Driving Through a Rainbow: The Truth Behind This Phenomenon

[A.T.]

There is only one direction for one wavelength. That is why the bow appears at all.

That's completely backwards. If all rays for one wavelength that hit your eye were parallel, you would not see a bow, bur rather a point in the distance.

Are you saying that you cannot justify your statement(s) in that quoted post?

Quote my statement, ask a question about it or explain how you understand it and give your objection to it.

 

[sophiecentaur]

This is tiresome. All you have to do is to justify (with a proper argument) what you wrote in the bold parts of the quote in post 29. I now know you are just being awkward. You understand perfectly well what I want. Please don't bother to respond until you have done that. Also, please respond to my points about parallax.
I couldn't make myself any clearer about what's needed. Are you, in fact, not able to respond with some Physics? Is there a human on the other end of this dialogue or just a annoybot?

 

[A.T.]

You understand perfectly well what I want.

Not really. You fail to point out what you think is wrong with my statements, or how you understand (or misunderstand) them. When I have a problem with something you said, I try to clarify it. For example:

That means that the 'red' rays (of around the same wavelength) are all parallel.

What exactly do you mean by "all" here?
- All 'red' rays that exist in the area?
- All 'red' rays that leave a single water droplet?
- All 'red' rays that hit the eye?
- Something else? (If yes, then what?)

 

[sophiecentaur]

I'm sure you know that I mean your statement that the rainbow 'is' where the drops are, is wrong. Have you not gleaned this from what I have already written several times? I have told you a good reason for saying that the image must be at infinity because of PARALLAX (that word you won't acknowledge). I am asking you for the grounds to justify that bit in bold, in the earlier post. You haven't given any valid reason for your statement. I can understand that you 'feel' it to be correct but that really isn't enough. If it is really that obvious, then you should be able to produce a diagram (which would happen to be at odds with all the others you can see on the Web) to show your reasoning.

Red rays of the same wavelength (say 800nm) are deviated exactly the same amount by all water drops. That means that they will all be parallel to each other. But you only needed to have read my post no.26, which says precisely that. Your eye, when in a given position, will receive 500nm light from just a few drops that are very close to each other along a radius of this cone. If you move your eye, you will pick up a different bunch of rays from a different group of drops. In both cases, the light will have been deviated exactly the same amount (All the 800nm light is deviated by All the drops in the cloud and by the same amount)
I am sure you will find an 'I' that I haven't dotted or a 'T' that I have not crossed in that statement but just make an effort with it and answer the substance.

The upshot of all that is that, as the angle doesn't change as you move your head, the direction of arrival (i.e. the direction in which the image of the bow appears to be) doesn't appear to change. Tell me how this could apply to a nearby object.

I ask again: do you know what parallax is and what it implies about our appreciation of distances in a scene? You can always read about it, you know. It's all over the Web.

 

[A.T.]

your statement that the rainbow 'is' where the drops are, is wrong.

That's just a semantical issue. When explaining the phenomenon that the OP described (apparently approaching the base of the rainbow) it is practical to consider the intersection of the volume that currently reflects light into the eye with the ground and other obstacles. If you don't like calling that volume the "rainbow" as I did for simplicity, that is fine with me. We can call it something else.

I have told you a good reason for saying that the image ...

You were not even able to say how you technically define "image" here. You didn't like the http://en.wikipedia.org/wiki/Virtual_image" but provided no alternative reference nor a diagram showing how it applies to a rainbow.

I am asking you for the grounds to justify that bit in bold, in the earlier post.

See above. The grounds is the intersection of the volume that currently reflects light into the eye with the ground and other obstacles.

Red rays of the same wavelength (say 800nm) are deviated exactly the same amount by all water drops.That means that they will all be parallel to each other.

No, that's not what it means. "Deviated exactly the same amount" means that they all have the same angle to the incoming sun rays. So for each drop they form a cone with the drop at the apex, that opens towards the sun. Rays forming a cone are not all parallel to each other.

 

[sophiecentaur]

Semantic? It's what you wrote and, if it were just a semantic issue, why didn't you mention it ten years ago?
I kept saying that it was other visual clues that 'told you' the rainbow was in the cloud. Now you seem to be agreeing. Why not agree earlier?
An Image is something you see. If its position can be identified rigorously (e.g. by parallax haha) then that's the Image Position - just like in a mirror. In the case of a rainbow, this will be at infinity (in the absence of other 'clues').
There was no point in suggesting an article other than that wiki one. It did not define an image, in any case. It was a simple set of ray diagrams for a couple of lenses, such as one draws in school.

