# FeaturedB Double Rainbows and the direction of their colors

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1. Aug 30, 2017

### lavinia

I just saw a double rainbow. On the brighter bow the color arcs were red on the left edge of the bow and shifted to blue on the right edge. On the dimmer bow the colors shifted from blue on the left edge to red on the right. Why were they separated oppositely like this?

2. Aug 30, 2017

### davenn

Have a read ... no point me requoting it all .... all about double rainbows and colour reversals

http://www.straightdope.com/columns/read/2279/what-causes-double-rainbows

Dave

3. Aug 30, 2017

### sophiecentaur

This diagram may help. It shows the primary and double bows are formed due to different paths through the drops. The angles aren't quite right with respect to the normals, to my view but the basic idea of multiple reflections is fairly clear.

4. Aug 31, 2017

5. Sep 4, 2017

### JeffJo

Most explanations of rainbows that you will find get something wrong (http://www.atoptics.co.uk/bows.htm is a very good exception). I'm not trying to criticize anybody here - we were all given very plausible-sounding, but incorrect, explanations by teachers we trusted. So pass them along when we become the teachers. What I'm tying to do, is change what gets taught.

For example, the drops that sophiecentaur's diagram shows reflecting violet light to the observer will reflect every other color of light to the same observer; but with less brightness. The drops that are shown reflecting red, reflect only red. Drops halfway in between reflect red, orange, yellow, and green; but the green is much brighter. The order of the colors is not directly a result which color in a single ray bends more.

A rainbow is not just an arc of a circle, it is a filled-in disk. If the sun were monochromatic, it would have a very bright rim that is 0.5° wide. (The sun's rays are not actually parallel since the sun appears to be this same 0.5° wide, and that's where this rim comes from.) This effect can be seen on hazy days near sunset, when all but the red light is filtered out by the atmosphere:

Other colors of light will have slightly smaller discs. Red's disk is about 42° wide, and violet's is about 40° wide. The order of the colors in a rainbow is caused by how these colors overlap.

One side effect of this is that the sky is brighter inside the colored bands, than outside of them. But people generally notice the darkness outside, so it has a name: Alexander's Dark Band. But in bright bows, the white inside of the bow becomes more apparent if you look for it:

So, why are the colors of the secondary rainbow reversed? It is not because the second reflection inverts the image - it doesn't. It is not directly because the colors from one ray come out above or below each other.

In the primary rainbow, these disks are centered on the anti-solar point (directly opposite the sun), and are about 40° (violet) to 42° (red) wide. So violet is on the inside of the colored rim, and red is on the outside.

In the secondary rainbow, these disks are centered on the sun itself, and are about 126° (violet) to 130° (red) wide. But since these disks are wider than 90°, they wrap around the top of the sky. They are seen, upside down, in the same direction as the primary. Violet is still on the inside of the colored rim, and red is still on the outside.

6. Sep 4, 2017

### sophiecentaur

I am interested why you would distinguish between the Sun and the anti solar point because are they not on the same line through your eye?
Isn't the 'upside down' a valid way to describe it?

7. Sep 4, 2017

8. Sep 4, 2017

### OmCheeto

No way am I getting involved in another rainbow discussion....

Me discussing rainbows at PF; "You are all crazy. I think I'll go finish reading my book, on evolution."
[ref]

I would recommend @anorlunda 's insight article: Rainbows are not Vampires

But stay away from the comments!
Rainbows may not be vampires, but they are surely zombies, as if you get too deep into them, they will consume your brain.

9. Sep 4, 2017

### JeffJo

Why would you not?

Maybe you missed my point. The rainbow is formed by light that hits the entire surface of the raindrop, not just a single ray. After internal reflection, that light is confined within a cone whose center depends on whether you are considering light that reflected an odd, or even, number of times inside the drop. Here's a comparison for the first and second rainbows (the colors are to help you follow the path of each incident ray; both are actually for red light):

Light that hits the center of the drop, and exists after one internal reflection, reflects directly back towards the sun. Technically, we should call that a 180° deflection, but since we observe rainbows by looking away from the sun along the line you described, it is easier to call it 0°. Here's a plot of this for three colors (and here, the color in the plot does indicate the color of the light):

See how it starts at a deflection angle of 0°, and increases to around 40° to 42°? The rainbow exists at all of these angles; but at most of them, all colors exist with about the same brightness, so we see white. Above 40°, some colors drop out, and we see colors. This is amplified by how the flatness of the curves near the maximum makes the colors brighter there.

