From 910f67af00f4dd5204cbdf89ecb744da4bec2c1d Mon Sep 17 00:00:00 2001 From: "Eugeniy E. Mikhailov" Date: Fri, 4 Sep 2020 15:47:39 -0400 Subject: added new edits from Irina --- blackbody_new.tex | 33 ++----- ediffract_new.tex | 259 +++++++++++++++++++++++++++++++++++++++++++++++++ interferometry_new.tex | 2 +- 3 files changed, 270 insertions(+), 24 deletions(-) create mode 100644 ediffract_new.tex diff --git a/blackbody_new.tex b/blackbody_new.tex index 56b3a35..1066c1a 100644 --- a/blackbody_new.tex +++ b/blackbody_new.tex @@ -70,10 +70,11 @@ where $c$ is the speed of light and $h=6.626076\times 10^{-34} J\cdot s$ is the \section*{Radiation sensor operation principle} -\includegraphics[height=2.5in]{./pdf_figs/thermopile} \\ - -Imagine a metal wire connected to a cold reservoir at one end and a hot reservoir at the other. Heat will flow between the ends of the wire, carried by the electrons in the conductor, which will tend to diffuse from the hot end to the cold end. Vibrations in the conductor's atomic lattice can also aid this process. This diffusion causes a potential difference between the two ends of the wire. The size of the potential difference depends on the temperature gradient and on details of the conductive material, but is typically in the few 10s of $\mu V/ K$. A thermocouple, shown on the left, consists of two different conductive materials joined together at one end and connected to a voltmeter at the other end. The potential is, of course, the same on either side of the joint, but the difference in material properties causes $\Delta V=V_1 - V_2 \neq 0$. This $\Delta V$ is measured by the voltmeter and is proportional to $\Delta T$. Your radiation sensor is a thermopile, simply a ``pile'' of thermocouples connected in series, as shown at the right. This is done to make the potential difference generated by the temperature gradient easier to detect. -\\ +Imagine a metal wire connected to a cold reservoir at one end and a hot reservoir at the other. Heat will flow between the ends of the wire, carried by the electrons in the conductor, which will tend to diffuse from the hot end to the cold end. Vibrations in the conductor's atomic lattice can also aid this process. This diffusion causes a potential difference between the two ends of the wire. The size of the potential difference depends on the temperature gradient and on details of the conductive material, but is typically in the few 10s of $\mu V/ K$. A thermocouple, shown on the left, consists of two different conductive materials joined together at one end and connected to a voltmeter at the other end. The potential is, of course, the same on either side of the joint, but the difference in material properties causes $\Delta V=V_1 - V_2 \neq 0$. This $\Delta V$ is measured by the voltmeter and is proportional to $\Delta T$. Your radiation sensor is a thermopile, simply a ``pile'' of thermocouples connected in series, as shown at the right. This is done to make the potential difference generated by the temperature gradient easier to detect. +\begin{figure} +\includegraphics[height=1.5in]{./pdf_figs/thermopile} +\caption{\label{sensor}\emph{Left}: thermocouple construction; \emph{right}: thermopile - an array of thermocouples connected in series.} +\end{figure} \\ \textbf{Important}: When using the thermal radiation sensor, make each reading quickly to keep the sensor from heating up. Use sheets of white isolating foam (with the silvered surface facing the lamp) to block the sensor between measurements. \\ \textbf{Sensor calibration}: To obtain the radiation sensor readings for radiated power per unit area $S$ in the correct units ($W/m^2$), you need to use the voltage-to-power conversion factor $22~V/W$, and the area of the sensor $2mm\times2mm$: @@ -90,7 +91,6 @@ Lamp, Power supply. Before starting actual experiment take some time to have fun with the thermal radiation sensor. Can you detect your lab partner? What about people across the room? Point the sensor in different directions and see what objects affect the readings. \textbf{These exercises are fun, but you will also gain important intuition about various factors which may affect the accuracy of the measurements!