```
\documentclass[a4paper]{article}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{esvect}
\usepackage{graphicx,caption}%
\usepackage[colorinlistoftodos]{todonotes}
\usepackage[]{comment}
\usepackage{booktabs}
% \usepackage{booktabs}
\usepackage{graphicx}
\title{Analyzing elliptically polarized light using Fresnel Rhomb}
\author{Experiment-1}
\date{17-01-2017}
\begin{document}
\maketitle
\begin{abstract}
The aim of the experiment is to study elliptically polarized light using a Fresnel rhomb and to determine the ratio of semi major axis to the semi minor axis of the various elliptically polarised light forms for different angles of plane polarised incidence at the rhomb
\end{abstract}
\section{Apparatus used:}
\label{sec:apparatus}
\begin{enumerate}
\item He-Ne Laser source
\item Fresnel Rhomb
\item Polarising and analysing Polaroids
\item Photo-diode
\item Micro ammeter
\item Suitable optical bench with uprights to hold the accessories
\end{enumerate}
\section{Theory}
\subsection{Polarisation}
Light wave is a transverse electromagnetic wave made up of mutually perpendicular, oscillating electric and magnetic fields perpendicular to each other.
\\
\\
When a beam of randomly polarised light is incident on the surface of an anisotropic crystal \footnote{An anisotropic crystal is one in which properties like conductivity, velocity of light, refractive index etc. depend on the direction along which the property is measured}, it separates into two rays- the Ordinary Ray(o-ray) and the Extraordinary Ray(e-ray). They are linearly polarised in mutually perpendicular directions. This phenomenon is called Birefringence or Double Refraction.
\\
\\
The e-ray and the o-ray undergo superposition after emerging from an anisotropic crystal. They combine to produce different states of polarisation depending upon their optical path difference:\\
\\
1. When the optical path difference is 0 or an even or odd multiple of $\lambda$ /2, the resultant light wave is linearly polarised.\\
2. When the optical path difference is $\lambda$ /4, and the wave amplitudes are equal, the resultant light wave is circularly polarised.\\
3. For any other path difference, the resultant light wave is elliptically polarised.\\
\begin{comment}
\subsection{Jones Vector}
The Jones vector describes the polarization of light in free space or another homogeneous isotropic non-attenuating medium, where the light can be properly described as transverse waves. Suppose that a monochromatic plane wave of light is travelling in the positive z-direction, with angular frequency $\omega$ and wavevector k = (0,0,k), where the wavenumber k = $\omega/c$. Then the electric and magnetic fields E and H are orthogonal to k at each point; they both lie in the plane "transverse" to the direction of motion. Furthermore, H is determined from E by 90-degree rotation and a fixed multiplier depending on the wave impedance of the medium. So the polarization of the light can be determined by studying E. The complex amplitude of E is written
\begin{figure}[h!]
\begin{center}
\includegraphics[width=100mm,scale=.6]{JV1.JPG}
\end{center}
\end{figure}
Note that the physical E field is the real part of this vector; the complex multiplier serves up the phase information. Here i is the imaginary unit with $i^{2}=-1 $.
The Jones vector is then
\begin{figure}[h!]
\begin{center}
\includegraphics[width=20mm,scale=.3]{JV2.JPG}
\end{center}
\end{figure}
\end{comment}
\subsection{Elliptical Polarisation}
Elliptical polarization is the polarization of electromagnetic radiation such that the tip of the electric field vector describes an ellipse in any fixed plane intersecting, and normal to, the direction of propagation. An elliptically polarized wave may be resolved into two linearly polarized waves, with their polarization planes at right angles to each other.
\begin{table}[h!]
\begin{center}
\includegraphics[width=85mm,scale=.6]{ellipse.png}
\end{center}
\end{table}
At a fixed point in space (or for fixed z), the electric vector $\vv{E}$ traces out an ellipse in the x-y plane. The semi-major and semi-minor axes of the ellipse have lengths a and b. While the maximum amplitude of the $ \vv{x}$ and $ \vv{y}$ components of $\vv{E}$ are $ E_{xo}$ and $E_{yo}$. The major axis of the ellipse is inclined at an angle $\alpha
$ with respect to $ \vv{x}$ .
