The Boon Companion has been experimenting with long exposure times and intentional camera movement, of late. She was just about to discard the motion-blurred cyclist above as a failed experiment when something about the image caught my eye.

In the thirtieth-of-a-second exposure, the bicycle wheel has rolled a short distance. But why do the spokes look curved? Why don’t the curves point towards the centre of the wheel? And why is the effect only visible in the lower half of the wheel?

So I sat down to figure out the trajectory of a bicycle spoke as the wheel rolls along the ground. As you do.

Any point on wheel rolling across a flat surface without slipping follows a curve called a trochoid (from Greek trochos, “wheel”). I won’t pester you with the relevant equations (they’re at the other end of the link above). Here’s what the trochoid curves look like for the two ends of a radial spoke, spanning the distance between a wheel hub and a thick wheel rim:

The shape of the wheel is plotted in grey dashes, with a single vertical spoke marked. As the wheel rolls left or right, the ends of the spoke follow the curved trochoid trajectories, with successive positions marked at 20º intervals.

But bicycle wheels don’t (usually) have radial spokes, and I felt obliged, going into the problem, to look at the position of real bicycle spokes. Here’s a very common pattern:

Thirty-six spokes, laid out in what’s called a “three cross” pattern. Fundamentally, there are only two different spoke alignments in this pattern—leading and trailing.

For a wheel rolling from right to left, the spoke I’ve highlighted in red is leading, and the blue spoke is trailing. They’re simply mirror images of each other. One side of the wheel has nine leading spokes, spaced 40º apart, and nine trailing spokes in the same pattern. The other side of the wheel is laid out exactly the same way, but with the pattern rotated by 20º. The final result is called a “three cross” pattern because each trailing spoke crosses three leading spokes on its side of the wheel (and vice versa).

In this pattern, the anchor point for the spoke at the hub is offset 60º relative to its attachment at the rim. So to see the trajectory of a representative bicycle spoke, I need to slide the trochoid curve for the hub 60º out of alignment with the rim curve. Here its, with the spoke drawn in at 20º intervals:

This is the trajectory of a trailing spoke for a wheel rolling right to left, and a leading spoke for a wheel rolling left to right.

We can already get a hint of why some sort of spoke pattern shows up in the lower half of the wheel, but not the upper. In the upper half, the spoke is moving rapidly sideways, as it pivots across the top of the wheel; in the lower half it performs a sort of dipping motion, arcing downwards towards the point at which the wheel contacts the road, and then arcing back up again.

Now, I figure a cyclist moving at a reasonable speed for a shared-use path will rotate the wheels through about 30º during a thirtieth-of-a-second exposure. Here’s a more detailed trajectory for a trailing spoke (for a bicycle moving right-to-left) during a 30º rotation:

You can see that the spoke slides along itself, to some extent—different parts of the spoke occupy the same spatial position at different times. These are the only parts of the spoke that will show up as a dense “shadow” during a prolonged exposure that blurs the other parts. In the final photograph we’ll therefore see something like the arc I’ve sketched in red, while the rest of the spoke is smeared into a blur:

Something similar happens for a leading spoke as it passes through the same position:

So, actually, while the leading/trailing distinction slightly changes the details, both kinds of spoke produce a curved shadow during a prolonged exposure.

If we catch a spoke that’s higher in its trajectory, we get another arc:

Now we can put these images together, showing the curved shadows of several spokes at once. Each of them will lie on a different set of trochoid arcs, shifted laterally according to how far the spoke lies from the lowest point of its trajectory. Like this:

I’ve marked the visual centre of the wheel, for reference. Notice how the partial shadow arcs formed by the lower spokes seem to point below the centre, while the arcs of the higher spokes point above the centre. It happens because none of the spokes are radial, and because the centre of the wheel is never stationary, but shifts horizontally as the spokes sketch out their curved arcs.

So. Although I was baffled by the photograph when I first saw it, I began to get an inkling of what was going on as I made the first trochoid sketches, and was then pleasantly surprised by how things began to fall neatly into place as I added more detail. I’m hoping you’re as pleasantly surprised as I am.

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Reflection: A transformation under which each point in a shape appears at an equal distance on the opposite side of a given line—the line of reflection.