If you ask me to draw a diagram showing how an image in a mirror appears where it does, I could do it. If you ask for one to show how and where a concave or convex lens produces an image, I could do it. I could even give an explanation of how a hologram works, with diagrams and how an image can be seen. If you shine a lamp on a piece of paper, I can also show how the paper appears to be where it is. No need for me to do this because the web is full of such diagrams. You have seen some. All I am asking is that you should be able to draw an equivalent, valid diagram to show how we 'see' a rainbow as being in a nearby rain shower / spray if we can't actually identify the position of the drops first. I don't need to see a mirror in order to identify where the reflected image is. Are you really determined not to acknowledge the existence of Parallax?

You are very choosy about which words of mine that you read (or remember). If you read carefully you will see that I referred to all drops on a radius from "that cone" deviate the 800nm rays by the same amount. This happens to be is true. Why state the obvious about conical rays rather than parsing my sentences with more care? I think you need to get the geometry of the situation straight - and then deliver a crushing blow to my ideas by producing that diagram. Without a diagram, there is no argument to support your view; it's just whimsy.
You can't 'gainsay' parallax. I don't think you can even say it.

 

[A.T.]

Semantic? It's what you wrote and, if it were just a semantic issue, why didn't you mention it ten years ago?

I described the volume that currently reflects light into the eye in post #3 already.

I kept saying that it was other visual clues that 'told you' the rainbow was in the cloud. Now you seem to be agreeing. Why not agree earlier?

If by "other visual clues" you mean obstacles (like the ground) intersecting the cone, see my post #5

An Image is something you see.

That is just vague gibberish. The "image of an object" as used in optics is a well defined geometrical construct. I don't see how the rainbow is a image of the sun. The sun could be a cube, and the rainbow would still be round.

No need for me to do this because the web is full of such diagrams.

Can then you post a link to a diagram that specifically shows how the rainbow is an image of the sun?

If you read carefully you will see that I referred to all drops on a radius from "that cone" deviate the 800nm rays by the same amount.

How is that relevant? You don't see the light coming from all the drops on a radius of the cone. You just see the light coming from all the drops on the surface of the cone. And those rays are not parallel.

deliver a crushing blow to my ideas

I don't even know what exactly your idea is, because you just throw around with red herrings, instead of explaining it. But given that you eventually manage to show how the rainbow is an image of the sun at infinity: How does this actually help to address the OP question, about the visual effect of approaching the rainbow?

 

[sophiecentaur]

How would you define an image? Would it have to be a 'perfect' image, such as you would see through some high quality binoculars or would you use the word Image to describe the blurred blobs that you see when you magnify the moons of Jupiter, seen through a cheap, Russian telescope? Does the Chromatic aberration in a poor camera lens make the result 'not an image'?
How bad does it have to be before you would insist that what you see is not an image?.
When you see a rainbow (oh God, I can see this is a total waste of time already), why are you not seeing a reflected image of the Sun with enormous chromatic aberration. The sunlight has passed through a refractive and dispersive medium to produce a visible pattern that relates to its light source. How does this differ, except in the actual extent of aberration, from a good image of the sun that you could see through a telescope with a filter in front of it? (No health and safety issues please) Both are 'images'.

What is "relevant" in my statement of all drops laying on a radius of a cone is that, if you move your eye a bit to the left, you intercept a different set of rays from a different set of drops. BUT the rays in the second case are parallel with the rays in the first case. They appear to come from the same direction.
Now, I feel pretty hamstrung at this point because you seem totally blind to the word 'parallax' (how can that be? Does it frighten you or are you just making a point of ignoring the one really appropriate word that clinches my argument). Anyway, I'll press on. When you move your eye, you see a new set of 800nm (along an arc) that comes from exactly the same direction as before. This even happens in a garden spray. How do you and your brain determine where this 'pattern' is? If there were a rain shower that extended from 50m to 5km in front of you, where would you assess the rainbow to 'be'? The only way to tell could be by assuming it was on the ground in front of you (i.e. as distant as possible from you but your brain would have to discount the possibility of it being underground - extra visual clues) But, for the parts of the bow that are above the skyline, you can only deduce that the position of this thing that you can see (or 'image') must be at a great distance (infinity). OR that it moves with you, as if connected on a frame, I suppose.

