2
$\begingroup$

I have been taught in my school (and read on many internet sites too) that frequency of light doesn't change by changing the medium.

Also I read that the colour we see depends on the energy of the photons (which depends on the frequency of the photon) reaching our eye in this answer. So we can conclude that when medium is changed colour of light doesn't change.


Now see this :

enter image description here

The heart shaped structure is actually a paperweight of glass. When I passed the green laser from it the colour of the light beam changed to red which means that the frequency changed (more specifically : the frequency decreased) on changing the medium.

Now this is what I want to know. How did the medium changed the colour of the light ?

Why is the picture in contradiction with the theory?

$\endgroup$
13
  • 1
    $\begingroup$ I would like to add that this is not happening at a particular angle. It is generally true for all angles of incidence $\endgroup$ – Ankit Jun 6 at 15:31
  • $\begingroup$ The red light is not a reflection of any other light .. I can even add pictures in complete darkness if needed. $\endgroup$ – Ankit Jun 6 at 16:03
  • 1
    $\begingroup$ Just a comment to give a simple intuitive feeling of the phenomenon. If you ever played with UV lights and fluorescent paints, or even just a plain old white t-shirt, you see that the UV light makes these materials glow. You are observing a very similar effect with the laser and the glass. The green light excites the material and it then radiates at higher wavelength. $\endgroup$ – José Andrade Jun 8 at 11:59
  • 1
    $\begingroup$ @shai horowitz trust me I can see them with my eyes.. but I don't know what can I do so that you guys can believe.. $\endgroup$ – Ankit Jul 11 at 13:26
  • 1
    $\begingroup$ We believe you. Were just pointing out hypothesis that can be easily shot down to narrow the playing field $\endgroup$ – shai horowitz Jul 11 at 14:25
2
$\begingroup$

Assuming everything is correct, and this red light is not a reflection of some other light source or something: one possible way of achieving this is that the medium is absorbing green light and emitting the red light through photoluminescence. So in this case it's technically not light changing frequency in the medium; it's "new" light produced by the medium, and it is characteristic for the particular medium. In short, if the medium absorbs green light and emits red in this process, it means that there is an allowed optical transition with energy corresponding to the green light, and then this excited state is relaxed to a lower-energy state. The latter can then relax to the ground state emitting a red photon. So technically, the spectrum of the photoluminescence contains information about the energy levels in the medium.

Note that in linear optics (single photon processes only) it can only go one-way (green to red in your example) because the energy of the green photon is higher. So the conservation of energy is not violated, and some energy is lost in the medium (e.g. on heating). This means that if you get a source of red light and shine its light on the same glass, it won't produce green light. In nonlinear optics, this is not true; for example, two photons with energies half of that of the green photon can be absorbed, and one red photon can be emitted, bit for that you need high intensity (laser light could work actually).

Of course, when I write "red" or "green" photon, I mean a photon with the energy corresponding to the wavelength of green or red light, respectively.

$\endgroup$
2
  • $\begingroup$ color perception could play a role too, hyperphysics.phy-astr.gsu.edu/hbase/vision/colper.html . i.e. the green giving rise to a cascade of new photons that add up to a perception of red. $\endgroup$ – anna v Jun 6 at 17:30
  • 2
    $\begingroup$ @annav not in this case. The picture clearly shows fluorescence. $\endgroup$ – José Andrade Jun 8 at 11:56

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.