I On Mixing Colors of Light

[DaveC426913]

I'm dying to ask about your pickup line intro.

 

[Charles Link]

I'm dying to ask about your pickup line intro.

I'm really too old to have much success with the younger generation, but if I have someone who I like, I might try a riddle on them that I got from the wife of one of my Saturday morning baseball players: "What type of fish does a bird like to sit on?" (You should be able to come up with the answer.) LOL

We got off topic here. Back to the physics=I'd still enjoy some feedback on the mathematics of the CIE map that I came up with=maybe people already have it figured out, but it took a google on my part, and one of the key things that I found in the google and applied it is that for any and all of the $(x,y,z)$ color coordinates $x+y+z=1$.

 

[Charles Link]

Just a brief comment or two about the CIE color chart: I had seen this chart 40+ years ago, but it isn't at all obvious=I found this by googling it last week, that there is a 3rd variable z that is part of the color coordinate description, and I figured out that the chart is a cross section in the plane $x+y+z=1$ of an $X,Y,Z$ vector space , where the $X$ is the red cone response or intensity of a light source, the $Y$ is the green cone intensity, and the $Z$ is the blue cone intensity. Any visible light source is characterized by these 3 terms, and the color is characterized by where the $X,Y,Z$ vector crosses the plane $x+y+z=1$. For a given $x$ and $y$, the $z$ is found by $z=1-x-y$.

They would really do well to mention/footnote about the 3rd variable when displaying a CIE chart. One other item is that ordinarily the direction of a vector is characterized by the direction cosines of its normalized unit vector. Instead, with the CIE system they characterize the direction of the vector by giving the coordinates $(x,y,z)$ where the vector (with its tail at the origin), crosses the plane $x+y+z=1$, (instead of where it crosses the unit sphere $x^2+y^2+z^2=1$).

I'm mostly repeating/ summarizing what I have previously posted in the above thread. I think there may be others who have also seen a CIE color chart, who also didn't have a clue that it represents 3 variables=many might think that it is just an x-y graph. It is actually a view from above of the cross section of a 3-D space that lies in the plane $x+y+z=1$.

See post 16 above for the CIE color chart with its x-y axes.

I am hoping that there are at least a couple of people that find this of interest. I welcome your feedback. :)

Edit: Just an additional comment=to see what this whole thread is about, you really only need to read the OP and this last post 53, where I summarized how to read the CIE color chart. (The chart can be found in post 16). One minor correction is that it seems that yellow light occurs in the narrow band from 580-590 nm, rather than at 600 nm, but in any case, as is seen by reading the CIE chart, light that appears yellow can indeed be made in a composite manner by the right combination of green at 550 nm and red at 650 nm, even though a prism spectrometer will be able to tell the difference with complete certainty where our eye cannot.

Edit 2: One detail that might be worth mentioning in reading the OP , especially for those not very familiar with optics, is how you can sample the output of a prism at a given angle. To do this, right after the prism you place a converging lens, and parallel rays will come to a focus in the focal plane of the lens. Parallel rays at a different angle will focus at a different location in the focal plane. In this way the rainbow of colors is generated in the focal plane of the lens.

 

[sophiecentaur]

It is actually a view from above of the cross section of a 3-D space that lies in the plane x+y+z=1.

The CIE chart only shows chrominance, which can actually produce very different perceived 'colours' when Luminance changes. Two areas with the same XY co ordinates can often look very different with different Z. For, high and low luminanceareas can look yellow and brown. Pale 'European' skins tend to have very similar chrominance values to darker skins because the dark pigment is a pretty good neutral grey. Hardly surprising really because the tissues , fat, blood etc. are much the same for all skins.

When the eye sees coloured light it also takes into account the other light sources around and 'integrates to grey' , which is the best it can do so the actual chrominances in a scene can look almost the same in bright light or dim light There are hundreds of made-up images (illusions) that totally confuse the eye because surrounding areas affect what you 'see'.

 

[Charles Link]

The CIE chart only shows chrominance, which can actually produce very different perceived 'colours' when Luminance changes. Two areas with the same XY co ordinates can often look very different with different Z. For, high and low luminanceareas can look yellow and brown. Pale 'European' skins tend to have very similar chrominance values to darker skins because the dark pigment is a pretty good neutral grey. Hardly surprising really because the tissues , fat, blood etc. are much the same for all skins.

When the eye sees coloured light it also takes into account the other light sources around and 'integrates to grey' , which is the best it can do so the actual chrominances in a scene can look almost the same in bright light or dim light There are hundreds of made-up images (illusions) that totally confuse the eye because surrounding areas affect what you 'see'.

I don't know that you have completely followed my explanation of the chart. There is the capital $(X,Y,Z)$, and the small $(x,y,z)$. The chart itself is small $(x,y,z)$, and for a given $(x,y)$, there is only one $z$ which is computed to be $z=1-x-y$.

