Driving Through a Rainbow: The Truth Behind This Phenomenon

[Mk]

Is it physically possible to drive through a rainbow? Why or why not?

 

[Mk]

From the Wiki:

A rainbow does not actually exist at a location in the sky, but is an optical phenomenon whose apparent position depends on the observer's location. All raindrops refract and reflect the sunlight in the same way, but only the light from some raindrops reaches the observer's eye. These raindrops are perceived to constitute the rainbow by that observer. Its position is always in the opposite direction of the sun with respect to the observer, and the interior is slightly brighter than the exterior. The bow is centred on the shadow of the observer's head, or more exactly at the antisolar point (which is below the horizon during the daytime), appearing at an angle of approximately 40°–42° to the line between the observer's head and its shadow (this means that if the sun is higher than 42° the rainbow is below the horizon and cannot be seen unless the observer is at the top of a mountain or a similar vantage point).

 

[A.T.]

As we got closer, we realized that we might actually approach, what appeared to be, the end of the rainbow.

When the rain is uniform, the rainbow that you see is a cone with your eye in the apex. So you will never have the impression of approaching the rainbow, because it always extends from your eye to very far away.

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.

The interior of the car became very bright, and it did, in fact, change colors all around us. I vividly remember my arm lighting up as though it were being shined upon by a sequence of colored lights.

When you get into light rain that sprays your windows with tiny water drops, with the sun low, you will see a lot of nice color effects. (However, this can be annoying when you are trying to land a glider, which don't have wipers.)

 

[Neveos]

When you get into light rain that sprays your windows with tiny water drops, with the sun low, you will see a lot of nice color effects. (However, this can be annoying when you are trying to land a glider, which don't have wipers.)

It was a bit more distinct than this to my memory, however I do not discount the possibility of wishful thinking playing a role in what I saw. I'm not willing to discard what I vividly saw, which was the colorization of the interior. There are actually a lot of videos surfacing on youtube of people getting close to the foot of rainbows, so it seems to have happened enough times to accrue a number of eyewitness accounts. A lot of other people are saying they witnessed a change in colors as well. However, from the videos, it does appear that it is less manifestation than I thought, so I am inclined to believe that I am wrong. However, being able to "pass" and "approach" a rainbow, to the degree that one is capable of is quite different than one would expect from messing with standard rainbows made in a mist.

 

[A.T.]

There are actually a lot of videos surfacing on youtube

Link some that are similar to what you saw.

of people getting close to the foot of rainbows,

That is no problem as I decribed above. In fact, once you drive into the rain, the foot of the rainbow (intersection of the cone and the ground) starts right outside of your car.

 

[sophiecentaur]

The above statements about the Physics of how a rainbow forms are correct, of course. A rainbow is a virtual image that is not, actually, anywhere . Mostly (when in the sky) it will appear to be at infinity. But it can often appear in front of a distant hill (your brain tells you that it can't be behind the hill!) or even in the grass of a nearby field.

If you spray a garden hose near you and the Sun is strong, you can produce a 'rainbow' which will show up vividly against a nearby dark surface. It is even possible to place your hand (which is not where you eye is) so that it appears to be 'in' the rainbow. To get this effect you really need things to be just right. You can also get rainbow-like effects from smears on a windscreen, too, with the colours appearing to be quite close.

However 'clearly' one remembers a really stunning rainbow, I think that the memory of your exact visual experience will very easily become distorted. A particularly vivid rainbow is a pretty overwhelming experience and I wouldn't necessarily trust even myself to make a highly accurate witness statement. Without a photograph of one of those experiences, I think you can't be absolutely sure of what you saw.

 

[A.T.]

A rainbow is a virtual image that is not, actually, anywhere.

The water drops that send light of a certain color to your eye have a well defined position. They form a cone surface with your eye in the apex, and an axis patallel to the sun rays.

Mostly (when in the sky) it will appear to be at infinity.

Not really. It will appear where the water drops are. It can extend from your position to a few kilometers away.

But it can often appear in front of a distant hill

It doesn't just appear to be in front of a distant hill. It is in front of it, if the rainy area is closer than the hill.

(your brain tells you that it can't be behind the hill!)

Whether it extends beyond the hill depends on whether it also rains behind the hill. So the brain might sometimes be right here.

 

[sophiecentaur]

The mirror that you gaze at yourself in also has a defined position, on the wall. That doesn't mean that your (virtual) image appears in that position. Is your image in a mirror 'in front' of the wall that the mirror is attached to? The mirror is not a projector screen and neither is a water droplet.

The only absolute position that you could justifiably assign to a rainbow would be the position of the Sun, from which the light originates. You said, yourself, that the 'rays come , via the water drops, into your eye; they start diverging when they leave the Sun and they keep diverging, putting their apparent source a long way away. The rays are never 'focused' onto a point, which they would have to be, for a real image to be produced at a certain point.

