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| %!TEX TS-program = xelatex | |
| %!TEX encoding = UTF-8 Unicode | |
| \documentclass[letterpaper, 12pt]{exam} | |
| \usepackage{amsmath,amsfonts,amssymb} | |
| \usepackage{boxedminipage} | |
| \def\letvar{\mbox{let }} | |
| \def\un #1{\mathrm{\;#1}} | |
| \def\ee #1{\ensuremath{\times 10^{#1}}} | |
| \def\vunit {\un{m\,s^{-1}}} | |
| \def\aunit {\un{m\,s^{-2}}} | |
| \def\iunit {\un{W\,m^{-2}}} | |
| \def\deg {\un{^\circ}} | |
| \title{\Large\sc polarization assessments} | |
| \author{Jonathan M. Sterling} | |
| \date{} | |
| \begin{document} | |
| \maketitle | |
| \thispagestyle{empty} | |
| \section{Notes} | |
| \subsection*{Assessment 11.5.1} | |
| \emph{Describe what is meant by polarized light.} | |
| Transverse waves can be polarized, which means that all oscillations are in the same plane. Another way to put this is that light generally is unpolarized: the direction of its oscillation is not correlated. In a wave, oscillation can occur in the $x$-direction, $y$-direction and $z$-direction. \textbf{When light is polarized, its vibrations are forced into the same direction.} | |
| \subsection*{Assessment 11.5.2} | |
| \emph{Describe polarization by reflection.} | |
| Reflected light is partially polarized such that oscillations are (to a certain degree) parallel to the surface of reflection, depending upon its nature. Light reflected off metallic surfaces is unpolarized. Light reflected off water and other non-metallic substances is polarized such that a large quantity of the wave oscillations occur parallel to the water-surface. Polarized sunglasses are used to block this partially-polarized light. | |
| \subsection*{Assessment 11.5.3} | |
| \emph{State and apply Brewster's law.} | |
| There is an angle of incidence such the reflected light is completely polarized, and that the reflected and refracted light are perpendicular to each other. The angle $\theta_i$ between the incident light and the normal to the boundary is equal to the angle between the reflected light and the normal to the boundary. | |
| \begin{align*} | |
| \theta_i + \theta_r = 90^\circ \implies \theta_r &= 90^\circ - \theta_i\\ | |
| n_i\sin\theta_i &= n_r\sin\theta_r\\ | |
| n_i\sin\theta_i &= n_r\sin(90^\circ - \theta_i)\\ | |
| n_i\sin\theta_i &= n_r\cos\theta_i\\ | |
| \tan\theta_i &= \frac{n_r}{n_i} | |
| \end{align*} | |
| \subsubsection*{Example} | |
| Determine the Brewster Angle for light that reflects against vegetable oil at $50\,^\circ C$ under water. | |
| \begin{align*} | |
| n_i &= 1.33\\ | |
| n_r &= 1.47\\ | |
| \theta_b &= \arctan\left(\frac{n_r}{n_i}\right)\\ | |
| &= \arctan\left(\frac{1.47}{1.33}\right)\\ | |
| \theta_b &\approx 47.9^\circ | |
| \end{align*} | |
| \subsection*{Assessment 11.5.4} | |
| \emph{Explain the terms \textsc{polarizer} and \textsc{analyzer}.} | |
| A \textsc{polarizer} is a device that takes an unpolarized wave (i.e. a wave with irregular oscillations) and forces its oscillations into the same plane. A polarizer could be a fine wire-grid. The transmitted intensity $I$ of light with initial intensity $I_0$ after passing through a polarizer is represented by Malus' law: | |
| \[ I = I_0 \cos^2\theta_i \] | |
| When two polarizers are placed in a row, the second one is called an \textsc{analyzer}. The intensity of the light that passes through the analyzer also follows Malus' law. | |
| \subsection*{Assessment 11.5.5} | |
| \emph{Calculate the intensity of a transmitted beam of polarized light using Malus' law.} | |
| A light wave polarized at $30^\circ$ with initial intensity $0.9\mathrm{cd}$ passes through a polarizer with polarization angle $50^\circ$. What is the intensity of the transmitted beam? | |
| \begin{align*} | |
| \theta_i &= 50^\circ - 30^\circ = 20^\circ\\ | |
| I &= 0.9\mathrm{cd} \cdot \cos^2 20^\circ\\ | |