When you say "the surface of the cone", I assume that you mean all the rays from 'front to back' of the rain shower in a particular direction. "All the drops" - your words. So where do you place the rainbow, in space? At the front and the back and everywhere in between? I don't think your brain could cope. But your arch enemy Parallax comes to your rescue. You use binocular vision or move your head and that tells you that what you are seeing is at infinity.

You ask for a diagram showing how the rainbow is an image of the sun. I can describe a similar situation, without a diagram, that doesn't involve chromatic aberration (dispersion). take a tiny fragment of mirror, some distance away and tilt it so that it shines the sun back into your eye. What you are seeing is an image of the sun. (Very degraded because the 'aperture' is small but still an image) Do the same with a lot of small mirrors. Depending on how you actually angle and place them all, you can make what is effectively a large corrugated, reflecting surface and produce a sum of all the sub-images: a new and perhaps distorted image of the Sun. That is still an image??
The raindrops are an equivalent to the mirrors but they happen to have dispersive properties and a peculiar geometry that introduces a more complicated rule to what happens to the light. Rings are produced, of different colours. If what is produced is not an image of the sun, then perhaps you could tell me at what stage in all of this, the light that you see stops being an image of the Sun.

 

[A.T.]

The raindrops are an equivalent to the mirrors but they happen to have dispersive properties and a peculiar geometry that introduces a more complicated rule to what happens to the light. Rings are produced, of different colours. If what is produced is not an image of the sun, then perhaps you could tell me at what stage in all of this, the light that you see stops being an image of the Sun.

Well, by that logic even a cloud is an "image of the sun": The water drops are just smaller and there is more of them so there is more dispersion, diffusion, more complex reflections. But if what is produced is not an image of the sun, then perhaps you could tell me at what water droplet size the light that you see stops being an image of the Sun.

 

[sophiecentaur]

@A_T
I really have to thank you for having made me think so hard about some of these matters. I know the OP wasn't concerned about what constitutes an Image and also that you have some very specific ideas about that. But you did not respond when I asked you where to draw the line. There is no "logic" that leads from what I said to saying a cloud could be an image of a sun, any more than a mirror is an image of my face. Rather than defending something to the last, I am actually after an answer to defining an Image. I think that, in order to include all images that you, A_T would have to acknowledge to be images, the definition would have to be something like: An image is an identifiable artifact of an optical system. So, when you look into a mirror (even a scratched and dirty one) you see (identify) an image of yourself that is not on the mirror itself. When you look through a hologram, you see an image which is definitely identifiable and which, again, is not 'on' the holographic plate. The same goes for good and bad images, seen with lens systems (both real and virtual): they seldom coincide with any of the actual lenses. Many of these images are deliberately placed at infinity, for comfort.
The essence of an Image is that it is identifiable and its apparent position is something that your eyes and brain need to assess. A rainbow ticks all these boxes. A cloud is not an image of the Sun, neither does it usually 'produce' an image of the Sun because there is nothing identifiable - just a generally white / grey pattern . You don't see a rainbow in a cloud, the image is too diffuse. You see a rainbow through Rain, which has 'appropriate' optical qualities. Clouds (water and ice) can produce identifiable 'images' in the form of halos and sundogs. Who would not call them images of the Sun?

Enough said on that topic. I hope you didn't perceive too many "red herrings" in all that.
I'm afraid that it was you who brought up the ultimate red herring in your fierce objection to my 'image of the Sun' remark.
What is more important is the position of this image. You do not seem to want to discuss this more important issue because you are not prepared to see how parallax is the basis of position placing. What is wrong with the term? I wish you would respond to that.

 

[Neveos]

Yeah, I've actually been in a number of long winded forum debates, and it is funny, but it is actually quite productive for a showdown to occur. It's the whole 'synthesis' from thesis vs. antithesis thing.

 

[sophiecentaur]

It can be entertaining reading when two guys get to talking as if the other is a complete idiot. The exasperation builds and politeness reduces. But, dammit, I don't bear a grudge. (Kicks passing cat.)

 

[sophiecentaur]

@ A_T
It's a shame that you never did provide an explanatory diagram that shows how an actual image is formed 'in' the water drops (i.e. show that, in the absence of any other visual information, that is where your eye/brain places it). I shall just have to accept that the parallax argument (which needs no diagram because it's so well known) must be right.