From what I can tell, I think the chart has been somewhat misunderstood or not fully understood by many. If we go to a different brightness level, we do scale up the $(X,Y,Z)$, but the $(x,y,z)$ stays the same. (The $(x,y,z)$ is found by where the vector $(X,Y,Z )$ crosses the plane $x+y+z=1$.) That of course is then what you are saying, that the $(x,y,z)$ for the scaled up case should really take on a different position on the chart=the chart assumes linearity of the system at all light levels, and that is only good to a very rough approximation at best. In any case, I find the chart to be based on some very good mathematics.

Edit: Computing where the $(X,Y,Z)$ crosses the plane $x+y+z=1$, it is given by $x=X/(X+Y+Z)$, $y=Y/(X+Y+Z)$, and $z=Z/(X+Y+Z)$. The $(x,y,z)$ are the color coordinates, where the $(x,y)$ is found on the chart, and it is simple arithmetic to compute the $z$ which is $z=1-x-y$. The chart is a map on the plane $x+y+z=1$ of the color that is found for each vector $(X,Y,Z)$ where each vector's color/direction is determined by where it crosses the plane $x+y+z=1$.

(They really IMO would do well to footnote that $z =1-x-y$ when showing the chart. The chart is actually a view from above of the plane $x+y+z=1$ that is colored in showing the colors of each $(x,y,z)$). .



Scaling up the $(X,Y,Z)$ gives the same $(x,y,z)$ on the color chart, which is really the best we can do with a linear model. It is not expected to be perfect, but it is reasonably accurate for many cases.

 

[sophiecentaur]

From what I can tell, I think the chart has been somewhat misunderstood or not fully understood by many.

Very true. The CIE chromaticity chart seems to be the result of several steps to produce a practical way into colourimetry without going into the details that you find in the Wiki Article, in which I (today) re-visited what I learned in the late 60s. (Not explicitly used since I moved departments soon afterwards.) Wiki can't always be relied upon but that article seems ok. The misunderstanding that you refer to is probably because it's not something you can reverse engineer - and it's much better than near enough for Jazz.
The experiments to determine the sensitivity curves are a great example of pulling yourself up by your own bootstraps but the end product (CIE chart) has proved itself to be useful. No lasers in those days.
The negative sensitivity values in the normalised sensitivity curves need thinking about!
The "similar threads" below are worth while looking through.

 

[pinball1970]

Good answer, but you reversed rods and cones.

Cones colour??

Operate in bright light and rods in dim, peripheral but 'grey' vision in the dark/low light.

 

[pinball1970]

Let's not forget that "yellow" is highly subjective. It's different for everyone, and even different depending on context.
View attachment 352556View attachment 352555

Anyone encountered this little test?
https://ismy.blue/
My score (today) is 172.

Pantone split them out and assign a hue, "lightness" and "C"
TPX is printed usually for textiles.
"C" is coated for paper.
The number make sense but the names as you can see are design centric madness.

 

[Charles Link]

The experiments to determine the sensitivity curves are a great example of pulling yourself up by your own bootstraps but the end product (CIE chart) has proved itself to be useful. No lasers in those days.
The negative sensitivity values in the normalised sensitivity curves need thinking about!

Yes, very good. I also thought the sensitivity curves that had some negative numbers were something that looked to be in error. That could be one of the reasons why, from what I could tell through the Wiki article, that they redid the original R,G,B response with X,Y,Z responses. I also "linked" the Wiki article in post 32 above.

and thank you very much for your interest in the topic. :)

 

[Charles Link]

Just an additional simple comment about the CIE map, e.g. as shown in post 37: I think in general they do get fairly accurate the wavelength numbers in the chart as they move around the tongue-shape from 380 nm to 750 nm. I do think though that on these charts we may be seeing some false coloring: e.g. yellow is supposed to occur between 580 nm and 590 nm and on my screen it looks very orange. It is my guess that it is very difficult to actually make an accurate color chart on our computer screens, since they are only working with 3 leds, and the entire upper left corner of the chart is also expected to then be somewhat inaccurate.

To make a more much accurate chart as far as viewing the colors, they would need to use perhaps 5 or 6 different led's spaced around the outer portion of the tongue of the CIE map. The led's also need to be near the edges rather than having color coordinates somewhere towards the interior of the map.

If you print it out with a color printer, it's likely to be even more approximate.

 

[sophiecentaur]

I do think though that on these charts we may be seeing some false coloring.

Yes; totally false /unreliable. I have a suspicion that the charts we see (the Google Results Page) are all copy and pastes of one original chart which was created way back when everything was different. We can ignore the printed charts, which can never be right - just an indication of what happens.

