Clouds changing color inside rainbow arch

[iLuv2Fly]

Took these pics yesterday evening and was wondering what causes the clouds color to change inside the rainbow arch. I don't recall seeing this before and search the web but no luck. I'd love to hear a scientific explanation

[Eise]

On 6/5/2018 at 2:19 PM, iLuv2Fly said:

Took these pics yesterday evening and was wondering what causes the clouds color to change inside the rainbow arch. I don't recall seeing this before and search the web but no luck.

The inside of a rainbow is always lighter than the outside. But these pictures were made when the sun was low above the horizon, so it was shining red. Therefore also the rainbow itself is mainly red. See this picture, that was obviously made when the sun was still higher above the horizon:

I just looked it up: the raindrops reflect both from the backside as from the frontside. The light that reflects from the backside of the raindrops makes the rainbow itself,  the light that reflects from the frontside are just reflected, and therefore has the same colour as the incoming light, mainly red in your case.

See here.

Edited June 8, 2018 by Eise

[koti]

On 5.06.2018 at 2:19 PM, iLuv2Fly said:

Took these pics yesterday evening and was wondering what causes the clouds color to change inside the rainbow arch. I don't recall seeing this before and search the web but no luck. I'd love to hear a scientific explanation

I will add to Eise’s post that the red color in your pictures is caused by the evening sun warm color temperature (you will have the same edffect just after the sunrise as well) The sunset/sunrise color temperature is about 2000K which is very a pleasant warm color. The „color temperature” concept comes from objects heated up and glowing - emitting a certain color of light depending on the temperature, like a steel rod will glow when heated different colors depending on the temperature. Heres a list of common sunlight temperetures depending on the angle at which the sunlight hits the earth at different times of the day:

 

Source:

https://www.eizo.com/library/basics/color_temperature_on_an_LCD_monitor/

Edited June 8, 2018 by koti

[JeffJo]

On 6/5/2018 at 8:19 AM, iLuv2Fly said:

Took these pics yesterday evening and was wondering what causes the clouds color to change inside the rainbow arch. I don't recall seeing this before and search the web but no luck. I'd love to hear a scientific explanation.

I just joined, and wanted to get my post count up . So I looked for some of my pet topics in old posts, and I think this one needs a better response in a science forum.

Most explanations of rainbows are satisfactory for middle schoolers, but get many details wrong if you go above that level of education. The problem is that it sounds so reasonable, that few students doubt the accuracy, and they repeat it when they become teachers.

The short explanation is that spherical raindrops do not act like prisms, they act like lenses (or the mirror in the back of a flashlight). The colors are not caused by the literal act of color separation, it is a secondary effect called color aberration. And the bands are caused by ... well, it isn't a simple analogy.

1. A spherical raindrop reflects light back toward the sun at every angle from 0° (straight back toward the sun) to about 40°. So the complete reflection is a beam of light, like from a wide-beam flashlight, that is about 80° wide.
2. But the beam does not have a constant brightness. In increases slowly as the incident sunlight moves off-center. And it actually becomes asymptotic at the maximum reflection angle. See the first figure below.
3. But the sun is actually about 0.5° wide, so this asymptote creates a bright band of light that is 0.5° wide.
4. The effect of diffraction is not to separate colors, it is to change this maximum angle. It's about 40° for violet light, and about 42° for red light.
5. So the bright bands for different colors are slightly offset, and the colors get less vivid as you move from red to violet. See the second figure (from https://atoptics.co.uk/bows.htm)
6. Finally, the area in the sky that is inside the bright bands still has some reflected light. This makes the sky there appear brighter, but what is usually noticed is that the sky outside is darker. It has been named Alexander's Dark Band for the man who first wrote about it. Third figure.
7. The same thing happens with the secondary bow. But the colors aren't "reversed," the bow is actually seen up-side down. Technically, the bright inside extends past the top of the sky and goes to the sun itself. You usually can't see this.

