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Does the color of an object have anything to do with its refractive index, and if so, what is the relation? I tried googling this and searching through the questions on stack exchange, but I haven't been able to find anything addressing this exact question.

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    $\begingroup$ Color is related to many many many different things, so you should constrain things a bit: do you mean a thick transparent solid plain object made of a pure mono-material with a smooth air/material interface ? $\endgroup$ – Fabrice NEYRET Oct 29 '15 at 17:04
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The color of an object is completely determined by the refractive index and the geometry of the object. The reason why it is not obvious is that one has to takes the imaginary part of the refractive index into account. The real part determines the wavelength of light in matter and the imaginary part describes absorption of light in matter.

When you ask what color has an object the physics question you ask is what light is reflected or transmitted by the object to your eye. Transmission and reflection are well described by the fresnel equations. They only depend on the the geometry of the problem, the refractive index of the object and the wavelength of the light. For the object to have color the refractive index has to change with wavelength, which we call dispersion.

There are now certain possibilities for the refractive index:

  1. Refractive index is purely real in the wavelength region of interest => no absorption (e.g. diamond, glass, air): Light is mainly transmitted. Few percent are typically reflected. Dispersion is only visible in certain geometries for example prism or rainbow.

  2. Refractive index has a low imaginary part (e.g. Rust, coal...): As in 1. few percent are reflected but in contrast to 1. the light transmitted into the substance is absorbed. Dispersion of the imaginary part determines the color. Rust has an absorption peak around 400 nm (blue) and therefore looks reddish.

  3. Refractive index has a high imaginary part (e.g. metals): Most light is reflected. Higher absorption of a color now means that it looks more like that color because higher imaginary part means higher reflectivity. For example copper has a high absorption in the red and low one in blue resulting in a reddish color.

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    $\begingroup$ Well, not entirely. Many things, including the famous blue jay feather, have an apparent color due to interference effects alone. The color of nontransmissive objects is due to absorption and reflection, and for all practical purposes there is no real part of the index of refraction. Even for transmissive objects, dispersion & prism can lead to the apparent color depending on the view angle. $\endgroup$ – Carl Witthoft Oct 27 '15 at 12:35
  • $\begingroup$ I don't see the contradiction to my answer. I stated prism as an example in 1. and i wrote "determined by refractive index and shape of the object" (maybe I should have written geometry). For interference clearly the geometry and refractive index matter. In 2. and 3. I explained the situation for nontransmittive objects. What I wanted to make clear is that the refractive index is the only material property you need to know to calculate optical response (i.e. apparent color), at least in the realm on linear optics. $\endgroup$ – Jannick Oct 27 '15 at 12:45
  • $\begingroup$ Ive always seen Fresnel equations applied to transparent media and metals. Would they explain the color of wood, plastic, etc? $\endgroup$ – jinawee Oct 27 '15 at 13:05
  • $\begingroup$ Yes they would in principle but maybe not quantitatively. The fresnel equations are strictly true only for a flat surface. For plastic the predicted reflectence should still be a very good approximation. The reflection is however diffuse because of surface roughness of the plastic. If you polish it you should see a clear reflection. For wood I think a flat surface is too much of an approximation. Still you only need the refractive index of wood to calculate the color but instead of the fresnel equations you now need to solve the maxwell equations directly (e.g. with a FDTD solver) $\endgroup$ – Jannick Oct 27 '15 at 13:52
  • $\begingroup$ @Jannick The contradiction is to the first sentence, "The color [...] is completely determined by the refractive index and the geometry". Instead of "completely", it's more like "mainly" determined by these. $\endgroup$ – Volker Siegel Oct 16 '18 at 12:22

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