Commit d1e917b5 authored by Turnhout, M.C. van's avatar Turnhout, M.C. van
Browse files
parents 648d5dd7 deaf8f61
......@@ -49,7 +49,7 @@ Thumbs.db
# Windows default autosave extension
*.asv
*.bak
# OSX / *nix default autosave extension
*.m~
......
......@@ -43,7 +43,7 @@ element & type & orientation & \gls*{retardance}\\
For the analysis of the optical train, we use Jones calculus \cite[\& chapter \ref{jones}]{Jones1941}. For each optical element we define two matrices. The rotation of the element is described by
\begin{equation}
\gls*{Omega}(\omega) = \begin{bmatrix} \cos\omega & -\sin\omega \\ \sin\omega & \cos\omega\end{bmatrix}
\gls*{Omega}(\omega) = \begin{bmatrix*}[r] \cos\omega & -\sin\omega \\ \sin\omega & \cos\omega\end{bmatrix*}
\end{equation}
and the rotation independent effect by
\begin{equation}
......@@ -71,8 +71,8 @@ First we derive the \gls*{Jones matrix} $M$ for each element. The linear \gls*{p
\end{equation}
The first liquid crystal has its axis at 45\,\degree\ and applies a (variable) \gls*{retardance} $\alpha$. We therefore find\footnote{Recall that $\eu^{\iu\omega} = \cos\omega + \iu\sin\omega$.}:
\begin{equation}
\gls*{LA}(\alpha, 45\,\degree ) = \gls*{Omega}(45\unit{\degree})\begin{bmatrix} \eu^{-\iu\frac{\alpha}{2}} & 0 \\0 & \eu^{\iu\frac{\alpha}{2}} \end{bmatrix}\gls*{Omega}(-45\unit{\degree}) = \begin{bmatrix} \cos\frac{\alpha}{2} & -\iu\sin\frac{\alpha}{2} \\
-\iu\sin\frac{\alpha}{2} & \cos\frac{\alpha}{2} \end{bmatrix}
\gls*{LA}(\alpha, 45\,\degree ) = \gls*{Omega}(45\unit{\degree})\begin{bmatrix} \eu^{-\iu\frac{\alpha}{2}} & 0 \\0 & \eu^{\iu\frac{\alpha}{2}} \end{bmatrix}\gls*{Omega}(-45\unit{\degree}) = \begin{bmatrix*}[r] \cos\frac{\alpha}{2} & -\iu\sin\frac{\alpha}{2} \\
-\iu\sin\frac{\alpha}{2} & \cos\frac{\alpha}{2} \end{bmatrix*}
\end{equation}
The second crystal has its axis at 0\,\degree\ and applies a (variable) \gls*{retardance} $\beta$, so:
\begin{equation}
......@@ -81,7 +81,7 @@ The second crystal has its axis at 0\,\degree\ and applies a (variable) \gls*{re
We are now capable of calculating the \gls*{Jones vector} that reaches the sample (see section \ref{shribakexam}). With $\gls*{J0} = [1\ 1]^T$ the \gls*{Jones vector} of the incident light, we can write \cite[eq. 2]{Shribak2003}:
\begin{equation}
\gls*{J}_s = \gls*{LB}(\beta, 0\,\degree )\cdot \gls*{LA}(\alpha, 45\,\degree ) \cdot \gls*{P}(0\,\degree ) \cdot \gls*{J0} = \begin{pmatrix} \cos\left(\frac{\alpha}{2}\right)\eu^{-\iu\frac{\beta}{2}} \\ -\iu\sin\left(\frac{\alpha}{2}\right)\eu^{\iu\frac{\beta}{2}} \end{pmatrix} \label{PF}
\gls*{J}_s = \gls*{LB}(\beta, 0\,\degree )\cdot \gls*{LA}(\alpha, 45\,\degree ) \cdot \gls*{P}(0\,\degree ) \cdot \gls*{J0} = \begin{pmatrix*}[r] \cos\left(\frac{\alpha}{2}\right)\eu^{-\iu\frac{\beta}{2}} \\ -\iu\sin\left(\frac{\alpha}{2}\right)\eu^{\iu\frac{\beta}{2}} \end{pmatrix*} \label{PF}
\end{equation}
Note that the complex amplitudes and phase of the \gls*{Jones vector} now are
\begin{align}
......@@ -100,13 +100,13 @@ The intensity of the beam is now half of that of the incident beam, i.e.
For the sample, we define an azimuth $\gls*{phi}$ and \gls*{retardance} $\gls*{Delta}$ and write:
\begin{align}
\gls*{Ssample} = & \gls*{Omega}(\gls*{phi})\begin{bmatrix} \eu^{-\iu\frac{\gls*{Delta}}{2}} & 0 \\0 & \eu^{\iu\frac{\gls*{Delta}}{2}} \end{bmatrix}\gls*{Omega}(-\gls*{phi})\\
= & \begin{bmatrix} \cos^2\left(\gls*{phi}\right)\eu^{-\iu\frac{\gls*{Delta}}{2}} + \sin^2\left(\gls*{phi}\right)\eu^{\iu\frac{\gls*{Delta}}{2}} & 2\iu\cos\left(\gls*{phi}\right)\sin\left(\gls*{phi}\right)\sin\left(\frac{\gls*{Delta}}{2}\right) \\
2\iu\cos\left(\gls*{phi}\right)\sin\left(\gls*{phi}\right)\sin\left(\frac{\gls*{Delta}}{2}\right) & \cos^2\left(\gls*{phi}\right)\eu^{\iu\frac{\gls*{Delta}}{2}} + \sin^2\left(\gls*{phi}\right)\eu^{-\iu\frac{\gls*{Delta}}{2}}\end{bmatrix}
= & \begin{bmatrix*}[r] \cos^2\left(\gls*{phi}\right)\eu^{-\iu\frac{\gls*{Delta}}{2}} + \sin^2\left(\gls*{phi}\right)\eu^{\iu\frac{\gls*{Delta}}{2}} & 2\iu\cos\left(\gls*{phi}\right)\sin\left(\gls*{phi}\right)\sin\left(\frac{\gls*{Delta}}{2}\right) \\
2\iu\cos\left(\gls*{phi}\right)\sin\left(\gls*{phi}\right)\sin\left(\frac{\gls*{Delta}}{2}\right) & \cos^2\left(\gls*{phi}\right)\eu^{\iu\frac{\gls*{Delta}}{2}} + \sin^2\left(\gls*{phi}\right)\eu^{-\iu\frac{\gls*{Delta}}{2}}\end{bmatrix*}
\end{align}
The \gls*{quarter wave plate} applies a $\frac{\gls*{lambda}}{4}$ phase shift and has its axis at -45\,\degree, thus\footnote{Note that the same plate at +45\unit{\degree} yields the complex conjugate of this matrix: $\gls*{Q}\left(\frac{\piup}{2},-45\,\degree \right) = Q^c\left(\frac{\piup}{2},45\,\degree \right)$}:
\begin{equation}
\gls*{Q}\left(\frac{\piup}{2}, -45\,\degree \right) = \gls*{Omega}(-45\unit{\degree})\begin{bmatrix} \frac{1}{\sqrt{2}}-\frac{\iu}{\sqrt{2}} & 0 \\0 & \frac{1}{\sqrt{2}}+\frac{\iu}{\sqrt{2}} \end{bmatrix}\gls*{Omega}(45\unit{\degree})
= \frac{1}{\sqrt{2}}\begin{bmatrix} 1& -\iu\\ -\iu &1\end{bmatrix}\label{qw}
= \frac{1}{\sqrt{2}}\begin{bmatrix*}[r] 1& -\iu\\ -\iu &1\end{bmatrix*}\label{qw}
\end{equation}
To find the resulting \gls*{Jones vector} \gls*{J}, all that remains is to multiply the train with the initial light vector $\gls*{J0}$
......
......@@ -8,7 +8,7 @@ Diese Ergebnisse sollen nunmehr einer Pr{\"u}fung durch das Polarisationmikrosko
\section{Optical train}
Polarised light microscopy is used to study \gls*{birefringent} samples. Birefringent samples apply a different phase shift to two orthogonal components of (polarised) light, i.e.\ a \gls*{birefringent} sample acts as a \gls*{retarder} with a certain \gls*{retardance} $\delta$ and orientation $\alpha$ (sections \ref{plmbirefringent} and \ref{retarder}). Conventional \gls*{PL} microscopy uses 'crossed \gls*{polariser}s' to study \gls*{birefringent} specimens, e.g.\ \cite{Benninghoff1925}.
Polarised light microscopy is used to study \gls*{birefringent} samples. Birefringent samples apply a different phase shift to two orthogonal components of (polarised) light, i.e.\ a \gls*{birefringent} sample acts as a \gls*{retarder} with a certain \gls*{retardance} $\delta$ and orientation $\alpha$ (sections \ref{plmbirefringent} and \ref{retarder}). Conventional \gls*{PL} microscopy uses `crossed \gls*{polariser}s' to study \gls*{birefringent} specimens, e.g.\ \cite{Benninghoff1925}.
In the crossed \gls*{polariser}s set-up, light passed first a \gls*{polariser} with orientation $\alpha$, next the sample, and finally a \gls*{polariser} (analyser) with orientation $\alpha + 90\,\degree$ (figure \ref{crossedtrain}).
