I think you mean "reflection" rainbow, not "reflected" in the way that I used those terms in the article. The reflected rainbow is seen on the lake surface. The reflection rainbow is seen in the sky.Being that the rainbow in the "What caused the Quadruple Rainbow?" thread looks suspiciously like your "mystery" rainbow, I'm going to guess that yours was a "reflected" rainbow.
I think you missed the point, as astronomer Berman said, "Rainbows are not 3D objects." I too struggled with what that means.Why denying virtual objects?
I am sure that, if rainbows didn't look so gorgeous, they wouldn't have got into folk lore and they wouldn't have that extra magical quality that seems to make people treat them differently from other optical phenomena.Maybe it is useful to add that your statement that two persons cannot see the same rainbow is a denial of the concept of virtual objects.
I don't wee why it has to be a laser beam. The same thing would apply to any object that happens to be obscured from one viewer and visible to the other. One observer is aware of the object and the other is not.The point may be clearer if we refer back to the article's analogy with a man pointing a perfectly collimated laser at my eye.
Because light reflected from an ordinary object goes in many directions. It is (at least partially) omnidirectional light.I don't wee why it has to be a laser beam.
I can hold my hand in front of me so that I can see an object with one eye but not the other. Is there any (relevant) difference. Colimation is a way over the top requirement for this explanation. Speaking as one who did all the basic Physics learning in the absence of handy laser pointers, I often find that people reach for a virtual laser to prove points when simple shadows can do just as well. What did the 19th century opticians do when they wanted to explain things?Because light reflected from an ordinary object goes in many directions. It is (at least partially) omnidirectional light.
Collimated light is unidirectional. A collimated beam can be aimed at your left eye and 0% of its light reaches your right eye.
So a man holding a paper with a red dot printed on it, is very different from a man holding a red laser pointer.
But in the case of the rainbow you are making a virtual image of the sun at infinity. Even if the different rays from this virtual image passes through different rain drops, it does not make them part of a different virtual image. When you construct virtual images of objects using lenses, the rays take different paths as well, passing through different parts of the lens. The important thing is that a virtual image is constructed. It is this virtual image which is reflected. Fine, the rays did not pass through the same rain drops, but the virtual image is not where the raindrops are.A big difference. Zero photons from the laser reach the other eye, with our without your hand.
You could likely get similar effects with the virtual images created by lenses by placing a mirror beyond the lens, clearly the virtual image is not going to be reflected.Now suppose that the rain just began so that the high drop is there but there is no low drop
I agree that rainbows are not 3D objects. Our difference is simply that I prefer to extrapolate the light rays back to infinity. For me raindrops are merely mirror particles at a finite distance, they are certainly not the location of the rainbow. Extrapolate the light rays back to infinity to find the virtual object. The celestial sky is the location of the rainbow. Everybody sees the rainbow at the same location in the celestial sky.I think you missed the point, as astronomer Berman said, "Rainbows are not 3D objects." I too struggled with what that means.
The imaging forming structure in a rainbow is different from what happens in a lens - it's more like a multiplicity of lenses, with each lens contributing within a narrow angle. It's a bit like what happens with a lenticular screen or a fresnel lens. I think it's a bit pointless to try to make the rainbow fit in with the more straightforward images that we see. Of course the image is not 'real' because the light behaves as if it comes from way behind the image forming structure. There is no parallax against distant objects so it can be classed as infinitely far away.When you construct virtual images of objects using lenses, the rays take different paths as well, passing through different parts of the lens.
I don't think they do, exactly. The distances are so large that it would be difficult to spot but when you move to the left, the bow moves to the left, with you. So it would be moving across the sky relative to the distant stars. The centre of the bow is in line with the Sun and the back of your head. But a rainbow at night? Weird idea! Perhaps it's an experiment you could do with the Moon - if you could arrange the rain to come at the right time of the day and month. But you would need to travel quite a distance sideways to see the effect against the moonscape as a background. (many km to observe a recognisable movement of a fuzzy thing like a rainbow.Everybody sees the rainbow at the same location in the celestial sky.
You are saying that the angle subtended by the Earth from the Sun is negligible. Yes, that sounds reasonable. But how does that relate to the fact that a rainbow moves against the Earth as you move? Ah - it moves by the distance you move, which means the apparent change in angle against the celestial sphere is zero.It is pure geometry. The Sun's celestial location is the same for everybody, at a given time. Hence the celestial locations of the antisolar point and the 42° circle around it are the same for everybody.