 

[A.T.]

Clouds (water and ice) can produce identifiable 'images' in the form of halos and sundogs. Who would not call them images of the Sun?

I personally wouldn't call halos "identifiable images of the Sun", because I cannot identify the sun in them. There is no clear correspondence between points of the object, and points of the image here. Otherwise, how do you define "identifiable". The transition to a diffuse cloud is smooth here.

What is more important is the position of this image. You do not seem to want to discuss this more important issue because you are not prepared to see how parallax is the basis of position placing.

I agree that when you move laterally, the rainbow behaves like it was an object at infinity. But is not the situation the Neveos was interested in (driving towards the base of the rainbow). It also doesn't help to explain the effects the Neveos was asking (approaching the base).

It's a shame that you never did provide an explanatory diagram that shows how an actual image is formed 'in' the water drops

I explained that this real image is not relevant for seeing the rainbow in the very same post I first mentioned it.

(i.e. show that, in the absence of any other visual information,)

Where did I ever say, that the effect of approaching the base of the rainbow works without obstacles? All my explanations (post #5, #7) refer to objects intersecting with the cone of the rainbow.PS: Next rainbow debate here:
https://www.physicsforums.com/showthread.php?t=534964

 

[sophiecentaur]

So, I repeat my question as to how aberrated can an image be before you would stop calling it an image. A halo or sundog are there because of the Sun. If the Sun is moved or removed, they will move or disappear too. Just like a rainbow. In the case of a rainbow, you could even point behind you accurately, to the position of the Sun, from the position of the bow. if they are not images by your estimation then how about the image of the Sun behind a thick cloud?
I really think you need to think this through better.

As to the actual image, I can't imagine what you mean with comments (some while ago), like

"Yes, assuming a spherical raindrop the real image for the reflection on the inner surface is in the raindrop. (I had a typo in my previous post, calling it "virtual"). But the key point I was making is, that there is no distinct virtual image here."

If you're not prepared to explain this with a diagram then it's meaningless. Where is a "real image" how does it form? Can you even do the correct diagram to show the formation of an image in a plane mirror?

Also your comments about "approaching the base" really do beg the question. They assume that the base really is somewhere you can reach. Any 'rainbow' that appears to be within arm's reach will appear that way because there is something very close, like a fence or bush. This will only work for a garden spray, which is not a good representation of a true rainbow - because the drops are not rain drops: more of a visible mist. More like a cloud than rain. You don't see the drops on a rain shower so you really have to rely on other objects

You say "But if there is only a stripe of rainy area, the rainbow is just a slice of a cone: an actual arc. When you approach the stripe of rain, you will see the near boundary of the arc on the ground coming closer (because it actually is). So even though the arc is not a fixed object, you can get closer to it, and drive trough its base.
"
This thing is conical, if you are are driving "through the base" then you would also have to be getting closer and closer to the other base where the other side of the bow is, too, because of symmetry. (You have to be on the axis of the cone at all times.) You drive at the base and it just moves to one side to avoid you. No chance of a pot of gold, I'm afraid, except in the case of your garden spray, when you could stretch your arm and touch the ground with your head still on the axis of the cone.

You have to look at more of the many diagrams about rainbow formation.

Each band reaches your eye from the whole depth of the rain shower, not the nearest bit. So where do you place it in your brain?

 

[Martha]

One direction for one wavelength.

What if a color has no single wavelength i.e., Magenta?

http://www.atmos.ucla.edu/~fovell/AS3/theory_of_color.html

What is the direction?

 

[sophiecentaur]

Is there a magenta ring in a rainbow? Perhaps in your world but not in mine.
There are other phenomena, such as oil films , that produice a different set of colours by interference but I think that would be better discussed in another thread. The geometry is very different.

 

[A.T.]

So, I repeat my question as to how aberrated can an image be before you would stop calling it an image.

That was actually my question to you: How do you define an "image of an object" objectively to include a halo, but not include a diffuse cloud? By "objectively" I mean based on in mathematical criteria, not subjective criteria like: "an image is something you can see / identify".

Where is a "real image" how does it form?