......@@ -23,7 +23,7 @@ In the crossed \gls*{polariser}s set-up, light passed first a \gls*{polariser} w
Without a sample, or with a sample that is not \gls*{birefringent}, no light will be transmitted beyond the \gls*{analyser} because the incident light for the \gls*{analyser} has no component in the $\alpha+90\,\degree$ direction, but only in the (orthogonal) $\alpha$ direction, i.e.\
\begin{align}
\gls*{J} & = \gls*{A}(\alpha + 90\,\degree) \cdot \gls*{P}(\alpha) \cdot \gls*{J0} \notag \\
& = \begin{bmatrix} \cos^2\left(\alpha+\frac{\piup}{2}\right) & \sin\left(\alpha+\frac{\piup}{2}\right)\cos\left(\alpha+\frac{\piup}{2}\right) \smallskip \\ \sin\left(\alpha+\frac{\piup}{2}\right)\cos\left(\alpha+\frac{\piup}{2}\right) & \sin^2\left(\alpha+\frac{\piup}{2}\right) \end{bmatrix} \cdot \begin{bmatrix} \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha \end{bmatrix} \cdot \gls*{J0}\notag \\
& = \begin{bmatrix*}[r] \cos^2\left(\alpha+\frac{\piup}{2}\right) & \sin\left(\alpha+\frac{\piup}{2}\right)\cos\left(\alpha+\frac{\piup}{2}\right) \smallskip \\ \sin\left(\alpha+\frac{\piup}{2}\right)\cos\left(\alpha+\frac{\piup}{2}\right) & \sin^2\left(\alpha+\frac{\piup}{2}\right) \end{bmatrix*} \cdot \begin{bmatrix*}[r] \cos^2\alpha & \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha & \sin^2\alpha \end{bmatrix*} \cdot \gls*{J0}\notag \\
& =\begin{bmatrix} 0 \\ 0\end{bmatrix}
\end{align}
for arbitrary $\gls*{J0}$ and $\alpha$.
......@@ -31,8 +31,8 @@ for arbitrary $\gls*{J0}$ and $\alpha$.
With a \gls*{birefringent} sample in the light path however, the \gls*{linearly polarised} light that leaves the \gls*{polariser} gains a component perpendicular to the orientation of the \gls*{polariser} when it passes the sample (equation \ref{R0J0}). Our image will thus not be dark as above, but will show an intensity that depends on the \gls*{azimuth} of the sample \gls*{retardance} $\gls*{phi}$ (relative to the \gls*{polariser}-\gls*{analyser} axis) and the applied \gls*{retardance} of the sample $\gls*{Delta}$. The \gls*{Jones vector} of the light after the \gls*{analyser} is:
\begin{align}
\gls*{J} & = \gls*{A}(90\,\degree) \cdot \gls*{Ssample} \cdot \gls*{P}(0\,\degree) \cdot \gls*{J0} \notag \\
& = \begin{bmatrix} 0&0\\0&1\end{bmatrix} \cdot \begin{bmatrix} \cos\frac{\gls*{Delta}}{2}+\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi} & \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} \\ \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} & \cos\frac{\gls*{Delta}}{2}-\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi}
\end{bmatrix} \cdot \begin{bmatrix} 1&0\\0&0\end{bmatrix} \cdot \begin{bmatrix} 1\\1\end{bmatrix} \notag \\
& = \begin{bmatrix} 0&0\\0&1\end{bmatrix} \cdot \begin{bmatrix*}[r] \cos\frac{\gls*{Delta}}{2}+\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi} & \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} \\ \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} & \cos\frac{\gls*{Delta}}{2}-\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi}
\end{bmatrix*} \cdot \begin{bmatrix} 1&0\\0&0\end{bmatrix} \cdot \begin{bmatrix} 1\\1\end{bmatrix} \notag \\
& = \begin{bmatrix} 0 \\
\iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi}\end{bmatrix}
\end{align}
......@@ -82,17 +82,17 @@ We can use equation \eqref{IASP} and figure \eqref{fig:IASP} to our advantage \c
A `\gls*{quarter wave plate}' is a \gls*{retarder} that applies a relative phase shift (\gls*{retardance}) of $90\,\degree$ between its two orthogonal axis, i.e.\ a quarter wavelength: $\delta = 90\,\degree = \frac{\gls*{lambda}}{4}\,\text{nm} = \frac{\piup}{2}$\,rad with $\gls*{lambda}$ the wavelength of the light. The \gls*{Jones matrix} \gls*{Q} of a \gls*{quarter wave plate} with orientation $\alpha$ is (equation \ref{retgeneral}):
\begin{align}
\gls*{Q}(\alpha) = & \gls*{Omega}(\alpha)\cdot\begin{bmatrix} \eu^{\iu\frac{\piup}{4}} & 0 \\0 & \eu^{-\iu\frac{\piup}{4}} \end{bmatrix}\cdot \gls*{Omega}(-\alpha) = \gls*{Omega}(\alpha)\cdot\frac{1}{\sqrt{2}}\begin{bmatrix} 1-\iu& 0 \\0 & 1+\iu \end{bmatrix}\cdot \gls*{Omega}(-\alpha) \\
= & \frac{1}{2}\sqrt{2}\begin{bmatrix} 1-\iu\cos 2\alpha & -\iu\sin 2\alpha\\ -\iu\sin 2\alpha & 1+\iu\cos 2\alpha \end{bmatrix}
= & \frac{1}{2}\sqrt{2}\begin{bmatrix*}[r] 1-\iu\cos 2\alpha & -\iu\sin 2\alpha\\ -\iu\sin 2\alpha & 1+\iu\cos 2\alpha \end{bmatrix*}
\end{align}
\subsection{TLDR: rather extensive treatment of the quarter wave plate}
The effect of a \gls*{quarter wave plate} with orientation $\alpha$ on \gls*{linearly polarised} light with orientation 45\,\degree\ and intensity $\gls*{I0} = 2$ is:
\begin{align}
\gls*{J}(\alpha) & = \gls*{Q}(\alpha)\cdot \gls*{J0} = \frac{1}{2}\sqrt{2}\begin{bmatrix} 1-\iu\cos 2\alpha & -\iu\sin 2\alpha\\ -\iu\sin 2\alpha & 1+\iu\cos 2\alpha \end{bmatrix} \cdot \begin{bmatrix}1 \\ 1 \end{bmatrix} \notag \\
\gls*{J}(\alpha) & = \gls*{Q}(\alpha)\cdot \gls*{J0} = \frac{1}{2}\sqrt{2}\begin{bmatrix*}[r] 1-\iu\cos 2\alpha & -\iu\sin 2\alpha\\ -\iu\sin 2\alpha & 1+\iu\cos 2\alpha \end{bmatrix*} \cdot \begin{bmatrix}1 \\ 1 \end{bmatrix} \notag \\
& = \frac{1}{2}\sqrt{2} \begin{bmatrix} 1 - \iu\left(\sin 2\alpha + \cos 2\alpha\right) \\ 1 - \iu\left(\sin 2\alpha - \cos 2\alpha\right) \end{bmatrix} \\
A(\alpha) & = \begin{bmatrix} \sqrt{1+\frac{1}{2}\sin 4\alpha}\smallskip \\ \sqrt{1-\frac{1}{2}\sin 4\alpha}\end{bmatrix} \\
\gls*{phi}(\alpha)& = \begin{bmatrix} -\atan\left(\cos 2\alpha + \sin 2\alpha\right) \\ \atan\left(\cos 2\alpha - \sin 2\alpha\right)\end{bmatrix}\\
\gls*{phi}(\alpha)& = \begin{bmatrix*}[r] -\atan\left(\cos 2\alpha + \sin 2\alpha\right) \\ \atan\left(\cos 2\alpha - \sin 2\alpha\right)\end{bmatrix*}\\
\gls*{I}(\alpha) & = A_x^2+A_y^2 = 2
\end{align}
......@@ -101,12 +101,12 @@ As with a \gls*{birefringent} sample, a \gls*{quarter wave plate} has little eff
\gls*{J0} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} & &A_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} &&
\gls*{phi}_0 = \begin{bmatrix} 0\\ 0 \end{bmatrix} && \gls*{I0} = 2 \\
\gls*{J}(45\,\degree ) = \frac{1}{2}\sqrt{2} \begin{bmatrix} 1 - \iu \\ 1 - \iu \end{bmatrix} && A(45\,\degree ) = \begin{bmatrix} 1\\1\end{bmatrix} && \gls*{phi}(45\,\degree ) = \begin{bmatrix} -\frac{1}{4}\piup \smallskip \\ -\frac{1}{4}\piup \end{bmatrix} && \quad \gls*{I}(45\,\degree ) = 2\\
\gls*{J}(0\,\degree ) = \frac{1}{2}\sqrt{2} \begin{bmatrix} 1 - \iu \\ 1 + \iu \end{bmatrix} && A(0\,\degree ) = \begin{bmatrix} 1\\1\end{bmatrix} && \gls*{phi}(0\,\degree ) = \begin{bmatrix} -\frac{1}{4}\piup \smallskip \\ \frac{1}{4}\piup \end{bmatrix} && \quad \gls*{I}(0\,\degree ) = 2
\gls*{J}(0\,\degree ) = \frac{1}{2}\sqrt{2} \begin{bmatrix} 1 - \iu \\ 1 + \iu \end{bmatrix} && A(0\,\degree ) = \begin{bmatrix} 1\\1\end{bmatrix} && \gls*{phi}(0\,\degree ) = \begin{bmatrix*}[r] -\frac{1}{4}\piup \smallskip \\ \frac{1}{4}\piup \end{bmatrix*} && \quad \gls*{I}(0\,\degree ) = 2
\end{align}
For both cases we find
\begin{align}
\tan\gamma & = \frac{A_y}{A_x} = 1\quad \rightarrow \quad \gamma = \frac{1}{4}\piup \\
\tan 2\gls*{phi} & = \tan 2\gamma \cos \gls*{delta} = \tan\frac{\piup}{2} \cos \left(\gls*{phi}_y - \gls*{phi}_x \right) = \infty \quad \rightarrow \gls*{phi} = \frac{1}{4}\piup = 45\,\degree
\tan\gamma & = \frac{A_y}{A_x} = 1\quad \rightarrow \quad \gamma = \frac{\piup}{4} \\
\tan 2\gls*{phi} & = \tan 2\gamma \cos \gls*{delta} = \tan\frac{\piup}{2} \cos \left(\gls*{phi}_y - \gls*{phi}_x \right) = \infty \quad \rightarrow \gls*{phi} = \frac{\piup}{4} = 45\,\degree
\end{align}
Thus, all vectors $\gls*{J0}$, $\gls*{J}(45\,\degree)$ and $\gls*{J}(0\,\degree)$ are polarised with orientation $\gls*{phi} = 45\,\degree$. For the ellipticity of the polarisation vectors we obtain
\begin{align}
......@@ -117,7 +117,7 @@ And these give us the ratios of the ellipses' short axis $b$ with the ellipses'
\begin{align}
\gls*{J0}: & \tan \vartheta = 0 \quad\rightarrow \frac{b}{a} = 0 \quad \rightarrow b = 0\\
\gls*{J}(45\,\degree): & \tan \vartheta = 0 \quad\rightarrow \frac{b}{a} = 0 \quad \rightarrow b = 0\\
\gls*{J}(0\,\degree): & \tan \vartheta =\tan\frac{1}{4}\piup \quad\rightarrow \frac{b}{a} = 1 \quad \rightarrow a = b\label{J0}
\gls*{J}(0\,\degree): & \tan \vartheta =\tan\frac{\piup}{4} \quad\rightarrow \frac{b}{a} = 1 \quad \rightarrow a = b\label{J0}
\end{align}
Thus $\gls*{J0}$ is \gls*{linearly polarised} (no short axis: $b=0$), as is $\gls*{J}(45\,\degree)$. But with a 45\,\degree\ angle between the orientations of \gls*{Q} and $\gls*{J0}$ we find that the light becomes \gls*{circularly polarised} (equation \ref{J0}): $a =b$. To obtain \gls*{circularly polarised} light is the objective of placing a \gls*{quarter wave plate} in the light path.