Here's the same plot for the secondary rainbow:

Since there are two reflections, light that hits the drop in the center technically deflects 360°, but we call it 180° when we look away from the sun. The white part of the rainbow is now at angles greater than 54°, as opposed to less that 40° for the primary rainbow. The colors "drop out" as the observation angle decreases below 54°.

One important point here is that absolutely nothing gets "inverted" by the second reflection inside the drop. In fact, if the indicies of refraction were higher, you could get a secondary rainbow that is greater that 180° from the sun, and so the colors would appear in the same order as the primary (which would also be higher in the sky).

Last edited: Sep 4, 2017
10. Sep 4, 2017

### A.T.

Maybe try less text, and instead a precise mathematical definition of "centered on the sun itself" vs "centered on anti-solar point".

11. Sep 4, 2017

### sophiecentaur

@JeffJo: I see what those two ray diagrams are trying to do but it confuses me that the incident rays are coloured before they enter the drop. I see that you have the problem of representing a range of entry points on the drop and the resulting caustic curves PLUS the dispersion. Quite a mess to represent to suit everybody. I also don't see how you appear to have total internal reflection with almost normal incidence on the far side of the drop. How can that happen if TIR only works beyond the critical angle? And again, the graphs are for different wavelengths but with what entry points?
I also see what you are saying about white light being the resultant of equal contributions over the drop except for a selective range of angles. I have always taken issue with people who seem to claim that the Rainbow Colours are 'pure' when, in fact, they are pretty desaturated.

I can still see no essential reason for choosing your reference angles differently for the two orders of rainbow. The only consequence of using the anti-solar direction would be that some of the angles from the secondary rainbow would be obtuse, which we could cope with, surely??

12. Sep 4, 2017

### collinsmark

Walter Lewin is known to give excellent undergraduate physics lectures on the topic of rainbows (as well as other things). Here is one lecture he gave on the subject:

13. Sep 4, 2017

### JeffJo

Did you read the part where I said "the colors are to help you follow the path of each incident ray; both are actually for red light"? They don't represent the color of the light, which is red in both plots.

Since TIR is not involved in the process of rainbow formation (and in fact is impossible in spherical raindrops - A good exercise for a beginning geometry student, who knows what an isosceles triangle is, is to prove it),
I did not intend to demonstrate it.

Did you look at the label of the horizontal axis?

[/quote]I can still see no essential reason for choosing your reference angles differently for the two orders of rainbow. You don't see what you apparently don't want to see. The reference comes from the formation process, not the observation process, as I am trying to explain. The process that determines the order of the colors, and so answers the original question.

Last edited by a moderator: Sep 4, 2017
14. Sep 5, 2017

### sophiecentaur

Great. I have been wrong all my life - but thanks.
They both start at zero, meaning that the reference for the angle of incidence is the same; they are parts of the same original ray and the emerging rays are 180 degrees apart. The graph shows that.
I think I could say the same thing here.
I think your claim that the two references are different is the thing that has been confusing me. Now I have looked 'properly' at the starting point of the graphs, they confirm what I am saying and I have no issue with them.

That movie with Walter Lewin gives the angle of change for a single reflection as given by 4r-2i and you can extend that to 6r-2i for two reflections. That can take you exactly the same way that Lewin goes for the secondary bow. The difference is that the secondary bow comes from incidence on the lower half of the drop. I imagine the angle i will be negative as the curve of the face goes the other way. The exaggeration of the ray paths through the drop do not help with understanding; there is actually very little difference between the overall deflections for the two overall paths.

15. Sep 5, 2017

### JeffJo

The label of the axis is "Angle of Incidence." It means the angle that a ray of light, which is incident upon the spherical raindrop, makes with the surface normal there. What you describe is the range of values the dependent variable can have. I call the variable A, and it does go from from 0° (light that is centered on the drop) to 90° (light that would graze an edge).

The formula for these plots is D(A,N,r) = 180°*(N-1) + 2*A - (2N+2)*B(A,r).
r is the index of refraction
N is the number of internal reflections (N=-1 coincidentally describes light that reflects externally)
A is the angle of incidence of the original sunlight. (Lewin calls it "i")
B(A,r) = asin(sin(A)/r) is the refracted angle (Lewin calls is "r")
D(A,N,r) is the angle where a viewer who is looking toward the anti-solar point would see the reflected light.​

Note that the reference to the observer's position is for convenience only. It is not a part of the actual process of rainbow formation. So your choice of using it is an arbitrary choice. It would be more general to leave the observer out, like The Calculus of Rainbows does, but I try to use more familiar values. But what I was trying to explain was why the colors are in a specific order. This is not a function of the observer's position.

