Light in other medium

informtohagrid

Hi..when light goes from one medium to other medium (like air to glass or glass to air) its velocity changes(either increases or decreases)..if velocity is changing then frequency or wavelength should also be changing..i want to know which one changes..frequency or wavelength or both...
I think frequency changes..is this correct..

cesiumfrog

nope, wavelength. *shrug*


...... I guess I could explain in more detail. Say cars enter a tunnel at a frequency of one per second. At some point, assume the traffic speed decreases (perhaps there is a highly visible radar trap): there must still be one car per second passing that point, otherwise the cars would keep piling up (until no more could fit into the tunnel), so the *length* between cars must decrease there..

Helios

No, the frequency f and hence the energy hf remain unchanged. The wavelength change is simply

[tex]\lambda[/tex] = n[tex]\lambda[/tex]'

where
n is the index of refraction
[tex]\lambda[/tex]' is the wavelength in the medium

informtohagrid

are there any theoritical or experimental proof that only wavelength changes when light goes from one medium to other medium?

MrXow

I agree with the wavelength changing, however I would not bet my life on it.

jtbell

The frequency has to be the same on both sides of the boundary between two media because the wave function has to be continuous at all times, that is, it can't "jump" discontinuously as it crosses the boundary.

For a classical analogue, consider a wave traveling down a stretched string which is fastened to a thick rope at one end. The speeds of the wave are different along the string and the rope. The string just before the junction has to oscillate at the same frequency as the rope just after the junction, in order for the string and the rope to stay connected at the junction.

informtohagrid

Any mathematical or theoritical derivation you have for the relation above...

Helios

Let v be the velocity of light in the medium

n = [tex]\frac{c}{v}[/tex] = [tex]\frac{f\lambda}{v}[/tex] = [tex]\frac{\lambda}{\lambda'}[/tex]

cesiumfrog

Strictly, Helios, you should have a ratio of two refractive indices there.