1
$\begingroup$

It makes sense to me that when looking through a sample (observer | sample | light), it should appear as the opposite of the light absorbed, but it does not make sense to me to expect the same when not looking through it, just standing by the same side as the light source (observer | light | sample, or light | observer | sample). In the first cenario, the observer sees light emitted - light absorbed (the transmitted, that barely interacts with the sample). In the second, the observer sees light reflected (to my understanding, light emitted from the de-excitations of excitations caused by the light source).

However, when I see a solution of $Cu^{2+}$, it looks the same when observed in both settings. Thereby implying that transmitted colour = reflected colour. Why is that?

Somewhat related question but without a satisfactory answer: Transmission, absorption, and reflection of light

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Since your experimental observation with copper salts negates your original hypothesis, it implies that the way you are trying to explain it is wrong. The main culprit and the source of all these problems is the color wheel which is taught in schools. Misconceptions persist for long. When you are looking at the copper solutions in two different settings, it is misleading to think that copper solution is reflecting blue light back to you. It is not.

In each case, copper(II) solution is absorbing a small portion of the red light from the visible spectrum and it appears to you like a pure blue solution.

Indeed it is our brain which has been created in such a way that when the visible spectrum has a certain red portion missing, it perceives the remaining spectrum as "blue".

Hint: Water in the ocean also appears blue? Water also very very weakly absorbs the red portion of the visible spectrum. You just need tons of water to perceive this effect.

There is a beautiful book by the name of The Physics and Chemistry of Color. The same author wrote an article "The fifteen causes of color: The physics and chemistry of color." It is certainly worth consulting. Article-behind paywall

The fifteen causes of color derived from a variety of physical and chemical mechanisms are summarized in five Croups in this article. Vibrations and simple excitations explain the colors of incandescence (e.g., flames), gas excitations (neon tube, aurora), and vibrations and rotations (blue ice and water). Ligand‐field‐effect colors are seen in transition‐metal compounds (turquoise, chrome‐oxide green) and impurities (ruby, emerald). Molecular orbitals explain the colors of organic compounds (indigo, chlorophyll) and charge‐transfer compounds (blue sapphire, lapis lazuli). Energy bands are involved in the colors of metals and alloys (gold, brass), of semiconductors (cadmium yellow, vermillion), doped semiconductors (blue and yellow diamond), and color centers (amethyst, topaz). Geometrical and physical optics are involved in the colors derived from dispersive refraction (rainbow, green flash), scattering(blue sky, blue eyes, red sunset), interference (soap bubbles, iridescent beetles), and diffraction (the corona aureole, opal).

Now you can imagine that all is not that simple.

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ (+1). Then there is human perception and such curious things as Impossible color and also en.m.wikipedia.org/wiki/Color_constancy. $\endgroup$
    – Ed V
    Jul 2, 2020 at 0:22
  • $\begingroup$ That's pretty nice, I'll be reading that article soon. However, this answer does not answer my question. The solution is absorbing red, nice, but why is the transmitted colour the same as the reflected? I thought the two process followed different mechanisms, with transmission being "barely interacting" and reflection being excitation -> de-excitation by photon emission. $\endgroup$
    – peruca3d
    Jul 2, 2020 at 1:39
  • $\begingroup$ Dear LocoAsker, as I said there is no reflection. You are still seeing transmitted light. Reflection is not de-excitation by photon emission. This topic which I tried to cover is at least one hour lecture. StackExchange is not a substitute to reading books. Your confusion is arising from mixing of several phenomena. Why don't you read the article first, and then we discuss further? $\endgroup$
    – AChem
    Jul 2, 2020 at 1:50
  • $\begingroup$ If you really want to see true reflected light from copper solution in a flat glass tumbler, shine white LED from an angle, do not look at the solution, but let the reflected light beam fall on a white wall or white paper. What color do you see? Don't forget to do a blank, i.e., try with water first. $\endgroup$
    – AChem
    Jul 2, 2020 at 2:22
  • $\begingroup$ @LocoAsker also see it this way : the light you see in both cases is that from the source minus what it is absorbed (plus scattering in general). Solid layered films, for instance a transparent colored material cast on a glass slide, can behave differently and similar to your expectation. $\endgroup$
    – Alchimista
    Jul 2, 2020 at 14:09

Your Answer

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

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