Viewing glass from an oblique angle

Question

When I view most glass from the side it's green which I've found out is due to impurities in the glass specifically from iron oxide. Why is it when I view the larger face from an oblique angle, it isn't nearly as green?

I cannot personally notice any different on the piece I have next to me even when I hold it at an angle that would be almost looking at the edge of the glass. It is pretty small (about 2.5" x 5" x .0625" or about 61mm x 127mm x 2mm, l x w x h) but I feel like it's big enough that I'd be looking through enough glass to get the green.

Answer 1

This might be surprising to some viewers. Feel free to down vote me, but please let me know if anything is wrong except the fact that I'm not relating the color of glass to impurities.

A "pure" piece of glass (without the iron impurity) can also appear green. This is not because of the color of metal ions. This can be explained by the classical drude model.

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Here is a rather concise derivation. I use the steps from the material cited with the link. The Drude model assumes that the electrons in a material is locally bounded by a "spring" force. The response (spatial displacement $\vec{r}$) of individual electrons to an external electric field $\vec{E}$ can be described by the force equation $$ m \frac{\partial ^2 \vec{r}}{\partial t^2} + m \Gamma \frac{\partial \vec{r}}{\partial t} + m \omega_0^2 \vec{r} = -q \vec{E} $$

Take the Fourier transform and get $$ \left ( -m \omega^2 - j\omega m \Gamma + m \omega_0^2 \right )\vec{r}\left ( \omega \right )= -q \vec{E}\left ( \omega \right ) $$

The dipole moment under field $\vec{E}$ is $$ \vec{\mu}\left ( \omega \right ) = -q\vec{r} = \left[ \frac{q^2}{m} \frac{1}{\omega_0^2 - \omega^2 - j\omega\Gamma} \right] \vec{E}\left ( \omega \right ) $$

The polarization of the material (containing $N$ dipoles per volume) is the sum of all dipoles $$ \vec{P}\left ( \omega \right ) = N \left< \vec{\mu}\left ( \omega \right ) \right> $$

We also define the susceptibility $\chi_e$ of the material as $$ \vec{P} \left ( \omega \right ) = \varepsilon_0 \chi_e\left ( \omega \right ) \vec{E}\left ( \omega \right ) $$

Therefore $$ \chi_e\left ( \omega \right ) = \frac{Nq^2}{\varepsilon_0 m}\frac{1}{\omega_0^2 - \omega^2 - j \omega \Gamma} $$

The complex dielectric constant (or truly, function vs frequency) is $$ \tilde{\varepsilon_r} \left ( \omega \right ) = 1+\chi_e\left ( \omega \right ) = {\varepsilon_r}'\left ( \omega \right ) + j{\varepsilon_r}''\left ( \omega \right ) $$

Now introduce the complex refractive index $$ \tilde{n}\left ( \omega \right ) = n\left ( \omega \right ) + jk\left ( \omega \right ) = \sqrt{\tilde{\varepsilon_r}} $$

What does this mean? For a planar wave incident on the glass, we have transmission, absorption and reflection. Particularly, the absorption is characterized by the imaginary part of the refractive index, $k$.

The transmission, absorption and reflection are dependent on the wave (light) frequency. Without showing the full expression of $k$ here, one can imagine there are absorption peaks at certain frequencies when the electrons resonant with the incident wave, so the wave attenuates when it travels through the glass.

So far I haven't stated whether or not the glass contains impurity iron oxide. The frequency response of the material is a first-principle model derived from classical physics. It predicts that, no matter if a piece of glass has impurity or not, it will have a frequency-dependent absorption!

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How does it compare to reality? Here is some data for soda-lime glass.

1. A rather unreliable source.
2. A reported research work.

Source #2 contains full-range data in terms of wavelength. It has both "clear" glass (with iron impurity) and "low-iron" glass (which I assume does not contain as much iron as the other one). I plotted a portion of their data. I mapped the wavelength to RGB color using this formula. You may also check this table for wavelength-color relationship as well.

Apparently, the "clear" glass with iron impurity has an absorption peak for red light. Other than that, both the "clear" glass and "low-iron" glass have similar trends in absorption vs wavelength. The longer the wavelength, the more the absorption. Although the deviation is tiny, it can still make a difference.

Since the glass absorbs a slightly larger portion of the red light, the "color" or what's left in the transmitted light seems to have a hint of green.

Also I suspect that the reason it looks "green" rather than "cyan" (missing red) is because human eyes are more sensitive to green lights. But I don't have reference here, so don't quote me on this.

The Drude model can also explain that glass is about 4% reflective. Think of the rear-view mirror in a car.

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To answer OP's question, when you look at a piece of glass from an oblique angle, the length of the path a beam of light has to travel through the glass before reaching your eyes becomes longer. According to general electromagnetic wave theory, in a lossy dielectric material, the longer the distance, the higher the exponential decay.

Edit: A more thorough but lengthy explanation of the electromagnetic wave decay in glass is here.

This also answers the question "The same happens for a thick glass block too. Why?"

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All references, including those used in the text and additional ones:

• Drude model:

  Lorentz and Drude models (Lecture notes, Instructor: R. Rumpf)

  Optical Properties of Materials (Lecture notes, Instructor: Y. Zhao)

  Section 7.5 "Frequency Dispersion Characteristics of Dielectrics, Conductors, and Plasmas", Classical Electrodynamics Second Edition, John David Jackson, Wiley, 1975

• Glass data:

  Soda–lime glass (Wikipedia)

  Optical properties of soda lime silica glasses (M.Rubin, Solar Energy Materials, 1985)

• Mapping Wavelength to Color:

  Spectral colors (Wikipedia)

  Answer to question "RGB values of visible spectrum"

• Backgrounds in Electromagnetic waves:

  Section 8.3: "Plane Waves in Lossy Media", Field and Wave Electromagnetics Second Edition, David Keun Cheng, Pearson, 1989 (I don't have access to it at the moment; chapter info may be inaccurate)

$$ k = \frac{1}{\sqrt{2}} \sqrt{ -{\varepsilon_r}' + \sqrt{{\varepsilon_r}'^2 + {\varepsilon_r}''^2} } \\ {\varepsilon_r}' = 1 + \frac{\omega_p^2 \left ( \omega_0^2 - \omega^2 \right )}{\left ( \omega_0^2 - \omega^2 \right )^2 + \left ( \omega \Gamma \right )^2} \\ {\varepsilon_r}'' = \frac{\omega_p^2 \omega \Gamma }{\left ( \omega_0^2 - \omega^2 \right )^2 + \left ( \omega \Gamma \right )^2} \\ \omega_p^2 = \frac{Nq^2}{\varepsilon_0 m} $$

Answer 2

As you stated, the degree of green is directly dependent on the thickness of glass you stare at (Beer-Lambert law). It actually comes from the absorption of the other wavelengths by the glass.

Due to refraction, even when you look at the glass from a grazing angle in the air, the light rays bend to a higher angle in the glass which makes the light path through the glass shorter (figure 2).

On the contrary, when you stare at the glass from the edge, total internal reflection makes the light rays travel through the whole length of the glass to your eye (figure 3).

Answer 3

Even when held at a glancing angle, the light within the glass follows a relatively short path. As light crosses the boundary between the air an the glass its angle changes (refraction). Light that leaves the glass at a glancing angle and enters your eye was traveling much closer to perpendicular to the surface when it was within the glass. It hasn't passed through a thick layer, hasn't encountered many impurities, and hence hasn't had its color changed much.