\subsection{Fresnel Rhomb}
A Fresnel Rhomb is a prism designed for converting linearly polarized light into elliptically polarized light. The device works based on the relative phase shift between the the two linearly polarized components of the incident light as a result of total internal reflection, two of which each contribute a nominal retardance \footnote[1]{Nominal Retardance - The difference in phase shift between two characteristic polarizations of light upon reflection from an interface.} of $45^{o}$ . The rhomb has interior angles of $54^{o}$ and $126^{o}$.
\begin{table}[h!]
\begin{center}
\includegraphics[width=55mm,scale=.6]{rho.png}
\end{center}
\end{table}
\begin{comment}
\subsection{Strokes Parameters}
The Stokes parameters are a set of values that describe the polarization state of electromagnetic radiation
The relationship of the Stokes parameters S0, S1, S2, S3 to intensity and polarization ellipse parameters is shown in the equations below and the figure at right.
\begin{table}[h!]
\begin{center}
\includegraphics[width=40mm,scale=.6]{ssss.JPG}
\end{center}
\end{table}
\begin{table}[h!]
\begin{center}
\includegraphics[width=55mm,scale=.6]{Polarisationellipse.png}
\end{center}
\end{table}
Here $I_{p},2\Psi$ and $2\chi$ are the spherical coordinates of the three-dimensional vector of cartesian coordinates $(S_{1},S_{2},S_{3})$. I is the total intensity of the beam, and p is the degree of polarization, constrained by $0 \leq p \leq 1$. The factor of two before $ \Psi$ represents the fact that any polarization ellipse is indistinguishable from one rotated by $180^{o}$, while the factor of two before $\chi$ indicates that an ellipse is indistinguishable from one with the semi-axis lengths swapped accompanied by a $90^{o}$ rotation. The phase information of the polarized light is not recorded in the stokes parameters. The four Stokes parameters are sometimes denoted I, Q, U and V, respectively.
If given the Stokes parameters one can solve for the spherical coordinates with the following equations:
\begin{table}[h!]
\begin{center}
\includegraphics[width=55mm,scale=.6]{I.JPG}
\end{center}
\end{table}
\end{comment}
\subsection{Production of Elliptical Polarisation }
Elliptically polarised light can be produced when two mutually perpendicular plane polarised waves of unequal amplitude and any phase difference are superimposed.
\\
\begin{table}[h!]
\begin{center}
\includegraphics[width=100mm,scale=.6]{set.JPG}
\end{center}
\end{table}
\\
As shown in the figure, laser is first passed through a polariser, the plane polarised light which comes out of the polariser is incident normally on the Fresnel rhomb in such a way that the incident light undergoes two internal reflections before transmitting perpendicularly on a parallel surface. The phase change for the internally reflected light happens to be a function of the angle of incidence. For the material used in Fresnel Rhomb (n=1.51), a phase change of $45^{o}$ happens at two angles of incidences $48^{o} 37' $ and $54^{o} 37'$. Fresnel constructed the rhomb such that both the internal reflections occur at an angle of $54^{o} 37'$ which contributes to a total phase change of $\pi /2$. Thus an incident linearly polarised light can be transmitted as an elliptically polarised light.
\\
\begin{table}[h!]
\begin{center}
\includegraphics[width=120mm,scale=.6]{EP.JPG}
\end{center}
\end{table}
\\
When this elliptically polarised light is made to fall on the analyser, it filters the beam such that components of only a specific polarization is transmitted. When the analyser is rotated about the transmitted axis, intensity of the transmitted light changes from maximum to minimum, according to whether the principal plane of the analyser is parallel or perpendicular to the major axis of the elliptical polarization. The light transmitted from the analyser is made to fall on a photocell which converts the optical energy into electrical energy and the current through the photocell gives a measure of the intensity transmitted by the analyser.
\\
\\
Thus the variations in the intensity of the light for different orientations of the analyser can be studied fairly accurately with the help of the photocell.
\subsection{Superposition of waves linearly polarised at right angles}
\begin{comment}
\end{}
We consider two light waves travelling in the same direction, z; one wave is polarised in the xy-plane and the other is polarised in the yz-plane. Let the two orthogonal waves be represented by
\begin{center}
\[E_x = E_{xo} cos(kz - \omega t)\]
\[E_y = E_{yo} cos(kz - \omega t + \delta)\]
\end{center}
The waves are of the same frequency, $\delta$ is the phase difference between the waves. At a given time t, the optical vectors $E_x$ and $E_y$ produce a resultant optical vector of magnitude, say A and a later time they give rise to a resultant vector of magnitude B which points in a different direction. With the progress of time, tip of the resultant optical vector moves along curve in the xy-plane. Applying the priciple of superposition to find the equation of the curve ttraced by the resultant of the two vectors.
