On dispersion , speed of wave is inversely proportional to its wavelength . Therefore, gamma has max speed. But then in visible light , violet has least wavelength still it has least speed in visible spectrum. How?


$n \propto \frac{1}{\lambda} $ , where n = index of refraction and $\lambda$ = wavelength of light

$n = \frac{c}{v} $ , where c = speed of light in vacuum and $v$ = speed of light in material or medium

$\therefore v \propto \lambda $ , so the speed of light in a material is proportional to the incident light's wavelength

In words: the index of refraction of a material is inversely proportional to the light's wavelength and is proportional to the reciprocal of the speed of light in a material. A bigger speed of light in a material or medium results in a smaller index of refraction, and vice versa. These together mean that speed of light in a medium is proportional to its wavelength. So the speed of light is not inversely proportional to its wavelength. See Material Dispersion in Optics.

The speed of light is always dependent on wavelength in a non-vacuum medium, so dispersion naturally happens anytime there are multiple modes (wavelengths) of light traveling in a medium. Higher energy light (blue) gets slowed down by the medium more than lower energy light (red) does. Then, gamma rays would actually be the slowest EM radiation in a medium, even though they have the highest energy. Also see What really causes light to appear slower in media?

The index of refraction of a material is set by the electromagnetic properties of the medium (electric permittivity and magnetic permeability), but these change slightly because of incident light. See this. Function for index of refraction

The variation of $n$ based on wavelength, $n(\lambda)$, can be quantified by the Abbe number and is useful in optics and lenses.

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    $\begingroup$ Good physics - but would you mind translating the first few sentences into algebra so we don't have to do that in our heads? It makes sense but it's really hard to read... $\endgroup$ – Oscar Bravo Feb 1 '19 at 7:06
  • $\begingroup$ Thanks - let me know if you think that helped or not. And it's my first time trying MathJax. $\endgroup$ – Alex Strasser Feb 1 '19 at 7:33
  • $\begingroup$ Looks great! I always think if you want to talk physics, you have speak maths since maths is the language of physics :-) $\endgroup$ – Oscar Bravo Feb 1 '19 at 12:54