The rainbow forms because, for N>=1, these functions have an extremum at A=asin(sqrt(1-(r^2-1)/(N^2+2N))).

If N is odd, this extremum is a maximum; if N is even, it is a minimum. This because if N is odd, D(A,N,r) starts at an angle equivalent to 0°, and increases. If it is even, it starts at an angle equivalent to 180° and decreases.

Red light has a lower index of refraction than violet light, so its max/min deviation is always further from this starting point than violet's max/min deviation. And its band in the corresponding rainbow is always further from the starting point (so it is quite literally "on the outside") than violet's. Whether you observe red to be higher, or lower, depends on which quadrant the max/min ends up in and whether it is a maximum or minimum.

For example, if the index of refraction were around 1.13 instead of 1.33 (I may have said this backwards earlier), then both rainbows would be seen in the order ROYGBIV from "top" to "bottom." The primary would be seen around 80° from the anti-solar point, instead of 40°. The secondary would technically be 200° from the sun, instead of 130°, but you could call that 20° from the anti-solar point after changing the reference and adding one more quadrant change.

This is why you have to understand how the rainbow is a full circle that is centered on 180°*(N-1) in order to explain the order of the colors. Whether the light enters the "upper" or "lower" half of the drop is not directly relevant - it just happens to coincide with the results when r=1.33.

16. Sep 5, 2017

### sophiecentaur

Thank you for bearing with me. I have been reading what you have written and my thinking / drawing have taken me to a Damascene moment. You cracked it when you pointed out that the light is never totally internally reflected. Also, you said
Looking at the two ray diagrams, way back, I was assuming they referred to two different light paths through the drop. BUT, of course, if they had been drawn to proper angular scale, the first and second order (and subsequent) rays start with the first ray as it enters the drop. that ray can bounce around and around inside the drop, releasing some light at each bounce. It is not necessary to have two different diagrams as what happens to a given wavelength could be shown in a single diagram, with the first and second exits (etc.) being at different angles. Usually the result from a white ray will be white rays emerging but, when those graphs happen to be near a minimum or maximum, there will be visible dispersion in just one direction.
The usual 'helpful' diagrams (the ones I posted and the ones you posted plus others) all give the message that there are different paths though the drop for the different orders, which distract the mind from what really is relevant. It is all about the maxima and minima that you describe, which determine the angles at which a colour will emerge.
It is worth noting that it was only when I drew the drop and rays without any reference to an observer that the mechanism became clear. Once one gets the pattern from a drop, you can then place your observer somewhere and do a transform to determine what he sees. (Again, you made that point earlier on.)

17. Sep 5, 2017

### Staff: Mentor

In the 1970s or 1980s, Scientific American had an article on this topic that included great graphics. If anyone can provide a link to that, I would be grateful.

18. Sep 5, 2017

### sophiecentaur

Some really appropriate graphics would be really helpful. I have in my head a sort of family of coloured, caustic curves and there's too much information in that picture to put it into 2D printing. I did find Walter Lewin's description of what goes on in the simple case very good. I bet some interactive graphic with sliders and a rotatable droplet could help. But, to be frank, I have got a lot out of this thread. Largely thanks to the patience of @JeffJo.
No wonder they stick to the simple diagrams for school kids.

19. Sep 5, 2017

### Staff: Mentor

It is deliciously complex. Especially considering cases of non spherical droplets (twinned rainbows)., supernumerary rainbows, and reflection rainbows all mentioned in the Insights article. It is enough to melt anyone's brain.

20. Sep 8, 2017

### JeffJo

21. Sep 8, 2017

### sophiecentaur

Re the 'Bows' link: Dew Bows !!!! Oh boy, I must get to see some of them. They would not be difficult to simulate with a garden spray and some nice long grass.
Some other good little pics on that page too.

22. Sep 11, 2017

### sophiecentaur

I'm afraid it is a lot more complicated than that. Look at all the links quoted in this thread.

23. Sep 11, 2017

### Staff: Mentor

No, not when the reflections are from a curved surface. Please re-read the earlier posts to learn more about what is going on. Thank you.

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