Jump to content

# Clouds changing color inside rainbow arch

## Recommended Posts

Then you missed the details of Snells law with chromatic dispersion and the subsequent changes of index due to changes in phase velocity of Light through a medium. Are you claiming the following doesn't occur

Obviously some materials are non dispersive, however a prism or water vapours are. The properties of the material and its sometimes dependency of frequency to index of refraction is the determining factor. Some materials are also birefringent though they depend on the polarization. (calcite for example).

Edited by Mordred
##### Share on other sites

On 6/24/2023 at 7:53 PM, JeffJo said:

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.

Perhaps you would prefer the phrase partial separation ?

3 hours ago, JeffJo said:
On 6/24/2023 at 9:44 PM, 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.

I would be interested in finding out how Snell's Law is applicable to that equation (which refers to reflection)

I wonder if you are trying to offer some form (possibly transverse) of chromatic aberration as the mechanism for the rainbow colours formation, as in the pictures from Wikipedia below.

Ring 3 perhaps on the first one ?

The second one shows purple fringing on the donkey's ears.

This discussion on Physics forums lends some support to that description, in some circumstance.

Edited by studiot
##### Share on other sites

Snells law of refraction

n1sinθ1=n2sinθ2

n_1 is the incident index, n_2 the refracted index, θ1 the incident angle,  θ2 the refracted angle.

you can get further details here

the article covers the phase velocities and how differences in phase velocities will change the index of refraction

however some examples for visible light apply this to red, yellow, violet in regards to a prism such as here

largely just a simplification.

also its often referred to as the angular dispersion with regards to the prism

Edited by Mordred
##### Share on other sites

25 minutes ago, Mordred said:

Snells law of refraction

n1sinθ1=n2sinθ2

n_1 is the incident index, n_2 the refracted index, θ1 the incident angle,  θ2 the refracted angle.

you can get further details here

the article covers the phase velocities and how differences in phase velocities will change the index of refraction

however some examples for visible light apply this to red, yellow, violet in regards to a prism such as here

largely just a simplification.

also its often referred to as the angular dispersion with regards to the prism

Hi Mordred, who were you addressing in this post ?

I think we are all aware of Snell's law, and that it applies only to refraction.

The equation I was querying refers to reflection.

##### Share on other sites

with water droplets the light enters and leaves the droplet so refraction does apply

here

obviously some light will simply reflect but you get refraction due to light entering and exiting the water droplets. Seriously if a water droplet were a perfect mirror then you would have 100 percent reflection. However a water droplet is semi transparent so naturally refraction will be involved. Anytime light enters a medium you apply refraction. With reflection its simply the surface or internal reflections in both cases Snell's law covers both. Angle of incident= outgoing angle for the reflected case

$\theta_i=\theta_O$

. with refraction obviously you have a much wider range.

Quite frankly I can't think of any reason for color separation via reflection as opposed to refraction.

I'm fairly positive your well aware of this though Studiot lol. Its still good info for other readers

Edited by Mordred
##### Share on other sites

On 6/25/2023 at 6:11 PM, studiot said:

Perhaps you would prefer the phrase partial separation ?

Why? There is nothing "partial" about separation. But the term applies to individual rays, not the entire reflection. It is total in individual rays. But at most observations angles where a reflection is observed, every color of light is observed in that reflection.

Quote

I would be interested in finding out how Snell's Law is applicable to that equation (which refers to reflection)

No, the equation refers to the angle of deflection. That deflection is caused by two transmissions (refraction accompanies transmission) and one reflection. And it references Snell's Law:

On 6/25/2023 at 2:40 PM, JeffJo said:

And I described [how Snell's Law is applied]. [Snell's Law] 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.

Quote

I wonder if you are trying to offer some form (possibly transverse) of chromatic aberration as the mechanism for the rainbow colours formation, as in the pictures from Wikipedia below.

Well, since I mentioned that explicitly, yeah. A cross-section of it is displayed in this diagram that I repeat:

The "red" reflection is a 42 degree wide, filled-in circle with a bright rim. The "violet" reflection is 40 degree wide, filled-in circle with a bright rim. Other colors do the same in-between. The 2 degree difference is caused by aberration, not separation. The specific angles are caused by the fact that the equation you didn't see Snell's Law represented in has a maximum at about A=60 degrees and D=40 to 42 degrees. Not because other colors "separated from" the violet light the eventually deflected 40 degrees. It is an example of dispersion, which means the equation evaluates differently for different colors, not the they spread out.

Edited by JeffJo
##### Share on other sites

On 6/27/2023 at 2:31 AM, JeffJo said:

Why? There is nothing "partial" about separation. But the term applies to individual rays, not the entire reflection. It is total in individual rays. But at most observations angles where a reflection is observed, every color of light is observed in that reflection.

No, the equation refers to the angle of deflection. That deflection is caused by two transmissions (refraction accompanies transmission) and one reflection. And it references Snell's Law:

Well, since I mentioned that explicitly, yeah. A cross-section of it is displayed in this diagram that I repeat:

The "red" reflection is a 42 degree wide, filled-in circle with a bright rim. The "violet" reflection is 40 degree wide, filled-in circle with a bright rim. Other colors do the same in-between. The 2 degree difference is caused by aberration, not separation. The specific angles are caused by the fact that the equation you didn't see Snell's Law represented in has a maximum at about A=60 degrees and D=40 to 42 degrees. Not because other colors "separated from" the violet light the eventually deflected 40 degrees. It is an example of dispersion, which means the equation evaluates differently for different colors, not the they spread out.

Dear oh dear.

Whilst I accept that much of what you are trying to say is in principle correct, you don't seem to be explaining it very well.

