Is there a way that I can make the overlapping points of these three glass objects clear, or would there be a way that I can control the colors of the overlapping points? I don't want to change the glass shaders of the objects, though, I want to be able to control the color of the glass in that area. This would be extremely helpful.
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asked Dec 29, 2019 at 1:11
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$\begingroup$ The whole point is about using glass for this. I want to be able to make the glass a certain color where it overlaps with the other colors. $\endgroup$– Nate_Sycro27Dec 29, 2019 at 19:43
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2 Answers
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What you are seeing corresponds with the way light behaves in the real world.
When you are dealing with filters you are subtracting colors from the light that goes through the filter.
If you are using primary colors (Red, Green and Blue), any filter with one of the primary colors will effectively cancel the other two out.
In other words: a Red filter will only transmit Red light, and will absorb Green and Blue.
Likewise a Green filter will subtract the Red and Blue out of white light and the result will be just green.
A Blue filter will only let Blue through, and absorb the other two colors.
If you then add a second filter then you will have no light left to go through.
Red light has no green content to go through, green light has no blue content to go trough, and blue has no red.
If you must use filters (such as glass), use Yellow, Magenta and Cyan instead.
(image source: http://hyperphysics.phy-astr.gsu.edu/hbase/vision/subcol.html)
Suggested reading to learn about additive or subtractive color mixing:
https://en.wikipedia.org/wiki/Additive_color
http://www.physics.wisc.edu/ingersollmuseum/exhibits/optics&color/subcolormix
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Are the 3 objects intersecting each other, or just in front of each other? If they're intersecting, you could use booleans to separate them into different objects and change the colors of those. If they're not, you could accomplish this with a pretty complicated node setup involving ray depth, which i'm not fully qualified to explain.
