\documentclass{article}
\usepackage[affil-it]{authblk}
\usepackage{graphicx}
\usepackage[space]{grffile}
\usepackage{latexsym}
\usepackage{textcomp}
\usepackage{longtable}
\usepackage{tabulary}
\usepackage{booktabs,array,multirow}
\usepackage{amsfonts,amsmath,amssymb}
\providecommand\citet{\cite}
\providecommand\citep{\cite}
\providecommand\citealt{\cite}
\usepackage{url}
\usepackage{hyperref}
\hypersetup{colorlinks=false,pdfborder={0 0 0}}
\usepackage{etoolbox}
\makeatletter
\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{%
  \errmessage{\noexpand\@combinedblfloats could not be patched}%
}%
\makeatother
% You can conditionalize code for latexml or normal latex using this.
\newif\iflatexml\latexmlfalse
\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}}

\usepackage[utf8]{inputenc}
\usepackage[english]{babel}

\iflatexml
\documentclass[prb,preprint]{revtex4-1} 
\fi
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{bm}% bold math

\usepackage{natbib}
\bibliographystyle{apsrev}



\begin{document}

\title{Measurement of the Faraday Effect}


\author{Ning Zhu}
\affil{Affiliation not available}

  
\author{Emily A Kaplan}
\affil{Affiliation not available}
  



\date{\today}

\maketitle




\section{Abstract}
We performed an experiment to measure the Faraday rotation of polarized light passing through a magnetic field, as well as measuring the Verdet constant of a glass tube with a length of 0.1 m. Linearly polarized light from a HeNe laser was sent through a magnetic field created by a solenoid. The light then passed through an adjustable polarizer and was detected by a photodiode. Using the photodiode, we measured how the intensity of the light changed when turning the polarizer through 360 degrees. We performed this experiment both with and without a glass tube inside the solenoid. We plotted the data as a function of voltage versus angle in plotly in order to find the angle where the voltage measured by the photodiode changed most rapidly. By keeping the polarizer at its most sensitive angle, we plotted voltage versus magnetic field and measured how the voltage changed with magnetic field. We used the lock-in amplifier to make the voltage measurement more accurate. 


\section{Aims} % Section titles are automatically converted to all-caps.
% Section numbering is automatic.
    To observe Faraday effect in this lab, which says that the rotation of plane of polarization of light changes when applying a magnetic field:
$$I=I_{0}cos2[\theta_{1}-\theta_{0}]$$
where $I_{0}$ is the original light intensity and I is the light intensity passing through two polarizers at angles $\theta_{1}$ and $\theta_{0}$. 
    To experimentally determine the Verdet constant of a glass tube, which describes the strength of Faraday effect for a particular material: 
$$\varphi_{B}=C_{v}BL$$
where $\varphi$ is the shift in polarization, B is the strength of applied magnetic field and L is the length of the glass tube.


\section{Introduction}



more introduction. Explanation of polarization? Transmission of originally unpolarized light through two polarizers?

Light intensity $I$ after passing through two polarizers at angles $\theta_1$ and $\theta_0$ is predicted to be 
\begin{equation}
I = I_0 \cos^2[\theta_1 - \theta_0]
\end{equation}
where $I_0$ is the original light intensity. 

An example for zero field is shown in Fig.~\ref{fig:PolarizationDataAtZeroField}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/PolarizationAngleForClass/PolarizationAngleForClass}
\caption{{\label{fig:PolarizationDataAtZeroField}
Something profound and explanatory about this figure, which I have named fig:PolarizationDataAtZeroField.%
}}
\end{center}
\end{figure}

The Faraday effect is ...

Following Melissinos and Napolitano \cite{Melissinos_2003} pages 204 - 210, we begin by assuming that the shift in polarization of the light $\phi(B)$ passing through a rod of material of length $L$ due to the application of an external co-axial magnetic field $B$  is linearly proportional to the applied magnetic field and the length of the material, so that 
\begin{equation}
\phi(B) = c_V B L
\end{equation}
where $c_V$ is a (wavelength-dependent) `constant' known as the Verdet constant. The value of the Verdet constant depends on the type of material and the wavelength of the polarized light. 