\\
\\
\[E = E_x + E_y\]
\[E = E_{xo} cos(kz - \omega t) + E_{yo} cos(kz - \omega t+\delta) \]
\\
\\
The equation of the curve may be found by eliminating 't' from the equations. Thus the equations can be written as \\
\[E_y = E_{yo}cos(kz - \omega t) cos \delta - E_{oy}sin(kz - \omega t)sin \delta\]
\[cos(kz - \omega t) = \frac{E_x}{E_{ox}}\] \\
\[E_y = E_{oy} \times \frac{E_x}{E_{ox}} cos \delta \pm \sqrt{1 - \frac{E_x^2}{E_{ox}^2}} E_{oy} sin \delta\]
Solving and rearranging the terms we obtain :
\[\frac{E_x^2}{E_{ox}^2} + \frac{E_y^2}{E_{oy}^2} - 2 \frac{E_xE_y}{E_{ox}E_{oy}} cos\delta = sin^2\delta\]
This is the general equation of an ellipse. Hence the tip of the resultant electric field vector traces an ellipse in the xy-plane. The major axis makes an angle $\alpha$ with the x axis.
\[tan(2\alpha) = 2E_{xo}E_{yo} \frac{cos\delta}{E_{xo}^2 - E_{yo}^2}\]
Let the elliptical vibration be represented by the equation,
\[\frac{{\left( {x} \right)^2 }}{{a^2 }} + \frac{{\left( {y } \right)^2 }}{{b^2 }} = 1\]
\end{comment}
Plane of incidence : x-z
\newline
Let the incident polarisation be inclined at an angle $\phi$ with the plane of incidence. Then,
\begin{equation}
\vec{E_{inc}} = E_{o}\cos\phi\cos\omega t \hat{i} + E_{o}\sin\phi\cos\omega t \hat{j}
\end{equation}\\
On passing through the Fresnel rhomb, a phase difference of $\frac{\pi}{2}$ is introduced between the two components of the incident polarisation vector. Hence, the transmitted electric field vector can be written as
\begin{equation}
\vec{E_{trans}} = E_{o}\cos\phi\cos\omega t \hat{i} + E_{o}\sin\phi\sin\omega t \hat{j}
\end{equation}
\begin{equation}
\vec{E_{trans}} = E_{x}\hat{i} + E_{y}\hat{j}
\end{equation}
where,
\begin{equation}
\vec{E_{x}} = E_{o}\cos\phi\cos\omega t
\end{equation}
\begin{equation}
\vec{E_{y}} = E_{o}\sin\phi\cos\omega t
\end{equation}
Thus,
\begin{equation}
\frac{E_{x}}{E_{o}\cos\phi}^2 + \frac{E_{y}}{E_{o}\sin\phi}^2 = 1
\end{equation}\\
The above is the equation of an ellipse, of the form
\begin{equation}
\frac{x}{a}^2 + \frac{y}{b}^2 = 1
\end{equation}
where,
\begin{equation}
x = \frac{E_{x}}{E_{o}} ,\space y = \frac{E_{y}}{E_{o}},\space a = \cos\phi
\space and\space b = \sin\phi
\end{equation}\\
Place an analyser at an angle $\theta$ with respect to x. Where $\hat{n}$ represents the axis of the analyser.