It's irrelevant for seeing in the rainbow as I said many times. I also posted the link already:
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/mirray.html#c3

You say "But if there is only a stripe of rainy area, the rainbow is just a slice of a cone: an actual arc. When you approach the stripe of rain, you will see the near boundary of the arc on the ground coming closer (because it actually is). So even though the arc is not a fixed object, you can get closer to it, and drive trough its base.
"
This thing is conical, if you are are driving "through the base" then you would also have to be getting closer and closer to the other base where the other side of the bow is, too, because of symmetry. (You have to be on the axis of the cone at all times.)

The rain boundary can be at some arbitrary angle to the cone axis, which breaks the symmetry. You eventually don't see the other base at all, until you get into the rain.

You drive at the base and it just moves to one side to avoid you.

If you keep the 42° angle to the sun it will not move to one side.

 

[sophiecentaur]

In order to see a bow or part of it, there need to be some drops present somewhere on this cone so that some deviared light can get to your eyes.
You can't "keep the 42degrees angle to the sun". The cone moves with you. Try drawing it out. Do you have a pencil and paper?

All this image definition stuff is irrelevant. A rainbow is an image and not an object. It is only there because of the sun and its position. I choose to refer to it as an image of the Sun for those reasons. Who's image it us doesn't affect its position.

 

[A.T.]

You can't "keep the 42degrees angle to the sun".

I meant: Move at 42° to the sun rays, which are almost horizontal. I see nothing that would stop me from doing so.

 

[DrZoidberg]

What determines the apparent distance of the rainbow?

 

[Martha]

SophieCentaur said:

"Is there a magenta ring in a rainbow? Perhaps in your world but not in mine."

Yes, apparently we are from different worlds.

Because in my world magenta (and purple) are very important and they ARE present in a rainbow.

"A warm magenta is barely visible on top of the red on a strong rainbow, it's stronger at the cool bottom of the rainbow. Also, yellow will appear again under the magenta on a strong rainbow."

http://realcolorwheel.com/rainbow.htm

If you need "visual" proof:

From “The mathematical colors of the rainbow using HSL” (Hue-Saturation-Luminosity)

Under these words, “A slice was taken through the original image and rotated. The slice at 400%”

See an actual picture of a rainbow in the link that follows

and note the explanation beneath which says: “Of interest in the image is the clear existence of colors noted above, including the equiangular colors such as cyan.

Leaving out "blends" such as red-orange and yellow-orange,

one can identify the presence of red, orange, yellow, green, cyan, blue,

purple, and magenta tones”


http://www.comfsm.fm/~dleeling/cis/hsl_rainbow.html

BTW...Pohnpei is an island. It is one of FOUR of the states of the Federated States of Micronesia. The picture of the rainbow was taken there.

Note the above link indicates a mathematically proposed eight color rainbow. ROYGCB

plus purple and magenta.

Given there are 8 particles in gluons that transmit the strong nuclear force from quark to quark containing 2 color charges, this number of "colors"/ "color particles" maybe indeed accurate.

"There are eight remaining *independent color states*, which correspond to the "eight types" or "eight colors" of gluons." Wikipedia

“Gluons differ from each other only in color, usually expressed as R, G, B, anti-R, anti-G, and anti-B. One would think that three by three possibilities would create nine gluons, but the math of the theory rules on combination out.”

http://washparkprophet.blogspot.com/2010/06/gluon-mass-qcd-developments-and-more.html

Yes, the author is not a scientist; he is a lawyer. Like me, he apparently has a tremendous interest in physics too.

"Color-octet scalars naturally appear in grand unified theories."

"The role of the ***color-octet mechanism*** in hadronic production of quarkonium (a flavorless meson whose constituents are a quark and its own antiquark)."

Leaving 2 "colors" out (magenta and purple) when discussing the math/geometry of a rainbow (which is actually circular when viewed from above) is a mistake, IMO.

Especially magenta which functions as a primary subtractive "color" and secondary additive color. It is very unique.

But I do "get it" i.e.,

"...digital imaging electronics vary greatly from that of the human eye."


Just because you can't see it, doesn't mean it isn't there and should thus not be factored in.

I am going to take this discussion to a different level for those who are interested/curious about a much a "bigger picture".

For those only interested in physics i.e., a "micro" view, and not the part it plays in the UNIVERSE, a "macro" view, you might want to stop reading now.

Just as a rainbow is "visual proof" of an endless existence (the rain WILL stop and sunlight will reappear), there is also "visual proof" in black holes which appear to look like laminin i.e., the "glue" that holds our cells together (a protein with alpha, beta and gamma chains).

