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The index of refraction can be written as $$n=\frac{\lambda_v}{\lambda_m}$$ where $\lambda_v$ is the wavelength in a vacuum and $\lambda_m$ is the wavelength in the medium. I’ve been told that since wavelength appears in the definition of an index of refraction, an index of refraction varies with wavelength. However, why would that be the case? The index of refraction is a ratio; if a wavelength of one wave is different from that of another wave passing through the same medium, the index of refraction should not be different for each wave, since they would have had different wavelengths in a vacuum too. So why is the index of refraction dependent on the wavelength?

marked as duplicate by ACuriousMind, Floris, Kyle Kanos, HDE 226868, Qmechanic Jul 30 '15 at 18:21

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You are mixing up two different things.

The refractive index is usually defined in terms of the velocity of light:

$$ n = \frac{c}{v} $$

where $v$ is the velocity in the medium. However the velocity is related to the frequency and wavelength by:

$$ v = \lambda f $$

so:

$$ n = \frac{\lambda_0 f_0}{\lambda f} $$

The frequency of the light, $f$, doesn't change as the light enters the medium, so $f_0 = f$ and the $f$s cancel to give:

$$ n = \frac{\lambda_0}{\lambda} $$

which is the equation you cite. The equation is not based upon any assumptions about the variation of $n$ with the frequency/wavelength of the light.

However the refractive index does change with frequency. This effect is called optical dispersion. The cause is the way the interaction of the light and the electrons in the medium change with frequency. See for example the questions Why do prisms work (why is refraction frequency dependent)? and Why does the refractive index depend on wavelength?.

I think you will have an easier time viewing the index of refraction from a speed-point of view.

Consider the following: The energy of a given photon is determined by its frequency (color): $E = h \nu $ (h being the Planck constant) Assuming the photon does not lose energy when entering the material, its frequency must be conserved. However, as light is an electromagnetic wave, its propagation in a material differs fundamentally between the vacuum and a crystalline material. In the material, the electric field of the photon will be harder to 'produce', since the crystal's electrons will react to it. The higher the frequency, the stronger this effect is - the light gets increasingly slower. The refractive index for photons with a certain wavelength is $n(\nu) = \frac{c_0}{c_m(\nu)}$ with $c_0$ being the speed of light in vacuum (equal for all wavelengths as far as we know) and $c_m(\nu)$ the speed of a photon of a certain frequency $\nu$ in the material.

Using $c = \lambda \nu$ (and assuming constant frequency) you will arrive at the expression you were using.

  • Not correct: there are materials with negative slopes to their dispersion curve; further most materials have a nonlinear relationship between input wavelength and index of refraction. – Carl Witthoft Jul 30 '15 at 12:03
  • To be fair this is true for most optically transparent materials at visible wavelengths. The refractive index becomes anomalous when the light energy reaches the band gap energy, and for most transparent materials the band gap energy is in the uv. – John Rennie Jul 30 '15 at 12:08
  • True - I was refering to 'classical', linear materials. No no nonlinear effects or effects found in metamaterials are accounted for. But if we would go along this path I would start with the more realistic complex valued refractive index - which would also mean the energy- and thus the frequency would not be constant for the photon alone. – Patrick Kraus Jul 30 '15 at 12:08

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