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In a vacuum, there is a one-to-one correspondence between light frequency ($\nu$) and wavelength ($\lambda$), ie. $\lambda=c/\nu$. But in a refractive medium, $\lambda=v/\nu$, so while the frequency may remain constant, the wavelength may not.

I've seen that in biology, wavelength is mostly used. But, I recall reading that human vision is sensitive to frequency. So, is human vision sensitive to wavelength or frequency?

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  • $\begingroup$ Huh. It's a question of what wavelengths of light will activate opsin proteins in the retina itself. If you changed the refractive index in the retina photons of the same energy(and therefore, frequency) should activate the same state transitions in proteins. $\endgroup$ – Resonating Oct 2 '14 at 6:16
  • $\begingroup$ My guess is therefore frequency instead of wavelength, but if anyone has empirical data I would love to read the methods of that. $\endgroup$ – Resonating Oct 2 '14 at 6:28
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It is not the wavelength or frequency that determines light absorption- it is the energy of the photon that matters. The energy of incident light should match the excitation energy of the chromophore. The medium itself can also affect light absorption by electronically interacting with the chromophore. Frequency is proportional to the energy (given by the relationship $E=h\nu$); while wavelength and velocity change in different media, frequency and energy remain constant.

The chromophore in the opsins (retinal) absorb energy to become active and you can say that they detect the frequency. However, sufficient number of opsins have to become active for the retinal cells to transmit the signal. So the overall light flux is also important. An interesting thing to note is that activation of the opsins hyperpolarizes the photoreceptor cell and reduces its tendency to fire.

Wavelength plays a role in diffraction and therefore diffraction is less in a medium with higher refractive index. Wavelength also determines scattering. So, while the opsins actually detect frequency, the wavelength can slightly affect the overall "vision" because of scattering/diffraction effects.

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    $\begingroup$ The energy of a single photon is proportional to its frequency. So if a single photon is detected, its frequency does determine the color. $\endgroup$ – Hans Sep 14 '16 at 19:49
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    $\begingroup$ The more important questions are: Does opsin detect single electron excitation by the photons, or integrate over a period of time and detect the total photon energy? An atom can absorb simultaneously two or multiple photons. I suppose, even if opsin can detect single electron excitation, it can not distinguish between single photon absorption and muti-photon simultaneous absorption. Is that correct? Both cases would imply the eye can not detect the true frequency of individual photons. $\endgroup$ – Hans Sep 14 '16 at 19:50
  • $\begingroup$ @Hans yes of course. Energy is proportional to frequency but I focussed on energy because the transactions (absorption/emission etc) are of the energy. Frequency is a sort of indicator. Regarding opsins: They detect energy of a photon (photon absorption causes a stereoisomeric change) but the retinal cells will respond only if a sufficient number of opsins are active. $\endgroup$ – WYSIWYG Sep 15 '16 at 5:16
  • $\begingroup$ Thanks for your answer. You say an opsin is able to detect a single photon. Is it able to absorb two or more photons simultaneously en.wikipedia.org/wiki/Two-photon_absorption#Two-photon_emission? Does a retinal cell fire only when the total number of active opsins presumably linked to it or the total energy of the absorbed photons exceeds a threshold? There is difference between the two. Either way, two different compositions of the incoming light can induce the same response from a retinal cell. Is that right? That would limit the resolving power of the retina regarding color. $\endgroup$ – Hans Sep 15 '16 at 8:02
  • $\begingroup$ @Hans as far as I know it is a single photon absorption. The firing of the retinal cells depends on total number of activated opsins but activation of opsins reduces the tendency to fire; there should be a threshold but I am not sure of how the threshold is established, at the moment. Usually different cells in the retina (rods and cones) express different opsins. Some residues in the opsin proteins can affect the absorption frequency of the chromophore (retinal). You can read more about cones/opsins on wikipedia or this linked post. $\endgroup$ – WYSIWYG Sep 15 '16 at 12:54
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Wavelength-frequency relation is $\lambda=\frac{c}{\nu}$ or $\nu\cdot \lambda=c$ in vacuum. In some refractive media you can substitute $c$ with speed of light at corresponding medium, essentially nothing changes. Speed determines the product of wavelength and frequency, not their ratio, as you suggested. Hence at different speeds frequency and wavelength will both decrease or increase, there is no reason to think that one of them stays constant and the other one changes.

If you still want to make a distinction and describe physical property of light with one of them, frequency $\nu$ would be more relevant as matter/light (photon/electron) interactions work as described by the formula $E=h\cdot \nu$

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  • $\begingroup$ A short comment: c is always the speed of light, there is no need to "substitute" it, when you have another medium. The speed depends on the medium, so you have to know the specific constant. $\endgroup$ – Chris Oct 2 '14 at 6:51
  • $\begingroup$ Frequency is supposed to remain constant: physics.stackexchange.com/questions/59469/… $\endgroup$ – jinawee Oct 2 '14 at 8:09
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As WYSIWYG wrote, energy is the answer, but to clarify the dependence of energy from momentum... sorry, the dependence of frequency from wavelength, have a look at the animation on this web page: http://www.acs.psu.edu/drussell/Demos/wave-x-t/wave-x-t.html, posted here in what I believe complies with fair use legislation:

enter image description here

It represents a sinusoidal wave, the basic component of any disturbance (as long as you can count on linearity, but that's another matter...).

Frequency tells you how fast the red dot goes up and down. This is what a receptor put in that specific position will sense: a disturbance (here it's the height of a string, in your case the amplitude of the electric field) changing $\nu$ times per second.

Wavelength, on the other hand, tells you how far the maxima and minima of the sinusoid are separated in the medium. In this case, two subsequent crests are separated by $\lambda$ meters. For a given disturbance frequency, this number depends on how fast the wave is allowed to travel in the medium. A slow wave will have a shorter wavelength (because it will travel a smaller distance in space during the time the red dot completes a cycle); conversely, a fast wave will have a longer wavelength.

Note: the speed can depend on the frequency of the disturbance - in that case the medium is said to be 'dispersive' and the relation between frequency and wavelength (or energy and momentum, or - if you are into quantum physics - between energy E and wavenumber k) is called a 'dispersion relation'.

In an isotropic homogeneous medium, the speed is the same in every point and in any direction and you can write

$$\lambda(\nu) = c(\nu) / \nu $$

The frequency is determined by the physical process in the source. It's very hard to change the frequency (i.e. the color, in the visible) of a light wave - you have to resort to nonlinear effects to do that. The wavelength is the result of the interaction with the medium, and changes all the time. But it's customary to express a sort of implied equivalence between frequency and wavelength based on the behavior the light wave would have in vacuum. Hence, when one says 550 nm photon, he is usually implying a photon with en energy corresponding to

$$E = h \ \nu = h \ c_0 /\lambda = 4.226 10^-19 \ J = 2.64 \ eV$$

Where $c_0$ is the speed of light in vacuo. (Note: the latter value is computed from the one before diving by the charge of the electron).

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In the context of human vision, wavelength and frequency are not mutually exclusive, but given by the first equation you pointed out in the question. That is, whenever the range of human vision is given they are not assuming that wavelength and frequency can change independently.

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  • $\begingroup$ Say you change the refractive index of the retina. Does the perceived color change? $\endgroup$ – jinawee Oct 2 '14 at 6:09
  • $\begingroup$ Also, I think there's a typo in your first equation. I don't have enough points to edit or comment. $v \lambda = c$ $\endgroup$ – General Zod Oct 2 '14 at 6:11

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