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It's quite a common fact that different types of glass have different refractive indices. Most sites I've found attribute these differences to variations in the 'density' of the glass, which is not very satisfying of an answer.

My questions:

  1. What is the underlying physical mechanism that determines the refractive index and its variations in glass?

  2. For other materials, such a Silicon\Gold\Silver\etc., does the refractive index have a fixed and known value (by model or in principle) under the assumption that the material in question is 'pure' (i.e. no foreign materials)?

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5 Answers 5

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Heavy glasses are mainly heavy because the use elements such as lead with many nuclei. As a result, these materials also have more electrons per atom, especially in the higher (more loosely bound) orbitals.

The looser bound the electrons, the more they will react to the electric field wave. How exactly this increases the refraction index I don't know, but I will provide some handwaving argument that is hopefully useful to you. The electrons getting dragged around (i.e. a plasmon) have an effective mass and thus the plasmon has kinetic energy, which takes away the energy of the actual photon and reduces propagation speed.

I hope this is not too inaccurate, but hopefully someone else can expand

Tables on refractive index and loss for metals. Metals mainly have a large loss term because of unbound electrons which couple to phonons and dissipate energy. When you dope or mix metals, surely their refractive index will change. Metals also have a real index of refraction, which is strongly frequency dependent as can be seen in the link. Unfortunately, the simple picture I draw here doesn't provide any clue as to why this is so.

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  • $\begingroup$ Question: How do you determine the index of refraction of a meta? Note: Refracted photons do not change in frequency or energy. My guess: Incoming photons are absorbed by the atoms of a transparent material and then stimulated to re-emission by following photons. Averaged over billions of photons, this slows the progress of a beam through the material. $\endgroup$
    – R.W. Bird
    Dec 27, 2021 at 14:58
  • $\begingroup$ @R.W.Bird I don't know if/how it is measured but I recall some tables listing real parts of the refractive index much higher than 1 (link added). About your note: Yeah, I also don't like that my provided mental picture "predicts" a drop in frequency (which doesn't occur) instead of wavelength. I also like your interpretation, but I can't say if it is more or less accurate than my own, so I will leave it for now. $\endgroup$
    – tobalt
    Dec 27, 2021 at 15:08
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Density of a substance doesn't determine it's optical density or refractive index. For example, kerosene oil has lower density that water (it floats on water) but has higher refractive index and is optically denser than water.

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The refractive index of a material is closely related to its dielectric susceptibility tensor. Both are a function of the wavelength (or frequency or angular frequency, depending on how you prefer to sort your calculations), and can be generally represented as:

$n^\dagger(\vec r, t)^2 = \epsilon_r(\vec r, t)\\ n^\dagger(\vec r, t)^2 = \epsilon_r(\vec r, t)\\ n^\dagger(\vec r, t)^2 = \epsilon_r(\vec r, t) $

Additionally, the refractive index can be a complex number generally represented as: $n^\dagger = n + i\kappa$.

In many books you will find that the dielectric function is referred to as "dielectric constant", but this is mostly because the dielectric function for low frequencies (typically for electronics or high frequency electronics, which have a lower frequency than optical frequencies, e.g. 5GHz typical high frequency signal e.g. WLAN vs 5.64 × $10^2$ THz for a 532 nm laser, for example).

The refractive index can be typically measured by measuring the reflection, transmission and absorption of a material using a photospectrometer that varies a light source in wavelength to give $R(\lambda), T(\lambda), A(\lambda)$. By using these measurements (and others such as ellipsometry which exploits a known polarization state of the incoming light to determine the polarization changes from an analyzed material) it is possible to obtain the refractive index of a material (without further proof of statement, but it is possible to find how these work e.g. 1963 paper explaining the principle.

Going further, why does the polarization matter to understand the refractive index?