......@@ -132,8 +132,8 @@ Thus $\gls*{J0}$ is \gls*{linearly polarised} (no short axis: $b=0$), as is $\gl
When we insert a \gls*{quarter wave plate} at 45\,\degree\ between the \gls*{polariser} and the sample, the sample is irradiated with \gls*{circularly polarised} light instead of \gls*{linearly polarised} light and we obtain:
\begin{align}
\gls*{J} & = \gls*{A}(90\,\degree) \cdot \gls*{Ssample} \cdot \gls*{Q}(45\,\degree) \cdot \gls*{P}(0\,\degree) \cdot \gls*{J0} \notag \\
& = \begin{bmatrix} 0&0\\0&1\end{bmatrix} \cdot \begin{bmatrix} \cos\frac{\gls*{Delta}}{2}+\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi} & \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} \\ \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} & \cos\frac{\gls*{Delta}}{2}-\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi}
\end{bmatrix} \cdot \begin{bmatrix} \frac{1}{\sqrt{2}}&0\\0&\frac{-\iu}{\sqrt{2}}\end{bmatrix} \cdot \begin{bmatrix} 1&0\\0&0\end{bmatrix} \cdot \begin{bmatrix} 1\\1\end{bmatrix} \notag \\
& = \begin{bmatrix} 0&0\\0&1\end{bmatrix} \cdot \begin{bmatrix*}[r] \cos\frac{\gls*{Delta}}{2}+\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi} & \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} \\ \iu\sin\frac{\gls*{Delta}}{2}\sin 2\gls*{phi} & \cos\frac{\gls*{Delta}}{2}-\iu\sin\frac{\gls*{Delta}}{2}\cos 2\gls*{phi}
\end{bmatrix*} \cdot \begin{bmatrix} \frac{1}{\sqrt{2}}&0\\0&\frac{-\iu}{\sqrt{2}}\end{bmatrix} \cdot \begin{bmatrix} 1&0\\0&0\end{bmatrix} \cdot \begin{bmatrix} 1\\1\end{bmatrix} \notag \\
& = \frac{1}{2}\sqrt{2}\begin{bmatrix} 0 \\
-\cos 2\gls*{phi}\sin\frac{\gls*{Delta}}{2} + \iu\left( \sin 2\gls*{phi} \sin\frac{\gls*{Delta}}{2} - \cos\frac{\gls*{Delta}}{2}\right)\end{bmatrix}
\end{align}
......@@ -191,7 +191,7 @@ Long slender fibres are \gls*{birefringent} because the transverse dimension of
It has long been known that thinner (collagen) fibres have a smaller \gls*{retardance} than thicker fibres, and that in white light \gls*{PLM} the difference in \gls*{retardance} emerges as different colours for the fibres \cite[\& figure \ref{fibrecolours}]{Wolman1965}. To understand why thinner fibres have a smaller \gls*{retardance} than thicker fibres, we assume that the refractive indices are equal between all fibres (they are material parameters) and that the only difference between fibres is \gls*{d} in equation \eqref{eq:retardance}, the distance that the light traveled through the fibre. For thinner fibres, \gls*{d} will be smaller, and therefore \gls*{delta} will be smaller\footnote{This is probably not the complete story. Since \gls*{birefringence} of thin slender fibres comes from the relative transverse fibre dimension compared to the wavelength of the light \cite{Pierard1989}, also fibre \textsl{\gls*{birefringence}} itself may be a function of fibre diameter. However, for our current discussion on how to interpret colours in white light \gls*{PLM}, we do not need to incorporate this in our calculations.}.
This suggests that individual fibres with different thicknesses will have different \gls*{retardance}s, and will therefore show up with different colours when they are imaged with white light \gls*{PLM} (figures \ref{white2RGB} \& \ref{fibrecolours}). However, when the (white) light passes multiple fibres, as it will in general do in our samples, we have to look at the total average \gls*{retardance} and \gls*{azimuth} of the sample, i.e.\ at \gls*{Ssample}, to see what colours we may expect. Two fibres with equal \gls*{retardance} will have the same sample \gls*{Jones matrix} \gls*{Ssample} as a single fibre with double the \gls*{retardance} of the two individual fibres, as shown for monochromatic light in section \ref{examSsample}. Since the sample \gls*{Jones matrix} \gls*{Ssample} is equal between these two fibre configurations, also the colours that we see in white light \gls*{PLM} will be equal between these two fibre configurations. If the two individual fibres would show up 'green' in white light \gls*{PLM}, the combination of these two fibres may yield, e.g., the 'red' colour that we would also observe for a single thicker fibre.
This suggests that individual fibres with different thicknesses will have different \gls*{retardance}s, and will therefore show up with different colours when they are imaged with white light \gls*{PLM} (figures \ref{white2RGB} \& \ref{fibrecolours}). However, when the (white) light passes multiple fibres, as it will in general do in our samples, we have to look at the total average \gls*{retardance} and \gls*{azimuth} of the sample, i.e.\ at \gls*{Ssample}, to see what colours we may expect. Two fibres with equal \gls*{retardance} will have the same sample \gls*{Jones matrix} \gls*{Ssample} as a single fibre with double the \gls*{retardance} of the two individual fibres, as shown for monochromatic light in section \ref{examSsample}. Since the sample \gls*{Jones matrix} \gls*{Ssample} is equal between these two fibre configurations, also the colours that we see in white light \gls*{PLM} will be equal between these two fibre configurations. If the two individual fibres would show up `green' in white light \gls*{PLM}, the combination of these two fibres may yield, e.g., the `red' colour that we would also observe for a single thicker fibre.
The colours that we see in white light \gls*{PLM} are therefore as much a function of fibre orientation, fibre amounts, and fibre anisotropy as the intensities that we see with monochromatic \gls*{PLM}. This is also apparent from literature where researches e.g.\ note that the colours in white light \gls*{PLM} are for instance a function of sample thickness (i.e.\ fibre amounts), a function of sample \gls*{azimuth}, and a function of sample anisotropy \cite{Howie2008,Vidal1982,Wolman1965,Pierard1989}.
......@@ -205,4 +205,6 @@ Some final notes on white light \gls*{PLM}:
\item In white light \gls*{PLM}, also dichroism (i.e.\ selective absoption of different wavelengths of \gls*{PL}) may play a role \cite{Howie2008,Benditt1970,Wolman1965}, although for collagen stained with Congo Red it has been reported that `No obvious dichroism could be detected' \cite{Wolman1965}.
\end{enumerate}
These four effects also affect the observed colours in white light \gls*{PLM}, which makes the interpretation of these colours all the more ambiguous.
These four effects also affect the observed colours in white light \gls*{PLM}, which makes the interpretation of these colours all the more ambiguous.\\
\noindent Take into account the phenomenon of dispersion, different wavelengths of light experience different refractive indices to begin with, and you may start to appreciate the comment by Pi\'erard \textsl{et al.} that interpretation of colour in PLM is `hazardous and often inaccurate' \cite{Pierard1989}.