Other answers did explain the pinkish hue correctly.

[swansont]

1 hour ago, JeffJo said:

The colors are not caused by the literal act of color separation, it is a secondary effect called color aberration.

Which is color separation, as a result of refraction and dispersion, which is what’s happening in a prism.

 

[JeffJo]

9 minutes ago, swansont said:

Which is color separation, as a result of refraction and dispersion, which is what’s happening in a prism.

 

Yes, the effect on any one incoming ray is the same as what you call "color separation." No, when seen as a whole, the effect on the resultant image is not due to that effect. Different colors from different parts of the original end up in the same place, not "separated" at all. And the reason I make this fine distinction is because many people believe, incorrectly, that the green band in a rainbow is completely "separated" from red. They believe that, because they attribute the rainbow to "color separation." And that is wrong.

What happens in prisms, only happens when there is a single ray of incoming white light (or parallel rays that extend in the same direction as the axis of the prism, so that the paths from different incoming rays experience the exact same deflections). What happens is that each color is deflected to a different place than any other color.

What happens in lenses and raindrops, where different rays do not experience the same deflections, is that the size of the resultant image varies by color.

[swansont]

1 hour ago, JeffJo said:

Yes, the effect on any one incoming ray is the same as what you call "color separation." No, when seen as a whole, the effect on the resultant image is not due to that effect. Different colors from different parts of the original end up in the same place, not "separated" at all. And the reason I make this fine distinction is because many people believe, incorrectly, that the green band in a rainbow is completely "separated" from red. They believe that, because they attribute the rainbow to "color separation." And that is wrong.

What happens in prisms, only happens when there is a single ray of incoming white light (or parallel rays that extend in the same direction as the axis of the prism, so that the paths from different incoming rays experience the exact same deflections). What happens is that each color is deflected to a different place than any other color.

What happens in lenses and raindrops, where different rays do not experience the same deflections, is that the size of the resultant image varies by color.

They are viewed differently. A rainbow is not projecting the light onto a screen as we do with a prism. i.e. we typically don’t look at the prism. If we did, from a similar distance as a rainbow, we would only see one color. But if we had a bunch of prisms, separated by an appropriate distance, we would see a rainbow.

The same principles of refraction and dispersion apply. With rainbows, there is also a reflection..

[exchemist]

7 hours ago, JeffJo said:

Yes, the effect on any one incoming ray is the same as what you call "color separation." No, when seen as a whole, the effect on the resultant image is not due to that effect. Different colors from different parts of the original end up in the same place, not "separated" at all. And the reason I make this fine distinction is because many people believe, incorrectly, that the green band in a rainbow is completely "separated" from red. They believe that, because they attribute the rainbow to "color separation." And that is wrong.

What happens in prisms, only happens when there is a single ray of incoming white light (or parallel rays that extend in the same direction as the axis of the prism, so that the paths from different incoming rays experience the exact same deflections). What happens is that each color is deflected to a different place than any other color.

What happens in lenses and raindrops, where different rays do not experience the same deflections, is that the size of the resultant image varies by color.

I must admit I don't see the difference. Surely the reason why the images for different wavelengths are different sizes is because refractive index is a function of wavelength, i.e., because of dispersion and thus refraction through different angles, just as it is for a prism?

The light is spread out into a spectrum by this process, i.e. the colours are separated, in both processes, surely?   

[Genady]

13 hours ago, JeffJo said:

• A spherical raindrop reflects light back toward the sun at every angle from 0° (straight back toward the sun) to about 40°. So the complete reflection is a beam of light, like from a wide-beam flashlight, that is about 80° wide.
• But the beam does not have a constant brightness. In increases slowly as the incident sunlight moves off-center. And it actually becomes asymptotic at the maximum reflection angle. See the first figure below.
• But the sun is actually about 0.5° wide, so this asymptote creates a bright band of light that is 0.5° wide.
• The effect of diffraction is not to separate colors, it is to change this maximum angle. It's about 40° for violet light, and about 42° for red light.
• So the bright bands for different colors are slightly offset

This explanation is symmetrical in regard to different colors. Perhaps I miss something. It would explain if the rainbow band were red on the edges and violet in the middle. What makes the violet to appear on the outside edge and the red on the inside edge of the rainbow band?

