# Why do high frequencies have a larger deflection through a prism than smaller frequencies? [duplicate]

The following is a proof that higher (angular) frequencies have a greater deflection than lower frequencies through a glass prism:

It was omitted in the image above, but $n_{\text{air}}$ is taken to be unity, which is the reason for the $1$ in Snell's law.

My problem is as follows; in a general dielectric (such as a glass prism) we have the following dispersion relation:

$$v_{\text{phase}}=\frac{\omega}{k}=\frac{1}{\sqrt{\mu_0\,\epsilon}}=\frac{c}{{\epsilon_r}^{1/2}}=\frac{c}{n}\tag{1}$$

so in short $$\frac{\omega}{k}=\frac{c}{n}$$

or $$n=\frac{k\,c}{\omega}$$

therefore $$n\propto \frac{1}{\omega}\tag{2}$$

Now looking at the writing in the image it says:

$v_{\text{phase}}$ decreases with $\omega$

According to $(1)$ $v_{\text{phase}}$ increases with $\omega$; since $v_{\text{phase}}\propto\omega$.

But that's not the only problem the text also says:

$n_{\text{glass}}$ increases with $\omega$

According to $(2)$ $n$ decreases with $\omega$. Another contradiction.

Is there anyone that can explain why $v_{\text{phase}}$ decreases with $\omega$ and why $n_{\text{glass}}$ increases with $\omega$?

## EDIT:

I'm voting to close this question now since I no longer care whether or not it it is a duplicate of this question as I already have accepted an answer to my question.

• There is no proof in general, since it all depends on the dispersion relationship of the particular glass used to make the prism. But, in general the index of refraction will change with frequency, so the matching conditions at the boundary change, requiring some angle of deflection. – Jon Custer Aug 9 '17 at 22:09
• You have implicitly assumed that the wave-number remains the same when crossing a boundary; only it doesn't. Understanding the relationship between $n$ and $\omega$ requires you to look at the microscopic physics on the wave interacting with the atoms of the material. A basic treatment can be found in any upper-division E&M text, where it makes up roughly a semester's worth of work. Treatments at the graduate level are both faster and more detailed. – dmckee --- ex-moderator kitten Aug 9 '17 at 22:24
• @JonCuster Okay then, I have edited my question. All I would like to know at this point is why those quotations in my question are correct. Could you please explain? – BLAZE Aug 9 '17 at 22:31
• Possible duplicate of Why do prisms work (why is refraction frequency dependent)? – Floris Aug 10 '17 at 0:44
• @BLAZE - see my updated answer. I think I should have cleared up your confusion. – Floris Aug 10 '17 at 1:24

higher (angular) frequencies have a greater deflection than lower frequencies through a glass prism

is false - materials have a refractive index that trends lower at higher frequencies (for X rays the refractive index is almost 1.0). But it is true for the limited frequencies that correspond to visible light, for a transparent glass prism (no absorption bands, i.e. colorless glass).

Considering a dielectric as made up from different damped harmonic oscillators, as you approach resonance you will initially get an increase in response amplitude and then a decrease, as you pass through resonance. This means that the refractive index as a function of frequency tends to follow a curve like this (after Hecht and Zajak, "Optics" (1973), figure 3.14)

If you have a material that has a positive slope in the range of frequencies of interest, your relationship holds - this is the "normal dispersion" range. But in general it doesn't (because of the "anomalous dispersion" that occurs around regions of resonance).

In general, the refractive index follows an equation of the form

$$\frac{n^2(\omega)-1}{n^2(\omega)+2} = \frac{Nq_e^2}{3\epsilon_0 m_e} \sum_j \frac{F_j}{\omega_{0j}^2 - \omega^2+i\gamma_j \omega}$$

(equation 3.38 from Hecht and Zajac (1973) "Optics"). You can see there are terms related to the individual resonances, $\omega_{0j}$, that correspond to absorption modes of the molecules that make up your medium. As the frequency increases, fewer and fewer of these modes will be excited, which is why the large scale trend is for the refractive index to decrease; but on a smaller scale, between characteristic frequencies, the trend is reversed. And it so happens that for a clear material like glass, the fact it is colorless implies that there can be no absorption in the visible range, and that therefore the entire visible range must have "normal" dispersion.

UPDATE

I was too hasty in my reply; from the comments, I realize there are specific parts of the derivation you are struggling with. Let's take these two in turn.

$v_{phase}$ decreases with $\omega$

This seems to be contradicted by the equation $v_{phase}=\frac{\omega}{k}$ which suggests that $v_{phase}$ should increase with frequency. But that ignores the fact that the wave number $k$ is also changing with frequency. That is why the more sensible expression is $v_{phase}=\frac{c}{n}$ - we know that $c$ is constant so we only have to worry about the effect of frequency on $n$. And we know that, over the range of visible light frequencies, the refractive index increases with frequency (decreases with wavelength, which is how it's shown in the image I reproduced from the answer to the duplicate question I marked).

The second statement,

$n_{glass}$ increases with $\omega$

is again not contradicted by your expression (2), because once again it contains $k(\omega)$ so you can't just state it will be proportional to $\frac{1}{\omega}$.

• @BLAZE I have edited my answer - hope this is clearer now, and more to the point of your question. – Floris Aug 10 '17 at 1:23
• @Floris Hi again, I have fixed errors in the quotations of your post (under the update) so that it matches the quotes in my question. Could you please revise your answer to see if your argument is still valid? Many thanks. – BLAZE Aug 10 '17 at 12:59
• @BLAZE sorry yes I don't know how I ended up writing $n$ instead of $\omega$ there... the argument I made only makes sense with $\omega$... – Floris Aug 10 '17 at 13:35

The phase velocity equation (1) is not relevant here. Yes, it relates all these properties at fixed frequency, but it ignores the fact that n is a function of frequency: $n(\omega)$, and hence, leads to confusion.

The index of refraction of any material is of course a material property--and in the prism, it increases with frequency. Generally, this is a result of the Kramer-Kronig relation, which relates the real part of the index of refraction (propagation) with the complex part (absorption). As you approach the absorption frequency from below, the index of refraction increases, and voila: dispersion.

Of course, now you can ask why there is absorption in the ultraviolet.

In the case of linear dispersive materials, two processes can lead to deviation of a composite light from its actual path- refraction, which leads to a change in phase velocity, and dispersion, which leads to frequency-dependent refractive index. The change in the magnitude of wavevector for a light in such a case is given by

$$k(\omega)=\frac{n(\omega)\omega}{c}=\frac{\omega}{v_p(\omega)}$$

Going with the proof given in the book, as frequency $\omega$ increases, the refractive index, $n(\omega)$, increases (Why?). This will naturally lead to a decrease in phase velocity, since

$$v_p(\omega)=\frac{c}{n(\omega)}$$

Now, for a given frequency $\omega$, the wavenumber decreases from the free-space value, when it undergoes refraction ($k(\omega)<k_0(\omega))$:

$$k(\omega)=\frac{2\pi}{\lambda(\omega)}=\frac{2\pi n(\omega)}{\lambda_0(\omega)}$$

Increasing the frequency will hence lead to an increase in refractive index as well as a decrease in wavelength. This will lead to an overall increase in $k$ with increase in $\omega$. This is what you are missing. You treated $k$ as a constant in saying $n\propto\omega^{-1}$, which is not true: $k(\omega)\propto\omega^2\implies n(\omega)\propto\omega$.