Is it physically possible to drive through a rainbow? Why or why not?
From the Wiki:
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.
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.
Link some that are similar to what you saw.
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.
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.
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.
Not really. 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.
Whether it extends beyond the hill depends on whether it also rains behind the hill. So the brain might sometimes be right here.
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.
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.
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.
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.
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.
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.
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.
Ouch - apologies for skipping over some previous posts... Mea culpa!
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'.
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.
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.
I can only refer to this and hundreds more sources.
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?
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.
There is no virtual image at all here. Concave mirrors don't create virtual images of objects that are outside of their focal length:
The sun is hardly inside the focal length of a raindrop.
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?)
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.
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.
The term "virtual image" that you originally used has a specific meaning which not applicable to a rainbow.
Wrong. The first occurrence of the word "image" in this thread is in your post:
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!
Too vague. Try this:
Show how it applies to a rainbow with a diagram analogous to the one shown there.
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.
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.
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