[JeffJo]

Again, the process known as "color separation" applies to individual rays of incoming light, not to the image as a whole. The green band contains lesser intensities of red, orange, and yellow light; but not blue, indigo, or violet. Citing "color separation" as the root cause of the rainbow does not explain this. Citing "color aberration" does.

But by citing  "color separation," it is easy to come to the conclusion that each color band comprises just one frequency of light (with allowances for the 0.5° width of the sun). And if you search the internet for explanations of the rainbow, you will find this incorrect conclusion explicitly mentioned in a many, maybe even a majority, of them. All I'm trying to do is separate (pun intended) the process as it applies to individual rays from the process that causes the rainbow.

This misconception is furthered when people incorrectly call Newton's "colors of the spectrum" the "colors of the rainbow." They are not the same; my second figure above shows this.

1 hour ago, Genady said:

This explanation is symmetrical in regard to different colors. Perhaps I miss something. It would explain if the rainbow band were red on the edges and violet in the middle. What makes the violet to appear on the outside edge and the red on the inside edge of the rainbow band?

This explanation is not symmetric in regard to different colors; I'm not sure why you claim that, or how you got those orders. The maximum angle of the deflection depends on the index of refraction, and I illustrated this in my first figure. I admit that I was trying to be brief, and so left the asymmetry to the interpretation of that figure, but it is there.

• The maximum for red light is about 42°, so the red band is seen reflected from drops that are about 42° from the sub-solar point (the shadow of your head). This is the outside, not the inside, of the rainbow.
• The maximum for green light is about 41°, so the green band is seen reflected from drops that are about 41° from the sub-solar point. This is the middle of the rainbow.
• The maximum for violet light is about 40°, so the violet band is seen reflected from drops that are about 40° from the sub-solar point. This is the inside, not the outside, of the rainbow.

Maybe this will help: The maximum of intensity is caused because the functional relationship between the incident angle and the deflection angle has a maximum around i=60° and d=40° to 42°. And the intensity of the light is inversely proportional to the slope of this function.

The intensity is illustrated by how the light rays "bunch up" near the maximum. This is exaggerated:

Yes, each colors separate in individual incoming rays. But not as that implies in the full set of outgoing rays.

+++++

[Edit] Swansont wrote:
 

Quote

 

They are viewed differently. A rainbow is not projecting the light onto a screen as we do with a prism. i.e. we typically don’t look at the prism. If we did, from a similar distance as a rainbow, we would only see one color. But if we had a bunch of prisms, separated by an appropriate distance, we would see a rainbow.

The same principles of refraction and dispersion apply. With rainbows, there is also a reflection..

 

There is no difference, except how the colors are formed. If we could create an array of small prisms suspended in the air, and all aligned the same way, we would see a linear band of spectrum (i.e., not quite the same as rainbow) colors across the array. The image project from one is the same as the image seen in the array.

Then say we have a screen, with a hole cut in it so that we can shine a bright beam of white light through it. Place a spherical flask of water in the beam. It will reflect a white circle of light onto the screen, with rainbow (i.e., not quite the same as spectrum) colors around the rim. Something like this:

The point is that the reflection from one, and the image from an (aligned) array, are the same.

Edited June 16, 2023 by JeffJo
Reached post limit

[Genady]

42 minutes ago, JeffJo said:

This explanation is not symmetric in regard to different colors

Got it. Thank you.

 

43 minutes ago, JeffJo said:

I'm not sure ... how you got those orders.

I looked, by mistake, on the secondary, upside-down rainbow.

[studiot]

On 6/15/2023 at 10:40 PM, JeffJo said:

A spherical raindrop reflects light back toward the sun..........................

