a rainbow requires an observer

Fraggle Rocker said:

If two people are a couple of miles apart, don't they see the rainbow in slightly different form or orientation?

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No, but one may see a more complete or intense rainbow. It shape and angular relationship to them will be exactly the same. This is true because the rainbow is not a physical thing that can be looked at from different angles or perspectives. It very definitely requires an observer to exist. It is a creation of the observer´s mind, is not a physical object:



As at Earth the sun´s rays are essentially parallel (0.5 or less degrees of angular divergence) any one water drop sphere will send to your eye only one color. Lets assume, just to be clear, that the drop of the illustration´s blue ray is entering your eye. (Its red ray is passing below your head. It might even strike the ground a mile in front of where you stand.) Another drop higher above it can send its red ray to your eye. I.e. coming to your eye are rays at 40 & 42 degree rays. You assume, correctly, that the 42 degree or red ray is from a higher source than the blue 40 degree ray. Thus you perceive the red color as coming from an arch larger than the blue color arc you perceive.

Every observer of every rainbow always sees the red arc 2 degree wider, higher, etc. than the blue arc – Thus the shape of all rainbows for all observers who see one is identical, but again, one may see only a small part of the rainbow arc if there are not water drops where they need to be for him. You can make a quite nice rainbow with fine mist garden hose, if you know how.* Also if you know where to look occasionally you can see the much weaker larger rainbow. It has the red arc inside the blue arc. I have seen it a dozen of so times.

*Even make the full 360 degree circle, but you will need help. I.e. someone else makes the mist for you in front you, when you are say looking out from a second story porch but sun must be coming from your back side. (House can not be blocking the sun.)
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This is same story for the 22 degree halo sometimes seen around the moon, but it is a little more interesting in its physic. If you shine a beam of monochromatic light thru a prism its direction of travel will change. Exactly how much depends upon the angle the incident ray strikes the first surface of the prism. If your “prism” were a hexagonal ice crystal, as your twist the crystal you will find that the beam is always bent by at least 22 degrees.

Thus in addition to the moon light going directly to your eyes, a part of the sky say 23 degrees from the direct line of sight with ice crystals can bend a little moonlight back, thru 23 degrees into your eye. That at 22 degrees can too, but that at 19 degrees cannot. I.e. the inter edge of the moon´s halo is quite sharp but the outer edge does not really exist – the halo light bent thru 25 degrees is just much weaker, etc.

Billy T said:

No, but one may see a more complete or intense rainbow. It shape and angular relationship to them will be exactly the same. This is true because the rainbow is not a physical thing that can be looked at from different angles or perspectives. It very definitely requires an observer to exist. It is a creation of the observer´s mind, is not a physical object:

As at Earth the sun´s rays are essentially parallel (0.5 or less degrees of angular divergence) any one water drop sphere will send to your eye only one color. Lets assume, just to be clear, that the drop of the illustration´s blue ray is entering your eye. (Its red ray is passing below your head. It might even strike the ground a mile in front of where you stand.) Another drop higher above it can send its red ray to your eye. I.e. coming to your eye are rays at 40 & 42 degree rays. You assume, correctly, that the 42 degree or red ray is from a higher source than the blue 40 degree ray. Thus you perceive the red color as coming from an arch larger than the blue color arc you perceive.

Every observer of every rainbow always sees the red arc 2 degree wider, higher, etc. than the blue arc – Thus the shape of all rainbows for all observes who see one is identical.

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This is same story for the 22 degree halo sometimes seen around the moon, but it is a little more interesting in its physic. If you shine a beam of monochromatic light thru a prism its direction of travel will change. Exactly how much depends upon the angle the incident ray strikes the first surface of the prism. If your “prism” were a hexagonal ice crystal, as your twist the crystal you will find that the beam is always bent by at least 22 degrees.

Thus in addition to the moon light going directly to your eyes, a part of the sky say 23 degrees from the direct line of sight with ice crystals can bend a little moonlight back, thru 23 degrees into your eye. That at 22 degrees can too, but that at 19 degrees cannot. I.e. the inter edge of the moon´s halo is quite sharp but the outer edge dose not really exist – the halo light bent thru 25 degrees is just much weaker, etc.

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Your perception of the rainbow can be differrent due to the things you spoke of earlier, but the scattered light that rainbow is made of is there whether you see it or not. It is not an artifact of your eyes or a hallucination due to a malfunction in sensory perception. Other animals can probably see them as well as long as there vision is sharp and sensitive, but they would be shades of grey rather then color, depending on the animal.

Believe said:

Can't win um all I guess

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Well at least you can learn - more that I can say for about half of the posters I correct.

Point I was trying to get you to understand was that no matter what set of wavelengths is present, if their intensities are adjusted to make the same relative neural activity ratios in the three different color receptor cells as when only single wavelength we call yellow is present, you perceive yellow EVEN IF NO WAVELENGTHS EVEN CLOSE TO YELLOW LIGHT ARE PRESENT.