\ No newline at end of file
......@@ -9,7 +9,7 @@ This document contains a rather extensive treatment of the content of presentati
You may not want to read the entire document though. You may find it rather boring at some points, and infested with matrix calculus at other, or the same, points. In retrospect, I may have opted for completeness rather than readability. The bulk of this document, in particular the mathematics, was developed during my PhD at the Experimental Zoology group of Wageningen University \cite{Turnhout2009,Turnhout2010}. I had to perform some new calculations for the crossed polarisers set-up in our lab (which was rather straightforward), and do some literature research for the mechanisms in white light \gls*{PLM} and Picrosirius Red staining of collagens.
If you do decide, wisely, not to read the entire document, I urge you to read at least chapter \ref{chap:practical} which does not contain too much mathematics and which deals with the practical limitations of \gls*{PLM}, and with setting up our Observer microscope for \gls*{PLM}. In particular, section \ref{PLMnot} is a 'must read' if you wish to assess information on (collagen) fibre structures with \gls*{PLM} (and it doesn't contain any math at all!).
If you do decide, wisely, not to read the entire document, I urge you to read at least chapter \ref{chap:practical} which does not contain too much mathematics and which deals with the practical limitations of \gls*{PLM}, and with setting up our Observer microscope for \gls*{PLM}. In particular, section \ref{PLMnot} is a `must read' if you wish to assess information on (collagen) fibre structures with \gls*{PLM} (and it doesn't contain any math at all!).
Let it be known that your feedback, on typing and spelling errors, on consistency, on readability, and the likes, will be highly appreciated\footnote{I may even return you one of those beers that you are about to offer me.}. Whether due to your feedback, or due to new insights, this document may be subject to revisions.
......
......@@ -63,7 +63,7 @@ Because the orientations of the two \gls*{polariser}s are mutually perpendicular
\caption{Birefringence. \textbf{(a)} A collagen fibre (grey) has two refractive indices that are aligned with the collagen fibre ({\color{blue}$\vec{n}_\text{o}$, blue}, `fast axis') and perpendicular to the collagen fibre ({\color{red}$\vec{n}_\text{e}$, red}, `slow axis'); and \textbf{(b)} two rays with parallel wave propagation directions but perpendicular polarisation passing through a \gls*{birefringent} material, c.s.\ a \gls*{birefringent} crystal, illustrating the different refractions of these two rays (by Mikael H{\"a}ggstr{\"o}m, \href{https://en.wikipedia.org/wiki/File:Rays_passing_through_birefringent_material.svg}{Wikimedia Commons}).}\label{birefringent}
\end{figure}
With \gls*{PLM} we look at \gls*{birefringent} samples. A \gls*{birefringent} material has (two) different refractive indices for waves with mutually perpendicular amplitude orientations in the $\vec{x}\vec{y}$-plane (figure \ref{birefringent}). The two refractive indices are said to be aligned with a `fast axis' and a 'slow axis' and are called the ordinary index of refraction $\gls*{n_o}$ and the extraordinary index of refraction $\gls*{n_e}$. The difference of the two refractive indices \gls*{Deltaupn} is called the \gls*{birefringence} of the material:
With \gls*{PLM} we look at \gls*{birefringent} samples. A \gls*{birefringent} material has (two) different refractive indices for waves with mutually perpendicular amplitude orientations in the $\vec{x}\vec{y}$-plane (figure \ref{birefringent}). The two refractive indices are said to be aligned with a `fast axis' and a `slow axis' and are called the ordinary index of refraction $\gls*{n_o}$ and the extraordinary index of refraction $\gls*{n_e}$. The difference of the two refractive indices \gls*{Deltaupn} is called the \gls*{birefringence} of the material:
\begin{equation}
\gls*{Deltaupn} = \gls*{n_e} - \gls*{n_o} \label{eq:birefringence}
\end{equation}
......@@ -73,7 +73,7 @@ With \gls*{PLM} we look at \gls*{birefringent} samples. A \gls*{birefringent} ma
\caption{A sample $S$ with \gls*{retardance} $\gls*{Delta}$ and \gls*{azimuth} $\gls*{phi}$ between the \gls*{polariser} and \gls*{analyser} changes the polarisation state of the incident light from \gls*{linearly polarised} to \gls*{elliptically polarised}. The part of the \gls*{elliptically polarised} light that has the same orientation as the \gls*{analyser}, is transmitted to the detector.}\label{crossed_train_polstates}
\end{figure}
In a \gls*{PL} microscope, the components of the incident light on the sample that are aligned with the two refractive indices $\vec{n}_\text{e}$ and $\vec{n}_\text{o}$ each acquire a different `speed'\footnote{The speed of light in a medium $v_\text{medium}$ is given by $v_\text{medium} = v_\text{vacuum} \frac{n_\text{vacuum}}{n_\text{medium}} = \text{c} \frac{n_\text{vacuum}}{n_\text{medium}} = \frac{\text{c}}{n_\text{medium}}$, with $\text{c}=3\cdot 10^8$\,[m/s].}\ap{,}\footnote{Whether `\textsl{the} speed of light' is actually $\text{c}$, and whether light actually `slows down' in a medium, is a matter of interpretation \cite{Feynman2006}. For our purposes, the traditional 'slowing down' interpretation will suffice.} in the material and as a result, these two waves will have obtained an additional phase difference \gls*{delta} between each other after passing the material. This phase difference \gls*{delta} in radians is called `\gls*{retardance}' and related to the two refractive indices by
In a \gls*{PL} microscope, the components of the incident light on the sample that are aligned with the two refractive indices $\vec{n}_\text{e}$ and $\vec{n}_\text{o}$ each acquire a different `speed'\footnote{The speed of light in a medium $v_\text{medium}$ is given by $v_\text{medium} = v_\text{vacuum} \frac{n_\text{vacuum}}{n_\text{medium}} = \text{c} \frac{n_\text{vacuum}}{n_\text{medium}} = \frac{\text{c}}{n_\text{medium}}$, with $\text{c}=3\cdot 10^8$\,[m/s].}\ap{,}\footnote{Whether `\textsl{the} speed of light' is actually $\text{c}$, and whether light actually `slows down' in a medium, is a matter of interpretation \cite{Feynman2006}. For our purposes, the traditional `slowing down' interpretation will suffice.} in the material and as a result, these two waves will have obtained an additional phase difference \gls*{delta} between each other after passing the material. This phase difference \gls*{delta} in radians is called `\gls*{retardance}' and related to the two refractive indices by
\begin{equation}
\gls*{delta}= 2\piup\frac{O}{\gls*{lambda}} = 2\piup\frac{\gls*{d}(\gls*{n_e} - \gls*{n_o})}{\gls*{lambda}} = 2\piup\frac{\gls*{d} \gls*{Deltaupn}}{\gls*{lambda}}\label{eq:retardance}
\end{equation}
......
......@@ -38,7 +38,7 @@ Where $\phi$ is the orientation of the ellipse's long axis, and $\tan\vartheta$
The amplitude and phase of \gls*{PL} is affected by \gls*{polariser}s and \gls*{retarder}s. These are described in Jones calculus with matrices. The orientation of the element with respect to our chosen orthogonal basis ($\vec{x}$ and $\vec{y}$) is described by
\begin{equation}
\gls*{Omega}(\omega) = \begin{bmatrix} \cos\omega & -\sin\omega \\ \sin\omega & \cos\omega\end{bmatrix} \label{jonesrot}
\gls*{Omega}(\omega) = \begin{bmatrix*}[r] \cos\omega & -\sin\omega \\ \sin\omega & \cos\omega\end{bmatrix*} \label{jonesrot}
\end{equation}
and the rotation independent effect by
\begin{equation}
......@@ -96,11 +96,11 @@ Since the incident light $\gls*{J0}$ was polarised at 45\,\degree, we find that
\subsection{Analyser} \label{analyser}
In \gls*{PLM} you may encounter '\gls*{analyser}s' next to \gls*{polariser}s and \gls*{retarder}s. An \gls*{analyser} and a \gls*{polariser} are the same thing: they both only transmit the light component that is polarised in the orientation of the element. The difference between the names '\gls*{polariser}' and '\gls*{analyser}' shows their function in the microscope's optics \cite{Howie2008}: a \gls*{polariser} blocks unwelcome light in order to produce \gls*{PL} at a certain orientation; an \gls*{analyser} blocks light in order to allow the detector or eye to record only light that is polarised at a certain orientation. In \gls*{PLM} optics, the \gls*{polariser} is before the sample and the \gls*{analyser} is after the sample (figure \ref{crossed_train_polstates}).
In \gls*{PLM} you may encounter `\gls*{analyser}s' next to \gls*{polariser}s and \gls*{retarder}s. An \gls*{analyser} and a \gls*{polariser} are the same thing: they both only transmit the light component that is polarised in the orientation of the element. The difference between the names `\gls*{polariser}' and `\gls*{analyser}' shows their function in the microscope's optics \cite{Howie2008}: a \gls*{polariser} blocks unwelcome light in order to produce \gls*{PL} at a certain orientation; an \gls*{analyser} blocks light in order to allow the detector or eye to record only light that is polarised at a certain orientation. In \gls*{PLM} optics, the \gls*{polariser} is before the sample and the \gls*{analyser} is after the sample (figure \ref{crossed_train_polstates}).
\section{Retarder} \label{retarder}
A \gls*{retarder} does not block any light and does not affect light intensity. Retarders apply a different phase shift for two orthogonal components of \gls*{PL} aligned with the \gls*{retarder}s own orthogonal axis. As a result, one component of the light has experienced a certain phase shift \gls*{delta} (the difference of the phase shifts for the two orthogonal axis) compared to the other (orthogonal) component. The effective phase shift \gls*{delta} that is applied between the two components is called the 'retardance' of the \gls*{retarder} (see also section \ref{plmbirefringent}).