Your brain tries to make what it can out of the light arriving at the eye. Optical illusions occur all over the place. The point of the scientific approach is to 'see' past these illusions and to try to explain them. To be successful, you need a certain amount of rigour. Note (Google) the difference between real and virtual images.

I wonder whether you have, in your mind some idea that the water droplets behave as if they had some sort of pigment in them. That would account for your idea that the rainbow is actually 'located' somewhere. But remember, as you move about, a droplet that was sending you red light could be a droplet that is now sending you yellow light.

 

[A.T.]

The mirror that you gaze at yourself in also has a defined position, on the wall. That doesn't mean that your (virtual) image appears in that position.

The term "virtual image" makes no sense in the case of a rainbow, because there is no focused virtual image of the sun, like in the case of a simple mirror.

The mirror is not a projector screen and neither is a water droplet.

And neither is a water droplet a simple mirror. And a single water droplet doesn't produce a rainbow, the collective of many water droplets does. And within that collective there is a subset for each each color which forms a cone. It's that cone determines where you see the rainbow, not some "virtual image" of the sun.

 

[sophiecentaur]

As you seem so sure of this, perhaps you would draw a simple ray diagram to show how this image of a rainbow forms on each water drop.
There is no focussing involved in order to see a virtual image.
Come on, this is all simple schoolboy stuff.

 

[A.T.]

There is no focussing involved in order to see a virtual image.

There is focussing involved in seeing anything. That is why your eyes have lenses.

- In the case of a single flat mirror the lenses are focused on the virtual image.

- In the case of a rainbow there is no distinct virtual image. Each water droplet produces a real image of the sun (for each color), within the droplet itself. But that image is quite irrelevant seeing the rainbow.

 

[sophiecentaur]

Of course your eyes focus becaus they form a real image of anything you see.
My issue is that you are claiming that there is an image 'in' the raindrops and that there is focussing involved in the formation of a virtual image there. This is nonsense which you will discover (using schoolboy ray tracing) when you try to draw a diagram of what goes on. When you have failed, take a look on Wiki or any of the other explanations you can find.

 

[xts]

You may make an easy experiment (hurry! Autumn is coming!): on a sunny day make a shower from garden sprinkler. You see mini-rainbow in it (it seems to be within a shower of droplets). Then move your head - you'll see that the rainbow moves accordingly - now your brain (trained to interprete perspective) changes the interpretation of the picture: you now see the rainbow very far away, and the shower is just a window, allowing you to watch it. Your brain may be only a bit confused by the shrubs in the background, as the perspective is such that rainbow is in bigger distance than shrubs, but shrubs do not block the view. That is something confusing to your brain, which is used to transparent windows and transparent showers, but not to transparent shrubs.

EDIT>
Ouch - apologies for skipping over some previous posts... Mea culpa!

 

[sophiecentaur]

Been there. Did it. See my earlier post. The illusion of the actual 'position' of that image can soon be resolved. The parallax as you move your head from side to side will show you that the image is, in fact, at the Sun' s distance. You are always at the centre of that cone. If the image really was where you say it is, the cone would no longer have its axis through your eye.
As I said earlier, draw the diagram, rigorously, and don't rely on what you think you see as 'evidence'.

 

[A.T.]

My issue is that you are claiming that there is an image 'in' the raindrops

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.

This is nonsense which you will discover (using schoolboy ray tracing) when you try to draw a diagram of what goes on. When you have failed, take a look on Wiki or any of the other explanations you can find.

What is nonsense, is your claim that a virtual image is somehow produced here. A concave mirror cannot produce a virtual image of an object which is placed outside of the mirrors focal length.

 

[sophiecentaur]

I can only refer to this and hundreds more sources.
http://rebeccapaton.net/rainbows/tir.htm"

This is how it is generally understood to work. Can you point our where, in the raindrop diagram, there is any focussing of a real image that you could possibly see? Rays of different wavelengths may cross over in the drop but your eye only intercepts a ray of one wavelength (say, red) from each drop. The guy next to you may see another ray (say yellow) from the same drop but that's not the yellow part of the image that you see.
As I keep saying, we all learned this in School and it's still right. (Most diagrams ignore the apparent focusing inside the drop because it is not relevant to what you experience.)
The problem with your interpretation of the effect is that it implies you can sit 'in' the rainbow. Why do you think the daft story about a pot of gold came from, if not because you can't actually get there?

Have you an answer for my 'parallax' objection to your idea about where the rainbow actually is situated? If you move your head from side to side and an image always moves as if it's a long way away then that image is a long way away. Wherever the drops happen to be, this is the effect that you see. I claim that, without any other visual clues, you could have no opinion whether the rainbow was formed by a nearby cloud or by one that is miles away.

btw, I don't claim that a virtual image is formed in the drop. Where did you deduce that from?