In our experiment, we measure light intensity by use of a photodiode. A photodiode is a current source with an output that is proportional to the light intensity (unless it saturates), so by placing a $1k\Omega$ resistor across the photodiode terminals and measuring the voltage across the resistor, we have a measure of light intensity. 

We did a lot of stuff to ensure we got a linear response, including ....



Suppose then that the original angle of polarization of the light emitted from the laser is $\theta_0$, the change in polarization angle  due the application of the magnetic field (as the light passes through a rod-shaped material) is $\Delta \theta(B) = C_v L B$, the angle of the polarization of the light after passing through the rod is $\theta_1 = \theta_0 + \Delta \theta$, and the angle of the second polarizer though which the polarized light passes before being detected by the photodiode (but after exiting the rod) is $\theta_2$. 

Then, as noted by \cite{Melissinos_2003} if the voltage output from the photodiode $V_{\textrm{ pd}}$ is proportional to the light intensity, the photodiode response is given by 
\begin{equation}
\label{eq:Diode_Response}
V_{\mathrm{pd}}(B) = V_0 \cos^2[\phi(B)]
\end{equation}
and the photodiode sensitivity $\eta$ = $dV_{pd}/dB$  is given by 
\begin{eqnarray}
\eta & = &  - 2 V_0 \cos[\phi]\sin[\phi] \frac{d\phi}{dB} \\
& = & -V_0 \sin[2\phi] \frac{d\phi}{dB} \\
& = & + c_V LV_0 \sin[2\phi] \\
\end{eqnarray}
since $\phi = \theta_{2} - \theta_{1} = \theta_{2} - \left(\theta_0 + \Delta \theta(B)\right)$. 
An example of a fit to Eq.~\ref{eq:Diode_Response} is shown in Fig.~\ref{fig:PolarizationDataAtZeroField}



Note that the sensitivity $\eta$  has a maximum value at $\phi = \pi /4$; at that orientation, 
\begin{equation}
\label{eq:MaxSensitivity}
\left.\frac{dV_{\mathrm{pd}}(B)}{dB}\right|_{\textrm{max}} =\left.\left( \frac{dV_{\mathrm{pd}}}{d\phi}\right)\right|_{\pi/4} \left(\frac{d\phi}{dB}\right)= -\left(V_0 \left.\sin[2\phi]\right)\right|_{\phi = \pi/4}\left(\frac{d\phi}{dB}\right).
\end{equation}

The derivation above (with some change in notation) is from \cite{Melissinos_2003}. See pages 205 - 210 for additional details. 

\section{Method}
Eqs.~\ref{eq:Diode_Response} and \ref{eq:MaxSensitivity} suggest 3 methods of measuring the Verdet constant $c_V$. 
\begin{enumerate}
\item Use Eq.~\ref{eq:Diode_Response} to Calculate $d\phi / dB$ and then $c_V$ from measurements of $V_{\mathrm{pd}}(B,\theta_2)$ as a function of  polarizer angle $\theta_2$ for two or more magnetic field strengths $B$ (one of which could be zero).
\item Use Eq.~\ref{eq:MaxSensitivity} to calculate $dV_{\mathrm{pd}}(B)/dB$, $d\phi / dB$ and $c_V$ from measurements of  $V_{\mathrm{pd}}(B)$ as a function of magnetic field $B$ (at the relative polarizer angle $\phi(0) = \pi/4$). 
\item Use Eq.~\ref{eq:MaxSensitivity} to calculate $d\phi / dB$ and $c_V$ from direct measurements of $dV_{\mathrm{pd}}(B)/dB$  (using a lock-in amplifier) for a modulated magnetic field $B = B_0 + B_1 \cos[2 \pi f t]$ (at the relative polarizer angle $\phi(0) = \pi/4$).  
\end{enumerate}

To find $c_V$ using  Eq.~\ref{eq:Diode_Response} , we need to fit a plot of our data to Eq.~\ref{eq:Diode_Response} to determine  $\theta_1(B)$ for each value of $B$. 