It’s not often I have occasion to shout at the television, but a recent episode of the BBC’s long-running television series QI precipitated just such an outburst. The cause of my vexation was their answer to the question that forms the title of this post. The offending episode was the R Series: Reflections, and the explanation was an excellent approximation to gibberish, involving as it did some business about “The mirror doesn’t flip things around; we flip things around,” intoned by Sandi Toksvig as she stood in front of a mirror fiddled with a bit of card with the word BOSS written on it (see above). To be fair to Toksvig, it probably wasn’t her idea, and she did manage to deliver the entire farrago while wearing the sort of anxious expression people wear when they’re not entirely convinced by their own argument.

The answer to the question is really that it is ill-posed. Mirrors actually don’t reverse left and right, for the simple reason that mirrors have no way of telling left from right. They have no left-right asymmetry, in other words. The only asymmetry they do possess is in the plane of reflection—stuff in front of the mirror is the real world; stuff “behind” the mirror is the reflected world.

That’s what’s being described in the definition at the head of this post, which refers to a two-dimensional reflection, like this:

Here we have a “line of reflection”, corresponding to the mirror; a letter “B” in front of the mirror, representing the real world; and a reflected letter “B” behind the mirror. Every point on the reflected “B” is the same distance from the mirror as the corresponding point on the original “B”. So because the spine of the “B” is the farthest part from the mirror, its reflection also lies farthest from the mirror. Conversely, the curved parts of the letter lie closer to the mirror on both sides. And it’s that preservation of “near” and “far” on either side of the mirror plane which causes the reflection to be a reversed image of the original. If we travel from one side of the mirror to the other, we encounter in turn spine-curves-mirror-curves-spine. So, actually (and pace Toksvig), the mirror very much does “flip things around”. Indeed, many introductory geometry texts gloss the word “reflection” as “a flip”.

The same thing happens when a three-dimensional person stands in front of a real mirror:

The reflected image’s left and right (and head and feet) are pointing in the same direction as the real person’s. What has been flipped in the mirror image is the direction in which the nose and toes are pointing. So the mirror has reversed front and back, not left and right. If you lie down with your feet pointing at the mirror, the reflection will also have its feet pointing at the mirror—so on this occasion the mirror leaves your front and back, and left and right, in the same positions, but reverses you top to bottom. Only if you stand sideways on to the mirror does it truly reverse your left and right—but that’s because you’ve chosen to place one side of your body close to the mirror and the other far from it, not because the mirror has some magical ability to tell left from right.

So why do we always think the mirror image has reversed left and right, no matter how we orientate ourselves before the mirror? The answer, I think, lies in the single plane of symmetry in our own bodies. Our fronts are very different from our backs, our heads are very different from our feet, but our left side is very similar to our right side. So it’s very difficult for us to see the mirror reflection as having reversed front and back—instead, we see it as another person who has turned around to face us. In which case, their right hand now moves when we move our left hand, and vice versa. No matter what the orientation of the reflection, we always interpret it as a left-right reversed person, because a head-foot reversed person or a front-back reversed person is harder to conceptualize.

Remove the left-right symmetry, and we stop talking about mirrors reversing left and right.

This barber’s pole lacks a clear left-right distinction. So instead, we find ourselves saying that it “spirals the opposite way”. We’re still, apparently, unable to discern that what has been reversed is the front and back, but now we can’t blame a left-right switch either.

So if anyone ever asks you the title question, permit yourself the slightest of headshakes and the faintest of smiles, and say: “But mirrors don’t reverse left and right; they preserve near and far.”

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In my last two posts about rainbows, I discussed the formation of the primary and secondary rainbows, respectively, tracing their origins to specific light paths through falling raindrops.

The primary rainbow ray follows a path like this:
For a raindrop at the apex of the rainbow arc, sunlight enters near the top of the drop, bounces once off the back, and then exits the bottom, descending towards the observer’s eye, making an angle of around 41.5º for the green ray shown.

For the secondary rainbow, sunlight enters near the bottom of the drop, is reflected internally twice, and then exits the front of the drop, descending towards the observer at an angle of around 51º.

These rainbow rays are special, representing the maximum or minimum angles of deflection of the incoming ray. And they are associated with a particular offset of the incoming ray in its interaction with the raindrop. I measured offset like this:The rainbow ray for the primary enters the raindrop at an offset of about 0.86; the secondary rainbow ray at about 0.95.

For more about these topics, in particular the importance of maximum and minimum deflections, I direct you back to my previous posts.