 

[A.T.]

Each water droplet produces a real image of the sun (for each color), within the droplet itself. But that image is quite irrelevant seeing the rainbow.

Can you point our where, in the raindrop diagram, there is any focussing of a real image that you could possibly see?

I made it pretty clear that the real image inside the raindrop is not relevant for seeing the rainbow. The only reason I even mentioned that image is because you claimed that we see a virtual image.

I don't claim that a virtual image is formed in the drop.

There is no virtual image at all here. Concave mirrors don't create virtual images of objects that are outside of their focal length:

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/mirray.html#c3
http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/mirray.html#c4

Concave Mirror Image

If the object is outside the focal length, a concave mirror will form a real, inverted image.

If an object is placed inside the focal length of a concave mirror, and enlarged virtual and erect image will be formed behind the mirror.

The sun is hardly inside the focal length of a raindrop.

 

[sophiecentaur]

Where is there a "concave mirror" that is, in any way, relevant to this? If you say the focussing of some rays inside the drop is not relevant then why should you be considering the formation of an image of the Sun 'inside the drop"? I certainly am not.

The easiest way to look at this is that one ray, of just one wavelength, per drop (sloppy terminology if you like but it can suffice) comes to your eye. This arrives because of total internal reflection at the back surface of the drop. The curvature of this surface is quite irrelevant because we are only considering a tiny arc of the circle; we can treat it as a plane mirror, in the same way that one can construct the usual image formed by an ordinary curved mirror - by breaking it down into a polyhedron. Only one face of this polyhedron is relevant in each drop and for each wavelength.

My problem with reading your posts is that you bring up ideas and then say they are not relevant. It's very hard to follow because of that. What I would like is for you to point out just how the image you 'see' of the rainbow appears to be at or in the drops. Can you really believe that my parallax (or even the binocular effect) can place the image anywhere other than at a great distance? And are you really dismissing the effect of other visual clues in the way the image 'position' is determined by the brain? (Do you understand what I am saying about Parallax?)

 

[A.T.]

My problem with reading your posts is that you bring up ideas and then say they are not relevant.

You brought up the concept of "virtual images" which is not relevant here. I just pointed out that the only image here is inside the drop and I said that it is not relevant right in the same post.

 

[sophiecentaur]

I'm afraid the whole point about a rainbow is that it is virtual. The light does not come from a point that is in line with your vision. It comes from the Sun, which is behind you. It is precisely the same as an image in a mirror. You have not proved it's not.

The light of different wavelengths just comes from different directions - not different points. How is that anything but a virtual image? It is you who first introduced the idea of focussing and image formation. What exactly is your modified stance on this matter? You seem to keep shifting.
And where is your rebuttal for my Parallax point? It all hangs on that. You have not yet said how you could actually tell where this cloud is positioned, without other clues. In fact you really haven't answered any of my points, I think. So, how about it? How can you construct an "image" inside the drop (image of what??), bearing in mind that the rainbow covers several degrees of your vision and a drop covers a fraction of a second of arc?
I think you are having a problem with the common definitions that we use in optics.

 

[A.T.]

I'm afraid the whole point about a rainbow is that it is virtual.

The term "virtual image" that you originally used has a specific meaning which not applicable to a rainbow.

It is you who first introduced the idea of focussing and image formation.

Wrong. The first occurrence of the word "image" in this thread is in your post:

A rainbow is a virtual image...

 

[sophiecentaur]

Did I mention "focussing" to produce an image? I don't think so - because I do not agree that any focussing is involved. Why do you 'focus' on trivial points rather than getting down to the real issue?

OK. I think we would both describe an image as 'something you can see'. You can see a rainbow so it is an image. Have you a problem with that? So what kind of image is a rainbow?

There are two classes of image, in optics: real and virtual. To be a real image, light from each part of the image must radiate from an identifiable point in space: the focus of a lens or curved mirror. You could put your finger at a point in space, after the image former and project that image onto your finger. Can you do that with a rainbow? No. Your finger will just be illuminated with white Sunlight, wherever you put it.

For virtual images, the light only 'appears' to come from a point. It is possible to make a virtual image appear to be anywhere you like, with lenses or mirrors but you can't put your finger there and project the image onto it. As there is nowhere that you can put your finger and illuminate it (project) with any colour of light, for a rainbow, I hold that it is a virtual image.
The only question remaining is where this image 'appears to be'.
It's just the same as any other elementary problem in basic optics. To determine the position of a virtual image, one has to use a parallax method. (look it up if you don't know what I'm talking about). If the angle at which a particular wavelength is deviated is constant, then no amount of movement of our head (the basis of parallax methods) can change the angle deviated (the green band will always form a circular arc with the direct line from the Sun at the centre). There is only one place for an image that behaves that way. That is at Infinity.