The dielectric function can be understood macroscopically through the polarization induced in a material by an external electric field (2 pg. 234, 6. Optical Properties I, 6.2. Macroscopic electrodynamics) in the form:

$P_i(\vec r', t') = \int \chi_{ij} (\vec r, \vec r', t, t')E_j (\vec r, t) d\vec r dt$

The tensor $\chi_{ij}$ is refered to as the electric susceptibility. The notation $\vec r', \vec r$ indicates that the polarization can apply to inhomogeneous, anisotropic, nonlinear materials (I have already posted a brief explanation of the difference between these properties, but said shortly, some materials can locally change the polarization state of an incoming light excitation. From here, it is possible to know the relationship between the electric susceptibility tensor and the dielectric function through the constitutive relation of the electric displacement vector, electric field and the electric susceptibility tensor (as well as its Fourier transform):

$ P_i(\vec q, \omega) = \chi_{ij} E_j(\vec q, \omega)\\ \epsilon_{ij}(\vec q, \omega) = 1 + 4\pi\chi_{ij}(\vec q, \omega) $

Here the polarization on the first equation is the fourier transform of the equation above it, and the equation next to it is from the constitutive relationship $\vec D = \vec E +4\pi \vec P$. The $4\pi$ term comes from gaussian units.

The dielectric function can be understood microscopically through semiconductor solid state physics (I recommend the Yu/Cardona book if you want to delve in deeper into the subject.

To your specific questions and with the framework established:

What is the underlying physical mechanism that determines the refractive index and its variations in glass?

As stated above, the refractive index and its variations on glass can be understood macroscopically through the polarization of light in the material due to an external electric field. This is also why it is possible to measure the refractive index macroscopically through optical properties such as reflection, transmission, absorption and ellipsometry.

For other materials, such a Silicon\Gold\Silver\etc., does the refractive index have a fixed and known value (by model or in principle) under the assumption that the material in question is 'pure' (i.e. no foreign materials)?

As I mentioned before, there is a much larger treatment of the refractive index for conductors and semiconductors that are their own field of study. The refractive index can be traced back from the crystal structure of the material and how the incoming light interacts with the material, with broadly 3 situations possible:

  • For transparent dielectrics, the light is not absorbed and passes through the material, meaning the light is not absorbed by it. Some portion of the light is reflected. This relation gives the refractive index of the material. The dielectric function is typically only real as there is (little to) no absorption of the material, except for specific wavelengths (Fused silica, for example, is a glass with a strong absorption band in the infrared (IR absorption of FS is caused by OH-chemical bond content, most relevant at wavelengths near the water absorption band at 2.7 µm but also critical at 940 nm, Source from Laseroptik).
  • For semiconductors there are specific wavelengths where there can be a direct absorption or emission of photons from the conduction and valence band, all which is its own field of study, but generally the same principle stands regarding the refractive index, besides at strong absorption wavelengths.
  • For metals the dielectric function is very high in visible and short wavelengths, meaning that light is strongly absorbed by the material (except, for example, at X-ray wavelengths which have a very short wavelength).

The refractive index can be both measured and predicted by ellipsometry, for example, or by analyzing the crystal structure of the material and the disposition of its conduction and valence bands. This way it is possible to predict the absorption ranges of the materials, and therefore, predict its refractive index.

I hope this was not waffling too much but the underlying concepts are indeed traceable, but it requires quite some context which is difficult to summarize in a single response.

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TL;DR

  1. Refractive index is determined by the way electrons interact with incoming electromagnetic waves. Impurity atoms in glass change the way they interact.

  2. Pure materials have principly known, frequency-dependent refractive indexes, but whether the material is a conductor affects the details of dependence.


There is a simple model that answers your question, i.e. why materials have the index of refraction they do.

Suppose each electron in the solid is bounded in place by hypothetical "springs", which mimic the forces exerted on the electron by the nuclei and other electrons. At the same time, the electron loses energy due to damping effects, which can be attributed to the electron being accelerated by electromagnetic waves.