But there will be a massive spread in regular TV displays because they are each busy producing the colours which their designers choose with the phosphors they've been given and with the white point that's been agreed on. Opinions about the "Yellow" we are all looking at (e.g. the post above) are not really valid - the viewing conditions for each of us are different, as well as the make of display. Go into a TV shop and look at the range of TV's (even within one make). Since we moved on from NTSC, TVs stopped having a Tint control but down in the depths of your preferences, you will find a set of options - one of which is 'Vibrant' and which is recommended for displays on the shop floor.

I already made my point about why the Y on the chart may not correspond to spectral yellow, the reason being that the original demo and 'conclusion' predates all real colour TV displays. If you were teaching the elements of colourimetry, in the 1930s, what would you have done to convince a class? R+G=Y may be in the realms of "Nature abhors a vacuum."

PS How many people reading this are viewing with a filament lamp at a known temperature? All your experimental conclusions are dodgy as you sit there today.

 

[Charles Link]

How many people reading this are viewing with a filament lamp at a known temperature? All your experimental conclusions are dodgy as you sit there today

I would like to see someone try the Gedanken experiment in the OP. I am pretty certain of the results though, if they could get the right proportion of green and red to generate the yellow. I don't currently have access to a prism or other experimental apparatus, but it could be a fun experiment to try. :)

Edit: and basically the tungsten filament lamp at 2500 K can be any filament lamp generating white light. They typically operate somewhere around that temperature.


Edit 2: and I thought I really had a good (Gedanken) experiment in the OP. I am surprised to date it has only received one "like" and that one came in within about 5 minutes after I first posted it.

 

[sophiecentaur]

"Gedanken" is 'all in yer head mate'. You can get any answer you want.

 

[Charles Link]

"Gedanken" is 'all in yer head mate'. You can get any answer you want.

Gedanken is German for thought. (I studied some German)=Gedanken experiment=thought experiment. I've got enough of an experimental background, especially with diffraction grating based spectrometers, as well as enough theoretical background in Optics, that I'm far from being just another beginner in these categories.

Meanwhile, although I didn't reference it to create the OP, the CIE color chart with its upper right border where $z \approx 0$ (with $x+y= 1$) =no blue content, on the pure spectral region from green at 550 nm to red at 650 nm lends a great deal of credence to the proposed experiment, even though no one has yet to say they tried it. The 580-590 yellow lies on the straight line connecting these two, so mixing the green and red in the right proportions should generate what appears as 580-590 nm.

I do think this one is good physics, rather than simply wishful thinking. Cheers. :)

 

[sophiecentaur]

I do think this one is good physics,

Except that no two spectral nor non-spectral colours can mix to produce a spectral colour. You can't draw a chord inside a curve and expect it to touch (/match). Every colour produced by a TV display has to lie within a triangle of whatever primaries you choose.

I'd say that colourimetry is not actual 'Physics' at all (however well it delivers good TV pictures) but a numerical description of subjective results.

so mixing the green and red in the right proportions should generate what appears as 580-590 nm.

The key word here is "appears". Thing is were pretty well never see spectral colours so what 'appears' to be a spectral colour could be very far away from that spectral curve. There are people who firmly believe that the 'blue' of the sky is spectral blue - have you seen the spectrum of or even looked at the RGB values on your favourite holiday pictures of a blue sky. Same goes for the 'pure' rainbow that people rave about.

 

[Charles Link]

Same goes for the 'pure' rainbow that people rave abou

I do think the rainbow is similar to a prism spectrometer=the light is separated into components=basically wavelength as a function of angle, but a good prism with the right wavelength dependent index of refraction will do a much better job. You do indeed get wavelength as a function of angle, and if you focus it with a lens to a focal plane, you will get wavelength as a function of position.

I think the creators of the CIE color chart did a good job when they put the pure spectral wavelengths on the border of the chart.

So far, I haven't gotten the best feedback from the posts I've put in this thread, but perhaps there are still some readers that can recognize what I think is good and also interesting physics. I think you do see some merit in it as well, but remain somewhat of a skeptic. :)

Edit: and you mention the blue sky=that is Rayleigh scattering of the white light, basically a T=6000 K blackbody spectrum from the sun, with scattering inversely proportional to the 4th power of the wavelength.

 

[sophiecentaur]

It does appear from the CIE chart, and its nearly straight line border that we should be able to generate what appears as an almost very precise yellow at 580-590 nm to the eye from wavelengths of green at 550 nm and red at 650 nm in the right combination.

