Why the Rainbow Sits at 42 Degrees

A rainbow is not a coloured object hanging over a particular field. It is a viewing geometry that follows the observer. This is the destination. The visible bow is an arc centred on the point opposite the sun, and its outer red edge sits about forty two degrees from that direction. We now have to explain why. Here is the observation to explain. To see a rainbow, the sun must be behind you. That is a fact about geometry, and it is our first clue. Stand with the sun at your back and follow the shadow of your head away from the sun. That line is the anti-solar direction, the centre line for every rainbow you see. A drop out here catches the sunlight and sends its red light back to your eye. The angle between that returning ray and the centre line is about forty two degrees. Rotate that direction right around the centre line and the eligible drops form a thin conical shell, whose two edges are the red lines of this side view. Now step sideways. The cone moves with your eye, and a different drop joins the new line of sight. The rainbow is fixed by an angle, not by a place in the shower. Raise the sun and the anti-solar direction tilts down. The whole bow then follows it below the horizon. Lower the sun and the bow rises again. So the question has two parts. Why is the angle about forty two degrees, and why does the light bunch up there instead of spreading evenly across the sky? Both answers are inside one raindrop.

Refraction is the bending of light as it crosses into water. An incoming ray and the centre of a spherical drop determine one flat slice where we can follow that bend. Seen head on, the slice is a circle. The same circle, centre, and ray will stay with us all the way to the answer. At the surface, the normal is the radius through the point of contact. The incoming ray makes the incidence angle i with that normal. At that surface the light turns toward the normal as it enters water, so the angle inside, r, is smaller than i. Snell's law gives the size of that bend. Sine i equals n times sine r, and water has n about one point three three. The ray reaches the back wall. Most light escapes and is lost, but a small fraction reflects. Symmetry makes the incoming and reflected angles there equal to the same r. At the front surface the reflected ray meets one more radius. It leaves water, opens from r back to i, and becomes the ray that can reach an eye. Each pair of angle marks appeared only for the event it named. The exit pair has now carried the same law to the last surface, so it can leave too. Now measure how far the ray turns at each event. At the entrance it turns through i minus r. The reflection is the large turn. Equal angles r leave a straight angle minus two r between the old direction and the new one. Leaving the drop adds another i minus r. All three local turns now stand beside the path: i minus r, one hundred eighty minus two r, and i minus r. Add those three turns. The deviation is one hundred eighty degrees plus two i minus four r. The angle D compares the original forward direction with the direction that actually leaves the drop. A ray through the centre has i and r both zero, so the expression gives a turn of one hundred eighty degrees, straight back. Snell's law also makes r the inverse sine of sine i over n, leaving i as the only free angle. Set the entry low on the drop, at twenty degrees. The ray returns almost the way it arrived, with a deviation of about one hundred sixty degrees. Slide the entry point upward. The deviation falls through one hundred fifty, one hundred forty five, and one hundred forty. Then it slows. There, it has stopped coming down. Keep sliding toward the rim. Only after the minimum does the curve climb again, and it keeps climbing as the entry point approaches the edge. The flat bottom makes the minimum important. These two rays enter at forty eight and seventy degrees, far apart on the face of the drop. Yet their outgoing directions differ by less than a quarter of a degree. A broad band of entry points therefore sends light back in nearly one direction, so the light piles up instead of spreading thin. That concentration is the rainbow. To locate it exactly, use the flat curve condition: at the minimum, the derivative of D with respect to i is zero. Solving that condition says d r by d i must equal one half. The angle inside must change half as fast as the incidence angle outside. Differentiate Snell's law. Cosine i equals n cosine r times d r by d i. Substituting one half gives cosine i equal to n cosine r over two. Now square both sides. Four cosine squared i equals n squared cosine squared r. Replace cosine squared r with one minus sine squared r. Snell's law then replaces n squared sine squared r with sine squared i. Finally, sine squared i is one minus cosine squared i. Collect the cosine terms. Cosine i is the square root of n squared minus one, divided by three. For water, i is fifty nine point four degrees and r is forty point two. Putting both into the turn formula gives a minimum deviation of about one hundred thirty eight degrees. Return to the special ray at the minimum. Its path is still the same path through the same drop; only the chosen entry point has moved. Carry the incoming direction through the exit point. Forward to backward is a straight angle of one hundred eighty degrees. The ray has already turned through D, about one hundred thirty eight degrees. Subtract that deviation from the straight angle. The remaining angle between the outgoing ray and straight back is forty two degrees. That is the promised number. Snell's law fixes the path inside one circle, the three turns create a minimum at one hundred thirty eight degrees, and the view back toward the sun leaves forty two degrees.

White sunlight contains many colours, and water bends each one by a slightly different amount. Red bends least; violet bends most. Follow the least-deviated red path through the drop, then the violet path. They enter close together, separate through the two refractions, and leave in different directions. For red, n is one point three three one and the viewing angle is forty two point three degrees. For violet, n is one point three four four and the viewing angle is forty point six degrees. A single drop therefore does not send an entire rainbow to one eye. Its red and violet leave along different lines, so one viewpoint receives at most one narrow part of that colour fan. Put two drops back in the shower. One is higher above the anti-solar line and one is lower, while the sunlight reaches both from behind the observer. The higher drop can send its red ray to the eye. Its violet ray passes over the observer's head and is missed. The lower drop can send violet to the same eye. Its red ray passes below the observer's feet. Red therefore arrives from the larger viewing angle and sits on the outside of the primary bow. Violet arrives from the smaller angle and sits inside. The anti-solar point is the centre of the geometry. The horizon lies above it whenever the sun is above the horizon. Rotate the primary viewing directions around that centre. Red traces the outer arc and violet traces the inner arc, with every point supplied by a different drop in the shower. The complete geometry is circular, but the ground hides the portion below the horizon. What remains visible is an arc, not a full ring. Look just inside the primary bow, then just outside it. The inside is brighter because a broad band of rays returns on that side of the minimum. The region just outside receives far fewer primary rays. A second internal reflection creates a secondary bow farther out. Its colour order reverses: violet is outside and red is inside. One drop supplies the path, the minimum deviation supplies the brightness, dispersion supplies the colours, and rotation around the anti-solar point supplies the arc. Together they make the rainbow in the sky.