It is well known that a prism can "split light" by separating different frequencies of light:

prism diagram

Many sources state that the reason this happens is that the index of refraction is different for different frequencies. This is known as dispersion.

My question is about why dispersion exists. Is frequency dependence for refraction a property fundamental to all waves? Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields? Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?

  • $\begingroup$ This related question physics.stackexchange.com/q/19422 provides dispersion as the answer but doesn't provide an answer to "why dispersion?". $\endgroup$ May 24, 2013 at 22:07
  • $\begingroup$ Pardon my laziness to just Google it, but isn't Newton credited with having discovered and studied this aspect of light? If I'm remembering correctly, then Newton wouldn't have neglected to postulate a corpuscular argument, which I'd be very interested in reading. $\endgroup$
    – David H
    May 24, 2013 at 22:52
  • 2
    $\begingroup$ Relevant en.wikipedia.org/wiki/Dispersion_relation. $\endgroup$ May 24, 2013 at 22:52
  • $\begingroup$ @joshphysics that article mentions "interaction of the waves with the transmitting medium". I'd like to understand if that's a special property of light or if all waves will have some sort of interaction. I assume no interaction implies an index of refraction of 1 (no refraction) so does that mean non-1 index of refractions always have dispersion regardless of the sort of waves? $\endgroup$ May 24, 2013 at 23:03
  • 2
    $\begingroup$ Ondřej and I have given brief answers, his with more depth on a certain aspect of the problem, mine with more breadth and less depth. For a full treatment, see the Feynman Lectures, I-31 and II-32. $\endgroup$
    – user4552
    May 26, 2013 at 17:28

4 Answers 4


Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field $\mathbf{E}$ (i.e., light), its movement can be described by the equation $$ m\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2}+m\gamma\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}+k\mathbf{x} = e\mathbf{E}. $$ The first and third terms on the LHS describe a classical harmonic oscillator, the second term adds damping, and the RHS gives the driving force.

If we assume that the incoming light is monochromatic, $\mathbf{E} = \mathbf{E}_0e^{-i\omega t}$ and we assume a similar response $\xi$, we get $$ \xi = \frac{e}{m}\mathbf{E}_0\frac{e^{-i\omega t}}{\Omega^2-\omega^2-i\gamma\omega}, $$ where $\Omega^2 = k/m$. Now we can play with this a bit, using the fact that for dielectric polarization we have $\mathbf{P} = \epsilon_0\chi\mathbf{E} = Ne\xi$ and for index of refraction we have $n^2 = 1+\chi$ to find out that $$ n^2 = 1+\frac{Ne^2}{\epsilon_0 m}\frac{\Omega^2-\omega^2+i\gamma\omega}{(\Omega^2-\omega^2)^2+\gamma^2\omega^2}. $$ Clearly, the refractive index is frequency dependent. Moreover, this dependence comes from the friction in the electron movement; if we assumed that there is no damping of the electron movement, $\gamma = 0$, there would be no frequency dependence.

There is another possible approach to this, using impulse method, that assumes that the dielectric polarization is given by convolution $$ \mathbf{P}(t) = \epsilon_0\int_{-\infty}^t\chi(t-t')\mathbf{E}(t')\mathrm{d}t'. $$ Using Fourier transform, we have $\mathbf{P}(\omega) = \epsilon_0\chi(\omega)\mathbf{E}(\omega)$. If the susceptibility $\chi$ is given by a Dirac-$\delta$-function, its Fourier transform is constant and does not depend on frequency. In reality, however, the medium has a finite response time and the susceptibility has a finite width. Therefore, its Fourier transform is not a constant but depends on frequency.

