• Beer-monster
However, there are some materials where absorption does not cause an increase in refractive index and in fact can lead to a decrease. This is due to the fact that the absorption causes the light to be scattered in many directions which diminishes the intensity of the light that reaches our eyes.

#### Beer-monster

Hi

I'm trying to get a better understanding of dispersion in light. Specifically, the relationship between refractive index and wavelength, and why the refractive index increases for decreasing wavelengths. However, I've been confused by all the various information I've come across.

Firstly, in class we were told that though the wavelength and speed change when light enters a different medium, the frequency remains constant. However, some other sources on the internet state that frequency changes due to the change in speed. The expression for the refractive index I was able to derive by treating the (motion of molecules in the medium as simple harmonic oscillators driven by the electric field of the incident wave) also depends on frequency.

Secondly, I get the impression that the increase of refractive index with decreased wavelength is specific to visible light. Is this correct?

In the end I would just like to know why refractive index increased with shorter wavelengths, and simply saying that its due to the light traveling slower in a different medium seems slightly circular reasoning.

Hope that's clear. Thanks.

The frequency stays constant. If the speed of light changes then the wavelength must change accordingly. The refractive index is not guaranteed to increase with wavelength (or vice-versa). Any such claims should be thought of as being specific to a certain material and within a certain bandwidth.

The actual relationship depends upon the physical properties of the medium. How the medium is structured (the lattice shape, spacings and types of bonds) and the type of atoms all affect the overall bulk behavior of the material. In the end, what we can do is to find the quantum mechanics that describe the medium when a photon passes through and use that to derive the dispersion relations (this is similar to your treatment the response of the atoms as oscillators). This is a very complex process so there are a myriad of ways to do more approximate techniques. You may want to take a look into the Debye relaxation. We can use the Debye relaxation to estimate the behavior of the material and use the mathematical relationships to find the behavior over a larger bandwidth. Associated with this method are such things like the Cole-Cole and Cole-Davidson models.

Basically the Debye relaxation is more accurate than just allowing the atoms to be oscillators because this would allow the atoms to perfectly correlate their movements with the applied fields. A more accurate picture is to realize that the inertia of the atoms and the bond forces causes a lag in their movements. The Debye relaxation considers such a lag in a heuristic model. There are certainly more complex models but I would say that Debye relaxation is a good start.

When our teacher brought this up, he showed some examples of refractive index in several glasses and the Sommelier relationship. He asked us to think about why the refractive index increases for shorter wavelengths, and dropped the hint that the relationship only applies for light in the visible region.

I've been giving this some thought and I can only guess that the difference has to do with the different atomic processes that occur when energy is absorbed i.e. vibrations for infrared and electronic transitions for the visible and ultraviolet.

I can imagine this making sense for the infrared, as (based on my picture of oscillating ions) this radiation would be closer to the resonance frequency of the oscillator which leads to anamolous dispersion. However, how would this apply to shorter visible and ultraviolet radiation as this would be far beyond the resonance frequency? Is refractive index related to absorption in general, so any strong absorption band would cause an increase in refractive index?

Beer-monster said:

When our teacher brought this up, he showed some examples of refractive index in several glasses and the Sommelier relationship. He asked us to think about why the refractive index increases for shorter wavelengths, and dropped the hint that the relationship only applies for light in the visible region.

I've been giving this some thought and I can only guess that the difference has to do with the different atomic processes that occur when energy is absorbed i.e. vibrations for infrared and electronic transitions for the visible and ultraviolet.

I can imagine this making sense for the infrared, as (based on my picture of oscillating ions) this radiation would be closer to the resonance frequency of the oscillator which leads to anamolous dispersion. However, how would this apply to shorter visible and ultraviolet radiation as this would be far beyond the resonance frequency? Is refractive index related to absorption in general, so any strong absorption band would cause an increase in refractive index?

Yes, refractive index (real part that is) and absorption of definitely interrelated. That is, the refractive index is defined by the permittivity and permeability of the material. Generally we talk about the real parts of the permittivity and permeability as defining the phase velocity and wavelength. The imaginary parts are responsible for the loss (that is the absorption) of the medium. Although, technically both the real and imaginary parts are used in the index of refraction though we typically only talk about the index of refraction for lossless (purely real) materials. The real and imaginary parts of the permittivity and permeability are related to each other by the Hilbert Transform (this is also known as the Kramers-Kronig Relation). Thus, if you know the real part (all of it, from 0 to infinity over the frequency) then you can find the imaginary part and vice-versa.

This relationship also means that we can predict some localized behavior of the two. For example, at the resonant frequency we can expect that the real part changes sign and the imaginary part has a peak. This means that near the resonant point the real part, the wavelength, will vary quickly and the medium will be very lossy. The fact that the real and imaginary parts are related by a Hilbert Transform is used in such formulations like the Cole-Cole equations that I mentioned previously.

But a material will have multiple resonances. The Debye relaxation technique is meant to be generally applicable to a simple inhomogeneous medium (say a light mixture of one substance in another) and over a small bandwidth. It only catches a single resonance but a material will have a very complex behavior over a large bandwidth (take a look at the absorption spectrum of water for example). There are also other reasons for resonances and absorption other than just simple phonon modes.