Are you sure you have got all your (ducks) (terms) in a row ?

I find your explanation a bit garbled.

Sorry.

 

In would read that statement to mean that the light bounces off the surface of the drop, which is largely untrue.

Strictly the light is only only reflected within the drop.

Refraction is the only surface effect operating.

Most of your other statements are true but very difficult to obtain a clear picture from.

 

[JeffJo]

2 hours ago, studiot said:

 

In would read that statement to mean that the light bounces off the surface of the drop, which is largely untrue.

Strictly the light is only only reflected within the drop.

Refraction i the only surface effect operating.

Most of your other statements are true but very difficult to obtain a clear picture from.

 

Some light always reflect off an air-water interface, in either direction. See Fresnel Equations. However, little light bounces off of the outside, and it scatters in many directions.

But strictly speaking, I didn't explicitly state whether the rainbow comes from light that reflects off of the inner, or the outer, surface. You read "outer" into what I said, ignoring the diagram that shows internal reflections.

All that I said is true. The only things making my statements difficult for you to see were my attempts to not go into detail, and your unwillingness to accept it (which I infer from the fact that you ignored my ray diagram showing internal reflections.)

But here's a more detailed explanation. See the ray diagram if you find it unclear:

1. Say a ray of light strikes the surface of the drop making angle A with the surface normal. 0°<=A<180°
   1. Both transmission and reflection will occur. But the reflected light is dim, and uninteresting.
   2. Most of the light will enter the drop with angle of refraction B=arcsin(sin(A)/k), where k is the index of refraction of water. 0°<=B<C, where C is the critical angle where total internal reflection occurs.
   3. The total deflection is A-B.
   4. Note that the line containing the surface normal passes thru the center of the drop, so it contains a radius.
2. This light will pass thru the drop, and make angle B with the surface normal at the back.
   1. It is angle B because the ray makes an isosceles triangle with two radii of the sphere.
   2. Many explanations for rainbows say total internal reflection occurs. It is impossible, since B<C.
   3. Both reflection and transmission occur, and again most transmits.
   4. The light that exits without ever reflecting does not form colored rainbow bands, but does make what is called the "zeroth-order glow" around the sun.
   5. The light that reflects is deflected 180°-2*B.
3. Step 2 repeats, in theory, indefinitely. But the amount of light that reflects becomes insignificant after a few internal reflections.
4. The light that exits has an angle of refraction of A.
   1. The total deflection is again A-B, and the sign of this angle is correct.
5. The light that exits after n internal reflections has a total deflection angle of 180*n+2*A-(2n+2)*B.

For the primary rainbow, it is customary to subtract this from 180 since you look away from the sun to see it. This is what I plotted above.

[swansont]

On 6/16/2023 at 8:24 AM, JeffJo said:

the process known as "color separation"

Can you provide a link or source that refers to this process? I see explanations that cite refraction, reflection and dispersion. I’m not aware of “color separation” as a physics phenomenon.

When I Googled ‘“color separation” rainbow’ the second hit was this thread, which leads me to think it’s not a common expression in rainbow explanations.

Quote

But by citing  "color separation," it is easy to come to the conclusion that each color band comprises just one frequency of light

Which is why dispersion is usually cited.

[studiot]

13 hours ago, JeffJo said:

But strictly speaking, I didn't explicitly state whether the rainbow comes from light that reflects off of the inner, or the outer, surface. You read "outer" into what I said, ignoring the diagram that shows internal reflections.

All that I said is true. The only things making my statements difficult for you to see were my attempts to not go into detail, and your unwillingness to accept it (which I infer from the fact that you ignored my ray diagram showing internal reflections.)

Thank you for you attempt to explain further, but I disagree that your explanation is totally correct in that.

1) You claimed diffraction occurs, but there can be no diffraction.

