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I tried to mimic the mechanism of typical screens to produce white color out of red, green and blue.

What I did is displayed the attached image on the screen, and moved far away as to let the diffraction effects take place, so that the three colors appear as if they're coming from the same point.

Nonetheless, quite paradoxically, what I have seen was black instead of white.

I don't know if this question fits this place, so excuse me.

enter image description here

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1  
It's not really diffraction -- more like lack of focus resolution. –  Carl Witthoft Jul 15 at 11:44
    
But HOW/WHY does this lack of focus occur? –  kalkanistovinko Jul 16 at 21:47
    
Your eye can only resolve spatial/angular separations down to a certain point. Beyond that, just as you can't read letters on a sign far away, you can't separate the color bars. –  Carl Witthoft Jul 17 at 11:43

4 Answers 4

up vote 41 down vote accepted

What you are seeing at a distance is not black. It is a darkish shade of gray, RGB gray 85,85,85. The reason you aren't seeing "white" is because each of those three rectangles has an HSV value of only 33% and you are seeing that merged square against a white background.

That merged square will appear to be whitish if you make the background black rather than white and view the screen in a very dark setting.

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Pity Jin's original proposal for a dark background on Physics.SE didn't make it! –  leftaroundabout Jul 15 at 18:49

This is an interesting question, but I think it concerns physiology rather than physics. As I'm a physiologist, I'll attempt an answer.

Our color vision begins with the different spectral sensitivities of our cone (photopic/bright light) receptors. These are often called red, green and blue cones, but in fact each is sensitive to an extent to a side range of wavelengths. But this does not account for many color vision phenomena.

Here's what does. The information conveyed by these cones is processed in the brain (including the retina, which embryologically is part of the brain) by what is called an "opponent" mechanism; i.e., this mechanism has three channels, one for luminosity (black/white B-W channel), one for blue or yellow (B-Y channel), and one red or green (R-G channel). Note that the latter two opponent channels in that the signal generated by a single wavelength of light is EITHER red or green, never both, or EITHER blue or yellow, never both.*

Now, if you add wavelenghts (as you did, as each stimulus is a mix of wavelengths) lights, what you see is the sum excitation of each channel. In order to perceive white you need to have a mix of wavelengths that stimulates relatively similarly B, Y, R and G. If you were far enough away that the stimuli were mixed (blurred) or better mixed the lights from each together with lenses I THINK they would look lightish red-blue. I say that because (i) your red stripe has some blue, the green stripe some yellow, and the blue lots of blue, for a B-Y channel B response, for blue; (ii) your red stripe has lots of red, the green stripe has only a little yelow, and the blue strip has some red as well, so the sum is R-G channel is R, red; and (iii) There will be activation of the non-color BW channel by wavelengths lights whose color aspects cancel out, so the light will be bright, or low hue saturation or "lightish."

If you want to understand better, you can see the classic paper on this on line (Hurvich and Jameson, 1957) on line.

http://jraissati.com/PHIL256/09_HurvichJameson_AnOpponentTheoryOfColorVision.pdf

See Figures 1&2 for the layout of the channles,and Fig 4 for the responses to lights of single wavelengths. Note that about 480 nm is RG = 0 and BY = B, so perceptually pure blue, 510 nm is BY = 0 and RG = G so pure green, 580 nm is RG= 0, BY = Y so pure yellow, and that no single wavelength is BY = 0 and RG = R, i.e. the is no spectral pure red. Think about that next time you see a rainbow, which is the only time most of us see an approxiamte dispyy of tjhe visible spectrum hat (picture and so are all mixes, not single wavelengths).

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While David Hamman's answer is correct, I wanted to expand a little bit on his answer:

When you use a CRT, you are looking at emitted light. In the case of emission, there is no "absolute" white - something will only look gray in comparison to something else with the same color ratio but brighter. When you turn up the brightness on your monitor, white is still white - just "whiter". In a dark room, even a CRT whose intensity is turned down a lot appears to render white as white.

By contrast when you have a sheet of paper with colors on it, what is really happening is that you have absorption of certain wavelengths / colors. A blue piece of paper absorbs everything except blue; ditto for red, and green. So what you have on your three rectangles is three pieces of paper that absorb 2/3 of the incident light:

      red  green  blue
red   100     0     0 
green   0   100     0
blue    0     0   100

When you illuminate the entire card with 100% white light, you get a red reflection from just 1/3 of the card - and the same for green and blue. If you have these three cards against a white background, the background will appear approximately 3x brighter as every part of the background reflects all three colors.

If you made the background gray (which is really just a "darker shade of white"), and shone a bright light onto the setup, you could actually persuade yourself that the background, and the three color sample, were white.

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A very good answer. While it is true that a blue paper absorbs other color other than blue, but why? –  luming Oct 19 at 15:04

To simplify the problem, I will use a source of light with only three different wavelengths. One for red, one for green and one for blue. If I pass "this light" through a red filter, only the red light will pass, and the green and blue components will be blocked. If I then put the green and/or blue filter on top of the red filter, no light will pass through, so the filters' area will look black. Since in "reality" the lights have a band of frequencies, some light will get through so the filters' area most likely will look "dark gay." Therefore, in your experiment, you should have expected black, not white.

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You are talking nonsense. Your filtering argument holds for any color including the white on your screen. –  kalkanistovinko Jul 18 at 11:42
    
@kalkanistovinko: it doesn’t. Guill’s red filter blocks spectral green and Guill’s green filter blocks spectral red. Which part of spectrum is a “white filter”☺ expected to block? In short, it is actually you who is talking nonsense, missing the difference between additive and subtractive color synthesis. –  Incnis Mrsi Oct 24 at 14:04
    
Actually, Ī misread the question. Guill argues about impossibility to synthesize red, green, and blue into white subtractively, whereas kalkanistovinko tried to observe their additive average. They are speaking about unrelated situations. Thanks to @Jim Ī realized my mistake by now. –  Incnis Mrsi Oct 24 at 15:22
    
@IncnisMrsi Don't mention it. Just trying to help however I can. –  Jim Oct 24 at 15:29

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