For example, we can have plasma modes. A metallic material allows its electrons to move fairly freely about the bulk. So when an electromagnetic wave strikes the material, the surface can behave like a plasma (where the ions are fixed but the electrons are free to roam). These surface plasma (or surface plasmons) can have their own resonance modes. In fact, there is a large area of reseach on designing nanoparticles that have surface plasmons that will allow us to achieve negative permeabilities and/or negative permittivities. This is done by impregnating a bulk with the nanoparticles so that they can create a resonance at a desired frequency (and like I mentioned above, a resonance can mean that the real part will change sign so there is a small bandwidth where it can be negative).

So I would advise caution in trying to extrapolate behavior beyond a bandwidth that you have knowledge about. You cannot take information over a finite bandwidth and use it to extrapolate what will happen outside of it. The physics can change (for example, with a plasma mode the waves cannot propagate below a certain resonant frequency but can above it) depending on your frequency.

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I just wanted to reply to one point of the original question, namely that the refractive index typically increases with frequency not only in the visible region. Only in small regions near absorption bands the refractive index decreases. One speaks of anomalous dispersion, then.

Thanks again.

I went through the maths again, working through a derivation of the complex refractive index and complex polarisation etc, and seeing how the imaginary parts describe loss. However, trying to thinks this out conceptually is still hurting my brain.

If I'm understanding the maths correctly, we can treat the system as a resonating oscillator. If we start at a frequency below the resonant frequency (approximating to one resonant mode for the moment) and increase towards resonance, the amplitude of the oscillating electrons will increase. This means that the energy of the oscillation should increase (and is dissipated by damping) but the EM wave should lose energy as it passes through the medium.

Now, I can't seem to reconcile this loss of energy with the refractive index. The wavelength of light shortens as it passes through a medium, which (unless I'm really in trouble) mean it has more energy. This also matches what I can work out using Planck's equation and the relationship between refractive index and the ratio of the wavelengths (in vacuum and in the medium).

Am I just getting confused between the energy of the oscillation and the EM wave? Does the wave not lose energy by making the electrons oscillate?

Beer-monster said:
Thanks again.

I went through the maths again, working through a derivation of the complex refractive index and complex polarisation etc, and seeing how the imaginary parts describe loss. However, trying to thinks this out conceptually is still hurting my brain.

If I'm understanding the maths correctly, we can treat the system as a resonating oscillator. If we start at a frequency below the resonant frequency (approximating to one resonant mode for the moment) and increase towards resonance, the amplitude of the oscillating electrons will increase. This means that the energy of the oscillation should increase (and is dissipated by damping) but the EM wave should lose energy as it passes through the medium.

Now, I can't seem to reconcile this loss of energy with the refractive index. The wavelength of light shortens as it passes through a medium, which (unless I'm really in trouble) mean it has more energy. This also matches what I can work out using Planck's equation and the relationship between refractive index and the ratio of the wavelengths (in vacuum and in the medium).

Am I just getting confused between the energy of the oscillation and the EM wave? Does the wave not lose energy by making the electrons oscillate?

Classically, the energy of the wave is dependent only upon the material parameters and the amplitude of the waves. Quantum mechanically, the energy of the photons that make up the electromagnetic wave is dependent only on the frequency. Wavelength does not affect the energy of the wave.

I think I get it now, after staring at my page of maths for ages.

Refractive index depends on permittivity which depends on polarisation. The polarisation of the electrons depend on their displacement i.e. amplitude. The amplitude is at a maximum at resonance. So refractive index increases as the frequency approaches the resonant frequency. However, in the region of resonance the imaginary part of the complex refractive index comes into play and leads to anomalous dispersion.

That sound right?

There is one point missing: You also have to consider the phase of the polarization relative to the exciting field. For frequencies well below the resonance, the two are in phase, while wide above the resonance, they are out of phase by 180 degrees. So the absolute value of polarization increases when you approach the resonance but its sign changes so that P always increases with the exception of a small region near the resonance.

Thanks. I think I have a better understanding now. Thanks to both of you.

## What is dispersion in optics?

Dispersion in optics refers to the phenomenon where different wavelengths of light have different speeds when passing through a medium. This results in the separation of light into its component colors, creating a rainbow-like effect.

## How does dispersion affect the appearance of objects?

Dispersion can cause objects to appear blurry or distorted, especially at the edges, due to the separation of colors. This is known as chromatic aberration and can be seen in images taken with a camera or telescope.

## What causes dispersion in optics?

Dispersion is caused by the variation in the refractive index of a medium with respect to the wavelength of light. The refractive index is a measure of how much a medium can bend light, and it is different for different colors of light.

## How is dispersion measured?

Dispersion is often measured by the dispersion coefficient, which is the rate of change of refractive index with respect to the wavelength of light. It can also be measured by the dispersion angle, which is the angle at which light of different colors is separated after passing through a prism.

## What are some practical applications of dispersion in optics?

Dispersion is essential in the design of optical devices such as lenses, prisms, and spectrometers. It is also used in telecommunications to transmit different wavelengths of light in optical fibers. Additionally, dispersion is utilized in spectroscopy to analyze the composition of materials based on the colors they absorb or emit.