On 6/15/2023 at 10:40 PM, JeffJo said:

The effect of diffraction is not to separate colors, it is to change this maximum angle. It's about 40° for violet light, and about 42° for red light.

The numbers 40 and 42 degrees are about correct for the blue/violet and the red/yellow respectively and these play an important role in why the observer sees a rainbow.

This is because of refraction alone and does not occur with the internal reflection either.

The internal reflection does invert the stacking order of the rays.

So you are correct in observing that the light reflected from the water droplet layer creating the return contains the full spectrum of the incoming sun's light.

But the double refraction that occurs as the light first enters then leaves the droplet accounts for the fact the the light leaving the droplet has been split into diverging blue and red rays with the blue above the red.

The result of this is that the blue rays from the higher drops arrive above the observer's head and outside the field of view, so he sees predominantly the red ones from the upper drops.

Conversely the red from the lower drops falls below his field of view so he sees predominately blue.

 

The light from a second total internal reflection within the droplet is inverted once again by the reflection so the blue is now on top and the second rainbow colours  appear inverted compared to the first.

 

On 6/15/2023 at 10:40 PM, JeffJo said:

Most explanations of rainbows are satisfactory for middle schoolers, but get many details wrong if you go above that level of education. The problem is that it sounds so reasonable, that few students doubt the accuracy, and they repeat it when they become teachers.

Actually I couldn't find any of these 'unsatisfactory' middle school explanations, perhaps because the rainbow is no longer taught in UK schools.

When it was taught it was on the old School Certificate syllabus for example .

Quote

Oxford Higher School Certificate

Give an account of the rainbow, explaining the appearances and formation of the primary and secondary bows.

Theory very similar to yours can be found in Physics texts of the 1940s and 1950s for example

Intermediate Physics by C J Smith

or

Higher Physics by Nightingale

 

But it is important to remember that the geometryand positioning  of the camera or eye that is also needed to play its part.

 

[JeffJo]

48 minutes ago, studiot said:

1) You claimed diffraction occurs, but there can be no diffraction
 

That was an obvious typo, and editing restrictions prevent me from changing it. It (quite obviously) should have been "refraction."

Quote

The internal reflection does invert the stacking order of the rays.

No, it does not. Reflections only invert the "depth" dimension. And if reflection did invert them, the colors would be seen in the same order.

The primary rainbow is a filled-in circle that is centered on the sub-solar point, with an angular radius of about 40° (violet, inside) to 42° (red, outside).

The secondary rainbow is a similar filled-in circle that is centered on the sun, not the sub-solar point. It has an angular radius of about 126° (violet, inside) to 130° (red, outside). This means that the image "wraps around" the top of the sky, and is seen about 10° above the primary rainbow. The colors are in the same order, inside-to-outside. But because it is seen upside-down, "inside" is higher than "outside."

This relative order is caused entirely by the magnitude of the deflection angles, and the alternating center (sub-solar point, sun), not the number of reflections. If the index of refraction were on the order of 2, the secondary rainbow would be seen the side of the sky with the sun, with red higher than violet.

I can show you the equations that lead to these results.

Quote

The light from a second total internal reflection within the droplet is inverted once again by the reflection so the blue is now on top and the second rainbow colours  appear inverted compared to the first.

There are no total internal reflections, so there can't be a second one. Perhaps this is a typo, too?

Quote

Actually I couldn't find any of these 'unsatisfactory' middle school explanations, perhaps because the rainbow is no longer taught in UK schools.

If you accept "total internal reflection," "reflections invert images", and "the rainbow is a spectrum," then I can see why you don't find them unacceptable. But it seems that Indian schools are quite fond of teaching rainbows, as many of the explanations you will find come from there. I'd say 90% get it wrong.

Maybe you should look at this site, if you think UK information is better: https://atoptics.co.uk/rainbows/primrays.htm . It is actually one of the leading sites that give entirely valid information.

Edited June 20, 2023 by JeffJo

[studiot]

1 hour ago, JeffJo said:

There are no total internal reflections, so there can't be a second one. Perhaps this is a typo, too?

