?

Why is rainbow light polarised?

Why can’t you see a rainbow when the Sun is high?

Does drop size change the rainbow angle?

FAQ

Why is a rainbow always 42 degrees?

Why is the secondary rainbow at 51 degrees?

Why is the sky darker between a double rainbow?

Why can’t you see a rainbow at midday?

Is the rainbow angle exactly 42 degrees?

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Sources