The COVID-19 lockdown, in my part of the world, has produced an outpouring of children’s rainbow art—often stuck up in people’s windows, but sometimes sketched on the pavements, too.
I’ve been struck by the generally good command of spectral colours on display, with red on the outside and an appropriate progression towards violet on the inside. I was amused by the one above, which is a pretty flawless piece of artistry, undermined only by the position of the sun.
It reminded me that I had been planning to write about rainbow colours for years, ever since I wrote about converging rainbows back in 2015.
That was when I posted this diagram, showing the relationship between the sun’s position and the rainbow arc:
The rainbow is a complete circle of coloured light, centred on the antisolar point—the direction exactly opposite the position of the sun, which is marked by the shadow of your head. We usually only see an upper arc, because the area below the horizon usually contains too few raindrops along the line of sight to generate bright colours.
Every raindrop 42½º away from the antisolar point reflects red light towards our eyes; every raindrop at 40½º reflects violet light in the same way. And between those extremes, raindrops reflect all the spectral colours in turn, changing their apparent colour as they fall. But quite why that happens is a little complicated, and that’s what I want to write about this time.
Here’s the route that a ray of green light takes when it passes through a rainbow-forming raindrop and bounces back towards our eyes. Let’s call this particular trajectory the “rainbow ray” for short:
If the raindrop is falling through the top of the rainbow arc, the light enters near the top of the drop, is refracted as it crosses the air-water interface, then reflected from the back of the drop, then exits through the bottom of the drop, being refracted again as it moves from water back to air. The angle between the incoming ray and the outgoing is about 41½º for green light. At the sides of the rainbow, you have to imagine the diagram above lying horizontal—the ray enters the outward-facing side of the drop, and is reflected towards you sideways. And, obviously, the rest of the rainbow is formed by light paths that are more or less tilted between the horizontal and the vertical.
Violet light is refracted more than green light, which is in turn refracted more than red light. So a ray of white light (drawn in black below) is split into a fan of coloured rays as it enters the raindrop. These follow slightly different courses within the drop, and exit at different angles (exaggerated here for clarity):
So we see red at 42½º, and violet at 40½º, with green between. Simple.
But why choose to consider just those rainbow rays? What about all the light that enters the drop closer to its rim, or closer to its centre? I’ll call this general group of light rays “reflected rays”, of which the rainbow ray is only one example.
If I plot a couple of examples, you can see that rays entering the drop farther out than the rainbow ray are refracted more strongly, and end up exiting the drop at an angle less than the rainbow ray. Those that enter the drop nearer the centre are refracted less, but bounce back at a narrower angle from the back of the drop, and also exit the drop at an angle less than the rainbow ray. So for a given colour of light, it turns out that the rainbow ray is the light path that results in the maximum deflection angle from the antisolar point. All other reflected rays exit at narrower angles, and so should appear within the visible coloured arc of the rainbow itself.
How does that help, though? Why is the rainbow ray’s role as the maximum angle of deflection important?
To show why, I’m going to make a plot of what happens to all the reflected rays, using a parameter I’ll call Offset*, which works like this:
It’s just the proportional distance from the centre of the raindrop at which the ray enters—a zero offset means that the ray spears straight into the centre of the drop; an offset of one means the ray just grazes the edge of the drop.
So here’s how red and violet reflected rays are deflected, for the full range of offsets:
Deflection peaks at an offset of about 0.86, the location of the rainbow ray, with lesser deflection occurring on either side, as shown in the diagram above. Red and violet are at their maximum separation at the peak, and the rounded peak of the curves means that a lot of rays close to the rainbow ray end up reflected in the same part of the sky as the rainbow ray. You can see from the chart that all the rays with offsets from 0.75 to 0.95 end up within two degrees of the rainbow ray; a similar span from 0.2 to 0.4 is spread over fifteen degrees.
