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Color Subtraction to the Eye Problem

  1. Jul 18, 2011 #1
    Yellow light shines on a sheet of paper containing a red pigment. We want to determine the color that we see. The obvious solution is this:

    Yellow can be though of as green and red light. Thus, green and red light can be though of striking the paper with red pigment. Red pigment absorbs its complementary color, which is cyan which can also be though of as blue and green. Thus, we are basically reduced to this problem:

    Green and red light strikes paper that absorbs blue and green light. Thus, only red light is reflected and thus the paper appears red.

    However, as far as I know, this is just a convention to simplify the subtractive color mixing process. However, the pigment only absorbs the frequency of cyan because the electrons vibrate to that frequency. I'd think that cyan can't be split to absorb two different frequencies instead of one. We only use this convention because our three kinds of cones on our retina overlap in receiving different wavelengths of light so that receiving red and green light is no different than receiving yellow light. So how can this be applied to pigments with flaw?
  2. jcsd
  3. Jul 19, 2011 #2
    I think the problem in your train of thought is that a pigment only absorbs a single frequency (such as cyan). In reality, http://www.uic.edu/classes/bios/bios100/lecturesf04am/absorption-spectrum.jpg" [Broken], because there are many possible vibrational, rotational, and shell transitions possible in its molecules.

    But getting to the core of your problem: if you shine green laser light mixed with blue laser light mixed with green laser light on an exotic material that only absorbs one frequency at cyan, we would see that surface as white. In other words, the qualitative color wheel methodologies would break down because they depend on broad-band light sources and material responses.
    Last edited by a moderator: May 5, 2017
  4. Jul 19, 2011 #3
    Yellow light is not red and green light, it is yellow light. There is a real difference between combining radiation frequencies and combining paint. If the red pigment is truly red, and the light is truly yellow (and only yellow), it will reflect none of the yellow light and the surface will appear black.
  5. Jul 19, 2011 #4
    To start from if you use red green and blue LEDs or lasers to illuminate a scene then there are infinite number of pigments that will look exactly the same. Any colour absorption/reflection phenomena in pigments that happen to fall between the spectral lengths of three lasers will be irrelevant because there is no radiation there from the source to expose it.

    This effect is called methamerism.

    Colour is complicated subject but in essence the "colour" makes only sense when you define all three: illuminant, target and detector (observer.) Human eye has three types of detectors so "colour" is 2D space (having discarded overall brightness.)

    To calculate response of three cone receptor types take three integrals (or sums) over spectrum range of e.g. 300-800nm:

    Y(i) = Sum( A * B * C(i)) over all wavelengths 300...800nm
    A - spectral density of illuminating source
    B - spectral reflectivity of target (pigment)
    C - spectral sensitivity of cone type "i"

    Notice how important is illuminant spectrum!

    In the end these three numbers Y(red) Y(green) Y(blue) will define what your brain sees as colour and brightness. They are what we call colour.

    This is just an essence. Cone sensitivities in general depend also on overall brightness level and even whether it is central or peripheral vision.

    P.S. There are few fundamental errors in original message. For example there is no "frequency of cyan" as what we see as cyan can only be created by mixing at least two monochromatic light sources with distinctive frequencies. In other words rainbows don't have cyan.
    Last edited: Jul 19, 2011
  6. Jul 19, 2011 #5
    Sorry, I may have confused people but I meant the paper to have a pure red pigment.

    I don't quite understand what you're saying. If we shine green laser light mixed with blue laser light mixed with green laser light that absorbs cyan, or green and blue, then it would absorb all the laser light shining on it which makes the surface black since there's an absence of light? Sorry if I'm misunderstanding you.

    Why would it reflect no light? Red pigment absorbs only its complement assuming it's a pure red pigment, which is cyan, which is green and blue. Yellow is "made" of green and red so only red is reflected since green is absorbed. So the paper is red?

    Yes, I understand. The brain is what sees the color and brightness, not the actual waves.