  • 1
    $\begingroup$ That's a good summary! Basically the resonance is given by $k$, but your answer does not really say what is this $k$ from. Dipole oscillations around the atom? $\endgroup$
    – fffred
    May 25, 2013 at 8:31
  • 1
    $\begingroup$ @fffred Yes, $k$ comes from the dipole oscillations of the electron. I thought that the basic meaning was clear from the text and I don't want to go into much detail concerning finding specific values of the constants. I am certain one can find more details in the literature; this is just to give the main idea. $\endgroup$ May 25, 2013 at 8:39
  • 1
    $\begingroup$ What about the atoms of the opaque materials?Oscillation is an assumption inconsistent to electronic cloud. $\endgroup$
    – elias2010
    Feb 28, 2018 at 12:09
  • $\begingroup$ Hey I had a doubt. The reason of frequency dependence of $\epsilon_0$ is rooted in the dipole oscillations of electron. For non dispersive medium, where refractive index $n$ is constant, do we have $\gamma$ = 0? But this is not possible as there must be some damping. $\endgroup$
    – Ruchi
    Feb 20, 2021 at 9:11
  • 1
    $\begingroup$ According to the equation above, by setting $\gamma = 0$ don't we still have a factor of $\Omega^2-\omega^2$ in the denominator? It would still seem to be frequency-dependent without the damping term. $\endgroup$
    – Jono94
    Oct 14, 2023 at 10:07

The simple explanation given in Hewitt's Conceptual Physics is that atoms in condensed matter have a high-frequency resonance, and the index of refraction for most substances is strongest at the blue end of the spectrum because that's the high-freqency end, which is closest to the resonance. The following is my attempt to flesh this out with a little more serious physics. It seems to capture some of the truth, but in some ways it's crude or wrong.

Index of refraction of silica glass as a function of wavelength (redrawn from Kitamura 2007).

Kitamura 2007 gives a summary of experimental data for silica glass over a wide range of wavelengths, along with a physical interpretation. The graph above is redrawn from Kitamura. What is observed is that the complex index of refraction has three prominent resonances with a shape that I think is referred to as a Lorentzian. At each resonance, the real part of $n$ swings low and then high, while the imaginary part has a peak, indicating absorption. They attribute each of these resonances to one or more qualitatively different physical phenomena. The visible spectrum lies between resonances at about 0.1 $\mu$m and 9 $\mu$m. The former is attributed to "interaction with electrons, absorption by impurities, and the presence of OH groups and point defects," the latter to "asymmetric stretching vibration of Si-O-Si bridges."

Although this is all pretty complicated, I think there's some fairly simple physics that can be extracted.

In the visible region, it looks like the decrease of the index of refraction with wavelength is due to a combination of two effects. This region of the graph picks up a negative slope from the 0.1 $\mu$m resonance on its left, and also a negative slope from the 9 $\mu$m on the right. This is a universal feature of any function formed by adding up a bunch of narrow Lorentzian resonances: far from resonances, it always has a negative slope. The bigger contribution to the slope seems to come from the resonance on the left, which is consistent with Hewitt's explanation.

Kitamura mentions several models that explain the resonances, of which the only one I'm familiar with is called the Lorentz model. In the Lorentz model, you take an electron to be a harmonic oscillator, like a little mass bound by a spring to a nucleus. The displacement of this driven harmonic oscillator (represented as a complex number to include its phase) is the Lorentzian $x=Af(\omega)$, where $f(\omega)= (\omega^2+i\gamma \omega-\omega_0^2)^{-1}$ and $A=(e/m)E$. As the electrons perform this oscillation in response to a plane wave, they generate their own coherenet plane wave. What is actually observed is the superposition of this wave with the incident wave. This superposition has two parts, a reflected wave and a transmitted one. In the limit of a low-density medium (such as a gas), the index of refraction is given by $n^2=1-\omega_p^2 f(\omega)$, where $\omega_p$, called the plasma frequency, is given by $\omega_p^2=Ne^2/m\epsilon_0$, where $N$ is the number density of electrons. The plasma frequency has an $e/m$ in it from the amplitude of the driven harmonic oscillator, and another factor of $e$ because the amplitude of the reemitted wave is proportional to the amount of charge oscillating. In the case of silica glass, I think the 0.1 $\mu$m resonance is probably what is described by the above mechanism, while the other resonances are similar mathematically but involve other effects than oscillation of bound electrons. E.g., the Si-O-Si bridges would resonate at a lower frequency due to the greater inertia of the nuclei compared to electrons.