This is an actual PHOTO from NASA:

An image of the core of the Whirlpool galaxy M51 (NGC 5149) taken by the Hubble Space Telescope. It shows an immense ring of dust and gas that is thought to surround and

hide a giant black hole in the center of the galaxy.

http://www.scienceclarified.com/Bi-Ca/Black-Hole.html

And yes, I am aware that black holes feed off of nearby stars and that black holes themselves create new stars...some of which ultimately form their own black holes.

And I am aware of the balance of clockwise and counter clockwise galaxies in our "observed universe". It is quite beautiful.

While you may chose to look this way (> (and that's okay and needed!)...

I'm trying to see this: (><) i.e. the micro and the macro view. How did we get from ONE "particle" to an entire never-ending universe?

I was studying light and sound (to heal) when I read in Discover magazine that Kent State math students (via massive computer help) had found "God's algorthim" to be 20. In other words, in 20 "moves" He created order out of chaos. This "number" is NOT God, it was His solution. That is astounding given the number of possibilities.

Using a "clock" drawing (time), I began to correlate numbers with their corresponding colors and charges and even to their corresponding elements. What I ended up with is a pattern that even surprised me. It directly matches an ancient Chinese drawing to explain "evolution".

Do you know 432Hz = the "frequency of light"? Square it. It correlates to "perfect A" (musically).

Do you know that SSS is the signal for a "Proton Flare"? Do you know what it represents translated to Morse Code? ... ... ...

Do you know that a tiny pulse of blue mimicks a computer's logic when it *carries* digital "ONE"?

Do you know in the Phoenician alphabet #1 = aleph which represents an "ox" and the last letter (#23) is Tau/tav and it represents a Tree? In olden times we used an ox to plow the earth. In doing so, we kept our eye on a distant Tree in order to plow a straight line.

There is so much significance/meaning to symbols that ancient civilization have left for us!

Another example:

^ = alpha/heaven/man
v = omega/earth/woman (think virgin)

When you use Greek symbols, think about their significance and be aware of the various forms they take.

Have you seen Carl Sagan explain the FOURth dimension (on YouTube)? It is FASCINATING. I wonder if his choice to use an apple was intentional? ;-)

There are two quotes that have driven my research and my attempt to understand how this all could be possible.

1. Gaudi (architect whose designs mimicked nature and are incredibly strong) said: "The straight line belongs to man. The curved line belongs to God."

2. "For behold, My imaged universe is ***mirrored*** to infinity;
it is repeated to the endless end;
yet there are but multiples of three in all My universe.
And again I say to thee,
***two of those very three are naught ***but My imaginings,

for My Trinity is but One.

(TSOL p. 138)

(TSOL refers to the “Secret of Light” by Walter Russell)

I DO believe in God and I DO believe in an Endless Universe. I DO believe that evolution/we DID have an "origin" and that He was/is the origin.

However, I also believe that evolution -> steady state universe.

It is very possible that our "Endless Universe" is that suggested by Paul Steinhardt i.e., two interconnected Klein bottle (shapes). Paul Steinhardt is a professor of physics at Princeton. I have his book on order.

It is my hope that someday science will prove religion is truth, not fiction. I hope that someday everyone will know that indeed God does exist and not just take His existence on "faith" alone.

Find the "God particle". It IS there and it IS within us.

Years ago, I was a maternity nurse in labor and delivery. Birth, life, is awesome.

Okay...I've probably overstayed my welcome in your world SophieCentaur. Bye.

 

[sophiecentaur]

A long and colourful reply! I LURVE Gaudi. I could end my days in Barcelona with pleasure.
Yes, the electronic analysis of light is a poor substitute for the real thing but, still, colour TV does a pretty damned good job,considering.

The "Magenta" issue.
This is a language problem. I thought that you were referring to the 'accepted' term Magenta. Magenta is not a spectral colour. The 'purple-ish" colours which you can see in a rainbow are spectral Indigo and Violet. Magenta is a specific, saturated, colour that is a mixture of Red and `blue primaries. In RGB terms, it is R=1,G=0,B=1. In colour-subtractive systems, magenta is 'Minus Green' There is a lot of confusion about 'formal' colourimetry terms and this particular region of colour is particularly confusing because it involves mainly the short wavelength receptors of the eye and is, subjectively, a bit uncertain. (When did anyone actually see indigo on TV?)

