2
$\begingroup$

From what I've studied about EM waves reflection over dielectrics, the reflected wave must have the same frequency than the incoming EM wave before reflection, else Maxwell equations aren't satisfied. So far so good.

However in practice I've noticed that for instance blue LED reflect as violet on white walls, pine needles, plastic and wooden floor to name a few.

I've also noticed the same similar shift from blue to violet while walking in the streets of a city in a thick haze night. When I far far away (farther than about 20 m) the lights appeared violet, while when I walked toward the lights, they were turning blue as I was advancing.

I couldn't find any explanation of how such a phenomenon could occur. I've thought about my eyes being tricked but I don't think this is the culprit. I had also thought about the material from which the light was being reflected off as absorbing better some frequencies than others, but since this occurs over almost any dielectric material I've seen so far, I also discarded this hypothesis.

I'd appreciate if someone could come up with an idea and also tell me whether he/she noticed a similar behavior.

Edit: I have just found the answer to this question by accident. My 2 years old son turned on a light bulb near a blue Christmas light. It turns out that when the two lights are on, the blue one appears almost violet, its reflection on the white wall is also almost violet. But when the light bulb is turned off and only the blue light remains, everything looks blue, no violet whatsoever!

So I conclude that it is really just perception. Incredible!!!

Improve this question
$\endgroup$
1
  • $\begingroup$ Fluorescence? From the UV. $\endgroup$
    – user137289
    Dec 25, 2016 at 18:56

2 Answers 2

Reset to default
3
$\begingroup$

What you are reporting is your perception of blue and your perception of violet. The frequency of electromagnetic radiation and the correlation to the color perception of the eye is not a one to one relation. A given frequency necessarily gives a color on the rainbow, but a given color can be a combinations of many frequencies.

So unless you have analyzed the frequency spectrum of what you see as "blue" and "violet" your question cannot be answered.

I suspect differential differences in the absorption of frequencies from the perceived blue spectrum in the case of mist, would change the balance in the diagram and give the perception of violet; of scattering off materials too.

color perc

Improve this answer
$\endgroup$
4
  • $\begingroup$ I see. If I take a picture with a digital camera and still notice a similar effect, would that overcome the problem of the human perception of the eye or is there a similar problem with digital cameras too? I didn't analyze the spectrum of the colors I saw, unfortunately. $\endgroup$
    – untreated_paramediensis_karnik
    Dec 25, 2016 at 19:06
  • $\begingroup$ I do not think digital cameras give frequencies either, they are geared to the human eye perception for accuracy. You need to measure the frequency spectrum . there exist spectrum analysers . maybe a lab you know has one? this is an advertisement as an example aaronia.com/products/spectrum-analyzers/… $\endgroup$
    – anna v
    Dec 25, 2016 at 19:52
  • $\begingroup$ I see, thanks. I don't know whether the lab has one, I might ask whenever I can. So the idea is to first determine whether the frequency of light really change as my eyes see, so that my question would make sense. Is that correct? $\endgroup$
    – untreated_paramediensis_karnik
    Dec 26, 2016 at 14:42
  • 1
    $\begingroup$ Yes. As I say in the answer I suspect that the recipe changes because of differences in absorption over the spectrum . It is a physiology question if so, not a physics one. $\endgroup$
    – anna v
    Dec 26, 2016 at 14:49
0
$\begingroup$

However in practice I've noticed that for instance blue LED reflect as violet on white walls, pine needles, plastic and wooden floor to name a few.

Try it in front of an actual mirror. Do you see the same shift then?

Improve this answer
$\endgroup$
1
  • $\begingroup$ Not sure why this was posted as an answer, it should be a comment. I already have the answer, as my edit suggest, and my approval of the already accepted answer... $\endgroup$
    – untreated_paramediensis_karnik
    Dec 22, 2018 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.