I finished my most recent post on this topic with a question: If a single internal reflection produces a primary rainbow, and two internal reflections produce a secondary rainbow, is there such a thing as a tertiary rainbow? And if so, where is it?

The path of the rainbow ray for a tertiary bow looks like this:

The three reflections carry it almost all the way around the raindrop, so that (in contrast to the primary and secondary bows) it leaves the drop heading away from the sun. This tells us that, to see a tertiary bow, we’d need to look towards the sun, rather than (as for the primary and secondary) towards the antisolar point.

Plotting my usual graph of light deflection and transmission against the full range of ray offsets with three internal reflections, I get this:

In contrast to my graphs for the primary and secondary, this one shows the deflection from the “solar point”—the position of the sun. The maximum occurs at an offset of about 0.97, reaching 42.5º for red light, and 37.7º for violet. The angular distance between red and violet is therefore about two-and-a-half times what we see in the primary bow. The light transmission is scaled to match my previous two graphs, and it shows that, because so much light is lost during multiple internal reflections, only about 1% of the tertiary rainbow ray survives to exit the drop. (But that’s equivalent to about 24% of the transmission for the primary rainbow ray, so not catastrophically dim.)

So it seems straightforward enough. We should see a broad, faint tertiary bow, about the same diameter as the primary but centred on the sun. Have you seen that? No, me neither.

The problem is that there’s a lot of light in the sky around the sun, particularly when rain is falling through the line of sight. Les Cowley at Atmospheric Optics calls this the “zero-order glow”, because it is formed from sunlight that passes through the raindrop without being reflected. So there’s always a directly transmitted, zero-order light ray parallel to the tertiary rainbow ray, like this:Much more light pours through the raindrop in the zero-order glow than survives through three internal reflections. This makes the tertiary rainbow very difficult to see.

But not impossible. There have been sporadic reports over the last few decades, by careful observers who knew what they were looking for. And finally, in 2011, a paper* appeared in the “Light And Color In The Open Air” edition of Applied Optics, entitled “Photographic evidence for the third-order rainbow”. The authors describe taking a photograph under favourable conditions (the sun obscured, a dark cloud in the region that would be occupied by the tertiary bow). The photographer could barely discern a hint of the tertiary rainbow—“only a faint trace of it at the limit of visibility for about 30 seconds”. But after a bit of image processing, a rainbow arc appeared in the resulting photograph. And, after careful analysis, the authors confirmed they had taken the first known photograph of a tertiary rainbow.

You’ll have discerned a pattern. Each successive rainbow (primary, secondary, tertiary) is produced by a rainbow ray which is increasingly offset from the centre of the drop, undergoing one more reflection that its predecessor. The increasing offset means that refraction is greater, with a larger difference between red and violet rays, leading to a broader rainbow. Each additional reflection along the light path means more light lost, and so a fainter rainbow.

Rather than bore you with light paths for higher-order rainbows, I’ll show you them all on one diagram. (The design is based on Jearl Walker’s “Amateur Scientist” column in the July 1977 issue of Scientific American—the calculations and drawing are all my own.)

The rosette shows the first twenty rainbows, produced by the first twenty internal reflections of light that enters the upper half of a water droplet. Light bounces around the drop clockwise, and the difference in deflection for successive rainbows quite quickly converges to settle at a little less than a right angle. So you can see that the primary (labelled “1” in the lower left quadrant) is followed by the secondary in the upper left quadrant, the tertiary in the upper right quadrant, and the quaternary in the lower right. The quinary rainbow (number five) brings the progression almost full circle, and appear just a little anticlockwise of the primary. I won’t bore you with the names for higher orders of reflection^†, but you should be able to pick out how they go around again, and again, and again, becoming successively wider and fainter.

Because violet is refracted more than red, all the rainbows have the same layout, with violet always lying clockwise of red.

Rainbows in the lower left quadrant are being reflected downwards, and back towards the light source. So an observer would look for them as circles around the antisolar point. Because the red light is descending more steeply than the violet, red will appear on the outer edges of all these rainbows. Rainbows in the lower right quadrant are formed from light that has been reflected downwards and away from the light source—so the observer must look towards the sun to see them. In this position, violet light is descending more steeply than red light, so all these rainbows have violet on their outer edges.