You wrote this, several posts ago and I have to take issue with it. Because of the way your colour vision processing works, you are always looking for a point that you can call the White of the illuminant. Only when you have decided on that point can you specify where B=0 and R and G = 1 for a given set of phosphors. Or, put it a different way, your eye is busy 'integrating to grey' and what it decides is a good yellow may lie in a wide range of places on the CIE chart. By extension, I'd suggest that spectral yellow could well 'look' very far from yellow in suitable background lighting (or the rest of the picture). Basically , the Illuminant makes a massive contribution to the subjective judgement of colours that we see.

Remember the the faces / complexions of women (in particular) in the cosmetic department of the old big stores, lit with massiv e fluorescent tubes. All the stuff they put on their faces produced that 'certain look' which the customers wanted to mimic. When they get (got?) home it's never quite the same. Remember how poorly people used to look under some street lamps? Same thing and that's not simple Physics.

 

[Charles Link]

It would be nice if we could get someone to conduct the proposed experiment. Otherwise, we are left with a lot of opinions, that may be a little biased, including mine. Cheers. :)

 

[sophiecentaur]

I do think the rainbow is similar to a prism spectrometer=the light is separated into components=basically wavelength as a function of angle,

The light from a rainbow arc is miles away from the spectral curve, it's very often very near to the white 'colour' of the Sun; it's usually very de-saturated just compare the next rainbow you see with what a prism can give you. Keep a prism handy in your pocket and do the experiment when you see a 'stunning' rainbow.

Look at the RGB values on your camera picture files. Prepare to be very surprised.

 

[sophiecentaur]

It would be nice if we could get someone to conduct the proposed experiment.

Perhaps you should try to define exactly how you would do this experiment (it has to be hardware). Your Gedanken is leaving our a lot of important parameters.

 

[Charles Link]

The light from a rainbow arc is miles away from the spectral curve, it's very often very near to the white 'colour' of the Sun; it's usually very de-saturated just compare the next rainbow you see with what a prism can give you. Keep a prism handy in your pocket and do the experiment when you see a 'stunning' rainbow.

Look at the RGB values on your camera picture files. Prepare to be very surprised.

The problem with the rainbow, if I understand it correctly, is some light will also reflect specularly off the front face of the droplets=so yes, on that one, I stand corrected. It's not the same as a prism. :)

 

[Charles Link]

Perhaps you should try to define exactly how you would do this experiment (it has to be hardware). Your Gedanken is leaving our a lot of important parameters.

The experiment is really a very simple one, for anyone with an Optics lab that has a couple of prisms of the type that are used in spectrometers.

See also from the bottom of post 53: Edit 2: One detail that might be worth mentioning in reading the OP , especially for those not very familiar with optics, is how you can sample the output of a prism at a given angle. To do this, right after the prism you place a converging lens, and parallel rays will come to a focus in the focal plane of the lens. Parallel rays at a different angle will focus at a different location in the focal plane. In this way the rainbow of colors is generated in the focal plane of the lens.

To get collimated/parallel rays onto the prism, you use a slit (or aperture) for the incident light and the slit is in the focal plane of a converging lens. The light will emerge from the converging lens with parallel rays incident onto the prism.

The light will emerge from the prism as parallel rays for each given wavelength, but the angle they emerge at will be wavelength dependent.

It's a simple matter to sample the green and red light with a couple slits in the focal plane of the exit lens from the prism and combine focused or even out of focus onto a white sheet of paper, (with other lights out during the experiment).

 

[sophiecentaur]

The problem with the rainbow, if I understand it correctly, is some light will also reflect specularly off the front face of the droplets

And what about all the desaturated blue that comes from all other directions in the sky. There is no 'net' dispersion of that light so it's a hemisphere's worth of solid angle of non-coloured light, scattered off the raindrops, to dilute those pretty rainbow colours.
But I can see that you can't come to terms with what I'm saying about the lack of specular spectral!!! Edit. colours for our vision to deal with. Just go outside with a prism / crystal drinking glass and see for yourself how desaturated.

I think the creators of the CIE color chart did a good job when they put the pure spectral wavelengths on the border of the chart.

The spectral colours are just on the limit of visibility. The inventors of the CIE chart didn't make any actual 'choice'. The space outside that closed (coloured) area just doesn't exist in visual terms. Look on it as a Mapping with an odd mapping rule. Forbidden spaces are not rare in mapping systems.

 

[Charles Link]

And what about all the desaturated blue that comes from all other directions in the sky. There is no 'net' dispersion of that light so it's a hemisphere's worth of solid angle of non-coloured light, scattered off the raindrops, to dilute those pretty rainbow colours.

For the rainbow, you are completely correct that there will be a lot of broad spectrum light mixed in with the rainbow.

For the prism, with the lenses at entrance and exit from the prism I have mentioned, which is what they use in a prism spectrometer, the light will come out as individual wavelengths at each location, with a small spread $\Delta \lambda$ (determined by the slit width that is used, as well as the $dn/d \lambda$ of the glass of the prism) that will typically be about 2-3 nm.

