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From https://en.wikipedia.org/wiki/Complementary_colors

The pairs of complementary colors vary depending upon whether the colors are physical (e.g. from pigments), or from light. These change the way in which the color is made, and therefore change the color model which applies. For pigments, subtractive colors apply, so the complementary/opposite color pairs, are red & green, yellow & violet, and blue & orange. In the RGB color model, which applies to colors created by light, such as on computer and television displays, the complementary/opposite pairs are red & cyan, green & magenta, and blue & yellow.

Since color printing ink does not produce color by pigmentation, but instead produces color by masking colors on a white background to reduce light that would otherwise be reflected, the same mix for producing black applies as for light producing white, i.e. the complementary/opposite pairs are red & cyan, green & magenta, and blue & yellow. The most clashing colors to the eye may still be as for painting.

Isn't the color of ink on a paper the color of light reflected by the ink, so same as color of pigments?

How does color printing ink not work the same way as pigmentation?

How does ink "produces color by masking colors on a white background to reduce light that would otherwise be reflected"? Thanks.

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  • $\begingroup$ Imagine an ad for lipstick. Now imagine that the color on the model's lips in the glossy magazine ad is subtly different under a certain lighting spectrum from the actual product under the same light. People in the color printing industry get fired for that. The ones who haven't been fired yet have knowledge of inks, pigments, papers and lighting that is both broad and deep; but most of those people are artists, not scientists. When they try to express what they know in words, it won't sound much like rigorous science. $\endgroup$ – Solomon Slow Mar 26 '15 at 19:50
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/14800/2451 , physics.stackexchange.com/q/23830/2451 and links therein. $\endgroup$ – Qmechanic Apr 18 '15 at 10:35

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