(2) Yes the wavelengths (and their relative intensities, plus 6 or more other factors) do determine the color perceived but their is no simple connection between wavelengths present and color perceived

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Interesting posts Billy T, thanks

Epictetus said:

Is it true that a rainbow requires an observer? Unlike a tree falling in a forest, which must certainly make a noise (IMHO),wouldn't the refracted light of a rainbow need to be seen to actually exist?

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Actually the tree falling in the forest makes a shock wave in the air that isn't sound until a brain processes it as such....

Diode-Man said:

Actually the tree falling in the forest makes a shock wave in the air that isn't sound until a brain processes it as such....

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I think the rainbow is the same. The conditions for a rainbow are there, but you do need an "observer". A camera will work. BTW - Polarized sunglasses may cancel the rainbow effect.

The physical data for rainbows and falling trees are present. Outside agencies are capable of detecting and interpreting such an event. The degree to which the data is interpreted is subject to the interpretive capabilities of the observer. Ergo, interpretation of the physical data can vary.

gmilam said:

... Polarized sunglasses may cancel the rainbow effect.

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An interesting comment, I had never thought of. Yes the light of the rainbow is probably somewhat (50%?) polarized, but different parts with different orientations. I.e. if view thru polarizer, some part should dim as the polarizer is rotated; which part that is will move along the arc.

Brewster´s angle effect is why:

Note there are 2 internal reflections inside the drop for EACH ray color:
but the only transmitted ray illustrated is the one that comes to your eye as a rainbow. Also the second internal reflections are not illustrated (for the normally seen rainbow - but the second would if figure were explaining the weaker, rarely seen, "reversed" rainbow as for it there are three internal reflections.)



where n1 is the refractive index of the initial medium through which the light propagates and n2 is the index of the other medium or the medium into which part of the light is transmitted. Air in the rainbow case. Normally, as shown in the illustration, n2 > n1 but not for the rainbow. I.e. the beam incident upon the reflecting surface is already inside the water drop and n2 is essentially 1.0 of air. After the sun light internally reflects off the “back side” of the drop the some what polarized reflected beam is still inside the drop, on its it third interaction with the air/water interface.

At that 3d interface the transmitted beam will become even more highly polarized than the internal beam striking the 3d interface. I´m too lazy to try to compute the two angles of incident for the 2n & 3d internal interface reflection, but if either of them, especially the 3d, is near the Brewster angle, then the rainbow rays entering your eye should be quite polarized.

Taking 1.5 as index for water, the Brewster angle IN RADIANS, is ~(2/3) where I have made slightly improper use of the fact that for small angles the arctan, in radians, is just the (n2/n1) ratio. As a radian is about 57 degrees, (as I recall) the Brewster angle is about 2x19 = 38 degrees.

The internal angle of incident in rain drop illustration looks to me to be a little larger, so effect would not be perfect polarization, but for the careful observer, probably noticeable if one rotates the polarizer – sees a slight dimming swing along the arc of the rainbow with that rotation.

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BTW 1, for the same EM wave scattering physics that makes the Brewster angle effect, the blue sky is highly polarized in a ring around the sun. I.e. that part of the sky where the sunlight has scattered thru 90 degrees to get to your eye, but once again what part you will see almost dark (thru polarizer) will change as you rotate the polarizer. This is a simple experiment all should do when you have a cloudless day. PS - Don´t look at the sun, even thru polarizer. The dark spot in the blue sky is far from the direction of the sun.

BTW 2, instruments exist that use Brewster angle effect to quickly and accurately measure the index of refraction of clear liquids. I used one in a summer job with a small oil company (Lion Oil). They had hired me to evaluate the maker´s claims for his new, tiny, curved, perforated metal chips as a distillation column packing material.

First day on the job I mixed 55 gallons of toluene with 55 of benzene and spent the rest of summer trying to separate them by a less than three meter tall distillation column filled with his packing material. My best separation gave that column 22 "theoretical plates," which is quit good for that length column. (The column performance depends upon how quickly and when heat is applied to the boiler tank - that is what I varied each day.)

I wrote my report, and headed back to Cornell, and never learned if they spent a million or so dollars buying his expensive metal chips based on my report. I measured the degree of separation with the instrument I mentioned as existing - it sort of looks like a microscope, but is hinged to allow a drop of liquid to be place between two glass plates.

Nobody on Sciforums would argue the interpretation of the double slit experiment that "matter does not exist unless it is observed".

This is a Dr. Fred Alan Wolf view. (he is Dr Quantum)
http://www.youtube.com/watch?v=DfPeprQ7oGc

Einstein and most Sciforum members will argue that the moon is there even if it is not viewed by anyone.

I have been in a position to measure and document recurring paranormal psychic hits beyond scientific explanations, and until I see a valid explanation I am open to a multitude of theories regarding Youngs experiment.

But a rainbow needing an observer would be the same argument as a tree needing an observer in my opinion.