Why an object has the color complementary to the color absorbed?

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SUMMARY

The discussion centers on the differences between color perception in solutions versus paints, specifically regarding complementary colors. When a solution absorbs a color, such as red, it transmits its complementary color, green, leading to the appearance of green. In contrast, mixing paints results in brown due to the subtractive color mixing process, where pigments absorb specific wavelengths of light. The key distinction lies in the principles of additive color mixing used by monitors versus the subtractive mixing of pigments in paints.

PREREQUISITES
  • Understanding of UV/VIS spectroscopy
  • Knowledge of additive and subtractive color mixing
  • Familiarity with color theory and complementary colors
  • Basic principles of light absorption and reflection in materials
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  • Research the principles of UV/VIS spectroscopy in detail
  • Explore the differences between additive and subtractive color mixing
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This discussion is beneficial for students of chemistry, artists working with paints, and anyone interested in the science of color perception and mixing.

blaisem
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Hello, I know from UV/VIS spectroscopy that for many solutions a single frequency from the visible spectrum is absorbed. The color of such a solution is the color complementary to the one absorbed.


If red is absorbed, then presumably it is no longer available to mix with its complementary green, so green is transmitted and the solution appears green. A colorless solution transmits all colors, so they are all subsequently still available to mix with each other to transmit the same white light that was incident on the solution. This leads me to believe that combinations of complementary colors make white. But this is inconsistent with what happens if I mix paint. Mixing two complementary colors makes brown, instead of white. If I mix all the primary and secondary colors together (as should occur with normal incident "white" light), I still get a mixture of browns.

So my question is:

Why is my solution colorless by mixing two complementary colors, but my paint brown?

I suppose another way to ask the question might be if I hypothetically labeled the colors with numerical values.

Colorless Solution:
If red has a value of 2 and green a value of 3, then in my solution 2 + 3 = 5 and 5 is equal to white light.

5 - 3 = 2 and corresponds to absorption of green leaving red behind.

Paint:
If I mix red and green together, 2 + 3 = 5, where 5 is no longer equal to white light, rather reflects brown. I have no idea how to explain this or what is "complementary" to brown such that brown would be reflected.

Thanks for any advice! I hope this question doesn't seem trivial or pointless, but the inconsistency between what happens in a solution and what is observed in paint makes it hard for me to wrap my head around the general concept.
 
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This is subtractive color. The pigments in the paint are designed to absorb a certain group of wavelengths. The rest gets scattered and reflected. So, if you mix pigment, incoming light will hit molecules of one pigment, get scattered around (because the paint is matte), and then hit molecules of another pigment, so both colors are absorbed.

Your monitor works on a different principle: additive color. In this case, the red emission from the red pixel is added to the green emission and the blue emission. That's because the monitor is the source of the light.
 
Thanks Khashishi, I had no idea there were two types of color mixing. That makes a lot more sense. Thanks for your reply :)
 

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