} - \begin{enumerate} \item \textbf{Before turning on the lamp}, measure the resistance of the filament of the Stefan-Boltzmann lamp at room temperature. Record the room temperature, visible on the wall thermostat. @@ -100,26 +100,13 @@ Before starting actual experiment take some time to have fun with the thermal ra % Resistance of filament (room temperature)=&$\underline{\hskip .7in}$ %\end{tabular} -\item Connect a multimeter as voltmeter to the output of the power supply. - {\bf Important:} make sure it is in the {\bf voltmeter mode}. - Compare voltage readings provided by the power supply and the multimeter and verify that their readings are reasonably close. - Which one is the correct one? Think about your measurement - uncertainties. - -\item Set up the equipment as shown in Fig. \ref{bbht}. VERY IMPORTANT: - make all connections to the lamp when the power is off. Turn the - power off before changing/removing connections. The voltmeter - should be directly connected to the binding posts of the - Stefan-Boltzmann lamp. In this case the multimeter voltmeter has - direct access to the voltage drop across the bulb, while the power - supply voltmeter reads an extra voltage due to finite resistance of - the current meter. Compare readings on the multimeter and the power - supply current meters. Are they reasonably close? +\item To indirectly measure the temperature of the filament, we will use the known dependence of its resistance on the temperature, given in Table.\ref{w_res:fig}. To ensure the accurate measurement, we will again use the four-point probe method (review the video on the course web site, if you need a refresher) by measuring the voltage drop across the lamp. VERY IMPORTANT: + make all connections to the lamp when the power is off, and ask the instructor to check your connections before proceeding. \item Place the thermal sensor at the same height as the filament, with the front face of the sensor approximately 5~cm away from the filament and fix their relative position. Make sure no other objects are viewed by the sensor other than the lamp. % \item Turn on the lamp power supply. Set the voltage, $V$, in steps of 1-2 - volt from 1-12 volts. At each $V$, record the current running through the lamp and the voltage from the radiation sensor. + volt from 1-6 volts. At each $V$, record the current running through the lamp and the voltage from the radiation sensor. Calculate the resistance of the lamp using Ohm's Law and determine the temperature $T$ of the lamp from the table shown in Fig. \ref{w_res:fig}. Don't forget to convert the measured temperatures to Kelvin scale: $T[K]=T[^oC]+273$. @@ -190,10 +177,10 @@ S(r)=\frac{P}{2\pi r^2} \newpage \section*{Universal thermometer} -Blackbody radiation gives us an ability to measure the temperature of remote objects. Have you ever asked yourself how do astronomer know the temperature of stars or other objects many light years away? The answer - by measuring the light they emit and analyzing it using the expressions for the blackbody radiation spectrum. Wein's law Eq.(\ref{weins}) links the wavelength at which the most radiation is emitted to the inverse of the object's temperature, thus the colder stars emit predominantly in red (hence the name ``red giants''), while emission pick for hot young stars is shifted to the blue, making them emit in all visible spectrum. +Blackbody radiation gives us an ability to measure the temperature of remote objects. Have you ever asked yourself how do astronomer know the temperature of stars or other objects many light years away? The answer - by measuring the light they emit and analyzing its spectrum conposition using the expressions for the blackbody radiation spectrum. Wein's law Eq.(\ref{weins}) links the wavelength at which the most radiation is emitted to the inverse of the object's temperature, thus the colder stars emit predominantly in red (hence the name ``red giants''), while emission pick for hot young stars is shifted to the blue, making them emit in all visible spectrum. \includegraphics[height=2.5in]{./pdf_figs/blackbody_radn_curves} -The human bodies, of course, are much cooler than stars, and emit in infrared range. This radiation is invisible for human eye, but using proper detection method it is possible to create thermal maps of the surroundings with the accuracy better than $1/10$th of a degree. Forward-looking infrared (FLIR) cameras have wide range of applications, from surveillance and military operations to building inspection and repairs, night-time navigation and hunting. As I write this in Fall 2020, in the middle of COVID19 pandemic, more and more locations use such sensors to measure visitors' temperature at the entrance of a building or a check points in the airports. +The human bodies, of course, are much cooler than stars, and emit in infrared range. This radiation is invisible for human eye, but using proper detection methods it is possible to create thermal maps of the surroundings with accuracy better than $1/10$th of a degree. Forward-looking infrared (FLIR) cameras have wide range of applications, from surveillance and military operations to building inspection and repairs, night-time navigation and hunting. As I write this in Fall 2020, in the middle of COVID19 pandemic, more and more