Add ray derivation

docs/src/ray_derivation.md

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+# Analytic Derivation of Vial Refraction
+
+If we remove the index matching bath in TVAM, the glass vial affects the ray propagation
+because of refraction at the glass and vial interface.
+Our ray optical backend [RadonKA.jl](https://github.com/roflmaostc/Radonka.jl) can handle non-parallel
+ray propagation if the intersection of the rays with the outer circle are calculated at entry and exit.
+
+
+In the following we derive analytical expressions for $y_i$ and $y_f$. 
+Given the circular shaped vial, the angle under which a ray hits the vial is $\alpha = \arcsin(y / R_o)$. Consequently, because of refraction we obtain $\beta=\arcsin(\sin(\alpha) / n_\text{vial})$.
+The orange segment $x$ is more inconvenient to derive but the law of cosines of triangles provides us with 
+
+$$R_i^2 = x^2 + R_o^2 - 2 \cdot x \cdot R_o \cos{\beta}.$$
+
+Solving the quadratic equation, the meaningful solution is
+
+$$x = R_o \cdot \cos{\beta} -  \sqrt{R_o^2 \cdot (\cos(\beta)^2 - 1) + R_i^2}.$$
+
+Again, applying law of cosine we can obtain an expression for
+
+$$\varepsilon = \arccos\left(\frac{x^2 + R_i^2 - R_o^2}{2 R_i x}\right) - \frac{\pi}{2}.$$
+
+Further, $\beta'=\mathrm{sign}(y) \cdot (\frac{\pi}{2}- \varepsilon)$ and because of refraction $\gamma=\arcsin(\frac{n_\text{vial} \cdot \sin(\beta')}{n_\text{resin}})$. 
+The total ray deflection angles are given by $\delta_1=\alpha - \beta$ and $\delta_2=\beta'-\gamma$ resulting in $\delta_\text{ges} = \delta_1 + \delta_2$. 
+
+
+```@raw html
+<img src="../assets/drawing_angle.png" alt="drawing of the angular geometry" width="500"/>
+```
+
+To calculate the height $y_i$ which describes the virtual height while entering the outer circle, we first need the distance $y' = R_i \cdot \sin(\gamma)$.
+Using the Pythagorean theorem we can derive
+
+$$p = \sqrt{R_o^2-y'^2}.$$
+
+Then, the angle $\eta$ is given by
+$$\eta =- \left(\arcsin\left(\frac{y'}{R_0}\right) - \text{sign}(y) \cdot \left(\frac{\pi}{2}-\delta_{\text{ges}}\right)\right)$$
+
+
+Then, the height of the ray at exiting the outer circle, is given by 
+$$y_f = R_o \cdot \sin(\eta).$$
+
+Because of the isosceles triangle, the height of the virtual ray entering the outer circle is given by
+
+$$y_i =  2\cdot p \cdot \sin(\delta_\text{ges}) + y_f$$
+
+