| I &\approx 0.8\mathrm{cd} | |
| \end{align*} | |
| \subsection*{Assessment 11.5.6} | |
| \emph{Describe what is meant by an optically active substance.} | |
| An optically active substance is one that can change the plane of polarization in a light-wave. This comes about from the molecular structure of the material. | |
| \subsection*{Assessment 11.5.7} | |
| \emph{Describe the use of polarization in the determination of the concentration of certain solutions.} | |
| The optical activity of a substance depends upon its molecular study. Therefore, measures of the polarization-properties of a solution can help to reveal the molecular structure of the solute. | |
| \subsection*{Assessment 11.5.8} | |
| \emph{Outline qualitatively how polarization may be used in stress analysis.} | |
| Stress analysis can sometimes be done geometrically, but some surfaces make this overly complex. An experimental method to analyze stress is called \textsc{photoelastic analysis}. | |
| When light enters the material, it experiences two different refractive indices (this property is called \textsc{birefringence}). The way that the light is divided and refracted creates colorful fringe patterns that we can see and analyze. | |
| \subsection*{Assessment 11.5.9} | |
| \emph{Outline qualitatively the action of liquid-crystal displays (LCDs).} | |
| A pixel of an LCD is a cell of aligned liquid crystal between two transparent plates, between a crossed polarizer and analyzer (that is, their axes of transmission are perpendicular to each other). When there is no electric field applied, the plane of polarization twists as the light passes through the crystals (they are arranged as a helix); when it emerges, it is polarized parallel to the transmission axes of the analyzer, and so exits the the cell appearing light. When an electric field is applied, the liquid crystals are aligned with each other such that when the light emerges, it is still polarized perpendicular to the analyzer, and thus is not transmitted. This creates a dark spot. | |
| %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% | |
| %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% | |
| \section{Problems} | |
| \begin{questions} | |
| \printanswers | |
| \question | |
| An unpolarized beam of light is incident on a Polaroid filter. The emerging beam then travels through a second Polaroid filter. The first filter has its transmission axis at $28\deg$ to the vertical, the second at $44\deg$ to the vertical. State the angle of polarization of the beam emerging from the second Polaroid, giving your answer in degrees to the vertical. | |
| \begin{solution} | |
| The light should emerge at the angle of transmission for the last polarizer, $\implies 44\deg$. | |
| \end{solution} | |
| \question | |
| A beam of polarized light of intensity $4.7\iunit$ is incident on a sheet of Polaroid. If the transmission axis of the Polaroid makes an angle of $20\deg$ with the plane of polarization of the incident beam, calculate the intensity in $\mathrm{W\,m^{-2}}$ of the transmitted beam. | |
| \begin{solution} | |
| Using Malus's Law: | |
| \begin{align*} | |
| I &= I_0\cos^2\theta_i\\ | |
| I_0 &= 4.7\iunit\\ | |
| \theta_i &= 20\deg\\ | |
| \therefore I &= (4.7\iunit)\cos^2(20\deg)\\ | |
| I &\approx 4.2\iunit | |
| \end{align*} | |
| \end{solution} | |
| \question | |
| Three Polaroid filters are lined up along the $x$-axis. The first has its transmission axis aligned in the $y$-direction, the second has its axis at $45\deg$ in the $y,z$ plane, and the third has its axis in the $z$-direction. A beam of light traveling in the $x$-direction has initial intensity $5.6\iunit$, and is polarized in the $y$-direction. The beam passes through each filter as it travels along the $x$-axis. | |
| \begin{parts} | |
