# Properties of light and refractive index of materials

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

• 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. – Jon Custer Feb 22 '17 at 18:07
• @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? – user_na Feb 22 '17 at 18:17
• 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. – peterh Feb 22 '17 at 18:31
• @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. – Jon Custer Feb 22 '17 at 18:55

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 $n=n(\lambda)$).
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