Light in the top half of the diagram is all spraying upwards—these rainbows would not be visible to an observer standing below the drop. But we’ve neglected the light that enters the lower half of the raindrop, and bounces anticlockwise. It produces its own rosette of rainbows, identical to the one above, except mirrored in the horizontal plane, like this:

The reversed light path means that violet always lies anticlockwise of red for this family of rainbows. So antisolar rainbows with this light path have violet as their outer colour, and solar rainbows have red outermost.

So the rainbows we see in the sky come from a mixture of the two sets of possible light paths in the two diagrams above. So let’s stack them together, and switch from a view centred on the water drop to a view centred on the observer:

The sky is full of overlapping rainbows! You should think of the upper and lower “copies” of each rainbow as being linked by a vertical circular arc sticking out at right angles to your screen, which the observer sees as a (potential) circular rainbow.

My diagram has the sun directly behind the observer, in which case the rainbows in the lower part of the diagram would be invisible under normal circumstances, superimposed on the ground, and each rainbow would form a semicircular arc against the sky. But the sun is usually some distance above the horizon—as it climbs higher, it pushes the antisolar rainbows lower in the sky, but carries the solar rainbows higher. So antisolar rainbows  generally form less than a semicircular arc, and when the sun is higher above the horizon than the radius of the rainbow they will drop entirely below the horizon. But solar rainbows will be lifted above the horizon by the sun, forming more than a semicircular arc, and at the extreme when the sun is higher above the horizon than the radius of the rainbow they can form complete circles. (Complete circles are also possible for antisolar rainbows, but only when the observer is looking down on water droplets suspended below the horizon, as from a plane flying over clouds.)

Are any of them visible to the naked eye? We know that the tertiary has been spotted very rarely. The quaternary sits right next to it, farther out in the zero-order glow, but additional light losses mean its rainbow ray transmission is just 15% of the primary. I don’t know of anyone who has seen it, but it has been photographed. In fact, the article^‡ entitled “Photographic observation of a natural fourth-order rainbow” appeared in the same themed edition of Applied Optics as the tertiary rainbow report. (Early reports of the tertiary photographs had inspired the author to search out the quaternary using image stacking.)

The quinary bow has also been photographed. With rainbow-ray transmission sitting at just 10% of the primary, it is also partially obscured by the brighter secondary bow. But its green, blue and violet portion sit within Alexander’s Dark Band, giving reasonable hope of detecting those colours. And Harald Edens managed to (just) pick out the green stripe of the quinary rainbow in a photograph taken in 2012.

The sixth-order rainbow sits within the bright sky at the inner edge of the primary bow, and seems like a poor candidate for detection. The seventh is better separated from the zero-order glow than the third and fourth, but its rainbow ray transmission is only 6% of the primary. However, it seems likely that some of the higher order bows will yield to the sort of photographic techniques commonly employed in astrophotography these days. Watch this space. Meanwhile, for your delectation, I’ve appended some basic descriptive data for the first twenty rainbows, as featured in my diagrams above.

* Großmann M, Schmidt E, Haußmann A. Applied Optics 50(28): F134-41
^† Oh, alright, I will. Beyond quinary, five, the sequence goes senary, septenary, octonary, nonary, denary, undenary, duodenary … at which point we need to start making up names until we get to vigenary, twenty.
^‡ Theusner M. Applied Optics 50(28): F129-F133

Note: All values, here and in previous posts, are calculated using a refractive index for red light (wavelength 700nm) of 1.33141, and for violet light (wavelength 400nm) of 1.34451. See the page dealing with refractive index on Philip Laven’s excellent website, The Optics Of A Water Drop, for further information and references.

* The order-11 and order-13 rainbows lie predominantly in the solar hemisphere, but extend slightly into the antisolar
^† The order-12 rainbow spans the solar point, and therefore overlaps itself—a small disc of blue-violet rainbow (radius 6.1º) is superimposed on a larger disc of green-yellow-red rainbow (radius 10.1º)

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In my previous post about rainbows, I described how the light of the rainbow was reflected back to our eyes by falling water droplets. For a raindrop at the top of the rainbow arc, light follows a path that enters near the top of the raindrop, bounces off the back, and then exits from the bottom:

The angle between the incoming light ray and the reflected ray ranges from 42.4º for red light to 40.5º for violet light. All other light rays, entering the drop either closer to its centre or closer to its edge, are reflected back at smaller angles—and it’s their smeared and superimposed light which accounts for the white glow visible within the arc of the rainbow, above. I used the name “Offset” for the parameter that measures how close to central the incoming light ray hits the droplet. It’s measured like this:

And by plotting the deflection of light rays at various offsets, along with light transmission along the reflected pathway, I showed how the rainbow forms at the point of maximum deflection, corresponding to an offset of about 0.86:

I called the ray that follows this maximum-deflection route the “rainbow ray”. (See my previous post for much more detail.)

