The Physics of Color Perception

[sophiecentaur]

Would you agree that the blue contains some green and the black is caused by a gap (no color at all) due to the refraction difference between red and blue?

I assume that's your question. There are many google links with the spectra of various primaries. Here's one. There is a small hump (linear scale) in the blue primary spectrum that takes it into the green. None of the primaries is (ever) monochromatic. It would just not be efficient enough.
As for the black stripe - it's just the logical result of the colours on either side and the spread of the components. The prism dispersion is consistent - just follow the rules.

Edit:
There is another factor which affects the outcome of display - camera - display chains. The spectral sensitivity of the three camera analysis filters. https://www.researchgate.net/figure/Ideal-camera-spectral-sensitivity-curves-for-a-Rec-709-b-Rec-2020_fig1_265550301has graphs of the analysis of colour cameras. They show a considerable amount of 'crosstalk' between the channels. The effect needs to be taken into account if you want to know the result of your setup. Remember, the TV system was never designed to be used for your experiment so, unsurprisingly, you got somewhat whacky results. Incidentally, the 'crosstalk' in the analysis is there deliberately to mimic the human optics and it's what allows our perception of colour with only three sensor analyses.

 

[SecularSanity]

OK. The very top left image is the most interesting because it's the nearest to ideal. It shows the nearest to a proper spectrum of the displayed white against a black background (ROYGB, at least). The fact that the black portion is now grey is because of unwanted light on other paths and it 'dilutes' the spectrum that's seen.

Top right you get Blue on the left and Cyan, next in where the blue and green mix. and the rightmost stripe is a sort of Magenta where the red mixes with the blue. Allowing for the imperfections of the experiment, it does what you'd expect.The other colours with a white line seem to do the 'right thing' but it's further and further away from ideal

To clear it up, you could do the same with red, green and blue lines on black to indicate the sort of spectrum the primaries have. But the same caveat applies because the TV system is not a spectrometer and just does its best to 'see things' the way you do.

IMO you have done as much as you can with a limited setup. You would need some proper monochromatic sources and a proper spectrometer if you want to confirm what we all know about refraction and dispersion.

Are you saying that in the blue square, on the left-hand side of the white line, the blue fades into the blue, and then disappears, and on the right-hand side of the white line the blue fades into the red and green creating magenta and cyan?

If so, then in the red square, on the left-hand side of the white line, the blue fades into the red creating magenta and on the right-hand side the red fades into the green creating yellow?

Is that correct?

 

[sophiecentaur]

Are you saying that in the blue square, on the left-hand side of the white line, the blue fades into the blue, and then disappears, and on the right-hand side of the white line the blue fades into the red and green creating magenta and cyan?

Sorry, I don't quite get what you are asking; specifics don't really get us anywhere. All I am saying is that the different components of the colours of the blocks are deviated by different amounts. The diagram of white on black shows by how much. If you take any colour of blocks, the final result will be a combination of what spills over from the next door block or stripe and what is shifter away from the edge of the target block. I went through the easiest example and the same principle applies everywhere.

It's 'and exercise for the student to complete'; I haven't the inclination to do it but if you are really interested, do it with strips of paper with RGB etc. written on them - to represent what the initial dispersion causes; one strip is the target and one strip is the shifted colour. Shift one against the other and it will show you what to expect when each element of the two rows of RGBs are added together. But the actual spectra of the primaries and the actual sensitivity curves will mess it up a bit so don't expect a perfect answer.

The whole idea of Science is to try to regularise the system and identify as few basic rules as possible; I have written sufficient for you to carry on independently,

 

[SecularSanity]

Sorry, I don't quite get what you are asking; specifics don't really get us anywhere. All I am saying is that the different components of the colours of the blocks are deviated by different amounts. The diagram of white on black shows by how much. If you take any colour of blocks, the final result will be a combination of what spills over from the next door block or stripe and what is shifter away from the edge of the target block. I went through the easiest example and the same principle applies everywhere.

It's 'and exercise for the student to complete'; I haven't the inclination to do it but if you are really interested, do it with strips of paper with RGB etc. written on them - to represent what the initial dispersion causes; one strip is the target and one strip is the shifted colour. Shift one against the other and it will show you what to expect when each element of the two rows of RGBs are added together. But the actual spectra of the primaries and the actual sensitivity curves will mess it up a bit so don't expect a perfect answer.

The whole idea of Science is to try to regularise the system and identify as few basic rules as possible; I have written sufficient for you to carry on independently,

Sorry to keep bringing this up, but no, you haven't been very clear.

I asked David Briggs who runs this website.

Here’s his response.

Thanks very much for your question, I must admit I was stumped for a while by that green colour! But if you look at my images through your prism you might see as I do that whereas the red dot is refracted cleanly, the blue dot appears doubled, with a green partner that is less strongly refracted. I had a look on fluxometer.com (fantastic site!) and sure enough on many (though not all) modern screens the blue phosphor has a more or less distinct second peak in green. (You can choose other devices on the right and then just click on the "B" to the right of 1200K). I was able to make a reasonable simulation of what I see through my prism by shifting the Green and Blue components of your image to the left and then adding a duplicate of the blue channel in dull green and aligned with the green image.



Thanks once again for your intriguing question! If you haven't already seen it, you might like to take a look at my new site of online resources on colour at https://sites.google.com/site/djcbriggs/colour-online.%5b/quote

I asked him the additional question that I posted above in post #28.

Here’s his reply.

In viewing your last image through a prism I do see the effect of the green stripe on the red and blue backgrounds, but the stripe appears yellow and cyan respectively because of additive mixing with the coincident red and blue light.

For anyone else who might want to ask questions, I sometimes get too busy to answer emails so if they're on Facebook they might have a better chance of getting an answer if they ask on my Dimensions of Colour Facebook page. https://www.facebook.com/DimensionsOfColour/%5b/quote

That sounds great, but the only problem is, is that I looked at a device that's showing no peak in the green, and I'm still seeing green. *shrug

 

[sophiecentaur]

That sounds great, but the only problem is, is that I looked at a device that's showing no peak in the green, and I'm still seeing green. *shrug

He has answered you absolutely, ifaics. He points out that you are getting Additive Mixing and that is producing a Green sensation in your eye. Perceived Green doesn't have to be spectral green wavelength.
I don't know what your "device" is but if it's a spectrometer of any sort, it won't mimic your colour vision. This point has been made / implied several times above. Perhaps you will see this now it's been pointed out explicitly.