Problem: Critical edge
Table of Important Numbers
| b\(_c\) | Relative mass | density (g/cm\(^3\)) | M (\(\mu_B\)) | |
|---|---|---|---|---|
| \(^1\)H | -3.7406 | 1.00794 | ||
| \(^2\)H | 6.671 | 2.0141 | ||
| O | 5.803 | 15.9994 | ||
| \(^1\)H\(_2\)O | -1.6482 | 18.01 | 1.0 | |
| Si | 4.1491 | 28.0855 | 2.33 | |
| SiO\(_2\) | 15.7551 | 60.0843 | 2.2 | |
| Co | 2.49 | 58.933 | 8.9 | 1.6 |
| Ru | 7.02 | 101.07 | 12.37 |
Calculate \(q_c\) for the following interfaces (left to right moves from the first medium to the second):
Question 1
air / silicon
The refractive index is given by this equation, \(q_c=\sqrt{16\pi\left(\rho_2\overline{b}_2-\rho_1\overline{b}_1\right)}\).
The exercise is therefore about calculating \(\rho\overline{b}_c\) for the different media. It is convenient to express \(q_c\) in Å\(^{-1}\). The numbers in the table above can be used to calculate \(\rho\overline{b}_c\) so long as they are expressed in Å.
Note that cm\(^{–3}\) = 10\(^{–24}\) Å\(^{–3}\)and \(\rho\) is a formula unit number density.
The conversion between mass density and number density is performed by:
\begin{align} \text{number density} = \frac{\text{mass density} \cdot \text{Avogadro's number}}{\text{atomic mass for the formula unit}} \end{align} For air: \(\rho\overline{b}_c\) = 0
For silicon: \begin{align} \rho\overline{b}_c &= \frac{2.33 \cdot 10^{-24}}{28.0855} \cdot 6.023 \cdot 10^{23} \cdot 4.1491 \cdot 10^{-5}\\ &= 2.0732 \cdot 10^{-6} \text{Å}^{-2} \end{align}
Thereby \(q_c\) for the air/silicon interface is:
\begin{align} q_c = \sqrt{16\pi \cdot 2.0732 \cdot 10^{-6} \text{Å}^{-2}} = 0.001 \text{Å}^{-1} \end{align}
Question 2
silicon / heavy water (\(^2\)H\(_2\)O)
For deuterated water (note that the number density for light and heavy water will be the same) \(\rho\overline{b}_c\) is:
\begin{align} \rho\overline{b}_c &= \frac{1 \cdot 10^{-24}}{18.01} \cdot 6.023 \cdot 10^{23} \cdot (5.803+2\cdot6.667) \cdot 10^{-5}\\ &= 3.576 \cdot 10^{-6} \text{Å}^{-2} \end{align}
Thereby \(q_c\) for the silicon/heavy water (\(^2\)H\(_2\)O) interface is:
\begin{align} q_c = \sqrt{16\pi \cdot (3.576-2.0732) \cdot 10^{-6} \text{Å}^{-2}} = 0.0087 \text{Å}^{-1} \end{align}
Question 3
silicon / light water (\(^1\)H\(_2\)O)
For protonated water (note that this is negative) \(\rho\overline{b}_c\) is:
\begin{align} \rho\overline{b}_c &= \frac{1 \cdot 10^{-24}}{18.01} \cdot 6.023 \cdot 10^{23} \cdot (5.803-2\cdot3.7406) \cdot 10^{-5}\\ &= -2.9719 \cdot 10^{-7} \text{Å}^{-2} \end{align}
Thereby \(q_c\) for the silicon/heavy water (\(^2\)H\(_2\)O) interface is:
\begin{align} q_c = \sqrt{16\pi \cdot (-2.029719-2.0732) \cdot 10^{-6} \text{Å}^{-2}} \end{align}
This is imaginary, i.e. there is no critical edge for a silicon / light water interface. This also comes from Snell's Law - the radiation is moving from a low refractive index to a high refractive index, so at the interface the wave is bent towards the surface normal.
Question 4
(38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O) / silicon
For (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O) \(\rho\overline{b}_c\) is: \begin{align} \rho\overline{b}_c &= (0.612 \cdot 3.576 - 0.388 \cdot 0.029719)\cdot 10^{-6}\\ &= -2.0734 \cdot 10^{-6} \text{Å}^{-2}\\ &\approx \text{the value for Si} \end{align}
Thereby \(q_c\) for the (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O)/siliconinterface is:
\begin{align} q_c = \sqrt{16\pi \cdot (2.0734-2.0732) \cdot 10^{-6}} \approx 0 \end{align}
Question 5
Will there be reflectivity in the case of question 3 or 4?
There is reflectivity in case silicon/light water (\(^1\)H\(_2\)O), as the two media have different refractive indices. The radiation sees a contrast between them. There is no critical edge, but there is still some reflection.
There is no reflectivity in case (38.8% \(^1\)H\(_2\)O + 61.2% \(^2\)H\(_2\)O)/silicon. The refractive indices for the two media are the same. As far as the radiation is concerned, it is still travelling in a homogeneous medium and it will not experience an interface.