Chromatic Dispersion, and Variability of Refractive Index on Wavelength

1. Feb 12, 2010

hbal9604@usyd

Hi I'm trying to sort this concept out in my head and have reached a stumbling block!!

1. ok so light travels through medium a and transmits through medium b and refracts. The angle or refraction is given by snell's law, and quantified by the refractive index of materials a and b.

2. But I now understand that the refractive index n of a material is not a static thing; rather, it varies depending on which wavelength light passes through it.

3. BUT ALSO, for light passing from material a to material b, its wavelength changes!!!!

According to this equation:

wavelength through medium = wavelength through vacuum/index of refcraction

4. frequency is constant (waves cannot be destroyed by a medium, right? can you elaborate on this by teh way????).

5. question: if we take a red beam of light passing through a vacuum (with it's characteristic wavelength), and then refract it through a prism, from (3) it would appear that it's wavelength changes, and so does this mean it's not red anymore? or is the quality we observe as colour purely governed by the frequency which is constant? Cause often enough colours are indexed by wavelength!!!!????

That is, does a red beam through a vacuum hit a prism, refract, have its wavelength changed and consequently appear to be yellow or something????

6. Please look at the following:

wavelength of light in some medium = w (not lamda)
wavelength of light in vacuum = w_0

refractive index:

n = c/v (i)

in some medium, the speed of light is given by

v = fw (ii)

in vacuum, the speed of light is given by

c = fw_0 (iii)

from (i) and (ii),

n = C/(fw) (iv)

BUT TAKE A BEAM OF LIGHT WITH VACUUM WAVELENGTH w_0, AND LET IT PASS
THROUGH A MEDIUM WITH INDEX OF REFRACTION n, THEN THE IT'S WAVELENGTH
CHANGES FROM w_0 to w according to the following:

w = w_0/n (v)

But now from iv and v, I get

n = w_0/w and n = c/(fw)

so

c/(fw) = w_0/w

and rearanging, c = fw_0

which is correct, but kind of circular. I don't know my head seems to be
in a mess!

7. What I'm trying to convey is my desire to understand the whole process: does the wavelength change when light crosses from material a to b BECAUSE of the refractive index which is a measure of how much slower the speed of light is in that medium? or is it the refractive index which responds to the particular specific wavelength in question (as required for dispersion). Because the two don't sit well TOGETHER in my head as that would seem to create a situation where a beam of light hits a medium, refracts because of an index of refraction, which is itself determined by the wavelength, which by the way changes once it's being transmitted because of the refractive index. DO YOU SEE MY CONFUSION??

THANK YOUUUUU!!!!

2. Feb 12, 2010

saunderson

Fundamental for your questions is the dispersion relation for electromagnetic waves

$$\omega^2 = c^2 \cdot \vec k ^2 \qquad \mathrm{or} \qquad \lambda = c \cdot T$$​

and Huygens' principle, that implies, that every point of an existing wave is the origin of a spheric wave with the same frequency / cycle duration!

The refraction index and the other things you have asked are consequences of Huygens' principle and the dispersion relation (as far as i noticed after i glanced through you post).

Before going further into detail you have to internalize these.

3. Feb 12, 2010

DrDu

The colour of light is given by its frequency, not wavelength. However, you are right that the frequency of light is often expressed via its wavelength in vacuum, which is not a problem as the n_0 of the vacuum does not depend on nor the wavelength neither the frequency.
First, let me stress that the wavelength dependence of the index of refraction is normally a tiny effect, often neglected, in contrast to the dependence on frequency. One speaks of spacial and temporal dispersion, respectively.
The most well known effect of a wavelength dependence of the refractive index is the phenomenon of optical activity, that is, the rotation of a the polarization plane of light in a medium. When the effect is stronger, it can lead to completely new phenomena, the most discussed of all maybe is the phenomenon of negative refraction.

4. Feb 12, 2010

hbal9604@usyd

so when you say that the frequency of light is the greater player, is this teh case:

that a beam of light of definite frequency hits a prism, which, due to teh inherent properties of that prism in conjunction with the specific frequency in question culminates in this case in a refractive index perculiar to this combination of frequency and refractive material (but which is minorly affected by the beam's wavelength also - not that it has much of an effect).

and now that this index of refraction is arrived at, the beam of light refracts accordingly at some angle given neatly by snell's law.

and the fact that white light is composed of all different frequencies superimposed, means that these various factors play out individually for each frequency and thus great a fanning out of all colours upon refraction, as each frequency has it's own (distinct) refractive index and consequently it's own (distinct) angle of refraction.

n = c/v = c/(f x wavelength)

so for greater frequency, smaller index of refraction, greater angle of refraction?

5. Feb 12, 2010

DrDu

All is correct but the interpretation of your last equation.
Take for example light in the vacuum: There f*wavelength=c which is a constant independent of frequency. Hence for the vacuum n=1 independently of frequency.
In materials n is a function of f, so wavelength=c/(f*n(f)). The latter equation describes especially how the wavelength changes when a light wave of fixed frequency f changes from a medium with refractive index n_1 to a medium with refractive index n_2.
In fact, the index of refraction increases with frequencies in the regions where the material is non-absorbing while in the regions where the material absorbs strongly it decreases (so-called anomalous dispersion).

6. Feb 12, 2010

nasu

The wavelength is changed as long as the light goes through that medium. When is exits the medium the wavelength is back to the original value (assuming that it exits back in the vacuum).
So the frequency is an intrinsic value of the wave whereas the wavelength depends on the medium.