 

[sophiecentaur]

I meant: Move at 42° to the sun rays, which are almost horizontal. I see nothing that would stop me from doing so.

You can move to the left or to the right. BUT, the line from the Sun, through your eye, will also shift. That moves the whole cone at the same time. You will still see the rainbow at the same angle to that new line. i.e the cone moves to the left or right by just the same amount as you moved.

Is this not the most elementary bit of geometry you could ever think of?

 

[Integral]

What determines the apparent distance of the rainbow?

A combination of the number of raindrops and the intensity of the sun. Once while trying to drive in a very intense thundershower I observed the end of the rainbow to be on my hood.

The amount of light from each raindrop should be pretty constant, so the more raindrops in the cone, the more light returned. Where the rainbow appears along the cone will be determined by the amount of light needed by the eye to resolve the colors. Anytime light changes direction your brain assumes that the source of the light lies somewhere along the projection of a line from your eye to the bend (read raindrop).

 

[sophiecentaur]

It's interesting (but not surprising, I suppose, because it correlates with the weather) that 'strange' experiences with rainbows seem always to be associated with driving. If the observer is actually driving at the time then the rainbow is (hopefully) not getting the full attention of that driver. This may add to the confusion as to what is actually being seen. Also, the drops / smears on the windscreen can affect what is obesrved.

The straightforward model of a rainbow is that light of one colour that is seen by the observer is produced at the same angle from all drops (near or distant). If the observer moves from side to side (or used binocular vision) this must be interpreted as if the image is at infinity. On the face of it, there is no way of locating a 'position' for the bow as being other than at infinity, if that's the only evidence available to the observer. This is not a very satisfactory conclusion for the brain to come to; we want things to be in a more definite position so our brains look for an improved location to place it in our perception space.
A rainbow is a very fuzzy object, for a start, so other clues are grabbed at. We are not disturbed by the minuscule de-focussing effect because we are focussed at nearer objects and so we 'place it' near us. We are not used to transparent objects (during our evolution, particularly) so, whenever we see a solid object in line with part of the rainbow (hill / wall / road / car bonnet) we 'reason' that the rainbow image must be in front of that object - even though there is strong parallax information which tells us otherwise. We just say that the rainbow moves. Contrast this with our assessment of what we see when observing distant mountains. We see exactly the same apparent movement but we don't believe they actually move as we drive through the countryside.

In the case of a garden spray, the spray itself is very visible and so we, again, are likely to 'see' the bow as being right next to us.

One thing that is pretty unarguable, however, is the direction that the bow appears with respect to the line of the Sun through our eye. Whatever else people may imagine they have seen, they cannot have moved to one side or the other of the rainbow, because that 'cone' we talk of, moves to prevent it. It always stays symmetrical, whether or not there is full or partial cover of the visual field by suitable rain drops and is locked to that axial line.

@A.T.
I think the following definition of an Image, one of several, taken from Websters, is reasonable:
"the optical counterpart of an object produced by an optical device (as a lens or mirror) or an electronic device."
(Nothing is implied here about quality or position.)

 

[A.T.]

You can move to the left or to the right. BUT, the line from the Sun, through your eye, will also shift. That moves the whole cone at the same time. You will still see the rainbow at the same angle to that new line. i.e the cone moves to the left or right by just the same amount as you moved.

Is this not the most elementary bit of geometry you could ever think of?

Sure, and what is your point here again?

 

[sophiecentaur]

My point is that, contrary to what you have written, you can't drive to the base of a rainbow. It isn't there.

 

[A.T.]

My point is that, contrary to what you have written, you can't drive to the base of a rainbow. It isn't there.

By "driving trough it" I meant driving into the rain, so you don't see the near boundary of the arc on the ground coming towards you anymore. This will create an effect close to what Neveos described.

But I thought you were trying to explain this here:

You drive at the base and it just moves to one side to avoid you.

I still don't know why this would happen, if the sun is quite low over the horizon, and you drive directly towards the arcs base (holding 42° to the anti-solar point).

 

[sophiecentaur]

So, when you steer (left, say) towards the 'base' (that's where the arc ends at the ground?), do you not realize that the 'base' just moves over to the left so that it remains at the same angle to the axis of the cone because the axis is still a line from the Sun through your head. You don't just leave the original axis behind. It's bolted to you.

As I have frequently written. You need to draw this out if you want to understand it. Your mental picture is just flawed.