When I look at Snell's law

$\frac{\sin\theta_2}{\sin\theta_1} = \frac{v_2}{v_1} = \frac{n_1}{n_2}$

I don't see any reference to wavelength.

If red and blue have the same speed in the same medium, why they refract differently? What am I missing?

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    $\begingroup$ Bear in mind that light does not posseses index of refraction. Usually this is attributed to materials $\endgroup$
    – amonk
    Commented Nov 12, 2020 at 9:31
  • 2
    $\begingroup$ @amonk thank you, but the Snell's law makes reference to the speed of light in a certain medium, thus my question $\endgroup$ Commented Nov 12, 2020 at 16:17

6 Answers 6


In general, red and blue light do not travel at the same speed in a non-vacuum medium, so they have different refractive indices and are refracted by different amounts. This phenomena is known as dispersion.

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    $\begingroup$ This is the only answer that actually address the OP's confusion. The speed is not the same for blue and red light in a medium. $\endgroup$
    – nasu
    Commented Nov 12, 2020 at 18:19
  • $\begingroup$ Sir can you tell why red and blue light do not travel at the same speed in a non-vacuum medium? And is there any role of Cauchy's dispersion formula? $\endgroup$
    – Aveer
    Commented Feb 10, 2022 at 7:59
  • $\begingroup$ @Aveer Because photons with different wavelengths/energies interact with the medium in different ways. See en.wikipedia.org/wiki/Dispersion_(optics). Cauchy's formula is an empirical relationship between refractive index and wavelength - see en.wikipedia.org/wiki/Cauchy%27s_equation. $\endgroup$
    – gandalf61
    Commented Feb 10, 2022 at 10:14

The refractive index is a function of wavelength. It has different values for different wavelengths. The way to show this in the mathematical notation is to write $$ n(\lambda) $$ just as you would write $f(x)$ for some function of $x$. So with this more complete notation Snell's law is written $$ n_1(\lambda_1) \sin (\theta_1) = n_2(\lambda_2) \sin (\theta_2). $$ The common version $$ n_1 \sin (\theta_1) = n_2 \sin (\theta_2) $$ is saying the same thing, once you remember that $n_1$ and $n_2$ here refer to the refractive index at the wavelength in question in the medium in question.

The wavelength will itself be different in the two media. This makes it slightly awkward to use the formulae in terms of wavelength as written above. However the frequency will be the same in the two media so one way to proceed is to find the index as a function of frequency and use that.

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    $\begingroup$ It's worth pointing out that while the wavelength will be different, the frequency remains the same (for intensities where the medium reacts linearly; isn't it great how physics is approximation on top of approximation on top of approximation ...) $\endgroup$
    – tobi_s
    Commented Nov 12, 2020 at 3:15
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    $\begingroup$ I think it would be better to re-phrase this answer in terms of frequency instead of wavelength. As (in the given wording) refraction index depends on wavelength, and wavelength in a medium changes depending on its refraction index, apparently creating a circular dependency. $\endgroup$ Commented Nov 12, 2020 at 8:06
  • $\begingroup$ @tobi_s Yes I thought about including that and your comment persuaded me to do so. I have now edited the answer a bit. I did not go into nonlinearity so your comment serves as a pointer to that. $\endgroup$ Commented Nov 12, 2020 at 9:27

Index of refraction actually does depend on the wavelength of the light in the medium. Typically this detail is left out of introductory physics classes, and the values of index of refraction given to students only applies to yellow light (at least this is what happened in my experience).

Of course, one obvious example of index of refraction being wavelength dependent is when white light moves through a prism. We end up with a spectrum coming out of the prism, indicating that the speed of light in the prism varies with wavelength of the light.


Optical dispersion is the property of a material to have a wavelength dependent index of refraction. A material with optical dispersion will have a different index of refraction for red vs. blue light meaning that red and blue light 1) have different velocities in the material and 2) refract at different angles at interfaces with the material.

If a material lacked dispersion (had constant index of refraction across wavelength) then red and blue light would have the same velocity and refract with the same angle. Most materials, however, do exhibit dispersion. For example glass does which is why a prism can be used to decompose white light into it's various wavelength components.


Velocity is equal to the product of a wave's frequency times its wavelength and, as the frequency of light remains the same as it changes media, the only thing that does differ is the velocity and the wavelength. Snell's law states that $\frac{v_2}{v_1}$ gives us the refractive index of light. As the wavelength of different colors of light varies, so do their speeds (in a non-vacuum medium). The greater the wavelength, the greater the speed, causing a difference in the refractive index. As there is a difference in the wavelength of red (about 700 nm) and blue (about 500 nm), they have different speeds in the same medium (other than vacuum/air) leading to different refractive indexes.

  • $\begingroup$ The last sentence is untrue. A difference in wavelength doesn't automatically different speeds in the same medium. $\endgroup$
    – Nayuki
    Commented Nov 13, 2020 at 6:34

Just as an addition to gandalf61's answer, which clarifies that blue and red light has different speeds in a medium:

When we say that red and blue light has different speed in medium we mean the speed of the wavefront. This speed of the wavefront is what actually differs for red and blue light.

Now individual photons, that make up the light, do travel at speed c in vacuum, inbetween the atoms of the medium, when measured locally.

But because there is interaction between the photons and the atoms in the medium, the wavefront slows down, this is what we mean when we say that light travels slower (relatively then in vacuum) in a medium.

Rayleigh scattering (/ˈreɪli/ RAY-lee), named after the nineteenth-century British physicist Lord Rayleigh (John William Strutt),[1] is the predominantly elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation. For light frequencies well below the resonance frequency of the scattering particle (normal dispersion regime), the amount of scattering is inversely proportional to the fourth power of the wavelength.


Since different wavelength photons scatter differently (because scattering is wavelength dependent) on the atoms of the medium, the wavefront will slow down differently in the case of different wavelength light. Blue wavelength photons have a shorter wavelength (relative to red wavelength) and so blue wavelength is a little bit closer to the size of the atoms in the medium, so blue will scatter more then red. The visible wavelength you are asking about is still much larger then the atoms' size in the medium, thus Rayleigh (elastic) scattering is the correspondent way, by the way this is the same reason the sky is blue.

So the answer to your question is that red and blue are refracted differently because ultimately they are made up of different wavelength photons, and though the individual photons all travel at speed c in vacuum inbetween the atoms of the medium (when measured locally), the interaction (scattering) between the photons and the atoms is wavelength dependent, and this causes the wavefront to slow down differently for different wavelength light.


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