I On Mixing Colors of Light

[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.