10
$\begingroup$

Wikipedia says that "in general, the refractive index of a glass increases with its density." And the refraction index of water vapor is less than ice, and even less than liquid water. Is there any simple explanation to that?

$\endgroup$
4

2 Answers 2

10
$\begingroup$

The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time.
The more this happens, the slower the effective average speed.
The denser the material, the more atoms there are in the way.

$\endgroup$
5
  • $\begingroup$ Great explanation. $\endgroup$
    – boyfarrell
    Commented Mar 29, 2014 at 6:04
  • 5
    $\begingroup$ The explanation is only apparent. In classical EM theory there is no time lag between interaction of external EM wave with charges and emission of secondary EM wave, both occur simultaneously. $\endgroup$ Commented Mar 29, 2014 at 8:02
  • 4
    $\begingroup$ By what mechanism is the light absorbed? Not an atomic electron transition, as surely re-emission would change frequency and hence color. $\endgroup$
    – xxyzzy
    Commented Jul 21, 2018 at 11:22
  • 1
    $\begingroup$ This explanation is wrong. See the Ewald–Oseen light extinction theorem for the correct explanation. $\endgroup$ Commented Aug 27, 2023 at 2:12
  • $\begingroup$ No, this answer is incorrect. In fact, the field of spectroscopy fundamentally relies on the principle that different materials interact narrowly with only specific wavelengths of light. Certain materials only interact with certain wavelengths, yet dense mediums like water slows light in a continuous, not discrete, fashion depending on wavelength. $\endgroup$
    – Zamicol
    Commented Dec 31, 2023 at 12:47
5
$\begingroup$

This is quite a subtle issue. The charges in the medium produce secondary spherical expanding EM waves when hit by the primary wave (external forces). There is immense number of these secondary waves. At any point of space, each secondary wave has slightly different wave vector. In a medium dense enough, these secondary waves add to the primary wave in such a way that the resulting wave has behaviour that is well described by a single macroscopic wave of the same frequency and (usually) same direction but (for most frequencies) with a reduced wavelength.

A common picture backed by successes of dispersion theory is that the relation $\mathbf j(t) = c\mathbf E(t-\Delta t)$ is valid, where $c, \Delta t$ are some medium property constants that depend on frequency of the wave, $\mathbf j$ is current density and $\mathbf E$ is total macroscopic electric field. With this assumption, Maxwell's equations imply that the resulting wave in the medium will have modified (in usual cases shorter) wavelength hence lower velocity (for a certain limited interval of frequencies it can have a longer wavelength and higher velocity).

$\endgroup$
10
  • 1
    $\begingroup$ This explanation cannot be correct. You state that "the resulting wave in the medium will have shorter wavelength hence lower velocity" but this is incorrect. If you change the wavelength (or the frequency, because frequency = c / wavelength) you only change the energy of the photon / light wave (and thus colour) but not the speed, which is fixed as long as it remains in the same type of medium. Indeed though as @arax notes, this fixed speed actually depends on the type of medium and can become smaller than c, but I do not know the correct explanation. $\endgroup$ Commented Aug 11, 2017 at 22:38
  • 1
    $\begingroup$ @PDiracDelta, the question is about index of refraction. Index of refraction is ratio of phase velocities of a wave in two media. Since frequency is the same in both media (linear media) and since $v = f \lambda$, decrease in wavelength implies decrease in phase velocity $v$. The energy of a photon or light wave has nothing to do with this explanation. $\endgroup$ Commented Aug 12, 2017 at 23:22
  • $\begingroup$ I'm trying to understand what you are saying, but if not light, then what kind of wave are you talking about? AFAIK a light and EM wave are the same thing. Also, to me your claim that the frequency remains the same is non-trivial. $\endgroup$ Commented Aug 14, 2017 at 11:03
  • $\begingroup$ I think I've found a good explanation here: physics.stackexchange.com/a/476/51901 (this question now apparently has been marked as a duplicate) $\endgroup$ Commented Aug 14, 2017 at 11:18
  • $\begingroup$ I am talking about an EM wave in material medium, which is a model of a light wave, but the important part in this explanation is relation of electric field to current density, not energy of the wave or the concept of a photon. $\endgroup$ Commented Aug 14, 2017 at 23:07

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