(Incidentally this same type of optical system, ( see post 72), often with spherical mirrors instead of lenses, is also used in diffraction grating type spectrometers. There may be those who ask, how do we first start out with an entrance slit for the spectrometer, and then fill the entire diffraction grating, typically 2" x 2" ? This answers how it is done, as well as how we sample the emerging light at a given angle from the grating (or prism)).

Once again, I do think I have what could be a very good experiment to try in the OP. I am hoping others find it of much interest, and I really don't see any well-founded reasons why it wouldn't work. It should be easy enough to do, if someone wants to try it and test it. :)

 

[Charles Link]

I also propose a very similar but slightly more sophisticated experiment from that of the OP: If you have a diffraction grating spectrometer available to you, as well as a Hg (mercury arc lamp) and a Na (sodium) lamp, tap off the (green) 546.1 nm spectral line from the Hg and mix it with a (red) 632.8 nm HeNe laser on a white sheet of paper, and it is my best assessment, if you get the proportions right, you won't be able to tell this apart from the (yellow) Na doublet at 589.0 nm (and 589.6 nm).

 

[sophiecentaur]

I do think I have what could be a very good experiment

You still haven't formally described the experiment

, if you get the proportions right

Thats the essence of additive colour mixing. No one would challenge that open statement as far as it goes. But that's a million miles from R=1,G=1 gives you a perfect match to spectral sodium lines. How would you define the positions of the three phosphors on the CIE chart?
Your 'experiment' takes place in every TV display that you watch (plus or minus a bit).

 

[Charles Link]

You still haven't formally described the experiment

Thats the essence of additive colour mixing. No one would challenge that open statement as far as it goes. But that's a million miles from R=1,G=1 gives you a perfect match to spectral sodium lines. How would you define the positions of the three phosphors on the CIE chart?
Your 'experiment' takes place in every TV display that you watch (plus or minus a bit).

I can't say I disagree with you, but you seem to be a little closed to something that could make a good laboratory demonstration. The experiment is really simple enough in either version, (post 1 or post 75), that I think I have already presented it in more than enough detail.

and yes, this same concept is used on TV and computer screens every day, but how many out there have any working knowledge of it, e.g. by doing a laboratory demo at the university or college? In some ways, I think you understand the concepts in enough depth that you might even find it unnecessary to do a Physics Forums post on the topic, but I think for some others, they may find it of interest. :)

 

[Charles Link]

See https://www.luxalight.eu/en/cie-convertor
I found this software interesting. For wavelength $\lambda$= 546 nm , you get (red cone) x=.27 and (green cone) y=.72 and for 633 nm you get x=.71 and y=.29. You can then let the capital X and Y be these values and combine (sum) them to get X=0.98 and Y=1.01 , which scales to x=.49 and y=.51 as the color coordinates, with z (blue cone) being very nearly zero in all cases. The x=.49 and y=.51 is found to appear the same as yellow at 577 nm.

See also post 75 where I used the same green and red wavelengths in what could make an interesting demo. Looks like it needs just a little more of the 633 nm in the combo to get it to appear nearly the same as the yellow Na doublet at 589 nm which has x=.57 and y=.43.

Edit: I think to determine proportions, in practice they begin by setting the Y's the same for the two vectors with different color coordinates, i.e. let each $Y=1.0$, and then scale up each vector accordingly, and then determine the ratio needed with an additional scaling factor on one of the vectors to get the proportion needed to generate the final desired color coordinates. It is the Y that they use to determine brightness.

 

[sophiecentaur]

in what could make an interesting demo

It might interest you in a private sort of way (quite fair enough and - enjoy) but you are mixing up mathematical formulae with subjective experience. The subjective experience would only be verifiable under highly controlled viewing conditions. You don't seem to have acknowledged the reality and difficulty of accurate colourimetry.
Colourimetry seldom involves spectral wavengths. The 'interesting' bit is how three coloured phosphors can add to produce a convincing match within that triangle. On the computer you are using now, you can install (free) basic photo software with control of RGB values for an area on the screen. It's simple to get a tolerable match against a coloured object AND to see how the real life illumination affects that match. Job done and you may not even need to get out of your chair.

 

[sophiecentaur]

I do think I have what could be a very good experiment

You still haven't formally described the experiment

 

[Charles Link]

You still haven't formally described the experiment

It's just a very simple demo, and seeing from your other post the extent of how much experimental work has been conducted in the field of colorimetry, what I have almost wouldn't even be in the category of experiment.