locations use such infrared sensors to measure visitors' temperature at the building entrances or the check points in airports. \end{document} diff --git a/ediffract_new.tex b/ediffract_new.tex new file mode 100644 index 0000000..8ace7bf --- /dev/null +++ b/ediffract_new.tex @@ -0,0 +1,259 @@ +\documentclass[./manual.tex]{subfiles} +\begin{document} + +\chapter{Electron Diffraction} + +\textbf{Experiment objectives}: observe diffraction of the beam of electrons on a graphitized carbon target and calculate the intra-atomic spacings in the graphite. + +\section*{History} + + A primary tenet of quantum mechanics is the wavelike + properties of matter. In 1924, graduate student Louis de + Broglie suggested in his dissertation that since light has + both particle-like {\bf and} wave-like properties, perhaps all + matter might also have wave-like properties. He postulated + that the wavelength of objects was given by $\lambda = h/p$, where $h$ is Planck's constant and $p = + mv$ is the momentum. {\it This was quite a revolutionary idea}, + since there was no evidence at the time that matter behaved + like waves. In 1927, however, Clinton Davisson and Lester + Germer discovered experimental proof of the wave-like + properties of matter --- particularly electrons. This discovery + was quite by mistake: while studying electron reflection + from a nickel target, they inadvertently crystallized their + target, while heating it, and discovered that the scattered + electron intensity as a function of scattering angle showed + maxima and minima. That is, electrons were ``diffracting'' from + the crystal planes much like light diffracts from a grating, + leading to constructive and destructive interference. Not only + was this discovery important for the foundation of quantum + mechanics (Davisson and Germer won the Nobel Prize for their + discovery), but electron diffraction is an extremely important + tool used to study new materials. In this lab you will study + electron diffraction from a graphite target, measuring the + spacing between the carbon atoms. + + +\begin{figure}[h] +\centering +\includegraphics[width=\textwidth]{./pdf_figs/ed1_new} \caption{\label{ed1}Electron +Diffraction from atomic layers in a crystal.} +\end{figure} +\section*{Theory} + Consider planes of atoms in a {\bf crystal} as shown in Fig.~\ref{ed1} +separated by distance $d$. Electron ``waves'' reflect from each of these planes. +Since the electron is wave-like, the combination of the reflections from each +interface produces to an interference pattern. This is completely analogous to +light interference, arising, for example, from different path lengths in the +Fabry-Perot or Michelson interferometers. The de Broglie wavelength for the +electron is given by $\lambda=h/p$, where $p$ can be calculated by knowing the +energy of the electrons when they leave the ``electron gun'': +\begin{equation}\label{Va} +\frac{1}{2}mv^2=\frac{p^2}{2m}=eV_a, +\end{equation} + where $V_a$ is the accelerating potential. The condition for constructive +interference is that the path length difference for the two waves +shown in Fig. \ref{ed1} be a multiple of a wavelength. This leads to Bragg's +Law: +\begin{equation}\label{bragg} +n\lambda=2d\sin\theta +\end{equation} +where $n = 1,2,\dots$ is integer. In this experiment, only the first order +diffraction $n=1$ is observed. Therefore, the intra-atomic distance in a +crystal can be calculated by measuring the angle of electron diffraction and +their wavelength (\emph{i.e.} their momentum): +\begin{equation}\label{bragg1} +d=\frac{\lambda}{2\sin\theta} = \frac{1}{2\sin\theta}\frac{h}{\sqrt{2em_eV_a}} +\end{equation} +\noindent +where $h$ is Planck's constant, $e$ is the electronic charge, $m_e$ is the +electron's mass, and $V_a$ is the accelerating voltage. +% +% +% +% +% Knowing $\lambda$ and the angles $\theta$ for which +%constructive interference occurs, the atomic spacing $d$ can be +%extracted. + +\section*{Experimental Procedure} + +\textbf{Equipment needed}: Electron diffraction apparatus and power supply, ruler and thin receipt paper or masking tape. + +\subsection*{Safety} +You will be working with high voltage. The power supply will be connected for you, +but inspect the apparatus when you arrive \textbf{before} turning the power on. +If any wires are unplugged, ask an instructor to reconnect them. Also, +\textbf{before} turning the power on, identify the high voltage contacts on the +electron diffraction tube, make sure these connections are well-protected and +cannot be touched by accident while taking measurements. +\begin{figure}[h] +\centering +\includegraphics[width=6in]{./pdf_figs/ed2} \caption{\label{ed2}Electron Diffraction Apparatus.