 

Did you make one when you posted this picture?

I have added an arrow to show the region of internal reflection.

 

 

 

If you wish to use the transmitted ray you have a problem since the sun and the observer need to be on the same side of the drops.

Here are the conditions for observing rainbows as published by the UK Met Office.

There have to be reflections, given the relative poitions of the bow, observer, sun and drops.

Quote

• The sun needs to be behind the viewer
• The sun needs to be low in the sky, at an angle of less than 42° above the horizon. The lower the sun in the sky the more of an arc of a rainbow the viewer will see
• Rain, fog or some other source of water droplets must be in front of the viewer

How are rainbows formed? - Met Office

 

 

 

1 hour ago, JeffJo said:

Maybe you should look at this site, if you think UK information is better: https://atoptics.co.uk/rainbows/primrays.htm . It is actually one of the leading sites that give entirely valid information.

Thank you for this link, it has some interesting sections, including the first none entitled 

Not all coloured patchesw in the sky are rainbows.

They go on to give examples of coloured patches due to transmissive and also of diffracted origin.

 

But when they come to rainbows themselves they offer the same explanation and caveats at I have ( and swansont did earlier)

[JeffJo]

16 hours ago, studiot said:

 

Did you make one when you posted this picture?

I have added an arrow to show the region of internal reflection.

Do you understand that "internal reflection" means only "reflection inside an object," and really has no special significance beyond the geometry? But that "total internal reflection" has a very specific meaning, and does not mean (although it often does) being inside an object?

When light encounters the boundary between two transparent media, some reflection always occurs, but transmission does not always. If the angle of incidence is greater than the critical angle, C=arcsin(1/k), no transmission occurs. This effect is responsible for fiber optics, the sparkle in diamonds, and the optics in binoculars. It's also why aerated water looks white, as total internal reflections occur on the outside of the bubbles.

I never said that internal reflections are not responsible for rainbows. I said that total internal reflection is not. But I usually try to say "reflections inside" so that I don't cause the kind of misunderstanding you seem to have. Even if it was just a habit to add "total" in front of "internal reflection," it is still wrong and teaches the wrong things.

Quote

If you wish to use the transmitted ray you have a problem since the sun and the observer need to be on the same side of the drops.

I have no idea what "problem" you think I have.

Quote

But when they come to rainbows themselves they offer the same explanation and caveats at I have ( and swansont did earlier)

And they offer the same explanations that I have. But not yours. They do not support "the light leaving the droplet has been split into diverging blue and red rays with the blue above the red" as the cause of a rainbow. It isn't the "diverging rays" that cause the rainbow, it is the fact ranges of the deflections are different, as a function of wavelength.

And that's what "dispersion" means. That ray tracing is a function of wavelength. For single rays, colors separate. For the entire image, aberration occurs.

Edited June 21, 2023 by JeffJo

[exchemist]

14 minutes ago, JeffJo said:

Do you understand that "internal reflection" means only "reflection inside an object," and really has no special significance beyond the geometry? But "that total internal reflection" has a very specific meaning?

When light encounters the boundary between two transparent media, some reflection always occurs, but transmission does not always. If the angle of incidence is greater than the critical angle, C=arcsin(1/k), no transmission occurs. This effect is responsible for fiber optics, the sparkle in diamonds, and the optics in binoculars. It's also why aerated water looks white, as total internal reflections occur on the outsideof the bubbles.

I never said that internal reflections are not responsible for rainbows. I said that total internal reflection are not. But I usually try to say "reflections inside" so that I don't cause the kind of misunderstanding you used. Even if it was just a habit to add "total" in front of "internal reflection," it is still wrong.

I have no idea what "problem" you think I have.

And they offer the same explanations that I have. But not yours. They do not support "the light leaving the droplet has been split into diverging blue and red rays with the blue above the red" as the cause of a rainbow. It isn't the "diverging rays" that cause the rainbow, it is the fact ranges of the deflections are different, as a function of wavelength.