From what I can tell, you may have even worked in this field for a while, so you could be light years ahead of where I am for experience in mixing a couple of colors of light together. I am used to working with what you might call pseudo monochromatic sources from the exit slit of a spectrometer, but I never tried mixing them together. I find the subject very interesting, but for this mixing colors topic, I am very much a beginner, even though I think I did get it right in the OP that you can mix spectral green with spectral red and get a yellow that is not spectral yellow, but could appear as spectral yellow. Cheers. :)

and the other concept mentioned in the OP is that when you mix them, you haven't changed the original physical form of the components of the light=the green and red still exist independently and can be separated out with a second prism. Something very elementary, and maybe very obvious to many, but I thought it worth mentioning when I wrote the OP a couple weeks ago.

 

[sophiecentaur]

you seem to be a little closed to something that could make a good laboratory demonstration.

You might be interested to know that when I was teaching, I gave a (year 10, iirc) demo to a Science class, in the School's small theatre. I got three spots with random RGB (ish) filters in them and from the lighting control booth, showed the results of very crude RGB mixing and also the effect of illuminant colour and shadows on the appearance of a person with various brightly coloured props.

No need for prisms or pure spectra; the basic idea is very easy to demonstrate, particularly in a dark theatre and with bright, large spots of coloured light. But. as I have already pointed out, you're only half way there to producing a synthesis that genuinely mimics the tristimulus sensitivity of the human eye in everyday life. Needless to say, they enjoyed it - it was not a real lesson but they mostly seemed to get the message.

If you have ever worked in a School, you will appreciate that the admin of setting up a demo with another department's equipment was really hard work and I never bothered to repeat it. I called in a lot of favours for that.

 

[Charles Link]

@sophiecentaur Very good=excellent. In some ways the Gedanken experiment I proposed in the OP might have been better if it were 50 or 60 years ago or more, but I still find the topic rather fascinating. I also think it would do well for a student these days to learn about the CIE chart, as is described in post 38 and other posts in this thread, even if it doesn't work completely in reality, because it assumes a linear response in the color cones in our eyes, which in actuality have a very non-linear character to them.

One concept that is present in the OP, which may not be entirely obvious, is that the light itself, with the different wavelengths does behave very linearly. When two sources are added together, their spectral content is completely linear. The silicon photodiode, used in cell phone cameras, etc. , is linear over several orders or magnitude or more, even though our eyes are not. Meanwhile if we add green and red light, it may look like yellow light, but it is still green and red light.

It may be worthwhile giving a "link" to a recent related post, where they discuss mixing of colors of paints. To me it is still fascinating that blue paint mixed with yellow will give you green. It really never gets old. See https://www.physicsforums.com/threa...each-contributing-hue-are-subtracted.1066237/ I was very much impressed with the answers that you @sophiecentaur and others gave in this thread. Cheers. :)

 

[sophiecentaur]

To me it is still fascinating that blue paint mixed with yellow will give you green.

Very different method here; it's subtractive. Firstly, the pigments used for three colour mixing have to reflect / pass a very broad spectrum of wavelengths so between the three primaries CMY, all wavelengths have to be reflected (or you won't see them).

The blue dye must reflect some green but no red. Also, the yellow dye must reflect some green but no red. The only colour band that will be reflected and not absorbed will be common to both dyes - i.e. green
If you want a paint with high reflectivity and a high saturation (as in bright cloths and logos etc.) you can't do it with CMY; you have to add in areas of spot colour where 'all' the wanted colour is reflected so the chemistry will have to be entirely different.
Mixing paints tends to give you brown / sludgy grey. All good paintboxes have loads of different pigments.

 

[Charles Link]

Thank you @sophiecentaur Very informative. I really hope this topic is something that the undergraduate physics students get shown in detail in their coursework, and that they don't miss it. Cheers. :)

 

[sophiecentaur]

I really hope this topic is something that the undergraduate physics students get shown in detail in their coursework

Problem is that it's very much a specialist set and doesn't link well to any other Physics field. When would you teach it?

Colourimetry is Psychophysics, really. It hangs on the tristimulus theory, rather than the science of producing filters and light sources. But all the big organisations must have experts on colourimetry so there would be a good career out there, waiting for a bright person to step in.

I left it all behind, years ago but, as with Catholicism, 'give me a colourimetry student for six months and he'll be into colourimetry all his life. (/her)

 

[Gleb1964]

The blue dye must reflect some green but no red. Also, the yellow dye must reflect some green but no red.

You want to say that yellow does not reflect blue.

 

[sophiecentaur]

You want to say that yellow does not reflect blue.

Yes; the Yellow dye must reflect wavelengths to and slightly beyond the boundaries of both the C and M filters so that must reject blues but pass orangey reds and some greens. Where the passbands actually meet and overlap must be difficult to choose. I guess the actual sources of materials would affect the choice.

It's easy for the CMY world to confuse a bear of little brain (I mean me). RGB mixing is easier to grasp, imo.