Assume the electron is constrained to the xy plane, and that a x-polarised electromagnetic wave propagates in the z direction. Now the motion of the electron can be modeled as a driven oscillator$^\dagger$: $$ m\ddot{\mathbf{x}}+m\gamma\dot{\mathbf{x}}+k\mathbf{x} = q\mathbf{E}_0\cos(\omega t). $$

This can be solved easily if we move to the complex plane. The solution is $$ \tilde{\mathbf{x}} = \frac{q}{m}\frac{E_0e^{-i\omega t}}{\omega_0^2-\omega^2-i\gamma\omega}\mathbf{\hat{x}}=\frac{q/m}{\omega_0^2-\omega^2-i\gamma\omega}\tilde{\mathbf{E}}_0, $$ such that $$ \mathbf{x}=\mathcal{R}[\tilde{\mathbf{x}}]. $$

We shall stay in the complex plane, and use the relation $$ \tilde{\mathbf{p}} = \tilde{\alpha}\tilde{\mathbf{E}} $$ to find $\tilde{\alpha}$, the complex polarisability. Here $\mathbf{p}$ is the complex dipole moment, which is simply $\tilde{\mathbf{p}}=q\tilde{\mathbf{x}}$, so $$ \tilde{\alpha} = \frac{q^2/m}{\omega_0^2-\omega^2-i\gamma\omega}. $$

This tells us the polarisability is a function of EM wave frequency. But there are different crystal sites in a material, where the $j$-th site is characterised by resonant frequency $\omega_j \equiv \sqrt{k_j/m}$ and damping $\gamma_j$. Different modes do not contribute equally, so they are weighted by a strength factor $h_j$. Therefore $\alpha$ in a dense material is $$ \tilde{\alpha} = \frac{q^2}{m}\sum_j\frac{h_j}{\omega_j^2-\omega^2-i\gamma_j\omega}. $$

By the Clausius-Mossotti relation, $\tilde{\alpha}$ in bulk materials is related to the complex electric susceptibility $\tilde{\chi}_\text{e}$ by $$ \frac{\tilde{\chi}_\text{e}}{3+\tilde{\chi}_\text{e}} = \frac{N\tilde{\alpha}}{3\epsilon_0}, $$

while the (complex) refractive index $\tilde{n}$ is$^\ddagger$ $$ \tilde{n}^2 = \frac{\tilde{\mu}\tilde{\epsilon}}{\mu_0\epsilon_0} \approx \frac{\tilde{\epsilon}}{\epsilon_0} = 1+\tilde{\chi}_\text{e}. $$

Therefore$^*$ $$ \frac{\tilde{n}^2-1}{\tilde{n}^2+2} = \frac{N\tilde{\alpha}}{3\epsilon_0}, $$ where $N$ is number density. The real part tells us the refractive index, while the imaginary part gives information about absorption. So far, the theory tells us that refractive index is frequency-dependent; impurities such as lead and boron can be accounted by considering the contribution of different components to total polarisation:

$$ \frac{\tilde{n}^2-1}{\tilde{n}^2+2} = \frac{1}{3\epsilon_0}\sum_i N_i\tilde{\alpha}_i, \quad \tilde{\alpha}_i = \frac{q^2}{m}\sum_j\frac{h_{ij}}{\omega_{ij}^2-\omega^2-i\gamma_{ij}\omega}. $$

Changing the proportion of different components of glass changes $N_i$ and $h_{ij}$, latter of which we assume to be frequency-independent, and so changes the refractive index. The same goes for other pure materials.

For pure metals such as gold and silver, however, a slight modification shall be made: optically, the effect of free electrons should overwhelm that of bound electrons, so metal can be approximated as a free electron gas. Then $\omega_j=0$ and $N|\tilde{\alpha}|/\epsilon_0 \ll 1$, so

$$ \tilde{n}^2 \approx 1 + \frac{q^2N}{\epsilon_0m}\frac{1}{-\omega^2-i\gamma\omega}. $$

From this we can see that the refractive index for pure metals is principly determined, and exhibits great variation with respect to frequency. For low frequency, $$ \tilde{n} = \sqrt{\frac{q^2N}{2\epsilon_0m\omega}}\left(1+i\right), $$ the imaginary part makes EM waves decay quickly and makes the metals practically opaque. For high frequency, $$ \tilde{n} \approx \sqrt{1-\frac{q^2N}{\epsilon_0m\omega^2}} \in \mathbb{R}, $$ so the metal becomes transparent, and the refractive index increases monotonically.