} +\end{figure} +\subsection*{Setup} + +The diagram of the apparatus is given in Fig.\ref{ed2}. An electron gun +(consisting of a heated filament to boil electrons off a cathode and an anode +to accelerate them, similar to the e/m experiment) ``shoots'' electrons at a +carbon (graphite) target. + +The electrons diffract from the carbon target and the resulting interference +pattern is viewed on a phosphorus screen. + + The graphitized carbon is not completely crystalline but consists of crystal +sheets in random orientations. Therefore, the constructive interference +patterns will be seen as bright circular rings. For the carbon target, two +rings (an outer and inner, corresponding to different crystal planes) will be +seen, corresponding to two spacings between atoms in the graphite arrangement +(see Fig.~\ref{ed3}). +\begin{figure} +\centering +\includegraphics[width=4in]{./pdf_figs/ed3} \caption{\label{ed3}Spacing of +carbon atoms. Here subscripts \textit{10} and \textit{11} correspond to the +crystallographic directions in the graphite crystal.} +\end{figure} + +\subsection*{Data acquisition} +Acceptable power supply settings: +\\\begin{tabular}{lll} +Filament Voltage& $V_F$&6.3 V ac/dc (8.0 V max.)\\ +Anode Voltage & $V_A$& 1500 - 5000 V dc\\ +Anode Current & $I_A$& 0.15 mA at 4000 V ( 0.20 mA max.) +\end{tabular} + +\begin{enumerate} +\item Switch on the heater and wait one minute for the oxide cathode to achieve thermal stability. +\item Slowly increase $V_a$ until you observe two rings appear around the direct beam. +Slowly change the voltage and determine the highest achievable accelerating +voltage, and the lowest voltage when the rings are visible. +\item Measure the diffraction angle $\theta$ for both inner and outer rings for 5-10 voltages from that range, +using the same thin receipt paper (see procedure below). Each lab partner should +repeat these measurements (using an individual length of the thin paper). +\item Calculate the average value of $\theta$ from the individual measurements for each +voltage $V_a$. Calculate the uncertainties for each $\theta$. +\end{enumerate} + +\textbf{Measurement procedure for the diffraction angle $\theta$} + +To determine the crystalline structure of the target, one needs to carefully +measure the diffraction angle $\theta$. It is easy to see (for example, from +Fig.~\ref{ed1}) that the diffraction angle $\theta$ is 1/2 of the angle +between the beam incident on the target and the diffracted beam to a ring, +hence the $2\theta$ appearing in Fig.~\ref{ed4}. You are going to determine +the diffraction angle $\theta$ for a given accelerated voltage from the +approximate geometrical ratio +\begin{equation} +L\sin{2\theta} = R\sin\phi, +\end{equation} +where the distance between the target and the screen $L = 0.130$~m is controlled during the production process to have an accuracy better than 2\%. {\it Note, this means that the electron tubes are not quite spherical.} + +The ratio between the arc length $s$ and the +radius of the curvature for the screen $R = 0.066$~m gives the angle $\phi$ in +radians: $\phi = s/2R$. To measure $\phi$ carefully place a piece of +thin receipt paper on the tube so that it crosses the ring along the diameter. +Mark the position of the ring for each accelerating voltage, and then +remove the paper and measure the arc length $s$ corresponding to +each ring. You can also make these markings on masking tape placed gently on the tube. + + +\begin{figure} +\centering +\includegraphics[width=4in]{./pdf_figs/edfig4} \caption{\label{ed4}Geometry of the experiment.} +\end{figure} + +\section*{Data analysis} + +Use the graphical method to find the average values for the distances between +the atomic planes in the graphite crystal $d_{11}$ (outer ring) and $d_{10}$ +(inner ring). To determine the combination of the experimental parameters that +is proportional to $d$, one needs to substitute the expression for the +electron's velocity Eq.(\ref{Va}) into the diffraction condition given by +Eq.(\ref{bragg}): +\begin{equation}\label{bragg.analysis} +\sin\theta=\frac{h}{2\sqrt{2m_ee}}\frac{1}{d}\frac{1}{\sqrt{V_a}} +\end{equation} + +Make a plot of $\sin\theta$ (y-axis, with uncertainty) vs $1/\sqrt{V_a}$ (x-axis) for the inner and outer rings +(both curves can be on the same graph). Fit the linear dependence and measure +the slope for both lines. From the values of the slopes find the distances +between atomic layers $d_{inner}$ and $d_{outer}$. + +Compare your measurements to the accepted values: $d_{inner}=d_{10} = .213$~nm +and $d_{outer}=d_{11}=0.123$~nm. + +\section*{Seeing with Electrons} + +The resolution of ordinary optical microscopes +is limited (the diffraction limit) by the wavelength of light ($\approx$ 400 +nm). This means that we cannot resolve anything smaller than this by looking at +it with light (even if we had no limitation on our optical instruments). Since +the electron wavelength is only a couple of angstroms ($10^{-10}$~m), with +electrons as your ``light source'' you can resolve features to the angstrom +scale. This is why ``scanning electron microscopes'' (SEMs) are used to look at +very small features. The SEM is very similar to an optical microscope, except +that ``light'' in SEMs is electrons and the lenses used are made of magnetic +fields, not glass. + +%\begin{tabular}{lll} +%Filament Voltage& $V_F$&6.3 V ac/dc (8.0 V max.)