    I guess what I'm just trying to say is why is that solution I presented valid? That solution is basically what is presented here at Example 2:
    http://www.physicsclassroom.com/Class/light/u12l2e.cfm" [Broken]
    You can do this for any type of problem like this right? My original post was questioning that method as to why it works because yellow light isn't really red and green light, but we make it that way because the brain knows no difference.
    Last edited by a moderator: May 5, 2017
  7. Jul 19, 2011 #6


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    Staff: Mentor

    Something that absorbs the wavelength of light that is Cyan will not absorb blue or green.

    Typically we say that something that is X color absorbs ALL other colors except for X color. I assume that they mean the paper is a pigment that reflects ONLY red light. Hence the yellow light is absorbed. Also, yellow LIGHT is NOT made of green and red. It is made of Yellow light only. The difference is that a COLOR is what we see, not the actual wavelengths of light. So while green and red light together CAN look yellow in color, yellow light itslef is NOT made of green and red.

    You can, but you have to be aware of the specifics of what you are doing and of the questions.
    Last edited by a moderator: May 5, 2017
  8. Jul 19, 2011 #7
    Thank you so much, that clears mostly everything but a few more questions:
    So can you give me an example when it is incorrect to use that method of reasoning (where we "split" wavelengths of light into the two wavelengths that make it up, forgive the misuse of vocabulary)?

    Why does the website I linked to use that kind of convention?

    Does all the secondary colors (cyan, magenta, and yellow) all have their own specific or range of wavelengths on the visible light spectrum?
    Last edited: Jul 19, 2011
  9. Jul 19, 2011 #8


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    Staff: Mentor

    The website is merely explaining what color subtraction and addition means. Hence saying that Yellow can be thought of as Green and Red only applies to the color yellow, not the wavelength of light. You cannot split a wavelength of light into two other wavelengths as you can split colors up. Blue light is ONLY blue. Red light is ONLY red. The site is not trying to say that light itself can be split, only that the colors can be.
  10. Jul 19, 2011 #9
    Oh, okay. But the methodology is correct? And under what circumstances may that be incorrect to use? Thank you by the way for clearing that up for me.
  11. Jul 19, 2011 #10


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    Staff: Mentor

    As long as you know which wavelengths of light are hitting an object, and which ones are being absorbed or reflected, then it works fine.
  12. Jul 19, 2011 #11
    Okay, so if the problem refers to the "color yellow", should I just assume that only the wavelength of yellow is referred to? Or whatever wavelengths that can stimulate the response the brain has for yellow including the yellow wavelength and the red plus green wavelengths since the problem never said yellow wavelength.
  13. Jul 19, 2011 #12


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    Staff: Mentor

    If it refers to the Color yellow, then keep in mind that the color is that way for 1 of a number of reasons. Either it is reflecting yellow light, or it is reflecting a combination of different colored light that LOOKS yellow. You'd have to come up with a way to figure out which wavelengths are being reflected to figure it out.
  14. Jul 20, 2011 #13
    There are infinite number of ways to create what you would call a "yellow" light.
    Same goes for "red" pigment.

    The problem you have referred to may actually have two solutions:
    1) "yellow" light with wide spectral distribution (incandescent lamp) shining onto a red pigment that only reflects narrow red -> pigment looks RED
    2) "yellow" light with narrow spectral distribution (LED) shining onto a red pigment with wide reflection spectrum -> pigment looks YELLOW
    Last edited by a moderator: May 5, 2017
  15. Jul 20, 2011 #14
    Surprise, surprise! Take a spectrodensitometer and try scanning stuff around. You will be amazed.
    I have one of these: http://www.xrite.com/product_overview.aspx?ID=278 so I can scan something if you wish me to.

    Even the most "deep" colours reflect an awful amount of other light.
    Monochromatic "yellow" light has wavelength about 590 nm.
    Have a look at this:
    Last edited: Jul 20, 2011
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