 

[Charles Link]

Yes; the Yellow dye must reflect wavelengths to and slightly beyond the boundaries of both the C and M filters so that must reject blues but pass orangey reds and some greens. Where the passbands actually meet and overlap must be difficult to choose. I guess the actual sources of materials would affect the choice.

It's easy for the CMY world to confuse a bear of little brain. RGB mixing is easier to grasp, imo.

I had to google this, and I see that CMY=cyan, magenta, yellow are the three colors that are used for printing. :) See https://blog.thenounproject.com/rgb...erence between the,, flyers or business cards).

 

[Charles Link]

Problem is that it's very much a specialist set and doesn't link well to any other Physics field. When would you teach it?

I think it would be good to introduce it to the physics majors in the third course of their first year/second year studies. They wouldn't need to cover it in great depth=perhaps about as much as we did here on the Physics Forums.

 

[sophiecentaur]

I think it would be good to introduce it to the physics majors

An interesting topic, of course but how would it fit into any existing scheme of work / syllabus? When I taught UK A level Physics, it was so content-loaded that fitting extra bits was difficult

 

[Charles Link]

An interesting topic, of course but how would it fit into any existing scheme of work / syllabus? When I taught UK A level Physics, it was so content-loaded that fitting extra bits was difficult

Then I guess it is all the better that we treated the topic of the mixing of colors of light on Physics Forums. I think most every physics major would benefit by having at least a little familiarity with the subject. :)

Edit: and perhaps worth mentioning again, the CIE theory in a nutshell is that the laws of vector addition apply for any two sources with their red=X, green=Y, and blue=Z components, so that $\vec{R}_{total}=\vec{R}_1+\vec{R}_2=(X_1+X_2) \hat{i}+(Y_1+Y_2) \hat{j}+(Z_1+Z_2) \hat{k}$.
Meanwhile the CIE color coordinates $(x,y,z)$ are found by where the vector $\vec{R}$ crosses the plane $x+y+z=1$, so that $x=X/(X+Y+Z)$, etc. (I'm repeating post 38).

 

[Charles Link]

and now for an observation of interest regarding the TV and computer screens that I have: On an older flat screen TV, perhaps 30 years old, I could clearly see the array of blue, green, and red LED's with my telescope eyepiece as a magnifying glass. With a newer TV and with this Chromebook computer screen, the LED array has much finer structure. I couldn't even tell for sure if they were using 3 LED's or possibly 5 or 6 or more to make the picture. Perhaps someone can supply some detail here=I could see with the eyepiece that there was some structure, but it was really too fine to totally resolve. It looked like there might even be a white LED in the structure, and perhaps even a yellow one, but I couldn't say for sure.

Edit: a google seems to suggest that they now might be using a white LED along with the red, green, and blue, but it wasn't real clear or definitive on the topic. Having a white one would provide an operating point near the center of the CIE chart, and make things easier for the other 3 LED's.

Edit 2: an additional google mentions things such as a white LED backlight, as well as things getting much smaller, (pixels, etc.), so that the older TV is actually much more interesting and definitive for viewing with the eyepiece from a telescope. Some of this nano type technology has gotten rather complex and much more difficult to figure out.

 

[Charles Link]

In reviewing this thread, I still encourage physics students to see posts 37,38, 46, and 55 (all on page 2) of this thread to learn about the CIE color map. From what we (@sophiecentaur and I ) concluded around post 90, the CIE color coordinates which came around 1931 is in many cases not being presented in the standard undergraduate physics curriculum, because there may be no place where it is a really good fit.

The CIE color map has withstood the test of time though, and should be considered good physics, and today's students would do well to have some familiarity with it. It should help students to understand how the right combination of red, green, and blue light can allow us to perceive most but not all of the colors that the human is able to recognize.

In the OP, a very simple experiment is presented that shows the yellow light you can get with the right mixture of green and red light. The physics student might also find that of interest. The yellow light can actually come in two forms as this experiment shows=pure yellow at around 580-590 nm (600 nm might be starting to be orange=a minor correction), and also from a mixture of green at 550 nm and red at 650 nm or thereabouts.

 

[sophiecentaur]

The yellow light can actually come in two forms as this experiment shows=pure yellow at around 580-590 nm (600 nm might be starting to be orange=a minor correction), and also from a mixture of green at 550 nm and red at 650 nm or thereabouts.

No one would ever use spectral line red or green; for good colour mixing, the primary sources (phosphors or equivalent) would never be monochromatic. In order to get a high output, efficiently, the primaries are broad band. You seem to be looking for something 'significant' about R + G = Y but we already know that the 'yellow' on a T V display is not formed by equal weights of the R and G phosphors. The yellow colour that is produced with that mixture is way off Sodium Yellow but Sodium Yellow is just 'any old colour' for a TV system. My point has always been that this particular mixing demo was first used long before there were TV primary phosphors and was used to 'make a point'. You surely can't think that the whole Colour TV system was invented around synthesising Sodium Yellow. You are putting the cart before the horse; the original choices of primary phosphors will (almost certainly) have been chosen for practical reasons so that a fair range of colours could be matched.