And that's what "dispersion" means. That ray tracing is a function of wavelength. For single rays, colors separate. For the entire image, aberration occurs.

Sure but the image is effectively split into an infinite series of images, one for each wavelength, is it not? So it does involve splitting into colours. That's why its full name is "chromatic" aberration, surely? 

[swansont]

4 hours ago, JeffJo said:

And they offer the same explanations that I have. But not yours. They do not support "the light leaving the droplet has been split into diverging blue and red rays with the blue above the red" as the cause of a rainbow. It isn't the "diverging rays" that cause the rainbow, it is the fact ranges of the deflections are different, as a function of wavelength. 

I’m not sure how “deflections are different” doesn’t mean that the rays diverge.

 

4 hours ago, JeffJo said:

And that's what "dispersion" means. That ray tracing is a function of wavelength. For single rays, colors separate. For the entire image, aberration occurs.

Dispersion necessarily means that the rays diverge.

[studiot]

On 6/15/2023 at 10:40 PM, JeffJo said:

The effect of diffraction is not to separate colors, it is to change this maximum angle. It's about 40° for violet light, and about 42° for red light.

 

23 hours ago, JeffJo said:

That was an obvious typo, and editing restrictions prevent me from changing it. It (quite obviously) should have been "refraction."

 

 

5 hours ago, JeffJo said:

Do you understand that "internal reflection" means only "reflection inside an object," and really has no special significance beyond the geometry? But that "total internal reflection" has a very specific meaning, and does not mean (although it often does) being inside an object?

When light encounters the boundary between two transparent media, some reflection always occurs, but transmission does not always. If the angle of incidence is greater than the critical angle, C=arcsin(1/k), no transmission occurs. This effect is responsible for fiber optics, the sparkle in diamonds, and the optics in binoculars. It's also why aerated water looks white, as total internal reflections occur on the outside of the bubbles.

I never said that internal reflections are not responsible for rainbows. I said that total internal reflection is not. But I usually try to say "reflections inside" so that I don't cause the kind of misunderstanding you seem to have. Even if it was just a habit to add "total" in front of "internal reflection," it is still wrong and teaches the wrong things.

 

 

You are right, after carefully only specifying reflections and distinguishing between total internal and internal and external an unwanted total slipped in.

 

However I think for everyone's benefit we should review some basic optics and its terminology since your excuse for claiming diffraction should read refraction, which you also seem claim does not occur.

However hopefully we are now agreed that the is no periodic structure that can cause diffraction.

So we can rule out diffraction.

Refraction is then the only mechanism that can bring about the observed colour separation.

In particular reflection of the light in your bubbles example is white just because reflection alone cannot bring about colour separation.

,

One difference between refraction and reflection is that for reflection the light remains in one medium, whilst refraction requires two.

 

Both the website I linked to and the one you linked state state quite clearly that the formation of a rainbow requires both refraction and reflection.

This is because the actual mechanism is more complicated.

If the colours were not split the observer would see only white light as with your air bubbles.

The split occurs at the first refraction, that occurs as the light enters the raindrop, somewhere on the incoming sunshine side.
Subsequently the red and blue etc paths are different in the passage of the now split light.
So when the red and blue rays strike the  surface on the opposing side of the drop, they will not strike at exactly the same place.
When they do strike, as you say, some will be reflected back into the liquid water, to pass through it a second time.

As you also said, this turning light away from where it was originally travelling to necessarily darkens that area.

Again some of the light will pass out of the drop on striking the surface a second time.

This also the difference spatial between the blue and red strike points will increase.

Refraction then occurs at the exit interface for both red and blue rays.

The upshot of this is that some of the blue light is directed out of the top of the observer's field of view and some of the red light out of the bottom.

So you are correct there is not the fully developed fan of pure colours as with a prism.

