# Why is there no dark band below a rainbow?

I've read articles about Descartes' theory of the rainbow and I can understand why there is a dark band of about 8 degrees (Alexander's band) between the primary and secondary bows: for all the water droplets in that area, no light rays are deflected toward the observer after 1 or 2 internal reflections.

What puzzles me is: why don't we see a similar dark area below the primary rainbow? In the pictures of rainbows one can see on the net, the area below the primary bow seems to be quite bright. Shouldn't it be dark too, since droplets in that area will also deflect no bright rays toward the observer?

The primary rainbow is at 40.7-42.4 degrees, where ~42 degrees is the critical angle for red light to reflect back to the observer, and 40 degrees for blue light. what this means is that for a droplet higher than 40 degrees (assume horizontal sun ray, 0 degrees), it can no longer reflect blue light to the observer. by the same logic, no red light can be reflected by any droplet higher than 42 degrees. since red light has the lowest refractive index of any visible light, this means no visible light at all can be reflected by water droplets higher than 42 degrees (this is what you already know as the dark area). at around 50.4-53.5 degrees we start seeing light again, but because of double total internal reflection rather than single TIR. because there are two relections occuring, the secondary rainbow is twice as thick.

note so far my use of the word critical angle. any droplet higher than the critical angle associated with a wavelength cannot reflect light of that wavelength to the observer. but below that angle, reflection is always possible. ie, the red band appears red because no other wavelength other than red can be reflected at 42 degrees. the yellow band appears yellow because no wavelengths other than red/yellow are reflected from that point, so on and so forth. as we move down the rainbow more and more wavelengths are being reflected--finally we reach the bright area under the rainbow itself (<40 degrees). here, all visible wavelengths are being reflected (white light instead of coloured light), resulting in this region being brighter than any other part of the sky. (as opposed to the rainbow itself where water droplets reflects fewer and fewer wavelengths going from 40-42 degreees, and finally no longer reflecting at all to the observer up to 50 degrees were reflections are again possible)

I can't comment yet, so I'll provide a different way to explain the answer. But the point I most want to get across is that the internal reflections are never total; in fact, symmetry makes it so that TIR is impossible after sunlight enters a spherical raindrop. By definition, the sunlight enters at less than the critical angle (what gregsan described is called the rainbow angle, not the critical angle). TIR happens when it tries to exit at angles greater than the critical angle. Since symmetry requires it to strike the back of the drop at the same angle it entered, TIR can't happen.

Light doesn't strike a raindrop at just one angle of incidence, it strikes at all angles from 0 to 90 deg. The light that strikes at 0 deg (dead-center on the drop) does not refract at all: it goes straight in, reflects 180 deg, and comes straight back out in the direction of the sun. Light that strikes at an angle A=5 deg off-center, ends up being deflected about D=5 deg. As you increase the off-center angle A, the deflection D increases at first, but the rate at which it increases slows until the off-center angle is about 60 deg and the deflection is about 40 deg (the rainbow angle). Here the deflection angle function D(A) has a maximum. D decreases as A increases past this point.

As a result, the sunlight striking the drop reflects in a pattern that is much like the beam of an 80-deg wide flashlight, aimed toward the sun. You see this light reflected from raindrops everywhere between 0 deg (measured from the original sun's rays) to the rainbow angle. But here's the catch: the intensity of light at each deflection angle D is inversely proportional to D'(A). And since D(A) has a maximum at the rainbow angle, D'(A) is zero there.

So the light is, in theory at least, infinitely bright in an infintessimally-small band at the edge of the flashlight beam. You can't really see this edge, but it is still VERY bright close to it. Rainbows happen because the beam for each color has a different width, making this bright edge appear at a different place for each color.

And the green band is not pure green. It is a mix of all colors from green to red, it's just that the green is brighter. The mixture makes it a paler green than you'd see in a true spectrum. The same holds for every color after red. And there is no "inside edge" to the rainbow - it simply fades from violet to gray when no color dominates the mixture. This is why the "inside" is not dark - it isn't really inside at all, it is the gray band.

In Alexander's Band, outside the (primary) rainbow, no raindrops reflect light to you this way until you reach the secondary bow. Its flashlight beam is aimed away from the sun, and is about 260 deg wide. It "wraps" around the sky so you can see it while looking away from the sun. The colors are actually in the same order (violet=inside, red=outside), but you see the entire bow backwards (inside=up, outside=down) because of the wrapping.