\begin{equation}
\hat{n} = cos\theta \hat{i} + sin\theta \hat{j}
\end{equation}\\
Component of transmitted Electric field is :
\begin{equation}
E_n = \vec{E_{transmitted}}.\hat{n}
= E_0cos\phi cos\theta coswt + E_0sin\phi sin\theta sinwt
\end{equation}
\begin{equation}
\begin{split}
{E_n}^2 & = {E_0}^2[cos^2\phi cos^2\theta cos^2wt + sin^2\phi sin^2\theta sin^2wt \\
& + 2cos\phi sin\phi sin\theta cos\theta coswt sinwt]
\end{split}
\end{equation}
\begin{equation}
I \propto < {E_n}^2 >
\end{equation}
\newline
Intensity is proportional to the time average of the $E_n$
\begin{equation}
<{E_n}^2> = \frac{{E_0}^2}{2} [cos^2\phi * cos^2\theta + sin^2\phi sin^2\theta]
\end{equation}
\begin{equation}
<{E_n}^2> = \frac{{E_0}^2}{2} [cos2\phi * cos^2\theta + sin^2\phi ]
\end{equation}
\begin{equation}
I = I_0 [cos2\phi * cos^2\theta + sin^2\phi ]
\end{equation}
\newline
Varying $\theta$ over 360º would produce a variation in intensity
\begin{equation}
I_{max} : \theta = 0,\pi
\end{equation}
\begin{equation}
I_{max} = I_0 [cos2\phi + sin^2\phi]
\end{equation}
\begin{equation}
I_{max} = I_0 [cos^2\phi - sin^2\phi + sin2\phi]
\end{equation}
\begin{equation}
I_{max} = I_0 cos^2\phi
\end{equation}
\begin{equation}
I_{min} : \theta = \pi/2, 3\pi/2
\end{equation}
\begin{equation}
I_{min} = I_0 sin^2\phi
\end{equation}
\begin{equation}
\frac{I_{min}}{I_{max}}=tan^2 \phi
\end{equation}
\begin{comment}
\end{}
In the parametric form of the ellipse, x=acos$\theta$ and y=bsin$\theta$
Let the axis of transmission PP through the analyser be inclined at an angle $\theta$ with the major axis of the ellipse as shown. Then the transmitted components will be $a cos\theta$ and $b \sin\theta$. Since the two components are $90^{o}$ out of phase.
\begin{center}
\[ I = a^2 cos ^2\theta + b^2 sin ^2\theta + 2 ab sin \theta cos \theta \cos 90 \]
\[ = a^2 cos ^2\theta + b^2 sin ^2\theta \]
\[ = a^2 cos ^2\theta + b^2 ( 1- cos ^2\theta ) \]
\[ = (a^2-b^2) cos ^2\theta + b^2 \]
\[I = Imax = a^2 \quad where \quad \theta = 0 \]
\[I = Imin = b^2 \quad where \quad \theta = \pi /2 \]
\\
\[\frac{Imax}{Imin} = (\frac{a}{b})^2 \]
\end{center}
\end{comment}
Thus, knowing $I_{max}$ and $I_{min}$ we can find the polarizer angles for the respective observation sets.
\section{Precautions}
\begin{enumerate}
\item Care should be taken to see that no external light other than the light from the analyser enters the photo-diode.
\item The height of the uprights should be properly adjusted so that the light from the source reaches the photo-diode after passing through the polariser, rhomb and the analyser.
\item While taking observations, the analyser should be rotated in one direction only to avoid backlash error.
\end{enumerate}
\section{Procedure}
\begin{enumerate}
\item Level the bed of the optical bench with the help of a spirit level and the screws provided. Switch on the laser source and adjust the height of the uprights so that the light from the source enters the polariser, rhomb, analyser and then the photo-diode.
\item Rotate the polariser and analyser to read zero on the respective circular scales. By rotating the analyser, see that the intensity of the transmitted light indicated by the micro ammeter reading varies between maximum and minimum on one full rotation, but is never zero.
\item Keeping the polariser at a fixed $\theta$, rotate the analyser from 0º to 360º in steps of 10º and note the corresponding reading in the digital microammeter.
\item Now, rotate the polariser and note the readings in the digital microammeter.
\item Repeat the observations for different values of polariser angles.
\end{enumerate}
\section{Data analysis}
\begin{enumerate}
\item Plot a polar graph between the orientation of the analyser and the corresponding microammeter reading for each orientation of the polariser.
\end{enumerate}
\begin{comment}
\section{Calcuations}
% Please add the following required packages to your document preamble:
% \usepackage{booktabs}
% \usepackage{graphicx}
\begin{table}[h!]
