Which electromagnetic radiation is faster in water, microwaves or light?

Question

Well I've been asked this question, but I haven't been able to come with an answer yet using books and some web searches. The point is as the title says, to answer the question with the whole phenomenon explanation if possible.

Answer 1

Electromagnetic radiation in a medium propagates according to the law $$ \mathbf E,\mathbf B \propto e^{\imath(\pm k_xx-\omega t)} $$ where $$ k_x^2 = \frac{n^2\omega^2}{c^2}\;. $$ The refractive index $n$ can also be complex, in which case its imaginary part describes the absorption of the EM wave in the medium. But the oscillating part is in any case $$ \propto e^{\imath ({\cal Re}\; n) x/ c} $$ where ${\cal Re}\; n$ is the real part of the refractive index. Thus the apparent speed of the EM is $$ c_{app} = \frac{c}{{\cal Re}\; n} $$ A diagram for the real part of the refractive index (online you can see it here) is the following:

(Optical) light has wavelengths in the range $400-700\;nm$, while microwaves have wavelengths in the range $1\; mm-1\: m$. From the diagram we see the $n$ is much smaller at ptical than micro wavelengths, thus optical EM waves travel faster than microwave EMs in water.

EDIT:

The above of course is about phase velocity. If you are interested in group velocity instead, $$ v_g = \frac{\partial \omega}{\partial k} $$ it can easily be checked from the same diagram that the same conclusion still applies.

Answer 2

This is basically about dispersion: EM waves with different frequencies travel at different speed in a medium because of interaction. Usually, as in a standard textbook experiment wherein a light beam is bent by a prism, higher frequency waves bends more. The more it bends, the more slowly it travels. So, lower frequency waves usually go faster. However, this is only partially true. Ultimately dispersion is determined by how strongly the EM waves interact with the medium. In regime of frequency ($\omega$) where the complex dielectric function $\varepsilon(\omega)$ varies very slowly, the speed is $c/\cos(\delta(\omega)/2)\sqrt{|\varepsilon(\omega)|}$, with $\delta(\omega)$ being the phase angle of $\varepsilon(\omega)$. According to water absorption spectra, the $\varepsilon(\omega)$ at frequencies of visible light is nearly a real quantity, i.e., with little absorption. The real part of $\varepsilon(\omega)$ has also to be small by virtue of Kramers-Kronig relation. While in microwave regime, there is a featureless absorption over a broad band. This means $|\varepsilon(\omega)|$ is indeed much bigger here. Thus, microwave is likely to travel much more slowly than visible light [this contradicts my previous conclusion as pointed out by Ben].

Answer 3

Provided that the electron & the atomic beams also exhibit refraction,it seems that this is a particle's property.Refraction index is proportional to particle's mass/size for specific medium.Photon behaves as particle in this effect.Mass is given by de Broglie equation:m=hv/c^2 , v=frequency.Thereby,light photons have more resistance in their motion inside water and going slower than microwaves.