3
$\begingroup$

"In the RGB color model, the light of two complementary colors, such as red and cyan, combined at full intensity, will make white light, since two complementary colors contain light with the full range of the spectrum." https://en.wikipedia.org/wiki/Complementary_colors#Colors_produced_by_light

Cyan light has wavelengths at around 495 nanometers, and red light about 700 nanometers. White light is light in which no color is seen. It looks like white, but actually consists of a combination of all visible colors.

The hue is determined by domination of the impression of one or two of the three types of cones. If the impressions are the same, we see white or gray, depending on the intensity.

Do I have to understand that because at 495nm (cyan) the S and M have a intersection point, by means that the S and M cones are impressed at the same level (at sensity 0.3) together with red (L) at the wavelength of 640nm also at sensity 0.3 (instead of 700nm). And this together creates white?

But with the two wavelengths 640 and 495nm can't be said that it contains light with the full range of the spectrum as is said in wikipedia. Fe if I have two lasers one with 495nm color and the other with 640nm and I mixe them will I see also white light? Or does it not work like that to create white light with complementary colors? enter image description here

$\endgroup$
3
  • 1
    $\begingroup$ I believe that this is physiological more than physical. $\endgroup$ Commented Feb 28, 2016 at 12:34
  • $\begingroup$ Peter: I disagree. Color is a subject that really needs both sides, and here the questions is alot related to spectrums of emission and absorption. Physiology & biology people would say "go ask to physics". $\endgroup$ Commented Feb 28, 2016 at 22:57
  • $\begingroup$ You don't create white light. You create the same sensation as the one produced by white light. This why is not just physics. A spectrometer won't "see" white light. The brain will. $\endgroup$
    – nasu
    Commented Dec 22, 2020 at 4:10

1 Answer 1

1
$\begingroup$

Your reasoning is well inspired, but take care about several mistakes:

  • Your curves are all normalized to [0,1],so you can't compare the absorption efficiency from these, or give any sense to an "intersection point".
  • Note also that the curves were not meant for laser-testing and their absorption curve might be drawn too smooth (but maybe it's ok).
  • The captor is not alone, you don't know the weighting retina and next layers will give, thus the interest of perceptive curves (but often what is presented as absorption is indeed what is perceived).
  • Of course, the relative intensity to the two lasers counts.

Besides, yes, the raw idea is there: you transform physical colors (i.e., spectrum) into perceived colors by integrating the tri stimulus (thus getting 3 values, that can be encoded on various way). For quantitative results, you should have a look ok XYZ color space and related stuff.

See also Color Metamerism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.