 

 

 

[Mordred]

Would also be useful to numerous readers to include Snells law.

[JeffJo]

On 6/21/2023 at 6:20 AM, exchemist said:

Sure but the image is effectively split into an infinite series of images, one for each wavelength, is it not? So it does involve splitting into colours. That's why its full name is "chromatic" aberration, surely? 

And those images overlap almost entirely. While "separation" - the word I used, and that is used most often - implies they don't.

 

On 6/21/2023 at 10:19 AM, swansont said:

I’m not sure how “deflections are different” doesn’t mean that the rays diverge.

Dispersion necessarily means that the rays diverge.

Diverge: "to move or extend in different directions from a common point." The vast majority of the reflection consists of white light, comprising all visible wavelengths. The overwhelming interpretation of rainbows (basis for this statement: Newton's "colors of the spectrum" are almost never called that, they are called "colors of the rainbow") is that each band is monochromatic. All I'm trying to get across, is that they are not. But those who learned it that way seem very reluctant to accept that it is wrong.

On 6/21/2023 at 12:33 PM, studiot said:

However I think for everyone's benefit we should review some basic optics and its terminology since your excuse for claiming diffraction should read refraction, which you also seem claim does not occur.

Never. I said that the result "separation of colors" is not observed. That is not synonymous with "refraction," which means that the ray tracing has to be handled separately.

This isn't that complicated, unless you intentionally try to misunderstand it.

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Refraction is then the only mechanism that can bring about the observed colour separation.

<Sigh.>

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Both the website I linked to and the one you linked state state quite clearly that the formation of a rainbow requires both refraction and reflection.

As does the process I described. I expressed it with Snell's Law.

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If the colours were not split the observer would see only white light as with your air bubbles

As they do in the rainbow reflection, inside the violet band, even after two refractions have affected the light.

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The split occurs at the first refraction, that occurs as the light enters the raindrop, somewhere on the incoming sunshine side.

Again: the split occurs to individual rays. Most recombine with the other colors, from incoming sun rays that hit the drop at a slightly different place.

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Subsequently the red and blue etc paths are different in the passage of the now split light.

And yet again: the split occurs to individual rays. Most recombine with the other colors, from incoming sun rays that hit the drop at a slightly different place.

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As you also said, this turning light away from where it was originally travelling to necessarily darkens that area.

No, in the area where the colors recombine, it brightens the sky. Alexander's Dark Band is actually unaffected.

On 6/22/2023 at 2:27 PM, Mordred said:

Would also be useful to numerous readers to include Snells law.

See step 1.2 of my outline of the process.

Edited June 24, 2023 by JeffJo

[Mordred]

1 hour ago, JeffJo said:

 

See step 1.2 of my outline of the process.

My meaning by applying Snells law is to demonstrate in the maths as to how it applies we both agree it does.  

[JeffJo]

21 hours ago, Mordred said:

My meaning by applying Snells law is to demonstrate in the maths as to how it applies we both agree it does.  

And I described it. It determines the angle B in the equation D(n)=n*180°+2*A-(2n+2)*B, where A is the angle of incidence of an incoming light ray, n is the number of (not total) internal reflections, and D is the total angle thru which that light is deflected. Since the primary and secondary rainbows are both seen when looking away from the sun, often this is subtracted from 180° and put in the range 0° to 180°. That's 4*B-2*A for the primary.

"Dispersion" does not mean colors separate, it means that each color's optics have to be handled separately. That is, each gets a different curve in plots like this:

The ranges of deflection angles that I labeled the "glow" is where no separation of colors is observed. (Yes, it still occurs to individual rays, but the colors recombine).

The Calculus of Rainbows shows how to determine the "rainbow angle" as a function of k, the index of refraction. For the primary, the angle of incidence for the incoming ray is A=arcsin(sqrt((4-k^2)/3)) = 59.48° for red light with k=1.3318. For this ray of red light, B=40.30°, and D=4B-2A=42.24°.