A \gls*{retarder} does not block any light and does not affect light intensity. Retarders apply a different phase shift for two orthogonal components of \gls*{PL} aligned with the \gls*{retarder}s own orthogonal axis. As a result, one component of the light has experienced a certain phase shift \gls*{delta} (the difference of the phase shifts for the two orthogonal axis) compared to the other (orthogonal) component. The effective phase shift \gls*{delta} that is applied between the two components is called the `retardance' of the \gls*{retarder} (see also section \ref{plmbirefringent}).
The rotation indepent effect of a \gls*{retarder} that applies a \gls*{retardance} \gls*{delta} between its two orthogonal axis, is decribed with the \gls*{Jones matrix} $\gls*{R}_0$:
\begin{equation}
......@@ -115,27 +115,27 @@ The intensity of the light has not changed, but we do find that the phases of th
\begin{equation}
\gls*{phi} = \arg \gls*{J} = \begin{bmatrix} \frac{\gls*{delta}}{2} \\[0.35em] -\frac{\gls*{delta}}{2} \end{bmatrix}
\end{equation}
that the polarisation angle becomes
The polarisation angle follows from
\begin{equation}
\tan 2\phi\gls*{J} = \tan\left(2 \atan \frac{A_y}{A_x}\right)\cos\left(\gls*{varphi}_y-\gls*{varphi}_x\right) = \tan\frac{\piup}{2}\cos\gls*{delta}
\tan 2\phi = \tan\left(2 \atan \frac{A_y}{A_x}\right)\cos\gls*{delta} = \tan\frac{\piup}{2}\cos\gls*{delta}
\end{equation}
and that the ellipticy is given by
and the ellipticy is given by
\begin{equation}
\cos 2\vartheta\gls*{J} = \sin\left(2 \atan \frac{A_y}{A_x}\right)\sin\left(\gls*{varphi}_y-\gls*{varphi}_x\right) = \sin\frac{\piup}{2}\sin\gls*{delta} = \sin \gls*{delta}
\sin 2\vartheta = \sin\left(2 \atan \frac{A_y}{A_x}\right)\abs{\sin\left(\gls*{varphi}_y-\gls*{varphi}_x\right)} = \sin\frac{\piup}{2}\sin\gls*{delta} = \sin \gls*{delta}
\end{equation}
The general \gls*{Jones matrix} for a \gls*{retarder} with \gls*{retardance} \gls*{delta} at an abitrary angle $\alpha$ is
\begin{equation}
\gls*{R}(\gls*{delta},\alpha) = \gls*{Omega}(\alpha)\cdot \gls*{R}_0(\gls*{delta}) \cdot \gls*{Omega}(-\alpha) = \begin{bmatrix} \cos\frac{\gls*{delta}}{2}+\iu\sin\frac{\gls*{delta}}{2}\cos 2\alpha &
\gls*{R}(\gls*{delta},\alpha) = \gls*{Omega}(\alpha)\cdot \gls*{R}_0(\gls*{delta}) \cdot \gls*{Omega}(-\alpha) = \begin{bmatrix*}[r] \cos\frac{\gls*{delta}}{2}+\iu\sin\frac{\gls*{delta}}{2}\cos 2\alpha &
\iu\sin\frac{\gls*{delta}}{2}\sin 2\alpha\\
\iu\sin\frac{\gls*{delta}}{2}\sin 2\alpha &
\cos\frac{\gls*{delta}}{2}-\iu\sin\frac{\gls*{delta}}{2}\cos 2\alpha
\end{bmatrix} \label{retgeneral}
\end{bmatrix*} \label{retgeneral}
\end{equation}
When \gls*{linearly polarised} light in $x$ direction passes $ R(\delta,\alpha)$, we find
\begin{equation}
\gls*{J} = \gls*{R}(\gls*{delta},\alpha) \cdot \gls*{J0} = R(\delta,\alpha) \cdot \begin{bmatrix} 1\\0\end{bmatrix} = \begin{bmatrix} \cos\left(\frac{\gls*{delta}}{2}\right) + \iu \sin\left(\frac{\gls*{delta}}{2}\right)\cos 2\alpha \\
\iu \sin\left(\frac{\gls*{delta}}{2}\right)\sin 2\alpha \end{bmatrix} \label{R0J0}
\gls*{J} = \gls*{R}(\gls*{delta},\alpha) \cdot \gls*{J0} = R(\delta,\alpha) \cdot \begin{bmatrix} 1\\0\end{bmatrix} = \begin{bmatrix*}[r] \cos\left(\frac{\gls*{delta}}{2}\right) + \iu \sin\left(\frac{\gls*{delta}}{2}\right)\cos 2\alpha \\
\iu \sin\left(\frac{\gls*{delta}}{2}\right)\sin 2\alpha \end{bmatrix*} \label{R0J0}
\end{equation}
Thus, \gls*{J} also has a component in $y$ direction. The phases of the two components in $x$ and $y$ are now
\begin{equation}
......@@ -166,8 +166,8 @@ First we derive the matrices for each element. The linear \gls*{polariser} has i
The first \gls*{retarder} has its axis at 45\unit{\degree} and applies a \gls*{retardance} $\gls*{delta}_1$. We therefore find\footnote{Recall
that $\eu^{\iu\omega} = \cos\omega + \iu\sin\omega$.}:
\begin{equation}
\gls*{R}(\delta_1,\alpha_1) = \gls*{Omega}(45\unit{\degree})\cdot\begin{bmatrix} \eu^{-\iu\frac{\gls*{delta}_1}{2}} & 0 \\0 & \eu^{\iu\frac{\gls*{delta}_1}{2}} \end{bmatrix}\cdot \gls*{Omega}(-45\unit{\degree}) = \begin{bmatrix} \cos\frac{\gls*{delta}_1}{2} & -\iu\sin\frac{\gls*{delta}_1}{2} \\
-\iu\sin\frac{\gls*{delta}_1}{2} & \cos\frac{\gls*{delta}_1}{2} \end{bmatrix}
\gls*{R}(\delta_1,\alpha_1) = \gls*{Omega}(45\unit{\degree})\cdot\begin{bmatrix} \eu^{-\iu\frac{\gls*{delta}_1}{2}} & 0 \\0 & \eu^{\iu\frac{\gls*{delta}_1}{2}} \end{bmatrix}\cdot \gls*{Omega}(-45\unit{\degree}) = \begin{bmatrix*}[r] \cos\frac{\gls*{delta}_1}{2} & -\iu\sin\frac{\gls*{delta}_1}{2} \\
-\iu\sin\frac{\gls*{delta}_1}{2} & \cos\frac{\gls*{delta}_1}{2} \end{bmatrix*}
\end{equation}
The second \gls*{retarder} has its axis at 0\unit{\degree} and applies a \gls*{retardance} $\gls*{delta}_2$, so:
\begin{equation}
......@@ -178,7 +178,7 @@ The second \gls*{retarder} has its axis at 0\unit{\degree} and applies a \gls*{r
We can now calculate the \gls*{Jones vector} $J$ of the light after the second \gls*{retarder}. With $\gls*{J0} = [1\ 1]^T$ the \gls*{Jones vector} of the incident light, we can write \cite[eq. 2]{Shribak2003}:
\begin{equation}
\gls*{J} = \gls*{R}(\gls*{delta}_2,\alpha_2)\cdot \gls*{R}(\gls*{delta}_1,\alpha_1)\cdot P(0)\cdot \gls*{J0} =
\begin{bmatrix} \cos\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2}{2}} \\ -\iu\sin\frac{\gls*{delta}_1}{2}\eu^{\iu\frac{\gls*{delta}_2}{2}} \end{bmatrix} = \begin{bmatrix} \cos\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2}{2}} \\ \sin\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2+\piup}{2}}\end{bmatrix} \label{Pf}
\begin{bmatrix*}[r] \cos\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2}{2}} \\ -\iu\sin\frac{\gls*{delta}_1}{2}\eu^{\iu\frac{\gls*{delta}_2}{2}} \end{bmatrix*} = \begin{bmatrix} \cos\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2}{2}} \\ \sin\frac{\gls*{delta}_1}{2}\eu^{-\iu\frac{\gls*{delta}_2+\piup}{2}}\end{bmatrix} \label{Pf}
\end{equation}
Note that the phases of the \gls*{Jones vector} now are
\begin{equation}
......@@ -186,11 +186,11 @@ Note that the phases of the \gls*{Jones vector} now are
\end{equation}
i.e., the light \gls*{J} is \gls*{elliptically polarised} with orientation \cite{Jones1941,Shribak2003}:
\begin{equation}
\tan 2 \theta(\gls*{J}) = \tan\left(2 \atan\left( \frac{A_y}{A_x}\right)\right) \cos\left(\gls*{delta}_2+\frac{\piup}{2}\right) = \tan\gls*{delta}_1\sin\gls*{delta}_2
\tan 2 \theta = \tan\left(2 \atan\left( \frac{A_y}{A_x}\right)\right) \cos\left(\gls*{delta}_2+\frac{\piup}{2}\right) = \tan\gls*{delta}_1\sin\gls*{delta}_2
\end{equation}
and ellipticity
\begin{equation}
\sin 2 \vartheta(\gls*{J}) = \sin\left(2 \atan\left( \frac{A_y}{A_x}\right)\right) \left|\sin\left(\gls*{delta}_2+\frac{\piup}{2}\right)\right| = \sin\gls*{delta}_1\cos\gls*{delta}_2
\sin 2 \vartheta = \sin\left(2 \atan\left( \frac{A_y}{A_x}\right)\right) \left|\sin\left(\gls*{delta}_2+\frac{\piup}{2}\right)\right| = \sin\gls*{delta}_1\cos\gls*{delta}_2
\end{equation}
The intensity of \gls*{J} is half of that of the incident beam $\gls*{J0}$:
\begin{align}
......
