Why does the refractive index of a material dependent on the wavelenght of light incident on it.

  • $\begingroup$ Because the electrons/phonons/polaritons/plasmons/etc. in the material all have different dispersion relationships based on solid state physics, so the material overall will react differently to different wavelengths. $\endgroup$
    – Jon Custer
    Commented Feb 22, 2017 at 18:07
  • $\begingroup$ @JonCuster isn't it rather the coupling of the light field and density of states of these parcticles/quasi-particles rather than their dispersion relation which is making the difference? $\endgroup$
    – user_na
    Commented Feb 22, 2017 at 18:17
  • $\begingroup$ JonCuster pretty well explained the reason, but it is too few to an answer. But it is not his fault, simply the question is too simple. If you want to know some deeper reason, or its mechanism, then ask for that, by extending your question. Do it fast, until it will be closed. $\endgroup$
    – peterh
    Commented Feb 22, 2017 at 18:31
  • $\begingroup$ @user_na - well, the dispersion relationships for each mode come directly in to the overlap integral, so I'm not sure I see how there is a real distinction to be made. If a dispersion relationship were different, you would get different coupling. $\endgroup$
    – Jon Custer
    Commented Feb 22, 2017 at 18:55

1 Answer 1


The refractive index is defined by the ratio $$n=\frac{v}{c},$$ between the speed of the light in the medium and the speed of the light in the vacuum. From the wave equation we read the velocity $v$ in terms of the permittivity and permeability of the medium, $$v=\frac{1}{\sqrt{\epsilon\mu}},$$ and, in particular, for the vacuum we have $$c=\frac{1}{\sqrt{\epsilon_0\mu_0}}.$$ From the three equations above and the definition of relative permittivity and relative permeability we can write the refraction index as $$n=\sqrt{\epsilon_r \mu_r}.$$

For simplicity, let us consider non magnetic materials (non metals would be enough), so that $$n=\sqrt{\epsilon_r}.$$ This relative permittivity depends on the frequency of the incident wave (or rather its wavelength), so $n=n(\omega)$ (or $n=n(\lambda)$).

To get an intuition on this frequency dependency we have to recall that from a microscopic point of view, the electronic permittivity actually measures how well an electronic cloud can be polarized by an electromagnetic field. The greater the distortion, the greater the dipole moment, the atomic polarizability, and the permittivity. The botton line is that we can understand how the refraction index depend on frequency if we understand how the electronic polarization depends on frequency.

A simple model to explain that consists on considering the centre of mass of the electronic cloud a driven oscillator. The equation of motion is $$\frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega_0^2x=qE,$$ where $qE$, the electric force, is the driven term. The constant $\omega_0$ is the cloud's natural frequency of oscillation and gamma is same damping. The amplitude of oscillation (for the steady solution) is frequency dependent, it follows curves like those in the figure

enter image description here

where the different curves are associated to different damping. As we can see, light with different wavelengths (different frequencies) put the electronic cloud to oscillate with different amplitudes leading to different refractive index. Of course this is a crude model, but it can predict refraction index that agrees with the experimental result up to the fourth decimal place.

  • $\begingroup$ But none of this looks like the answer to my question.My question is towards the behaviour of light of different wavelegth in the same medium.The refractive index increases on decreasing the wavelenth of the incident light. What causes this to happen and why not light's velocity decrease by the same amount when incidented in the same medium.And this property is so vivid that it causes dispersion and so what makes all of this stuff , this light's behaviour to change on the change in its wavelength. $\endgroup$ Commented Feb 22, 2017 at 22:59
  • $\begingroup$ @akashagarwal But that is exactly what I answered. Just replace frequency by inverse of wavelength in my answer. Different wavelengths of light interacting with the same atom put the electronic cloud to oscillate with different amplitudes. $\endgroup$
    – Diracology
    Commented Feb 22, 2017 at 23:03

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