What determines the arc of a rainbow?

The rainbow that you see is for your eyes only. Someone standing off to your left or right will be seeing an entirely different rainbow in a slightly different position in the sky.

The rainbow that you see is actually built, so to speak, around the axis line extending through the sun and your head. The shadow of your head, if you could see it, would always fall right at the center point of the arc of the rainbow (i.e. if the arc had been drawn by a giant compass, the point where that pivot point of the compass would have been). The primary rainbow will appear to lie about 40 degrees off of that sun/head axis. Because that axis line will be different for all viewers, each one effectively gets their own rainbow.

Also, because that axis line will vary when the sun is higher or lower in the sky, you’ll see more or less of the arc at different times of the day. If the sun is more than 40 degrees above the horizon, then you won’t see any of the arc at all (assuming flat terrain).

Great answer. I knew it was different for each viewer, but did not know my head was the center point. Would the arc form a circle? Also, is the arc shape going to be the same at various times of day, knowing the sun will change relative to my head?

The arc would form a circle of the ground didn’t get in the way. Airplane passengers could, under ideal conditions, get a full circle rainbow (as for time of day, see my edited post above).

Your edit made my response seem ignorant, but I can live with that. Thanks for all your rainbow expertise. I wondered because I saw a huge double rainbow once and noticed they were the same arc and the center appeared inline with the sun. I didn’t think about my head being along that line too, but that makes total sense. I thought it maybe had to do with the curvature of the Earth, or maybe atmospheric conditions that influenced the arc.

Finally, why is 40 degrees a critical number? I’m assuming it was to do with the dispersion of the light spectrum or something? Like above that number, the light won’t appear to separate into its components? Or is it just that you’d see so little of it on the horizon?

@cockswain The 40 degrees comes from the “refractive index” of water, i.e. how much water bends light. If all of the wavelengths of light were bent by exactly the same amount, rainbows would appear as a thin white arc.

But each wavelength gets bent to a slightly different degree, so the rainbow has width and distinct bands of color.

Excellent answers! I know red has a longer wavelength than purple. I’m trying to understand now why red is on the outside and the shorter wavelengths move closer to the axis. Is red being bent more or less than purple?

Violet is bent more than red, which would seem to mean that violet would appear on the outside of the arc; but the reverse is true. That’s because you’re actually seeing the reflection of the light coming off the inside rear wall of the water drop, so it’s a mirror image of the expected order.

Note that the secondary rainbow in a double rainbow has a reverse order of colors; that’s because the secondary is the result of yet another internal reflection inside the water drop, which reverses the order one more time.

I hadn’t noticed that about the double rainbow actually. Without asking, you’ve explained that phenomena as well. You know your physics, congrats.

Ah, I did think of another question: what determines the distance between a double rainbow?

Strictly speaking, it’s not a matter of “distance”, but of angle. That extra bounce inside the sphere of the drop kicks the light out another 10 degrees or so, putting the secondary bow out at around 51 degrees from the axis.

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