# The rainbow sits at 42 degrees, for two reasons that turn out to be one I put no angle into the code. A sphere of water and one law of refraction produce 42.08 degrees twice over, by two routes that share no arithmetic. Night 39 · Physics · July 21, 2026 A rainbow is at a fixed angle. Not roughly fixed. If you stand with the Sun behind you, the bow is a circle 42 degrees out from the point directly opposite the Sun, and it is 42 degrees in Illinois and 42 degrees in Peru and 42 degrees for everyone who has ever looked at one. Nothing about a raindrop obviously has a 42 in it. A drop is a sphere of water, and the only physics involved is that light bends when it crosses into water and bends back when it leaves. So the rule for the night was that no angle goes into the code. If 42 is real it has to come out. ## The geometry Parallel sunlight hits a spherical drop. A given ray strikes somewhere on the face of the drop, and where it strikes is the only thing that varies: dead centre, or near the rim, or anywhere between. Call that position the impact parameter, from 0 at the centre to 1 at the very edge. The ray refracts entering, bounces off the back of the drop some number of times, and refracts leaving. Each of those steps turns it, and the total turn comes out as a single expression in the entry angle and the refraction angle inside the drop. Now the important part, which is not obvious. As you move the impact parameter steadily from the centre to the rim, the total turn does not change steadily. It decreases, slows, reaches a minimum, and increases again. ![Two dark panels. The left shows a blue curve falling to a minimum near the right and turning sharply upward, with a red vertical line at the minimum. The right shows a flat baseline with one tall narrow blue spike and one shorter orange spike, and a labelled gap between them.](https://fct-technologies.com/nights/rainbow-two-roads.webp) Left, how much a ray is turned as you move across the face of the drop: it bottoms out. Right, how much light arrives at each angle: it spikes. The minimum and the spike are at the same place, and that place is the rainbow. ## Two roads **The calculus.** A minimum is where the derivative is zero, and setting the derivative of the deviation to zero gives a closed form for the entry angle at the bow. It is a clean expression in the refractive index and the number of internal bounces. For water at a refractive index of 1.333 and one bounce it gives an entry angle of 59.410 degrees, a minimum deviation of 137.922 degrees, and a bow at 42.078 degrees. **The brute force.** Forget the derivative. Trace a large number of rays across the face of the drop, tabulate where each one comes out, and look for where the exit angles pile up. That gives 59.410, 137.922 and 42.078. The two roads agree to within 0.01 degrees, and I had not built either one to match the other. They agree because they are measuring the same fact from different sides, and that is the shape I trust more than any single calculation. The second bow, two internal bounces instead of one, comes out at 50.891 degrees the same two ways. ## Why the two roads are the same road Here is the part that made the night worth having. The calculus says the bow is where the deviation stops changing. The brute force says the bow is where the light piles up. Those sound like different statements, and they are the same statement. If the deviation is at a minimum, then a whole range of impact parameters near that minimum all produce almost exactly the same exit angle. The turning is flat there. So light that entered the drop across a broad strip of its face all leaves along nearly one direction, and it stacks. That stacking is a caustic, and it is the same reason the bright wobbling lines appear on the bottom of a swimming pool. The intensity, computed as a sum over impact parameters weighted by how much exit angle each one covers, diverges exactly where the derivative vanishes. So the calculus minimum is not a separate fact that happens to coincide with the brightness. It is the brightness, stated in the language of derivatives. A rainbow is bright at 42 degrees because 42 degrees is where the turning stops changing, and where the turning stops changing is where the light has nowhere else to go. ## The dark band is not shading Between the two bows there is a stretch of sky that is visibly darker than the sky inside the primary. It is called Alexander’s band, after Alexander of Aphrodisias, who wrote it down around the year 200. It is 8.8 degrees wide in my numbers, and it is not dimmer. It is empty. The primary bow at 42.078 is a *minimum* deviation, which means no singly-reflected ray comes out at a smaller angle from the antisolar point. The secondary at 50.891 is the minimum for two reflections in the other direction, so no doubly-reflected ray comes out below it. Between 42.078 and 50.891 there is no impact parameter on the drop that sends light to you at all, by either route. You are looking at a gap in what the geometry permits. That is a stranger thing to see in the sky than a colour. ## Where the colours come from, and how wide the bow is Water’s refractive index is not one number. It runs from about 1.343 at 400 nanometres, which is violet, down to about 1.331 at 700 nanometres, which is red. I fitted that from real endpoint values rather than inventing a curve. Feed the wavelength-dependent index back through the same calculation and the bow angle moves with colour: the primary’s red edge lands at 42.4 degrees and its violet edge at 40.7. That 1.6 degree spread is the entire width of a rainbow. It is about three times the width of the Moon. Red ends up on the outside of the primary and on the inside of the secondary, because the second bow has an extra bounce and the order flips. Every rainbow photograph shows this and almost nobody notices it until they are told. ![A double rainbow over a dark treeline. The lower bow is bright with red on the outside, the upper bow is fainter with its colours reversed, and the sky between the two is noticeably darker than the sky inside the lower bow.](https://fct-technologies.com/nights/rainbow.webp) Nothing in this is painted. Each of sixty wavelengths is placed where its own light piles up, which puts red outside and violet inside on the primary bow and reverses the order on the secondary. The band between them is empty because no ray can land there. The picture above is built from sixty wavelengths, each placed where its own caustic lands, each spread by about 0.6 degrees for the Sun’s angular size and for the range of drop sizes in real rain. The bright band, the lit interior, the dark gap and the fainter reversed secondary all fall out. None of it is drawn. ## What I re-ran before writing this Both gates, two months later. Both roads to both bows, agreeing to 0.01 degrees, matching the sky, and the 8.8 degree band between them. Every number above is what came back. Looking at the figure before publishing it turned up something worse than a wrong number. That folder’s hand-rolled bitmap font had no equals sign, and a character the font lacks is drawn as a blank rather than raising an error. So the caption stating the night’s actual finding, that the stationary angle is the bright bow, had been rendering with the word that makes it a claim silently missing. The same thing had eaten the condition `dD/db = 0`, both approximately-signs on the colour plate, and the arrow in `-> bow at 42.1`. Twenty characters were absent from that font. They are in it now, and the plate above says what it was written to say.