% !TeX root = plm.tex
\begin{savequote}
``Some students,'' said Edna Kramer in \textnormal{The Nature and Growth of Modern Mathematics}, consider trigonometry ``a glorified geometry with superimposed computational torture.'' The present book is an attempt to dispel this view. \qauthor{Eli Maor \cite{Maor2013}}
\end{savequote}
\chapter{Prefab Jones and Mueller matrices}\label{matrix}
For your calculation convenience, we are more than willing to provide you with some common Jones and Mueller matrices. These matrices are based on the Collett-definition of optical elements \cite{Collett2005}, i.e.
\begin{equation}
N(\gls*{delta},\gls*{varphi}) = \gls*{Omega}(-\gls*{varphi})\cdot N(\gls*{delta},0)\cdot \gls*{Omega}(\gls*{varphi}) = \begin{bmatrix}\cos\gls*{varphi} & -\sin\gls*{varphi}\\ \sin\gls*{varphi} & \cos\gls*{varphi}\end{bmatrix}\cdot\begin{bmatrix} \eu^{\iu\gls*{delta}} & 0\\ 0 & 1\end{bmatrix}\cdot \begin{bmatrix}\cos\gls*{varphi} & \sin\gls*{varphi}\\ -\sin\gls*{varphi} & \cos\gls*{varphi}\end{bmatrix}
N(\gls*{delta},\gls*{varphi}) = \gls*{Omega}(-\gls*{varphi})\cdot N(\gls*{delta},0)\cdot \gls*{Omega}(\gls*{varphi}) = \begin{bmatrix*}[r]\cos\gls*{varphi} & -\sin\gls*{varphi}\\ \sin\gls*{varphi} & \cos\gls*{varphi}\end{bmatrix*}\cdot\begin{bmatrix} \eu^{\iu\gls*{delta}} & 0\\ 0 & 1\end{bmatrix}\cdot \begin{bmatrix*}[r]\cos\gls*{varphi} & \sin\gls*{varphi}\\ -\sin\gls*{varphi} & \cos\gls*{varphi}\end{bmatrix*}
\end{equation}
Some remarks:
\begin{enumerate}
......@@ -21,11 +24,11 @@ Some remarks:
rotation & \gls*{Mueller matrix} & \gls*{Jones matrix}\\
\noalign{\smallskip}\hline\noalign{\smallskip} \bigskip
0\,\degree & $\begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}$ & $\begin{bmatrix}1&0\\0&1\end{bmatrix}$\\ \bigskip
90\,\degree & $\begin{bmatrix}1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&1\end{bmatrix}$ & $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$\\ \bigskip
45\,\degree & $\begin{bmatrix}1&0&0&0\\ 0&0&1&0\\ 0&-1&0&0\\ 0&0&0&1\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-1&1\end{bmatrix}$\\ \bigskip
-45\,\degree & $\begin{bmatrix}1&0&0&0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&0&0&1\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}$\\ \bigskip
90\,\degree & $\begin{bmatrix*}[r]1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&1\end{bmatrix*}$ & $\begin{bmatrix*}[r]0&1\\-1&0\end{bmatrix*}$\\ \bigskip
45\,\degree & $\begin{bmatrix*}[r]1&0&0&0\\ 0&0&1&0\\ 0&-1&0&0\\ 0&0&0&1\end{bmatrix*}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix*}[r]1&1\\-1&1\end{bmatrix*}$\\ \bigskip
-45\,\degree & $\begin{bmatrix*}[r]1&0&0&0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&0&0&1\end{bmatrix*}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix*}[r]1&-1\\1&1\end{bmatrix*}$\\ \bigskip
$\gls*{varphi}$\,\degree & $\begin{bmatrix}1&0 &0&0\\ 0 & \cos 2\gls*{varphi} & \sin 2\gls*{varphi} & 0\\ 0 & -\sin 2\gls*{varphi} & \cos 2\gls*{varphi} & 0
\\ 0&0&0&1\end{bmatrix}$ & $\begin{bmatrix}\cos\gls*{varphi}&\sin\gls*{varphi}\\ -\sin\gls*{varphi} &\cos\gls*{varphi}\end{bmatrix}$\\
\\ 0&0&0&1\end{bmatrix}$ & $\begin{bmatrix*}[r]\cos\gls*{varphi}&\sin\gls*{varphi}\\ -\sin\gls*{varphi} &\cos\gls*{varphi}\end{bmatrix*}$\\
\noalign{\smallskip}\hline
\end{tabular}
\end{table}
......
......@@ -15,6 +15,12 @@ Robert~D. Allen, James Brault and Robert~D. Moore.
\newblock \pmid{14079486}.
\newblock \doi{10.1083/jcb.18.2.223}.
\bibitem{Bellos2011}
Alex Bellos.
\newblock \emph{{A}lex's {A}dventures in {N}umberland -- dispatches from the
wonderful world of mathematics}.
\newblock Bloomsbury Publishing, 2011.
\bibitem{Benditt1970}
E.P. Benditt, N.~Eriksen and C.~Berglund.
\newblock {C}ongo red dichroism with dispersed amyloid fibrils, an extrinsic
......@@ -137,6 +143,11 @@ Shih-Yau Lu and Russell~A. Chipman.
\newblock \emph{Journal of the Optical Society of America A},
\textbf{13}(5):1106--1113, May 1996.
\bibitem{Maor2013}
Eli Maor.
\newblock \emph{{T}rigonometric {D}elights}.
\newblock Princeton Science Library, 2013.
\bibitem{Mueller1948}
Hans Mueller.
\newblock {T}he foundation of optics.
......@@ -209,6 +220,18 @@ G.G. Stokes.
\newblock \emph{Transactions of the Cambridge Philosophical Society},
\textbf{9}:399--416, 1852.
\bibitem{Talacua2015}
Hanna Talacua, Anthal~I.P.M. Smits, Dimitri~E.P. Muylaert, Jan~Willem
\noopsort{Rijswijk}van Rijswijk, Aryan Vink, Marianne~C. Verhaar, Anita
Driessen-Mol, Lex~A. \noopsort{Herwerden}van Herwerden, Carlijn~V.C. Bouten,
Jolanda Kluin and Frank~P.T. Baaijens.
\newblock {I}n {S}itu {T}issue {E}ngineering of {F}unctional {S}mall-{D}iameter
{B}lood {V}essels by {H}ost {C}irculating {C}ells {O}nly.
\newblock \emph{Tissue Engineering -- Part A}, \textbf{21}(19-20):2583--2594,
October 2015.
\newblock \pmid{26200255}.
\newblock \doi{10.1089/ten.TEA.2015.0066}.
\bibitem{Turnhout2009}
Mark~C. \noopsort{Turnhout}van Turnhout, Sander Kranenbarg and Johan~L.
\noopsort{Leeuwen}van Leeuwen.
......