To find  $c_V$ using Eq.~\ref{eq:MaxSensitivity}, note that at that maximum sensitivity setting, $\phi(B) = \pi/4 - c_V L B$ and $V_{\mathrm{pd}}(0) =V_0 / 2 $. 






\section{Results}\label{sec:Results}
%\label{DispEqSection} You can label sections for reference




\begin{center}
Figure one
\end{center}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/Voltage-vs-Angle-for-Faraday-Rotation/Voltage-vs-Angle-for-Faraday-Rotation}
\caption{{Figure one shows the values of voltages obtained by changing the rotation angles over a 360 degrees range with a step size of 15 degrees. (0) means there is no magnetic field applied, and (-3) means there is a magnetic field applied, and the current goes through the solenoid is -3A.%
}}
\end{center}
\end{figure}

\centering
Figure two\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/V-vs-B/V-vs-B}
\caption{{Figure two shows how photodiode voltage changes when changing the strength of magnetic field. Current being put into solenoid ranges from -3A to 3A with a step size of 0.5A.%
}}
\end{center}
\end{figure}

\section{Analysis}
1. Direct Fit
$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}$$
%$$\frac{\Delta\theta}{\Delta B}=\frac{\Delta V}{\Delta B} \times \frac{\Delta \theta}{\Delta V}$$
$$\theta_{B}=104 degrees; \theta_{0}=108 degrees$$
$$\Delta\theta=\theta_{B}-\theta_{0}=-0.069 radians$$
$$\Delta B=-3A\times 11.1\frac{mT}{A}=-33.3mT$$
$$\frac{\Delta\theta}{\Delta B}=\frac{-0.069radians}{-33.3mT}=0.00207\frac{radians}{mT}=2.07\frac{radians}{T}$$
$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1m}\times 2.07\frac{radians}{T}=20.7\frac{radians}{T*m}$$

2. Slope Fit\newline
As shown in Figure 2:
$$V=0.000443B+0.127$$
Therefore:
$$\frac{\Delta V}{\Delta B}=0.000443\frac{V}{mT}=0.443\frac{V}{T}$$
As shown in Figure 1: 
$$V=\frac{0.021\times (1+cos(\frac{2\pi\times(\theta-162)}{180}))}{2}$$
Taking the derivative at $\theta=117$ degrees:
$$\frac{\Delta V}{\Delta \theta}=\frac{0.21}{2}\times(-sin(2\frac{\pi}{180}(-45)))=0.21V$$
$$\frac{\Delta\theta}{\Delta B}=\frac{\Delta V}{\Delta B} \times \frac{\Delta \theta}{\Delta V}=0.443\frac{V}{T}\times\frac{1 radian}{0.21V}=2.1095\frac{radians}{T}$$
$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1}\times2.1095\frac{radians}{T}=21.095\frac{radians}{T*m}$$

3. Lock-in Method
$$I_{rms}=0.6137\times V_{rms}$$
$$B=11.1\times\frac{mT}{A}\times I$$
$$Gain=G_{preamplifier}\times G_{filter}\times G_{lock-in-amplifier}\times G_{lowpass filter}=500\times2\times10\times1=10000$$
$$V_{Photodiode}=\frac{V_{output}\times1.11}{10000}$$
The main method is to change the magnetic field and then to observe the corresponding voltage output from photodiode.30mV, 40mV, 50mV and 60mV have been used to supply the magnetic field, and these are all rms values. Calculation method is shown above, and detailed numeric calculation is shown in the appendix.

\section{Discussion}

your discussion here

\section{Conclusion}

your conclusion here. One is that my textbook\cite{Melissinos_2003} is really helpful and I should consult it more often. 

\section{Conclusion}

time for a conclusion

\appendix %\appendix*   % Omit the * if there's more than one appendix.

\section{Where to find data files, figures, tables, and notes}
where is everything?

\begin{acknowledgments}
who are your thankful for?
\end{acknowledgments}

\selectlanguage{english}
\FloatBarrier
\bibliographystyle{plain}
\bibliography{bibliography/converted_to_latex.bib%
}

\end{document}

