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Note: Energy= h*f where h is a Planck's constant and f is frequency. Thus, the color is determined by frequency. – Vlad K Nov 17 at 16:02


Outside your eye, the frequency stays constant, and the wavelength changes according to the medium, so I would say the frequency is what counts more. This explains why objects' colour don't change when we look at them under (transparent) water ($n=1.33$) or in air ($n=1$).

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I think this is right and is the only answer I've upvoted because it talks about the detection mechanism. Arguments based on what color things appear in swimming pools etc. are pointless because the light passes into the medium of our eyeball before detection. The correct experiment is to change the refractive index in your eyeball and see if the color changes. – Rob Jeffries Nov 17 at 20:58
@Rob On the other hand, this perpetuates the misconception that the energy in light is purely a function of frequency, and not of the number of photons (a.k.a. "intensity"). There's very little, really, to be gained from the mention of energy; the frequency/wavelength correspondence is confusing enough without dragging even more related variables into the mix. – Emilio Pisanty Nov 17 at 21:44


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How is the "experimental fact" that "refraction does not change color" found? – Rob Jeffries Nov 17 at 16:49



Wikipedia has a lot of good further info on this, but it is scattered among several articles. Probably is the best starting point. For something much more thorough and technical, see Poynton's excellent Color FAQ at

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Building on prior answers, the facts are: Color is determined by the energy of the EM Wave that reaches your eyeball. Energy is defined as $E = hf$, where $h$ is Planck's constant and $f$ is the light's frequency.

Thus, the color of an EM Wave is defined by its frequency. In other words, measuring the frequency of an EM Wave is sufficient to identifying the color of light or the type of EM Wave that it is. This is opposed to measuring the wavelength, which would knowledge of what the refractive index of the medium that the wavelength was measured in is to determine what color of light or type of EM wave the EM wave is.

Note: Although $f$ can be defined by $v/l$, where $v$ is the speed of an EM wave in a medium and $l$ is the the wavelength in a medium, upon changing mediums the only constant is the frequency of the wave.

An example of why frequency is the defining factor: When you throw a red brick in a pool, the wavelength of the EM Wave carrying the color of the object varries. If you were to measure the wavelength carrying the color of the brick, that information would be useless or misleading in identifying the color of the brick unless you knew the refractive index of (speed of EM Waves in) the medium that you were measuring in. On the other hand - measuring the frequency of the EM Wave carrying the color of the brick anywhere would be sufficient in determining the color of the brick is Red, for it does not change regardless of what medium the EM Wave is found in.

From this, we can conclude that the color we see is dependent on the frequency of the EM Wave. (The Wave just happens to have an certain wavelength at that speed of EM wave determined by the medium the wave is in.)

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There is no "thus" about your last sentence, since you could equally have said "$E = hc/\lambda$, thus ... defined by its wavelength." – Rob Jeffries Nov 17 at 16:48
The speed of the wave and wavelength both vary depending on what medium the em wave is in - what does not varry is the frequency of the wave (v/lambda) – Vlad K Nov 17 at 16:59
Exactly, so perhaps add that the the answer. – Rob Jeffries Nov 17 at 18:52
The wavelength of the wave you "see" is the wavelength of the light in your eyeball, and does not depend on what (non-absorbing) media it has travelled through prior to that. So though I'm sure it is frequency that is the important factor, your argument does not lead to that conclusion. – Rob Jeffries Nov 17 at 20:50
According to David Z's answer we do not see a wavelength; rather, it is the energy of photons that fall on sensors the in our eyes that we "see" - which is determined by the frequency. – Vlad K Nov 17 at 21:15

Here's my addition. Many of the answers above use the erroneous argument that frequency is the determining quantity, on the basis that the same object viewed in different media appears to be the same colour.

This is meaningless, since the light has to travel through the vitreous humor (with refractive index 1.33) immediately prior to reaching the retina. Thus light of a given frequency will also reach the retina with exactly the same wavelength whatever medium that light has travelled through to get there.

No: the answer must be based on the physiology of the receptors. However, I do offer one obvious experiment in favour of frequency rather than wavelength. During a vitrectomy, the vitreous humor is temporarily replaced with other substances, often air or other gases with a completely different refractive index. In none of the few articles I have read, often for the patient's benefit (e.g. here) does it mention drastic changes in colour perception as one of the temporary side effects.

Therefore I deduce that since the frequency of the light is invariant, but its wavelength as it reaches the retina could be changed by 30%, that it must be frequency that determines colour perception.

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