From what we (@sophiecentaur and I ) concluded

You are overstating the claim that we concluded anything. Sorry but I really do not want to be associated with your personal version of colourimetry. Some of what you say is fair enough but does it make total sense when you look at the spectra of three example phosphors? On a plasma screen Red is a difficult one and they are all very wide band .

The only place where they are three 'dots' on a CIE chart.


Plasma Screen Phosphors

 

[Charles Link]

@sophiecentaur Very good. You are supplying much needed detail on the state of the art. I'm also trying to address what may be an audience of freshman undergraduate physics students, but I don't even know for sure because I'm getting very little feedback. From what I can tell, Physics Forums doesn't get the wide audience like it did five or ten years ago or more, because the google search engines aren't directing very many to our posts, but that is another subject. In any case, your feedback is very good, and maybe we really didn't conclude anything, but I wonder how many students get exposed to the color mixing of light topic and the CIE chart as part of their curriculum. Cheers. :)

 

[Klystron]

As the OP requests feedback, kudos for an interesting thread on color mixing and colorimetry. Back when I taught electronics and radar science, students became fascinated with the electromagnetic spectrum with visible light but a small, though important, segment of the continuum.

Physics students at uni put much effort into veiwing and understanding interference patterns derived from interferometers and crystal diffactrion images as much as visible light spectra, supporting the OP's thesis that this subject may be underappreciated or undertaught in physics departments. One found art students, particularly painters, fascinated by color charts, color mixing, pigments, dyes and paint constituents with interest in human perception of visible light.

 

[sophiecentaur]

One found art students, particularly painters, fascinated by color charts, color mixing, pigments, dyes and paint constituents

That is virtually the life blood of most artists and there is a lot of discussion and information; With a few exceptions, artists are dealing with subtractive colour mixing and the CIE chart wouldn't be of any use. With additive mixing, the primaries don't 'tread on each other's toes' but three broad band pigment primaries inherently cause crosstalk. That calls for a different approach. Too hard guv'.

I wonder how many students get exposed to the color mixing of light topic and the CIE chart

Which ten hour university course could include one hour on colourimetry? Who would even teach it? There are so few obvious slots in mainstream science threads and, of course, it's more psychology than Physics.

I could imagine an entertaining and informative extra curricular presentation Physics Club meeting?) working very well. Colour projectors could provide three 'primaries' for mixing and comparison with a laser monochromatic source. Boy, but the preparation time for such a presentation (and clearing up afterwards).

However, apart from the guys I worked with, briefly and many years ago, I never met anyone with a burning ambition to spread the gospel and with sufficient depth of knowledge.

 

[Charles Link]

Just a couple additional comments to post 95 from @sophiecentaur : It may be in practice that most sources have somewhat complex spectral characteristics, but part of the purpose of showing how two monochromatic sources add together, as vectors in the CIE 3-D color space with their coordinates defining the direction of the vector, is it shows the fundamentals of how the CIE system works. The complex spectral sources then have their color coordinates found by performing integrals, but those are nothing more than linear summations of many monochromatic sources. I didn't present the integrals in the above posts, but they can be found in the Wikipedia link of post 32. $X=\int x(\lambda) I(\lambda) \, d \lambda$, and similarly for $Y$ and $Z$.

Even though the human perception ultimately is not entirely linear and perhaps even very non-linear, the linear system that the CIE color coordinates provide can give some very good estimates to what the individual will see for these more complex sources.

 

[sophiecentaur]

but part of the purpose of showing how two monochromatic sources add together, as vectors in the CIE 3-D color space with their coordinates defining the direction of the vector, is it shows the fundamentals of how the CIE system works

I don't know where this is taking you. The analysis of the eye gives three signals, each of which is the correlation of the admitted light and the three analysis curves. What the eye does with monochromatic light is just by the by. We didn't evolve to deal with monochromatic sources so it's no surprise that we cannot actually identify them. Worse than that, when we look at a reasonably saturated colour our brain sees them as much more saturated than they are - hence my comments about the subjective effect of a rainbow; we fool ourselves.

Before the Enlightenment no one ever saw pure spectral sources so why would be ever needed to spot one?

I think you should look at the whole thing again and take on board the fact that both analysis and synthesis are involved. If you want a well matched TV picture then we need to include both in our channel. I think you are a bit too obsessed with spectral colours. Can you apply your personal theory to explain how we see all those non-spectral colours that lie between the straight portion of the CIE chart and the central White Point? Nothing spectral in that area.