An interesting feature of the graph is that there are broad plateaus, and as we go up in wavelength, each plateau is successively higher than the preceding one. This is explained by the Lorentz theory. In the limit the response of a driven harmonic oscillator approaches zero in the limit $\omega\gg\omega_0$, but approaches a constant (with reversed phase) for $\omega\ll\omega_0$. Adding the contributions from the various resonances produces an ascending staircase as observed.

Is frequency dependence for refraction a property fundamental to all waves?

The above does seem to suggest that there's some very universal behavior going on in the interaction of EM waves with matter.

Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields?

No, I think it's basically the linear response of a driven harmonic oscillator.

Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?

I'm sure this would be a holy grail for people doing optics. AFAIK, the best way of getting rid of dispersion in real devices seems to be combining two materials so that the dispersion cancels out. Silica glass does seem to have a relatively constant $n$, and this would be because the visible spectrum is relatively far from the two nearby resonances. To get less dispersion in the visible spectrum, I guess you would want a substance in which the resonant frequency that glass has at 0.1 $\mu$m was displaced higher.

Kitamura, http://www.seas.ucla.edu/~pilon/Publications/AO2007-1.pdf

  • 1
    $\begingroup$ Does this imply that dispersion is somewhat temperature dependent? If we cooled a prism down near absolute zero would we changes in the dispersion? Edit: I just noticed the paper has "at room temperature" in the title. It seems it is temp dependent. $\endgroup$ May 24, 2013 at 23:26
  • 1
    $\begingroup$ @BrandonEnright: To me, the Lorentz model seems to imply that $n$ would be roughly temperature-independent, except for maybe a slight change in $N$ due to thermal expansion...? I think the papers I came across on the web seemed to indicate that the temperature dependence was quite small, and hard to measure. $\endgroup$
    – user4552
    May 24, 2013 at 23:30
  • 1
    $\begingroup$ Also, it seems the motion of the electrons really matters. I thought refraction of light was based on the strength of the electric field but it looks like there is more to it than that. Hypothetically, if we replaced all electrons in a glass with muons instead then that material would have completely different refractive and dispersive properties? $\endgroup$ May 24, 2013 at 23:33
  • 2
    $\begingroup$ The temperature dependence is straightforward as Ben surmised a matter of thermal expansion. Refractive index is more or less proportional to density (=electron density, but there are some exceptions, eg glasses containing titanium). If one looks at closely related substances only eg organics, the deviations of that proportionality contain some structural information, look for "Parachor". The temperature dependence is easily measured, as everybody knows who ever worked with a refractometer. $\endgroup$
    – Georg
    Jul 16, 2013 at 11:01
  • $\begingroup$ I don’t think this answers the question completely, but I think it’s the best one so far. I also appreciate the IoF refraction graph you posted. Do you know where I can find more? Those graphs were actually what I was searching for when I happened upon this question and answer. Thanks. $\endgroup$
    – Digcoal
    Oct 26, 2021 at 21:37

I will hand wave here, looking at the problem a photon at a time.

We know from the double slit experiment that even individual photons impinging on the double slit geometry display an interference pattern, characteristic of the frequency/energy of the photon and the geometry of the slits.

One can think of a crystal as a very large number of three dimensional obstacles/slits ( 10^23 molecules in a mole give a huge number even for a one centimeter crystal in the path of your illustration).

crystal lattice

A photon impinging on the surface of the lattice, finds not two slits , but a depth of slits all the way through. The observed effect of the different angular distribution according to the impinging frequency of the photon must be the result of the quantum mechanical interference of the photon, which must be constructive in the angle of refraction given by its frequency and index of refraction and destructive everywhere else, otherwise we would be seeing interference fringes ( actually we do get a second rainbow, but that is a different story :) , though should be similar).

