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From my understanding of fluorescence, a "fluorescent yellow material" (like in highlighters) is a material that contains yellow dyes and fluorescent dyes absorbing green to give yellow. Then, the material appears "more yellow" than usual objects because it has two sources of yellow.


If the above is right, it appears possible to create a "fluorescent white material" simply by having :

  • three base colors (blue green red)
  • three fluorescent dyes with their corresponding quantities to counter absorption from other dyes (2blue 2green 1red)

Does it exist ? Or am I making a mistake there ?

EDIT :

  • The source is supposed to be sun light
  • "white" is seen from the observer point of view, I don't mean all the visible spectrum, just 3 colors are enough.
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  • $\begingroup$ This is a good, interesting question. No good answer yet. Give it some time though and if you don't get a definitive answer, try the chemistry stack exchange. $\endgroup$ – docscience Sep 8 '16 at 13:45
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The phosphor on ordinary fluorescent tubes and also in white leds would seem to qualify; they take ultraviolet light and convert it to white(ish) light, using a blend of substances to get the color balance right.

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  • $\begingroup$ Interesting ! +1 :) However I can't accept the answer since it doesn't take sun light as energy source. $\endgroup$ – Ghislain Bugnicourt Sep 8 '16 at 16:33
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    $\begingroup$ The envelope of a fluorescent tube is UV-blocking for safety. You can break the tube open to expose the phosphor to sunlight, though. $\endgroup$ – Whit3rd Sep 9 '16 at 0:50
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The one thing you can be pretty sure of is, a surface cannot send out more light than is coming in, unless it has a heat source and is hot. Fluorescent white surfaces would violate energy conservation, unless there was an ultraviolet source. We do see that, it's called a "black light" and can make white clothing look fluorescent.

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  • $\begingroup$ I'm not talking about emiting more light than what the material receives, just giving you this impression by taking energy from the UV (the source is supposed to be sun light). $\endgroup$ – Ghislain Bugnicourt Sep 8 '16 at 13:41

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