So, in the vicinity of the rainbow ray, there’s a lot of light in a small area of sky, and the spectral colours are well separated. Farther from the rainbow ray, the deflected light is smeared over a large area of sky within the rainbow arc, and the spectral colours are not well separated—all these other rays average out to a patch of white light filling the curve of the rainbow, with no colour separation. This bright area within the rainbow can often be strikingly visible, if the rainbow has dark clouds behind it.
One other thing contributes to the colour intensity of the rainbow—oddly, that’s that fact that some light is lost from the reflected rays every time they interact with an air/water or water/air interface. Here’s a diagram of how much light goes missing from the green rainbow ray as it passes through the raindrop:
The large proportion of light that shoots straight out the back of the drop, doubly refracted but without being internally reflected, creates a bright patch around the sun that appears whenever the solar disc is viewed through falling rain. Les Cowley at the excellent Atmospheric Optics site has dubbed this the “zero-order glow”.
Interestingly, for a given ray each interaction with an interface results in exactly the same ratio of reflection to transmission (though not quite in my diagram, which features rounded figures). This is unexpected (at least to me), because the reflective properties of a water/air interface are generally different from those of an air/water interface; the former features the phenomenon of total internal reflection, for instance. But it turns out that the first passage through the air/water interface changes the angle of the ray just enough to make it interact with subsequent water/air interfaces in exactly the same way as its initial air/water encounter.
If I plot the final amount of transmitted light for all the different offset rays, and add it to my previous graph, it looks like this:
The transmission data are in brown, and refer to the new, brown axis on the right side of the chart. You can see that transmission starts to ramp up just as we get into the vicinity of the rainbow ray, boosting the brightness in the rainbow’s part of the sky. The peak of transmission does occur at very high offsets, beyond the rainbow ray, but in that region the angle of deflection changes very rapidly with slight changes in offset, which diffuses that light over a large arc.
The calculations I did to produce the transmission graph above involved Fresnel’s equations, so I had to track two different polarizations of light independently. For light reflecting from a surface between two transparent mediums, there’s a critical angle of incidence called Brewster’s angle, at which the reflected light becomes totally polarized. At that angle, the reflected light is entirely s-polarized; light polarized at right angles to this (p-polarized) is completely transmitted through the reflective surface. (Your polarizing sunglasses are designed to filter out s-polarized light reflected from horizontal surfaces, to reduce glare.)
The Brewster angle for an air/water interface is around 53º; for water/air it is about 37º. And it turns out that any light entering the water at an angle of incidence of 53º has its angle changed to 37º by refraction. So in the case of our raindrop, a ray that strikes the surface of the drop at 53º (corresponding to an offset of about 0.8) will continue through the drop and strike the water/air interface at 37º—it hits two Brewster angles in succession! This means that p-polarized light that hits the drop at Brewster’s angle is entirely refracted into the drop—none of it escapes by reflection from the air/water interface. But then it hits the back of the drop, and now none of it is re-reflected—it is all transmitted, again. So at an offset of 0.8, no p-polarized light gets into the reflected ray—it is all lost out the back of the raindrop.
So now if I mark up the total amount of transmitted light with its s- and p-polarized components, you can see that the light making up the rainbow will be strongly s-polarized, because the rainbow rays are pretty close to Brewster’s angle:
The resulting polarization follows the curve of the rainbow. Your polarizing sunglasses will largely block the light coming from the top curve of the rainbow, but will let light through from the sides. However, if you tilt your head, you’ll remove light from the sides of the rainbow, and bring the upper curve into view.
Here’s a nice short little YouTube video, by James Sheils, demonstrating how to make the lower curve of a rainbow appear and disappear using a polarizing filter:
And that’s it, for now. Some time in the future I’ll get around to discussing the secondary rainbow.
* The measure I’ve called Offset is sometimes called the “impact parameter”, a term borrowed from nuclear physics. While the analogy is strong, if you know its original application, I’m not sure the phrase itself helps with visualization, so I’m sticking with Offset in this and subsequent posts.