Then the problem is reduced to explaining the frequency dependence. I will hand wave again and say that the smaller the frequency the larger the distances in the interference pattern of the probability wave ; the photon will see the lattice gaps differently

enter image description here

according to its wavelength, as is true for the double slit experiment, so a fanning out is to be expected .

  • 2
    $\begingroup$ What you're describing is a mechanism for x-ray diffraction, not refraction. $\endgroup$
    – user4552
    May 25, 2013 at 15:13
  • $\begingroup$ @BenCrowell It certainly is an unusual explanation for dispersion but I guess it could still work. You'd just need to do the calculations in detail to check it. And there is one more thing - it explains dispersion only in crystals, not in, e.g., glass. $\endgroup$ May 25, 2013 at 15:59
  • $\begingroup$ @OndřejČernotík as it HAS to work a photon at a time, i.e. single photons of the correct frequency should fall in the correct band of color, I do not see how it could be something else than a probability interference distribution pattern for photons one at a time. To get the color spectrum a calculation would include the distance of the unit cells of the crystal. In general the classical wave should merge with the photon ensemble frame except that calculating the latter would be too complex. $\endgroup$
    – anna v
    May 25, 2013 at 19:46
  • $\begingroup$ @BenCrowell X-rays have wavelengths much smaller than the distances of the unit of the crystal. They build up an interference pattern that can be fourier transformed to the crystal lattice. Here I am talking of larger wavelengths. My only argument is, as I said above, that it HAS to work a photon at a time, as the double slit experiment does too. $\endgroup$
    – anna v
    May 25, 2013 at 19:50
  • $\begingroup$ @BenCrowell I believe the diffraction Anna describes is the main one at work in opals. So "prismatic" (light splitting) behaviour is not only owing to refraction. $\endgroup$ Jul 16, 2013 at 11:38

Provided that the electron & the atomic beams also exhibit refraction,it seems that this is a particle's property.Velocity and deflection angle depends on particle's mass/size for specific medium.Photon behaves as particle in this effect.Mass is given by de Broglie equation:m=hv/c^2 , v=frequency

  • 3
    $\begingroup$ Dear Nikos Binis. It is usually frown upon to directly copy-paste identical answers. (The problem is if everybody start to copy-paste identical answers en mass.) In general in such situations, please consider one of the following options: (i) Delete one of your answers. (ii) Flag for duplicate posts and delete one of your answers. (iii) If you think the three posts are not duplicates, then personalize each answer to address the three different specific questions. $\endgroup$
    – Qmechanic
    Dec 7, 2014 at 16:21
  • $\begingroup$ You have right,forgive me.I ask for your compensation.I just want the science community to rethink about photon mass. $\endgroup$
    – elias2010
    Dec 11, 2014 at 10:56
  • 1
    $\begingroup$ @elias2010 They are not going to rethink photon's mass because no experiment up to now indicates that the photon has mass. If you have any way to verify it, you've got a Nobel prize wating for you. Although that would be hard 1because the upper bound of the photon's mass is around $10^{-27} \mathrm{eV/c^2}$. $\endgroup$
    – jinawee
    Dec 28, 2014 at 18:00
  • $\begingroup$ Here is an experiment:Schmitt,Weitz,Vewinger,Klaers "Bose-Einstein condensation of photons in an optical microcavity”,Nature 468(7323):545-548 $\endgroup$
    – elias2010
    Dec 30, 2014 at 8:41
  • $\begingroup$ There is no experiment for the opposite.Moreover,virtual photon has mass. $\endgroup$
    – elias2010
    Dec 30, 2014 at 8:52

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.