| \part Calculate the intensity in $\mathrm{W\,m^{-2}}$ of the transmitted beam emerging from the third filter. | |
| \begin{solution} | |
| \begin{align*} | |
| I_0 &= 5.6\iunit\\ | |
| I_1 &= I_0\cos^2(0\deg) &= 5.6\iunit\\ | |
| I_2 &= I_1\cos^2(45\deg - 0\deg) &= 2.8\iunit\\ | |
| I_3 &= I_2\cos^2(90\deg - 45\deg) &= 1.4\iunit | |
| \end{align*} | |
| The final intensity is $1.4\iunit$. | |
| \end{solution} | |
| \part In which direction is the emerging beam polarized? | |
| \begin{choices} | |
| \choice In the $y$-direction. | |
| \choice In the $x$-direction. | |
| \CorrectChoice In the $z$-direction. | |
| \choice At $45\deg$ in the $y,z$ plane. | |
| \end{choices} | |
| \end{parts} | |
| \question | |
| Calculate the angle required between the incident plane of polarization and the transmission axis of a sheet of Polaroid to reduce the intensity of a beam of polarized light from $7.2\iunit$ to $3.9\iunit$. | |
| \begin{solution} | |
| \begin{align*} | |
| I = I_0\cos^2\theta_i \implies \theta_i &= \arccos\sqrt{\frac{I}{I_0}}\\ | |
| &= \arccos\sqrt{\frac{3.9}{7.2}}\\ | |
| \theta_i &\approx 43\deg | |
| \end{align*} | |
| \end{solution} | |
| \question | |
| The polarizing angle for a particular glass is $58.2\deg$. Calculate the refractive index of the glass. | |
| \begin{solution} | |
| \begin{align*} | |
| \theta_B &= 58.2\deg, \quad n_1 = 1\\ | |
| \theta_B = \arctan\left(\frac{n_2}{n_1}\right) \implies n_2 &= n_1\tan\theta_B\\ | |
| &= \tan58.2\deg\\ | |
| n_2 &\approx 1.61 | |
| \end{align*} | |
| \end{solution} | |
| \question | |
| Calculate the polarizing angle for a beam of light traveling in water ($n_w = 1.33$) incident on a block of leaded glass ($n_g = 1.54$). | |
| \begin{solution} | |
| \begin{align*} | |
| \theta_B &= \arctan\left(\frac{n_g}{n_w}\right)\\ | |
| &= \arctan\left(\frac{1.54}{1.33}\right)\\ | |
| \theta_B &\approx 49.2\deg | |
| \end{align*} | |
| \end{solution} | |
| \question | |
| Light traveling in air is partially reflected from a glass block. The reflected light is found to be 100\% polarized when the angle of incidence is $57.5\deg$. | |
| \begin{parts} | |
| \part State the magnitude of the angle between the reflected and refracted beams, in degrees. | |
| \begin{solution} | |
| The angle between the reflected and refracted beams must be $90\deg$. | |
| \end{solution} | |
| \part Calculate the refractive index of the block. | |
| \begin{solution} | |
| \begin{align*} | |
| \theta_B = \arctan\left(\frac{n_2}{n_1}\right)\implies | |
| n_2 &= n_1\tan\theta_B\\ | |
| &= \tan57.5\deg\\ | |
| n_2 &\approx 1.57 | |
| \end{align*} | |
| The refractive index of the block is $1.57$. | |
| \end{solution} | |
| \end{parts} | |
| \question | |
| A beam of light is 100\% polarized by reflection from the surface of a swimming pool. The reflected beam is then passed through a Polaroid. The transmission axis of the Polaroid makes an angle of $34.5\deg$ with the normal to the plane of polarization of the reflected beam. If the reflected beam has intensity $6.75\iunit$, calculate the intensity of the beam transmitted by the Polaroid. | |
| \begin{solution} | |
| \begin{align*} | |
| \theta_i = 90\deg - 34.5\deg &= 55.5\deg\\ | |
| I &= I_0\cos^2\theta_i\\ | |
| &= (6.75\iunit)\cos^2 55.5\deg\\ | |
| I &\approx 2.16\iunit | |
| \end{align*} | |
| \end{solution} | |
| \end{questions} | |
| \newpage | |
| \section{Made-up Problem} | |
| There is a beam of light following the $x$-direction, polarized in the $y$-direction, passing through one polarizer every $5\un{cm}$. The transmission angles of the polarizers on the $y,z$ plane are represented by the sequence $\{\frac{\pi}{3},0,\frac{\pi}{3},0,\cdots\}$. Hence, deduce a general, non-recursive solution to the intensity of the light after the $n$th polarizer (where $n$ starts at $1$). Assume that the initial intensity was $1.0\iunit$. | |
