Fix some small issues

sections/Auswertung.tex

@@ -9,10 +9,10 @@
 First the resolution limit has to be measured:
 
 \begin{align*}
-    B_{1} & = 0,3 \cdot 10⁻³ m \\
+    B_{1} & = 0,3 \cdot 10^-3 m \\
     B_{2} & = 0,6 \cdot 10^-3 m \\
     B &= (0,45 \pm 0,15) \cdot 10^-3 m
-  \end{align*}
+\end{align*}
 
 Using the eqation $\ref{7}$ $\epsilon$ is calculated:
 
@@ -27,7 +27,7 @@ Using the eqation $\ref{7}$ $\epsilon$ is calculated:
     &= 0,31 \circ \\
     \Delta \epsilon &= 0,05 \circ \\
     \epsilon &= (0,26 \pm 0,05) \circ
-  \end{align*}
+\end{align*}
 
 Since the diffraction index n for air is equal to one, the numercal aperture is equal to sin($\epsilon$) (see $\ref{8.1}$):
 

sections/Physical-Principals.tex

@@ -22,7 +22,7 @@ The relationship between B and G or b and g is described in $\ref{1}$
 \label{1}
 \begin{align*}
     \frac{B}{G} & = \frac{b}{g} \tag{1}
-  \end{align*}
+\end{align*}
 
 \subsection{Magnification}
 
@@ -38,7 +38,7 @@ of an optical intrument. The proportion of B and G can be calcuated thus:
 \label{2}
 \begin{align*}
     \gamma & = \frac{tan(\alpha_{2})}{tan(\alpha_{1})} \tag{1}
-  \end{align*}
+\end{align*}
 
 
 \label{B}
@@ -58,7 +58,7 @@ The magnification, based on the equation $\ref{2}$, can be calculated thus:
     \gamma_{magnifying} & = \frac{tan(\alpha-{2})}{tan(\alpha_{1})} \\
     & = \frac{G/f}{G/a_{0}} \\
     & = \frac{a_{0}}{f}      \tag{3}
-  \end{align*}
+\end{align*}
 
 \subsection{Microscope}
 
@@ -92,17 +92,17 @@ Based on the equation $\ref{3}$ the calculation of the magnification of the ocul
 
 \label{3.1}
 \begin{align*}
-  \gamma_{ok} &= frac{a_{0}}{g} \\
+  \gamma_{ok} &= \frac{a_{0}}{g} \\
   &= \frac{a_0}{f_{ok}} \tag{3.1}
-  \end{align*}
+\end{align*}
 
 For $\gamma_{ok}$ at the distance of $a_{0}$ ($\gamma_{ok}(a_{0})$) the equation $\ref{6}$ has to be taken into account:
 
 \label{6}
 \begin{align*}
-  frac{1}{f} &= frac{1}{g} + frac{1}{b} \tag{6} \\
-  frac{1}{g} &= frac{1}{f} + frac{1}{a_{0}} 
-  \end{align*}
+  \frac{1}{f} &= \frac{1}{g} + \frac{1}{b} \tag{6} \\
+  \frac{1}{g} &= \frac{1}{f} + \frac{1}{a_{0}} 
+\end{align*}
 
 The final formula for $\gamma_{ok}$ is then as follows:
 
@@ -135,9 +135,9 @@ The minimum distance $d_{min}$ is then calculated thus, A being the numerical ap
 
 \label{8}
 \begin{align*}
-  d_{min} &= frac{\lambda}{A} + frac{1}{b} \tag{8} \\
+  d_{min} &= \frac{\lambda}{A} + \frac{1}{b} \tag{8} \\
   A &= n \cdot sin(\epsilon) \tag{8.1} 
-  \end{align*}
+\end{align*}
 
 n is the refraction index and $\lambda$ refers to the wavelength of the light being used.