From c066f70fdfee3b8f0f26def3f29acfbfed4ef63e Mon Sep 17 00:00:00 2001
From: Eugeniy Mikhailov <evgmik@gmail.com>
Date: Fri, 13 Sep 2013 21:15:50 -0400
Subject: chapters now compiles separately as well via subfiles package

unused chapters moved to separate folder
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+\documentclass[./manual.tex]{subfiles}
+\begin{document}
+
+\chapter{Atomic Spectroscopy}
+
+ \textbf{Experiment objectives}: test a diffraction grating-based spectrometer, study the energy spectrum of atomic hydrogen (H) and a hydrogen-like atomic sodium (Na), determine values of quantum defects of low angular momentum states of Na, and measure fine splitting using Na yellow doublet.
+
+\subsection*{History}
+
+    The observation of discrete lines in the emission spectra of
+    atomic gases gives insight into the quantum nature of
+    atoms. Classical electrodynamics cannot explain the existence
+    of these discrete lines, whose energy (or wavelengths) are
+    given by characteristic values for specific atoms. These
+    emission lines are so fundamental that they are used to
+    identify atomic elements in objects, such as in identifying
+    the constituents of stars in our universe. When Niels Bohr
+    postulated that electrons can exist only in orbits of discrete
+    energies, the explanation for the discrete atomic lines became
+    clear.  In this laboratory you will measure the wavelengths of
+    the discrete emission lines from two elements - hydrogen and sodium -
+    to determine the energy levels in the hydrogen-like atoms.
+
+\section*{Hydrogen Spectrum}
+
+    The hydrogen atom is the simplest atom: it consists of a single proton and a single
+electron. Since it is so simple, it is possible to calculate the energy spectrum of the electron
+bound states - in fact you will do that in your Quantum Mechanics course. From this calculations it
+follows that the electron energy is quantized, and these energies are given by the expression:
+\begin{equation}\label{Hlevels_inf}
+E_n=- \frac{2\pi^2m_ee^4}{(4\pi\epsilon_0)^2h^2n^2}
+ = -hcRy\frac{1}{n^2}
+\end{equation}
+%
+where  $n$  is the  {\bf principal quantum number} of the atomic level, and
+\begin{equation} \label{Ry}
+Ry=\frac{2\pi m_ee^4}{(4\pi\epsilon_0)^2ch^3}
+\end{equation}
+is a fundamental physical constant called the {\bf Rydberg constant} (here $m_e$ is the electron
+mass).  Numerically, $Ry = 1.0974 \times  10^5 cm^{-1}$ and $hcRy = 13.605 eV$.
+
+Because the allowed energies of an electron in a hydrogen atom, the electron can change its state
+only by making a transition ("jump") from an one state of energy  $E_1$ to another state of lower
+energy $E_2$ by emitting a photon of energy $h\nu  = E_1 - E_2$ that carries away the excess energy.
+Thus, by exciting atoms into high-energy states using a discharge and then measuring the  frequencies
+of  emission  one can figure out the energy separation between various energy levels. Since it is
+more convenient to use a wavelength of light $\lambda$ rather than its frequency $\nu$, and using the
+standard connection between the wavelength and the frequency $h\nu = hc/\lambda$, we can write the
+wavelength of a photon emitted by electron jumping between the initial state $n_1$ and the final
+state $n_2$ as follows:
+\begin{equation} \label{Hlines_inf}
+\frac{1}{\lambda_{12}}=\frac{2\pi^2m_ee^4}{(4\pi\epsilon_0)^2ch^3}
+\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]= Ry \left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]
+\end{equation}
+%Based on this description it is clear that by measuring the frequencies (or
+%wavelengths) of photons emitted by an excited atomic system, we can glean
+%important information about allowed electron energies in atoms.