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{@{}cccccc@{}}
\toprule
\multicolumn{6}{c}{\textbf{Calculation}} \\ \midrule
\textbf{} & & & & & \\
\textbf{\begin{tabular}[c]{@{}c@{}}Polariser Angle - Observation\\ (Deg)\end{tabular}} & 50 & 78 & 109 & 226 & 260 \\
\textbf{Imax (μA)} & 210 & 169.2 & 86.4 & 31.6 & 192.6 \\
\textbf{Imin (μA)} & 85.3 & 12.8 & 5.9 & 24.2 & 8.9 \\
\textbf{} & & & & & \\
\textbf{\begin{tabular}[c]{@{}c@{}}Polariser Angle by ratio of\\ intensities (Rad)\end{tabular}} & 0.567 & 0.268 & 0.256 & 0.719 & 0.212 \\
\textbf{\begin{tabular}[c]{@{}c@{}}Polariser Angle by ratio of\\ intensities (Deg)\end{tabular}} & 32.512 & 15.379 & 14.645 & 41.191 & 12.132 \\
& & & & & \\
\textbf{\begin{tabular}[c]{@{}c@{}}Polariser angle after\\ correction (Deg)\end{tabular}} & 57.488 & 74.621 & 104.645 & 221.191 & 257.868 \\ \bottomrule
\end{tabular}%
}
\end{table}
\end{table}
\end{comment}
\section{Result}
The elliptically polarized light from the Fresnel's Rhomb was studied for different polarizing angles and the difference between polarizing angles was found using the ratio of the minimum and maximum intensities was calculated.
\begin{comment}
Here the ratio of a/b and the eccentricities of the ellipses of polarizer angles 50º, 72º, 109º and 226º are tabulated
\begin{table}[!hbt]
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
Polarizer Angle & From gnu && From $I_{max}$ and $I_{min}$ &\\
\hline
& a/b & $\epsilon$ & a/b & $\epsilon$ \\
\hline
50º & 1.059 & 0.328 & 1.569 & 0.771\\
\hline
78º & 2.865 & 0.937 & 3.636 & 0.961\\
\hline
109º & 3.103 & 0.947 & 3.827 & 0.965\\
\hline
226º & 1.009 & 0.136 & 1.143 & 0.484\\
\hline
\end{tabular}
\end{center}
\end{table}
\section{Discussion - Data analysis}
\subsection{Documentation - gunplot, Fitting an ellipse to the data set}
The ellipse was constructed using nonlinear least-squares Marquardt-Levenberg algorithm via gnuplot. The theory of non-linear least-squares (NLLS) is generally described in terms of a normal distribution of errors, that is, the input data is assumed to be a sample from a population having a given mean and a Gaussian (normal) distribution about the mean with a given standard deviation. Here an iteration creates an ellipse who's parameters are redefined based on the chisquare test. In a nonlinear system, the derivatives used in minimizing the least square distance are functions of both the independent variable and the parameters, so these gradient equations do not have a closed solution. Instead, initial values must be chosen for the parameters. Then, the parameters are refined iteratively. For a sufficiently large size, and knowing the population standard deviation, one can use the statistics of the chisquare distribution to describe a "goodness of fit".\\
Here, it is sufficient to say that a reduced chisquare (chisquare/degrees of freedom, where degrees of freedom is the number of datapoints less the number of parameters being fitted) of 1.0 is an indication that the weighted sum of squared deviations between the fitted function and the data points is the same as that expected for a random sample from a population characterized by the function with the current value of the parameters and the given standard deviations. \\
\subsection{ Asymptotic Standard error}
Asymptotic standard error is an approximation to the standard error of the sample-mean's estimate of parameter (a,b), based on mathematical simplification.
The Central Limit Theorem states that the mean of n samples taken from independent identically distributed random numbers with finite variance converges in the limit n tends to infinity, to a normal distribution. The theorem doesn't guarantee that the means of a finite sample are normally distributed, but we calculate the standard error of the mean under the simplifying assumption that the means are normally distributed hence the term "Asymptotic Standard Error".
\end{comment}
\section{Sources of Error}
\begin{enumerate}
\item The Fresnel Rhomb used in the experiment section had a narrow range over which a transmitted ray underwent two internal reflections, when the light beam was incident normal to the plane. Thus, there was a huge loss in the maximum intensity derivable from this set-up because the rhomb was not able to accommodate the entire laser cross section. Thus the laser was incident at an angle to the normal of the plane in a way that it was able to transmit maximum intensity. This would lead to an error, as it deviates from the analysis employed which requires normal incidence.
\item The ability to accurately measure the rotation of the analyzer. The analyzer could slip back and forth in its holder such that even when a certain value’s tick mark was lined up with the reference mark the actual tilt could be several degrees different.
\item The error due to improper alignment of the apparatus.
\end{enumerate}
\end{document}
```