Could you please respond to that lot (and to what I wrote about distracting visual clues) because that's the whole thing in a nutshell?

Show me any significant error in the above and you will deserve a cream bun!

 

[A.T.]

an image as 'something you can see'.

Too vague. Try this:
http://en.wikipedia.org/wiki/Virtual_image

Show how it applies to a rainbow with a diagram analogous to the one shown there.

 

[sophiecentaur]

I don't like the very first sentence. Rays from all objects diverge, for a start.
But that definition accords with my statement that the rainbow appears behind (miles behind) the drops. Your link doesn't discuss an image at infinity. Strange, because that is where we usually set the image when using binoculars.
You could try the "ray tracing" link on your link. It could help you to work out a proper answer instead of, yet again, picking small holes in my statements.

But if an image isn't something you can see, what is it? Does your link contradict that statement? (Or any others, for that matter) "Vague" doesn't imply 'wrong' in any case.

You are just putting off the task of showing, in simple terms, how the rainbow appears nearby when there are no other clues. Get back to me when you can do that. A suitable ray trace diagram should do it. I challenge you.

 

[A.T.]

"Vague" doesn't imply 'wrong' in any case.

It implies 'not even wrong'. Get back to me when you have drawn the diagram showing how a virtual image is formed for a rainbow.

 

[sophiecentaur]

I actually asked you first to show me a proper diagram showing your argument so you should produce your diagram first, I think: if you can. If you're really that convinced, it should be a mere trifle for you to achieve it.
Actually, any of the diagrams you can already see in dozens of links demonstrate what I say. They all show that light of the same wavelength is deflected at a constant angle. That means that the 'red' rays (of around the same wavelength) are all parallel. If that doesn't imply a spource at infinity then I don't know what does. Add to that my parallax argument and my version is pretty compelling. But I think that you can't understand what my parallax argument implies. You have not yet responded to that particular idea.

 

[A.T.]

I actually asked you first to show me a proper diagram showing your argument

You continuously ask me to to show some things, that I never claimed. If you have a question about something that I actually wrote, quote my statement, explain how you understand it and your objection to it. (Just as I do with your statements)

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

Every water droplet creates a cone of reflected 'red' rays, which are not all parallel. The red rays that form the red arc in the eye also obviously not parallel, because have to cross in the eye. So which red rays are "all parallel" again?

 

[sophiecentaur]

I do really wonder sometimes, you know (should I try the Turing test?). Yes, the different colours from each droplet diverge but the rays of the same colour from adjacent drops are parallel. This is just the same as the rays that come from a very distant object. For a nearby object / image, the angle changes as you move about (which is the basis of stereoscopic vision, rangefinders and birds of prey that move their heads from side to side). So the nearby drops (being an optical system) produce light that enters the eye which mimics what you'd get if you had a massive illuminated 'rainbow' picture situated at a great distance.

I could remind you that what I originally asked you for - the only important thing I have, in fact asked you for - was a diagram which shows how the rays of different wavelengths are produced and then travel towards your eye in such a way that the colours appear to be in or at least near to the drops. This would have to involve the angles changing for adjacent drops. Have you seen a description of what goes on inside a drop which suggest that this is possible? My explanation of parallax is a total clincher, as far as I'm concerned but you seem to be ignoring the word 'parallax' for some reason. Do you really not grasp the significance of it?

You wanted a quote from one of your posts: Part of Post 7:

"
The water drops that send light of a certain color to your eye have a well defined position (DIRECTION, ACTUALLY). They form a cone surface with your eye in the apex, and an axis patallel to the sun rays.
. . . . . .
It will appear where the water drops are. It can extend from your position to a few kilometers away.
.. . . .

It doesn't just appear to be in front of a distant hill. It is in front of it, if the rainy area is closer than the hill. "

That's what I wanted you to justify. I think you knew that, though. Please don't tell me it was another typo.

 

[A.T.]

the rays of the same colour from adjacent drops are parallel.

No. Not even the rays reflected by one single drop of the same color are all parallel. They form a cone with the drop in the apex and the line drop-sun as axis.

You wanted a quote from one of your posts: Part of Post 7:

"
The water drops that send light of a certain color to your eye have a well defined position (DIRECTION, ACTUALLY). They form a cone surface with your eye in the apex, and an axis patallel to the sun rays.
. . . . . .
It will appear where the water drops are. It can extend from your position to a few kilometers away.
.. . . .

It doesn't just appear to be in front of a distant hill. It is in front of it, if the rainy area is closer than the hill. "

That's what I wanted you to justify.

Justify what exactly? You forgot your specific questions under the individual quotes.

 

[sophiecentaur]

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

Are you saying that you cannot justify your statement(s) in that quoted post? All of that quote implies the same thing about where the bow appears to be.
P.s. Parallax?

 

[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.