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Fix some small issues

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Eamonn Travers 2024-11-17 14:46:07 +01:00
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2 changed files with 15 additions and 15 deletions
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6
sections/Auswertung.tex
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@ -9,10 +9,10 @@
First the resolution limit has to be measured: First the resolution limit has to be measured:
\begin{align*} \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_{2} & = 0,6 \cdot 10^-3 m \\
B &= (0,45 \pm 0,15) \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: Using the eqation $\ref{7}$ $\epsilon$ is calculated:
@ -27,7 +27,7 @@ Using the eqation $\ref{7}$ $\epsilon$ is calculated:
&= 0,31 \circ \\ &= 0,31 \circ \\
\Delta \epsilon &= 0,05 \circ \\ \Delta \epsilon &= 0,05 \circ \\
\epsilon &= (0,26 \pm 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 $\ref{8.1}$): Since the diffraction index n for air is equal to one, the numercal aperture is equal to sin($\epsilon$) (see $\ref{8.1}$):

20
sections/Physical-Principals.tex
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@ -22,7 +22,7 @@ The relationship between B and G or b and g is described in $\ref{1}$
\label{1} \label{1}
\begin{align*} \begin{align*}
\frac{B}{G} & = \frac{b}{g} \tag{1} \frac{B}{G} & = \frac{b}{g} \tag{1}
\end{align*} \end{align*}
\subsection{Magnification} \subsection{Magnification}
@ -38,7 +38,7 @@ of an optical intrument. The proportion of B and G can be calcuated thus:
\label{2} \label{2}
\begin{align*} \begin{align*}
\gamma & = \frac{tan(\alpha_{2})}{tan(\alpha_{1})} \tag{1} \gamma & = \frac{tan(\alpha_{2})}{tan(\alpha_{1})} \tag{1}
\end{align*} \end{align*}
\label{B} \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})} \\ \gamma_{magnifying} & = \frac{tan(\alpha-{2})}{tan(\alpha_{1})} \\
& = \frac{G/f}{G/a_{0}} \\ & = \frac{G/f}{G/a_{0}} \\
& = \frac{a_{0}}{f} \tag{3} & = \frac{a_{0}}{f} \tag{3}
\end{align*} \end{align*}
\subsection{Microscope} \subsection{Microscope}
@ -92,17 +92,17 @@ Based on the equation $\ref{3}$ the calculation of the magnification of the ocul
\label{3.1} \label{3.1}
\begin{align*} \begin{align*}
\gamma_{ok} &= frac{a_{0}}{g} \\ \gamma_{ok} &= \frac{a_{0}}{g} \\
&= \frac{a_0}{f_{ok}} \tag{3.1} &= \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: 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} \label{6}
\begin{align*} \begin{align*}
frac{1}{f} &= frac{1}{g} + frac{1}{b} \tag{6} \\ \frac{1}{f} &= \frac{1}{g} + \frac{1}{b} \tag{6} \\
frac{1}{g} &= frac{1}{f} + frac{1}{a_{0}} \frac{1}{g} &= \frac{1}{f} + \frac{1}{a_{0}}
\end{align*} \end{align*}
The final formula for $\gamma_{ok}$ is then as follows: 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} \label{8}
\begin{align*} \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} 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. n is the refraction index and $\lambda$ refers to the wavelength of the light being used.