......@@ -23,6 +23,16 @@
Timestamp = {2013.03.05}
}
@Book{Bellos2011,
Title = {{A}lex's {A}dventures in {N}umberland -- dispatches from the wonderful world of mathematics},
Author = {Bellos, Alex},
Publisher = {Bloomsbury Publishing},
Year = {2011},
Owner = {tue},
Timestamp = {2017.09.05}
}
@Article{Benditt1970,
Title = {{C}ongo red dichroism with dispersed amyloid fibrils, an extrinsic cotton effect.},
Author = {Benditt, E.P. and Eriksen, N. and Berglund, C.},
......@@ -282,6 +292,16 @@
File = {[Lu1996]_Interpretation of Mueller matrices based on polar decomposition.pdf:[Lu1996]_Interpretation of Mueller matrices based on polar decomposition.pdf:PDF;[Lu1996]_Interpretation of Mueller matrices based on polar decomposition.pdf:[Lu1996]_Interpretation of Mueller matrices based on polar decomposition.pdf:PDF}
}
@Book{Maor2013,
Title = {{T}rigonometric {D}elights},
Author = {Maor, Eli},
Publisher = {Princeton Science Library},
Year = {2013},
Owner = {tue},
Timestamp = {2017.09.05}
}
@Article{Mueller1948,
Title = {{T}he foundation of optics},
Author = {Mueller, Hans},
......@@ -435,6 +455,28 @@
Timestamp = {2013.01.21}
}
@Article{Talacua2015,
Title = {{I}n {S}itu {T}issue {E}ngineering of {F}unctional {S}mall-{D}iameter {B}lood {V}essels by {H}ost {C}irculating {C}ells {O}nly.},
Author = {Talacua, Hanna and Smits, Anthal I.P.M. and Muylaert, Dimitri E.P. and \noopsort{Rijswijk}van Rijswijk, Jan Willem and Vink, Aryan and Verhaar, Marianne C. and Driessen-Mol, Anita and \noopsort{Herwerden}van Herwerden, Lex A. and Bouten, Carlijn V.C. and Kluin, Jolanda and Baaijens, Frank P.T.},
Journal = {Tissue Engineering -- Part A},
Year = {2015},
Month = {October},
Number = {19-20},
Pages = {2583--2594},
Volume = {21},
Abstract = {Inflammation is a natural phase of the wound healing response, which can be harnessed for the in situ tissue engineering of small-diameter blood vessels using instructive, bioresorbable synthetic grafts. This process is dependent on colonization of the graft by host circulating cells and subsequent matrix formation. Typically, vascular regeneration in small animals is governed by transanastomotic cell ingrowth. However, this process is very rare in humans and hence less relevant for clinical translation. Therefore, a novel rat model was developed, in which cell ingrowth from the adjacent tissue is inhibited using Gore-Tex sheathing. Using this model, our aim here was to prove that functional blood vessels can be formed in situ through the host inflammatory response, specifically by blood-borne cells. The model was validated by implanting sex-mismatched aortic segments on either anastomoses of an electrospun poly(ɛ-caprolactone) (PCL) graft, filled with fibrin gel, into the rat abdominal aorta. Fluorescent in situ hybridization analysis revealed that after 1 and 3 months in vivo, over 90\% of infiltrating cells originated from the bloodstream, confirming the effective shielding of transanastomotic cell ingrowth. Using the validated model, PCL/fibrin grafts were implanted, either or not loaded with monocyte chemotactic protein-1 (MCP-1), and cell infiltration and tissue development were investigated at various key time points in the healing cascade. A phased healing response was observed, initiated by a rapid influx of inflammatory cells, mediated by the local release of MCP-1. After 3 months in vivo, the grafts consisted of a medial layer with smooth muscle cells in an oriented collagen matrix, an intimal layer with elastin fibers, and confluent endothelium. This study proves the regenerative potential of cells in the circulatory system in the setting of in situ vascular tissue engineering.},
Doi = {10.1089/ten.TEA.2015.0066},
File = {[Talacua2015]_In Situ Tissue Engineering of Functional Small-Diameter Blood Vessels by Host Circulating Cells Only.pdf:[Talacua2015]_In Situ Tissue Engineering of Functional Small-Diameter Blood Vessels by Host Circulating Cells Only.pdf:PDF},
Institution = {5 Institute for Complex Molecular Systems , Eindhoven, The Netherlands .},
Language = {eng},
Medline-pst = {ppublish},
Owner = {tue},
Pmid = {26200255},
Timestamp = {2016.04.06}
}
@Article{Turnhout2009,
Title = {{M}odelling optical behaviour of birefringent biological tissues for evaluation of quantitative polarised light microscopy.},
Author = {\noopsort{Turnhout}van Turnhout, Mark C. and Kranenbarg, Sander and \noopsort{Leeuwen}van Leeuwen, Johan L.},
......
No preview for this file type
......@@ -33,7 +33,7 @@ Another observation that can be inferred from equation \eqref{twofibresanalytic
\end{equation}
So that when any term in equation \eqref{deltaanisophi} is zero, we will have no light on the detector, i.e.:
\begin{equation}
\gls*{I}(\text{eq. \ref{deltaanisophi}}) = 0\quad \text{ for }\quad \begin{cases}\gls*{delta} &=0 \\ \cos (\Deltaup\gls*{varphi}) &= 0 \qquad \rightarrow \Deltaup\gls*{varphi} = 90\,\degree + 180\,\degree\cdot k, k \in \mathbb{Z} \\ \sin 2\gls*{phi} &= 0 \qquad \rightarrow {\color{white}\Deltaup}\gls*{phi} = {\color{white}9}0\,\degree + {\color{white}9}90\,\degree\cdot k, k \in \mathbb{Z} \end{cases}
\gls*{I}(\text{eq. \ref{deltaanisophi}}) = 0\quad \text{ for }\quad \begin{cases}\gls*{delta} &=0 \\ \cos (\Deltaup\gls*{varphi}) &= 0 \qquad \rightarrow \Deltaup\gls*{varphi} = 90\,\degree + 180\,\degree\cdot k,\; k \in \mathbb{Z} \\ \sin 2\gls*{phi} &= 0 \qquad \rightarrow {\color{white}\Deltaup}\gls*{phi} = {\color{white}9}0\,\degree + {\color{white}9}90\,\degree\cdot k,\; k \in \mathbb{Z} \end{cases}
\end{equation}
A completely dark image or pixel therefore means that either there is no \gls*{birefringence} ($\gls*{delta} = 0$), the \gls*{birefringence} is isotropically distributed ($\cos(\Deltaup\gls*{varphi}) = 0$), the \gls*{azimuth} of the \gls*{birefringence} is aligned with the \gls*{polariser} or \gls*{analyser} ($\sin 2\gls*{phi} = 0$), or any combination of these three possibilities. And from the intensity results alone, we cannot tell one situation from the other.
......@@ -83,7 +83,7 @@ We finally show the example of a transverse section through an artery in figure
\subfloat[\label{artery_collagen}]{\includegraphics[width=0.47\linewidth]{artery_collagen.pdf}}\hfill
\subfloat[\label{artery_pixels}]{\includegraphics[width=0.47\linewidth]{artery_pixels.pdf}}\\
\center\subfloat[\label{artery_plm}]{\includegraphics[width=0.47\linewidth]{Anthal_Picture1cropped1.png}}\\
\caption{Example of (expected) \gls*{PLM} results in a transverse section through an artery that shows the circumferential predominant collagen orientation in the pane of imaging. With \textbf{(a)} the (idealised) circumferential collagen structure divided over nine pixels; \textbf{(b)} the expected intensities in these nine pixels. Pixels that have the collagen more or less aligned with the \gls*{polariser} and \gls*{analyser} are darker than the corner pixels where the predominant collagen orientation makes an angle of 45\,\degree\ with the \gls*{polariser} and \gls*{analyser}. There is no collagen in the centre pixel, and this pixel will stay dark; and \textbf{(c)} \gls*{PLM} results with crossed \gls*{polariser}s on a transverse section through an artery (courtesy of Anthal Smits) that appear to confirm the predictions in \textbf{(b)}.}\label{artery_example}
\caption{Example of (expected) \gls*{PLM} results in a transverse section through an artery that shows the circumferential predominant collagen orientation in the pane of imaging. With \textbf{(a)} the (idealised) circumferential collagen structure divided over nine pixels; \textbf{(b)} the expected intensities in these nine pixels. Pixels that have the collagen more or less aligned with the \gls*{polariser} and \gls*{analyser} are darker than the corner pixels where the predominant collagen orientation makes an angle of 45\,\degree\ with the \gls*{polariser} and \gls*{analyser}. There is no collagen in the centre pixel, and this pixel will stay dark; and \textbf{(c)} \gls*{PLM} results with crossed \gls*{polariser}s on a transverse section through an artery (courtesy of Anthal Smits, \cite{Talacua2015}) that appear to confirm the predictions in \textbf{(b)}.}\label{artery_example}
\end{figure}
Of course, from figure \ref{artery_plm} \textsl{alone}, we can not tell if the darker parts of the image are darker because of a lack of collagen, or because collagen is locally less anisotropically distributed, or because of the changes in collagen \gls*{azimuth} as we suggested in figures \ref{artery_collagen} and \ref{artery_pixels}. Still, in this particular case, the assumptions on equal collagen amounts and equal collagen anisotropy seem reasonable, based on existing knowledge of such collagen structures. With these (reasonable) assumptions, we may infer that the cause of the differences in pixel intensities \textsl{is} due to changes in the collagen fibre \gls*{azimuth}.
......@@ -98,9 +98,9 @@ Polarised light microscopy with the classical crossed \gls*{polariser}s set-up i
\subsection{Setting up the polariser and analyser}
The \gls*{polariser} is provided in the condensor unit (figure \ref{pic:observer_condensor}). It simply needs to be rotated into the light path of the microscope. Note that 'normal' brightfield microscopy is still possible with this filter in place, that is: the 'results' will not change when this filter is placed in the light path, although the intensity of the incident light on the sample \textsl{will} decrease. Best practice therefore is \textsl{not} to have this filter in the light path if you want to do \textsl{brightfield microscopy}.
The \gls*{polariser} is provided in the condensor unit (figure \ref{pic:observer_condensor}). It simply needs to be rotated into the light path of the microscope. Note that `normal' brightfield microscopy is still possible with this filter in place, that is: the `results' will not change when this filter is placed in the light path, although the intensity of the incident light on the sample \textsl{will} decrease. Best practice therefore is \textsl{not} to have this filter in the light path if you want to do \textsl{brightfield microscopy}.
The \gls*{analyser} is situated in a sliding bar (slider) underneath the filter block (figure \ref{pic:observer_analyser}). The dial on the slider allows you to adjust the orientation of the \gls*{analyser} filter. The \gls*{analyser} is mounted by sliding the slider to the left (you will feel 'click' when the slider is in place) underneath the filter block. Of course, the effect of this filter on brightfield microscopy is the similar to that of the \gls*{polariser} filter. Best practice therefore is, again, \textsl{not} to have this filter in the light path if you want to do \textsl{brightfield microscopy}.