$^\dagger$ The Lorentz force due to the magnetic component of the EM wave always points in the z direction, therefore has no effect on the motion since we assumed the motion occurs in the xy plane.

$^\ddagger$ In the context of most materials the permeability $\mu$ is a constant to a good approximation, i.e. $\mu(\omega) \approx \mu_0(\omega)$.

$^*$ This is known as the Lorentz–Lorenz equation.

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Refractive index defines the change in speed for propagating the light through the material relative to propagating the light in vacuum.

$$ n \equiv \frac{c}{v}$$

Higher values indicate slower velocities $v$ relative to the speed of light $c$. Although the speed of light decreases, the frequency $\nu$ does not. This maintains the light at the same energy $E = h\nu$, where $h$ is Planck's constant. Instead, as $n$ index of refraction increases above $1$ for vacuum, the speed of light decreases with constant frequency, and the wavelength also decreases.

$$\lambda = \frac{v}{\nu} = \frac{c}{n\ \nu}$$

The oscillating electric field of light interacts with solid materials in four ways.

  • It oscillates the electrons in atomic or molecular orbitals relative to the (statically placed) nucleus. This is the electronic mode.
  • It oscillates dipolar bonds through vibration or bending. This is the molecular mode.
  • It oscillates cation+anion ionic bonds through vibrations. This is the ionic mode.
  • It oscillates (sloshes around) free electrons relative to the (static) atom background. This is the plasmon mode.

These are not absorption modes, these are modes that slow down the speed of light as it travels through the material. A simple but useful analogy is to imagine running over two different surfaces, a flat, hard tarmac versus a flat but sandy beach. When you exert the same energy to run in both cases, you will travel at a slower speed across the sand than across the tarmac. Light exerts the same energy $h\nu$. With higher $n$, it travels slower. The analogy becomes even more interesting when you have a set of rows of a marching band travel across the two surfaces. You can envision that the spacing between the rows of the marching band decreases as the band transitions from tarmac to sand, the rows travel at a slower speed, yet you see the same frequency of rows passing you at a fixed point either on the tarmac or the sand.

Ceramics such as glass have only electronic and ionic modes -- they have no dipolar bonds or free electrons. Interactions with the electronic oscillation mode increase as atomic or molecular size increases. We observe this by tracking the polarizability of atoms or molecules as a function of their size (diameter). Ionic polarizability increases as the ionic bond force constant decreases or alternatively as the charges on the cation or anion increase. An alternative way to consider this is to appreciate that the polarization $P$ (C/m$^2$) of any dipole system (molecular or ionic in this case) is also expressed as a number density of dipoles $\rho_{N,D}$ (number / m$^3$) times the dipole moment of the dipoles $\mu_D$ (C m).

$$P = \rho_{N,D}\ \mu_D $$

The statement that refractive index increases with the "density" of the glass is a (sloppy) shorthand way of saying that, as you make the glass denser, you put more dipoles in the same volume of the material, making it more susceptible to slowing down light being transported through it.

Finally, across the other materials ... Metals have electronic and plasmon modes. Polymers have electronic and sometimes molecular modes. Semiconductors have electronic and (weaker) plasmon modes. Each mode (electronic ... plasmon) has its own resonance frequency. As long as light is below the resonance, it will sense the mode. Once the light frequency is above the resonance of a given mode, the light no longer senses that mode. The material then becomes equivalent to a vacuum in that it no longer slows down the speed (or changes the wavelength) as the light travels through.

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