| \printanswers | |
| \begin{solution} | |
| The first step is to deduce a sequence $\theta_n = \{\frac{\pi}{3},0,\frac{\pi}{3},0,\cdots\}$: | |
| \[ \theta_n = -\frac{\pi[(-1)^n - 1]}{6} \] | |
| The intensity of the beam after $n$ polarizers is a recursive function: | |
| \[ I_n = I_{n-1}\cdot\cos^2(\theta_n - \theta_{n-1}) \] | |
| This can be rewritten as | |
| \[ I_n = I_{n-2}\cdot\cos^2(\theta_{n-1} - \theta_{n-1})\cdot \cos^2(\theta_n - \theta_{n-2}) \] | |
| Essentially, the intensity after $n$ polarizers is the product (with index $i$) of the squares of the cosine of the difference between $\theta_i$ and $\theta_{i-1}$: | |
| \begin{align*} | |
| I_n &= \prod_{i=1}^n\cos^2(\theta_i - \theta_{i-1})\\ | |
| &= \prod_{i=1}^n\cos^2\left(\frac{-\pi([(-1)^i - 1] - [(-1)^{i-1} - 1]}{6}\right)\\ | |
| &= \prod_{i=1}^n\cos^2\left(\frac{-\pi[(-1)^i - (-1)^{i-1}]}{6}\right)\\ | |
| I_n &= \prod_{i=1}^n\cos^2\left(\frac{\pi[(-1)^{i-1} - (-1)^i]}{6}\right), \mbox{where } n\in\mathbb{Z}:n>0 | |
| \end{align*} | |
| \end{solution} | |
| \end{document} |
Describe what is meant by polarized light.
Transverse waves can be polarized, which means that all oscillations are in the same plane. Another way to put this is that light generally is unpolarized: the direction of its oscillation is not correlated.
In a wave, oscillation can occur in the x-direction, y-direction and z-direction. When light is polarized, its vibrations are forced into the same direction.
Describe polarization by reflection.
Reflected light is partially polarized such that oscillations are (to a certain degree) parallel to the surface of reflection, depending upon its nature. Light reflected off metallic surfaces is unpolarized.
Light reflected off water and other non-metallic substances is polarized such that a large quantity of the wave oscillations occur parallel to the water-surface. Polarized sunglasses are used to block this partially-polarized light.
State and apply Brewster's law.
There is an angle of incidence such the reflected light is completely polarized, and that the reflected and refracted light are perpendicular to each other.
The angle $\theta_i$ between the incident light and the normal to the boundary
is equal to the angle between the reflected light and the normal to the boundary.
\begin{align*}
\theta_i + \theta_r = 90^\circ \implies \theta_r &= 90^\circ - \theta_i\\
n_i\sin\theta_i &= n_r\sin\theta_r\\
n_i\sin\theta_i &= n_r\sin(90^\circ - \theta_i)\\
n_i\sin\theta_i &= n_r\cos\theta_i\\
\tan\theta_i &= \frac{n_r}{n_i}
\end{align*}Determine the Brewster Angle for light that reflects against vegetable oil at 50\,^\circ C under water.
\begin{align*}
n_i &= 1.33\\
n_r &= 1.47\\
\theta_b &= \arctan\left(\frac{n_r}{n_i}\right)\\
&= \arctan\left(\frac{1.47}{1.33}\right)\\
\theta_b &\approx 47.9^\circ
\end{align*}Explain the terms polarizer and analyzer.
A polarizer is a device that takes an unpolarized wave (i.e. a wave with irregular oscillations) and forces its oscillations into the same plane. A polarizer could be a fine wire-grid. The transmitted intensity $I$ of light with initial intensity $I_0$ after passing through a polarizer is represented by Malus' law:
\[ I = I_0 \cos^2\theta_i \]When two polarizers are placed in a row, the second one is called an analyzer. The intensity of the light that passes through the analyzer also follows Malus' law.
Calculate the intensity of a transmitted beam of polarized light using Malus' law.
A light wave polarized at $30^\circ$ with initial intensity $0.9\mathrm{cd}$ passes through a polarizer with polarization angle $50^\circ$. What is the intensity of the transmitted beam?
\begin{align*}
\theta_i &= 50^\circ - 30^\circ = 20^\circ\\
I &= 0.9\mathrm{cd} \cdot \cos^2 20^\circ\\
I &\approx 0.8\mathrm{cd}
\end{align*}Describe what is meant by an optically active substance.