\\ +%Anode Voltage & $V_A$& 2500 - 5000 V dc\\ +%Anode Current & $I_A$& 0.15 mA at 4000 V ( 0.20 mA max.) +%\end{tabular} +% +%\newpage +%\noindent +%Section: $\underline{\hskip 1in}$\\ +%Name: $\underline{\hskip 1in}$\\ +%Partners: $\underline{\hskip 1in}$\\ +%$\hskip 1in\underline{\hskip 1in}$\\ +%\section*{Electron Diffraction} +%Step through the electron accelerating potential from 5 kV to 2 kV in +%steps of 0.5 kV. Record $V_a$ (from needle reading on power supply) and +%measure the diameter of the two rings. In measuring the diameter, +%either try to pick the middle of the ring, or measure the inside and +%outside of that ring +%and average. Each lab partner should measure the diameter, and then average +%the result. {\bf Make sure you give units in the following tables} +% +%\begin{center} +%{\bf Final data -- fill out preliminary chart on next page first} +%\end{center} +% +%\begin{tabular}{|p{20mm}|p{20mm}|p{20mm}|p{20mm}|p{20mm}|}\hline +%$V_a$&(inner s)&$d_{inner}$&outer s &$d_{outer}$\\\hline +%&&&&\\\hline +%&&&&\\\hline +%&&&&\\\hline +%&&&&\\\hline +%&&&&\\\hline +%&&&&\\\hline +%\multicolumn{2}{|l|}{Average $d_{inner}\pm\sigma=$} +%&\multicolumn{2}{|l}{Average $d_{outer}\pm\sigma=$} +%&\multicolumn{1}{l|}{}\\\hline +%\end{tabular} +% +% +% +% +%Make a plot of $1/sqrt{V_a}$ versus $s$ for the inner and outer rings +%{both curves can be on the same graph). +% +%Compare your measurements to the known spacing below.$d_{outer}=$\\ +%True values are: $d_{inner}=.213 nm$ and $d_{outer}=0.123 nm $. +% +%\begin{boxedminipage}{\linewidth} +% +% +%Turn in this whole stapled report, including your data tables. Attach your +%plot(s) at the end. +%\end{boxedminipage} +%\newpage +%\begin{boxedminipage}{\linewidth} +%Note: start with the External Bias at 30 V. Decrease the bias if you need +%to, to increase the intensity of the rings. DO NOT EXCEED 0.2 mA on ammeter! +%\end{boxedminipage} + +%\newpage + +\end{document} + diff --git a/interferometry_new.tex b/interferometry_new.tex index 10bbbea..27d40b7 100644 --- a/interferometry_new.tex +++ b/interferometry_new.tex @@ -36,7 +36,7 @@ Assuming that the initial optical field is described by $E_0\cos(kz-\omega t)$, \begin{equation} E_{total} = \frac12 E_0 \cos(kz-\omega t + kL_1) + \frac12 E_0 \cos(kz-\omega t + kL_1) = E_0 \cos(k\Delta L/2) \cos(kz-\omega t + k(L_1+L_2)), \end{equation} -where $L_{1,2}$ are the round-trip optical path for each interferometer arm, and $Delta L = L_1-L_2$ is the optical path difference for the two beams. You may notice that the final expression describes a laser fields with the amplitude $E_0 \cos(k\Delta L/2)$. As a result, the intensity we are going to see on the screen will be $I_0 \cos^2(k\Delta L/2)$, where $I_0$ is the total intensity of the initial laser beam. +where $L_{1,2}$ are the round-trip optical path for each interferometer arm, and $\Delta L = L_1-L_2$ is the optical path difference for the two beams. You may notice that the final expression describes a laser fields with the amplitude $E_0 \cos(k\Delta L/2)$. As a result, the intensity we are going to see on the screen will be $I_0 \cos^2(k\Delta L/2)$, where $I_0$ is the total intensity of the initial laser beam. Mirror $M_1$ is mounted on a precision traveling platform, and thus we can adjust its position (by turning the micrometer screw) with great precision to make the distances traversed on both arms exactly the same. Because the thickness of the compensator plate and the beamsplitter are the same, both wavefronts pass through the same amount of glass and air, so the path length of the light beams in both interferometer arms is exactly the same. Therefore, the two fields will arrive in phase to the observer, their amplitudes will add up constructively, and we should see all the original laser beam to go toward the viewing screen. However, if now you turn the micrometer to offset the length of one arm by a quarter of the wavelength, $\Delta L = \lambda/2$, the two beams arriving to the beam splitter will be out of phase $k L_1-kL_2 = \pm pi$, resulting in the total cancellation (destructive interference). -- cgit v1.2.3