I also produced a little diagram of how light is lost from the rainbow ray each time it encounters a surface at which it is either reflected or transmitted, like this:

So the rainbow ray arrives at your eye containing only about 4.5% of the light that entered the water drop.

What I want to talk about this time is the fate of the light that undergoes a second internal reflection (labelled “0.5%” in the diagram above).

It’s possible for light from this second reflection to exit the drop when it next encounters the water-air interface, like this:

The second reflection takes the light to the front of the raindrop, and it exits on an upward course, crossing the path of the incoming ray. For this ray to reach the eyes of an observer looking towards the antisolar point (the centre of the rainbow arc), we have to flip the geometry upside-down, like this:

So now the incoming light enters near the bottom of the raindrop, and the reflected light is deflected downwards, towards an observer on the ground.

This light is the source of the secondary rainbow, which is larger and fainter than the primary rainbow I’ve been describing so far. (A secondary rainbow is dimly and partially visible in the photograph at the head of this post.)

Like the singly reflected light that forms the primary bow, the doubly reflected light that forms the secondary bow has its own characteristic angle of deflection, but this time (because of the flipped light-path described above) it’s the minimum angle of deflection from the antisolar point where light is concentrated to form the secondary rainbow ray:

I’ve kept the scale of the transmission curve the same, to allow comparison with that of the primary rainbow. And the range of the angle-of-deflection axis is the same, but it spans from 45º to 90º this time, rather than the 0º to 45º of the primary plot.

The minimum deflection occurs at an offset of about 0.95. The deflection is 50.4º for red light, and 53.8º for violet.

So there are several things going on with this secondary rainbow. Firstly, because the light enters closer to the edge of the raindrop, the refraction is greater, and that causes the separation between red and violet light to be greater—so the secondary rainbow has a width of 3.4º, compared to just 1.9º for the primary rainbow. Secondly, the fact the light-path is flipped over compared to the primary reverses the colour sequence—red is on the inside of the secondary bow, but on the outside of the primary. Thirdly, the additional reflection means more light is lost from the rainbow ray as it passes through the drop. This is partially offset by the fact that the rainbow-ray offset is closer to the offset of maximum light transmission—so the light transmitted into the secondary bow works out to be about 43% of that transmitted into the primary. Finally, because the rainbow ray occurs at a minimum deflection from the antisolar point, the rest of the sunlight entering the drop is reflected to greater angles than the rainbow ray—it lights up the sky outside the secondary bow.

So if we imagine a raindrop falling vertically towards the antisolar point, it at first sends a doubly reflected mixture of light, appearing white, towards the observer’s eye. When it has fallen to 53.8º from the antisolar point, it sends a relatively pure, doubly reflected violet light to the observer—the top of the secondary bow. As it falls from 53.8º to 50.4º, it reflects all the spectral colours in sequence, ending with red light at the inner rim of the secondary bow. Then, it quite literally goes dark. It is too low to send any doubly reflected light in the observer’s direction, but too high to send any singly reflected light. The region of sky between the secondary and primary bow is therefore noticeably darker than either the region outside the secondary, or inside the primary.

This dark region is called Alexander’s Dark Band*, in honour of the Greek philosopher Alexander of Aphrodisias, who described it in about 200AD.

After passing through the Dark Band, the raindrop lights up with singly reflected red wavelengths when it reaches 42.4º, and then runs through the spectral colours until it reaches violet at 40.5º. Below that point, it reflects a mixture of wavelengths (white light again) until it hits the ground.

The whole sequence looks like this:

The obvious question now is this: if the primary rainbow forms from singly reflected light, and the secondary from doubly reflected light, what happens to triply reflected light? Is there a tertiary rainbow? And if so, where is it?

That’s the topic for next time.

* Not to be confused with the 1911 Irving Berlin hit, Alexander’s Ragtime Band.

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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.

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