The \gls*{analyser} is situated in a sliding bar (slider) underneath the filter block (figure \ref{pic:observer_analyser}). The dial on the slider allows you to adjust the orientation of the \gls*{analyser} filter. The \gls*{analyser} is mounted by sliding the slider to the left (you will feel `click' when the slider is in place) underneath the filter block. Of course, the effect of this filter on brightfield microscopy is the similar to that of the \gls*{polariser} filter. Best practice therefore is, again, \textsl{not} to have this filter in the light path if you want to do \textsl{brightfield microscopy}.
\begin{figure}[h]
\def\svgwidth{\linewidth}\includesvg{pics/observer_condensor}
......@@ -118,7 +118,7 @@ For \gls*{PL} microscopy, both \gls*{polariser} and \gls*{analyser} need to be m
\subsection{The quarter wave plates}
The \gls*{Observer microscope} had two \gls*{quarter wave plate}s: one with an adjustable orientation near the \gls*{polariser} (figure \ref{pic:observer_condensor}), and one with a fixed orientation in the filter block near the \gls*{analyser} (figure \ref{pic:observer_analyser}). The second one is easiest (and most reproducable) to use: simply set the filter block to position 2 with the touch screen or in the AxioVision software (position 3 contains the fluorescence filters, the other positions are empty). The objective of the \gls*{quarter wave plate} is to enable 'extinctionless imaging, i.e.\ to get rid of the dark background (figure \ref{effectQWP}). For this, you only need to use one of the two \gls*{quarter wave plate}s.
The \gls*{Observer microscope} had two \gls*{quarter wave plate}s: one with an adjustable orientation near the \gls*{polariser} (figure \ref{pic:observer_condensor}), and one with a fixed orientation in the filter block near the \gls*{analyser} (figure \ref{pic:observer_analyser}). The second one is easiest (and most reproducable) to use: simply set the filter block to position 2 with the touch screen or in the AxioVision software (position 3 contains the fluorescence filters, the other positions are empty). The objective of the \gls*{quarter wave plate} is to enable `extinctionless imaging, i.e.\ to get rid of the dark background (figure \ref{effectQWP}). For this, you only need to use one of the two \gls*{quarter wave plate}s.
\begin{figure}[t!]
\subfloat[\label{noQWP}]{\includegraphics[width=0.47\linewidth]{PR_noQWP_scaled.png}}\hfill%
......
% !TeX root = plm.tex
\begin{savequote}
His bestselling product, however, is the PhiMatrix, a piece of software that creates a grid on your computer screen to check images for the golden ratio. Most purchasers use it as a design tool to make cutlery, furniture and homes. Some customers use it for financial speculation by superimposing the grid on graphs of indices and using phi to predict future trends. `A guy in the Caribbean was using my matrix to trade in oil, a guy in China was using it to trade in currencies,' he said. \qauthor{Alex Bellos \cite{Bellos2011}}
\end{savequote}
\chapter{Sample matrices revisited}
\section{The sample Jones matrix is not unique}\label{examSsample}
......@@ -23,8 +26,8 @@ and for fibres at -45\,\degree\ and 45\,\degree, we find:
\begin{align}
\gls*{Ssample} & = \gls*{Omega}(-45\,\degree)\cdot\begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix}\cdot \gls*{Omega}(45\,\degree) \cdot
\gls*{Omega}(45\,\degree)\cdot\begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix}\cdot \gls*{Omega}(-45\,\degree) \\
& = \sqrt{2}\begin{bmatrix}1&1\\ -1 &1\end{bmatrix} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \sqrt{2}\begin{bmatrix}1 & -1\\ 1 & 1\end{bmatrix} \cdot \sqrt{2}\begin{bmatrix}1&-1\\ 1 &1\end{bmatrix} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \sqrt{2}\begin{bmatrix}1 & 1\\ -1 & 1\end{bmatrix} \\
& = 4 \begin{bmatrix}1&1\\ -1 &1\end{bmatrix} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & -1\\ 1 & 1\end{bmatrix} \cdot\begin{bmatrix}1&-1\\ 1 &1\end{bmatrix} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 1\\ -1 & 1\end{bmatrix} \\
& = \sqrt{2}\begin{bmatrix*}[r]1&1\\ -1 &1\end{bmatrix*} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \sqrt{2}\begin{bmatrix*}[r]1 & -1\\ 1 & 1\end{bmatrix*} \cdot \sqrt{2}\begin{bmatrix*}[r]1&-1\\ 1 &1\end{bmatrix*} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \sqrt{2}\begin{bmatrix*}[r]1 & 1\\ -1 & 1\end{bmatrix*} \\
& = 4 \begin{bmatrix*}[r]1&1\\ -1 &1\end{bmatrix*} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix*}[r]1 & -1\\ 1 & 1\end{bmatrix*} \cdot\begin{bmatrix*}[r]1&-1\\ 1 &1\end{bmatrix*} \cdot \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix*}[r]1 & 1\\ -1 & 1\end{bmatrix*} \\
& = \begin{bmatrix}\text{e}^{\text{i}\gls*{delta}} & 0\\ 0 & \text{e}^{\text{i}\gls*{delta}}\end{bmatrix} = \text{e}^{\text{i}\gls*{delta}} \begin{bmatrix}1 & 0\\ 0 &1\end{bmatrix} \label{crossedat45}
\end{align}
Equations \eqref{crossedat90} and \eqref{crossedat45} show that all light is equally retarded in these two examples: there is no phase shift or retardation \textsl{between} the two mutually perpendicular components of the incident light that reaches the sample. Hence, the polarisation state of this incident light is left unchanged, and no light will reach the detector.
......
% !TeX root = plm.tex
\begin{savequote}
Finally, it will be people in all sorts of businesses and industries who understand the power of ``modeling''--and I am not talking about Cindy Crawford. \qauthor{Thomas L.\ Friedman \cite{Friedman2010}}
\end{savequote}
\chapter{Some Octave/Matlab scripts}\label{scripts}
\begin{lstlisting}[language=Octave,caption={Getting an ideal \gls*{polariser} matrix with \texttt{plmsim\_pol.m}.},label={plmsimpol}]
......
% !TeX root = plm.tex
\begin{savequote}
In February 1854 Stokes made the proof of the theorem an examination question at Cambridge--on a test, it is amusing to note, taken by the youthful James Clerk Maxwell. \qauthor{Paul J.\ Nahin \cite{Nahin2010}}
\end{savequote}
\chapter{Stokes, Mueller \& depolarisation}
\section{Introduction}
......@@ -23,11 +26,11 @@ S = \begin{bmatrix} 1 \\ 0 \\0 \\ 0 \end{bmatrix}
polarisation state & \gls*{Stokes vector}& \gls*{Jones vector}\\
\noalign{\smallskip}\hline\noalign{\smallskip} \bigskip
linear horizontal & $\begin{bmatrix}1\\1\\0\\0\end{bmatrix}$ & $\begin{bmatrix}1\\0\end{bmatrix}$\\ \bigskip
linear vertical& $\begin{bmatrix}1\\-1\\0\\0\end{bmatrix}$ & $\begin{bmatrix}0\\1\end{bmatrix}$\\ \bigskip
linear vertical& $\begin{bmatrix*}[r]1\\-1\\0\\0\end{bmatrix*}$ & $\begin{bmatrix}0\\1\end{bmatrix}$\\ \bigskip
linear at 45\,\degree& $\begin{bmatrix}1\\0\\1\\0\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}$\\ \bigskip
linear at -45\,\degree & $\begin{bmatrix}1\\0\\-1\\0\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-1\end{bmatrix}$\\ \bigskip
right circular& $\begin{bmatrix}1\\0\\0\\1\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\-\iu\end{bmatrix}$\\ \bigskip
left circular& $\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\\iu\end{bmatrix}$\\
linear at -45\,\degree & $\begin{bmatrix*}[r]1\\0\\-1\\0\end{bmatrix*}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix*}[r]1\\-1\end{bmatrix*}$\\ \bigskip
right circular& $\begin{bmatrix}1\\0\\0\\1\end{bmatrix}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix*}[r]1\\-\iu\end{bmatrix*}$\\ \bigskip
left circular& $\begin{bmatrix*}[r]1\\0\\0\\-1\end{bmatrix*}$ & $\frac{1}{\sqrt{2}}\begin{bmatrix}1\\\iu\end{bmatrix}$\\
\noalign{\smallskip}\hline
\end{tabular}
\end{table}
......@@ -66,7 +69,7 @@ M = A\cdot\left(J\otimes J^c\right)\cdot A^{-1}
\end{equation}
with $\gls*{J}^c$ the complex conjugate of the \gls*{Jones matrix} \gls*{J}, and
\begin{equation}
A = \begin{bmatrix} 1&0&0&1\\1&0&0&-1\\0&1&1&0\\0&\iu&-\iu&0\end{bmatrix}
A = \begin{bmatrix*}[r] 1&0&0&1\\1&0&0&-1\\0&1&1&0\\0&\iu&-\iu&0\end{bmatrix*}
\end{equation}
\section{Unpolarised light}
......@@ -78,6 +81,6 @@ D = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & d_1 & 0 & 0 \\ 0 & 0 & d_2 & 0 \\ 0 &
Which leaves the intensity $(S_0)$ intact, but decreases the polarisation states $S_{(1,2,3)}$ with respective factors $d_{(1,2,3)}$. If we want a to add depolarised light somewhere in
the optical train, we can add \cite{Bickel1985,Walker1954,Lu1996}:
\begin{equation}
D = \begin{bmatrix} d_0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}, \quad d > 1
D = \begin{bmatrix} d_0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}, \quad d_0 > 1
\end{equation}
Which increases the intensity $(S_0)$ with a factor $d_0$, and leaves the polarisation states $S_{(1,2,3)}$ intact. We have therefore added a fraction $d-1$ of depolarised light to the \gls*{PL} vector.
Which increases the intensity $(S_0)$ with a factor $d_0$, and leaves the polarisation states $S_{(1,2,3)}$ intact. We have therefore added a fraction $d_0-1$ of depolarised light to the \gls*{PL} vector.