I'm sorry but several members from different English speaking countries on diverse continents have all told you that they understand that when rays of different colours that are initially combined into a single ray, start going in different directions from some point or other in their path separation is said to occur from that point.
Yes I agree you could also say they diverge or use yet other terms, but you cannot discount all of us who use the term separation.

The equation you quoted (which I agree with)

D(n) = n*180+2A - 2(n+2)B is purely geometric and does not rely on Snell's law.

It is true that you can substitute for B as a function of A but this remains true regardless of whether that function obeys Snell's law or not.

In other words the above equation is geometry, Snell's law is physics.

Swansont nailed it when he said to you

On 6/16/2023 at 1:24 PM, JeffJo said:

 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.

On 6/20/2023 at 6:36 PM, studiot said:
• 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

Please note this  picture you posted breaks the condition I already poste as in the quote below yours.

The image on the screen is on the same side of the simulated water drop as the light source. (that is the wrong side for a rainbow).

In particular it is a real image.

The sky image of a rainbow is a virtual image.

Your screen corresponds to the retina of the eye or the plate of a camera system.

The real image on the retina or camera is caused by the optics of the eye or camera focusing the virtual image of the rainbow itself.

The real image is not the rainbow.

As I said in my first post it is very complicated, and this is still not the whole story because the red and blue etc do not come from the same drops.

So in that sense we do not see a split of the red and blue from the same drop, we see red from a drop at a certain altitude and blue from a drop below it, brought to the same focus on the retina.

##### Share on other sites

• 1 month later...

A couple of associated observations.

First one of my own on a recent visit to the National Aquarium in Plymouth.
Here they have many viewing 'portholes' into various tanks.
Some of these portholes have a hemispherical window to give a fisheye view of the interior.

So the setup is similar to the glass ball @JeffJo posted and earlier picture to.
However the lighting is inside the tank so on the opposite side from the viewer.

I looked very carefully but could discern no rainbow halo.

This is in accordance with Met Office stipulation that the viewer has to be between the light source and the reflecting/refracting ball.

Secondly here is an interesting associated phenomenon as recorded by the joint French - Spanish survey to measure the shape of the Earth.

Quote

Between gales, Pambarmarca's ghostly forts produced for Bouger a memorable spectacle. The Breton hydrographer was amazed by a phenomenon that he was later credited with being the first to analyse. It occurred on mornings when,  ' a very brilliant rising sun' , projected his image on to a nearby bank of cloud. Of particular fascination was the rainbow coloured halo around his head. and the fact that others could see their own image too, yet not those of the companions they were standing beside.

##### Share on other sites

On 8/10/2023 at 1:00 PM, studiot said:

A couple of associated observations.

First one of my own on a recent visit to the National Aquarium in Plymouth.
Here they have many viewing 'portholes' into various tanks.
Some of these portholes have a hemispherical window to give a fisheye view of the interior.

So the setup is similar to the glass ball @JeffJo posted and earlier picture to.
However the lighting is inside the tank so on the opposite side from the viewer.

I looked very carefully but could discern no rainbow halo.

This is in accordance with Met Office stipulation that the viewer has to be between the light source and the reflecting/refracting ball.

Secondly here is an interesting associated phenomenon as recorded by the joint French - Spanish survey to measure the shape of the Earth.

The effect you describe is probably what is called a glory, or spectre. See https://en.wikipedia.org/wiki/Glory_(optical_phenomenon)

To understand rainbows and many similar (but often unrelated) effects, start at https://atoptics.co.uk/rainbows/primrays.htm . If you move the slider, you can see how the white incoming ray separates between red and violet. But also and how that separation has nothing to do with the location of the "angle of minimum deviation" or "rainbow angle" as described in this authoritative text.

But I'm getting a little tired of the people here who are less interested in a scientific approach, as much as conforming to what they learned. No matter how clearly it is contradicted by actual facts. LIke http://www.trishock.com/academic/rainbows.shtml

On 6/29/2023 at 3:30 PM, studiot said:

The equation you quoted (which I agree with)

D(n) = n*180+2A - 2(n+2)B is purely geometric and does not rely on Snell's law.

It is true that you can substitute for B as a function of A but this remains true regardless of whether that function obeys Snell's law or not.

I'm glad you agree. But this isn't a vote, it is fact. And yes, the path description holds when angles A and B apply, whether or not the function obeys Snell's Law. But another fact is that when it applies to rainbows, it always does obey Snell's Law. So such speculation is moot. The correct deviation depends on Snell's Law.

##### Share on other sites

13 minutes ago, JeffJo said:

The effect you describe is probably what is called a glory, or spectre. See https://en.wikipedia.org/wiki/Glory_(optical_phenomenon)

To understand rainbows and many similar (but often unrelated) effects, start at https://atoptics.co.uk/rainbows/primrays.htm . If you move the slider, you can see how the white incoming ray separates between red and violet. But also and how that separation has nothing to do with the location of the "angle of minimum deviation" or "rainbow angle" as described in this authoritative text.

But I'm getting a little tired of the people here who are less interested in a scientific approach, as much as conforming to what they learned. No matter how clearly it is contradicted by actual facts. LIke http://www.trishock.com/academic/rainbows.shtml

thank you for this useful information.

I have learned something.

14 minutes ago, JeffJo said:

I'm glad you agree. But this isn't a vote, it is fact. And yes, the path description holds when angles A and B apply, whether or not the function obeys Snell's Law. But another fact is that when it applies to rainbows, it always does obey Snell's Law. So such speculation is moot. The correct deviation depends on Snell's Law.

But sadly,  if you continue to talk down to other folks in this manner you will become pretty lonely here.

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account

## Sign in

Already have an account? Sign in here.

Sign In Now
×

• #### Activity

• Leaderboard
×
• Create New...

## Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.