An optically active substance is one that can change the plane of polarization in a light-wave. This comes about from the molecular structure of the material.
Describe the use of polarization in the determination of the concentration of certain solutions.
The optical activity of a substance depends upon its molecular study. Therefore, measures of the polarization-properties of a solution can help to reveal the molecular structure of the solute.
Outline qualitatively how polarization may be used in stress analysis.
Stress analysis can sometimes be done geometrically, but some surfaces make this overly complex. An experimental method to analyze stress is called photoelastic analysis.
When light enters the material, it experiences two different refractive indices (this property is called birefringence). The way that the light is divided and refracted creates colorful fringe patterns that we can see and analyze.
Outline qualitatively the action of liquid-crystal displays (LCDs).
A pixel of an LCD is a cell of aligned liquid crystal between two transparent plates, betwee a crossed polarizer and analyzer (that is, their axes of transmission are perpendicular to each other). When there is no electric field applied, the plane of polarization twists as the light passes through the crystals (they are arranged as a helix); when it emerges, it is polarized parallel to the transmission axes of the analyzer, and so exits the the cell appearing light. When an electric field is applied, the liquid crystals are aligned with each other such that when the light emerges, it is still polarized perpendicular to the analyzer, and thus is not trasmitted. This creates a dark spot.
| %!TEX TS-program = xelatex | |
| %!TEX encoding = UTF-8 Unicode | |
| \documentclass[letterpaper, 12pt]{exam} | |
| \usepackage{amsmath,amsfonts,amssymb} | |
| \usepackage{boxedminipage} | |
| \def\letvar{\mbox{let }} | |
| \def\un #1{\mathrm{\;#1}} | |
| \def\ee #1{\ensuremath{\times 10^{#1}}} | |
| \def\vunit {\un{m\,s^{-1}}} | |
| \def\aunit {\un{m\,s^{-2}}} | |
| \def\iunit {\un{W\,m^{-2}}} | |
| \def\deg {\un{^\circ}} | |
| \title{\Large\sc made-up problem} | |
| \author{Jonathan M. Sterling} | |
| \date{} | |
| \begin{document} | |
| \maketitle | |
| \thispagestyle{empty} | |
| \begin{questions} | |
| \printanswers | |
| \question | |
| There is a beam of light following the $x$-direction, polarized in the $y$-direction, passing through one polarizer every $5\un{cm}$. The transmission angles of the polarizers on the $y,z$ plane are represented by the sequence $\{\frac{\pi}{3},0,\frac{\pi}{3},0,\cdots\}$. Hence, deduce a general, non-recursive solution to the intensity of the light after the $n$th polarizer (where $n$ starts at $1$). Assume that the initial intensity was $1.0\iunit$. | |
| \begin{solution} | |
| The first step is to deduce a sequence $\theta_n = \{\frac{\pi}{3},0,\frac{\pi}{3},0,\cdots\}$: | |
| \[ \theta_n = -\frac{\pi[(-1)^n - 1]}{6} \] | |
| The intensity of the beam after $n$ polarizers is a recursive function: | |
| \[ I_n = I_{n-1}\cdot\cos^2(\theta_n - \theta_{n-1}) \] | |
| This can be rewritten as | |
| \[ I_n = I_{n-2}\cdot\cos^2(\theta_{n-1} - \theta_{n-1})\cdot \cos^2(\theta_n - \theta_{n-2}) \] | |
| Essentially, the intensity after $n$ polarizers is the product (with index $i$) of the squares of the cosine of the difference between $\theta_i$ and $\theta_{i-1}$: | |
| \begin{align*} | |
| I_n &= \prod_{i=1}^n\cos^2(\theta_i - \theta_{i-1})\\ | |
| &= \prod_{i=1}^n\cos^2\left(\frac{-\pi([(-1)^i - 1] - [(-1)^{i-1} - 1]}{6}\right)\\ | |
| &= \prod_{i=1}^n\cos^2\left(\frac{-\pi[(-1)^i - (-1)^{i-1}]}{6}\right)\\ | |
| I_n &= \prod_{i=1}^n\cos^2\left(\frac{\pi[(-1)^{i-1} - (-1)^i]}{6}\right), \mbox{where } n\in\mathbb{Z}:n>0 | |
| \end{align*} | |
| \end{solution} | |
| \end{questions} | |
| \end{document} |