0
$\begingroup$

Does the speed of Electromagnetic wave depend on its wave length? For vacuum I'm aware that it's a constant $c=\frac{1}{\sqrt{\mu_o\epsilon_o}}$. Similarly can we say speed of light in any medium is a constant and is equal to $v=\frac{1}{\sqrt{\mu\epsilon}}$.

If yes, then refraction shouldn't be possible right ?

If no, then does $\frac{1}{\sqrt{\mu\epsilon}}$ have any significance?

$\endgroup$
3
$\begingroup$

$\epsilon$ - and, to a lesser extent $\mu$ - depends on the frequency of the light. The atoms of the medium have a particular excitation frequency, or frequencies, and the polarisability of the atom rises as the EM frequency rises to a peak when it equals the atomic excitation frequency, and then falls off. It's standard forced SHM https://en.wikipedia.org/wiki/Harmonic_oscillator#Sinusoidal_driving_force . So different frequencies, and hence different wavelengths, have different velocities.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ does $v=\frac{1}{\sqrt{\mu\epsilon}}$ have any significance? $\endgroup$ – Vilvanesh Mar 22 at 16:23
  • $\begingroup$ @Vilvanesh, yes it's the speed of light in that medium. If there was no dependence on wavelength, then a prism wouldn't separate white light into its color components. $\endgroup$ – The Photon Mar 22 at 16:37
0
$\begingroup$

The speed of light in a material does in general depend on wavelength, and there's no better demonstration of this fact than a prism. Since the index of refraction is

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

(where $c$ is the speed of light in vacuum and $v$ is the speed in the medium) and the angle of refraction depends on $n$, the fact that light splits into colors when going through a prism tells us that $n$ depends on the wavelength, and then so does $v$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ does $\frac{1}{\sqrt{\mu\epsilon}}$ have any significance? $\endgroup$ – Vilvanesh Mar 22 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.