+
+%To make more accurate calculation of the Hydrogen spectrum, we need to take
+%into account that a hydrogen nucleus has a large, but finite mass, M=AMp (mass
+%number A=1 and Mp = mass of proton)\footnote{This might give you the notion
+%that the mass of any nucleus of mass number $A$ is equal to $AM_p$. This is not
+%very accurate, but it is a good first order approximation.} such that the
+%electron and the nucleus orbit a common center of mass.  For this two-mass
+%system the reduced mass is given by $\mu=m_e/(1+m_e/AM_p)$. We can take this
+%into account by modifying the above expression (\ref{Hlines_inf}) for
+%1/$\lambda$ as follows:
+%\begin{equation}\label{Hlines_arb}
+%\frac{1}{\lambda_A}=R_AZ^2\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] \mbox{
+%where } R_A=\frac{R_{\infty}}{1+\frac{m_e}{AM_p}}
+%\end{equation}
+%In particular, for the hydrogen case of  ($Z=1$; $M=M_p$) we have:
+%\begin{equation}\label{Hlines_H}
+%\frac{1}{\lambda_H}=R_H\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]
+%\end{equation}
+%Notice that the value of the Rydberg constant will change slightly for
+%different elements. However, these corrections are small since nucleus is
+%typically several orders of magnitude heavier then the electron.
+
+
+Fig.~\ref{Hspec.fig} shows the energy levels of hydrogen, and indicates a large number of observed
+transitions, grouped into series with the common electron final state.  Emitted photon frequencies
+span the spectrum from the UV (UltraViolet) to the IR (InfraRed). Among all the series only the
+Balmer series, corresponding to $n_2$ = 2, has its transitions in visible part of the spectrum.
+Moreover, from all the possible transition, we will be able to to observe and measure only the
+following four lines: $n_1=6 \rightarrow  2$, $5 \rightarrow  2$, $4 \rightarrow 2$,
+ and $3 \rightarrow 2$.
+
+\begin{figure}
+\includegraphics[width=0.7\linewidth]{./pdf_figs/spec}
+\caption{\label{Hspec.fig}Spectrum of Hydrogen. The numbers on the left show the energies of the
+hydrogen levels with different principle quantum numbers $n$ in $eV$. The wavelength of emitted
+photon in {\AA}  are shown next to each electron transition. }
+\end{figure}
+
+%In this lab, the light from the hydrogen gas is broken up into its spectral
+%components by a diffraction grating.  You will measure the angle at which each
+%line occurs on the left ($\theta_L$) and ($\theta_R$) right sides for as many
+%diffraction orders $m$ as possible, and use Eq.(\ref{mlambda}) to calculate
+%$\lambda$, using the following expression, derived in the Appendix.
+%\begin{equation}\label{mlambda}
+%m\lambda = \frac{h}{2}\left(\sin\theta_L+\sin\theta_R\right)
+%\end{equation}
+% Then the same
+%expression will be used to check/calibrate the groove spacing $h$ by making
+%similar measurements for a sodium spectral lines with known wavelengths.
+%
+%We will approach the data in this experiment both with an eye to confirming
+%    Bohr's theory and from Balmer's early perspective of someone
+%    trying to establish an integer power series linking the
+%    wavelength of these four lines.
+
+\section*{Sodium spectrum}
+Sodium (Na) belongs to the chemical group of \emph{alkali metals} [together with lithium (Li),
+potassium (K), rubidium (Rb), cesium (Cs) and Francium (Fr)]. All these elements consist of a closed
+electron shell with one extra unbound electron. Not surprisingly, the energy level structure for this
+free electron is very similar to that of hydrogen. For example, a Na atom has 11 electrons, and its
+electronic configuration is $1s^22s^22p^63s$.  Ten closed-shell electrons effectively screen the
+nuclear charge number ($Z=11$) to an effective charge $Z^*\approx 1$, so that the $3s$ valent
+electron experiences the electric field potential similar to that of a hydrogen atom, given by the
+Eq.~\ref{Hlevels_inf}.
+
+However, there is an important variation of the energy spectrum of alkali metals, related to the
+electron angular momentum $l$. In hydrogen the energy levels with same principle quantum number $n$
+but different electron angular momentum $l=0, 1, \cdots (n-1)$ are degenerate. For Na and others the
+levels with different values of $l$ are shifted with respect to each other. This is mainly due to the
+interaction of the unpaired electrons with the electrons of the closed shells.  For example, the
+orbits of the  electron with large angular momentum value $l$ is far above closed shell, and thus
+their energies are basically the same as for the hydrogen atom. An electron with smaller $l$ spends
+more time closer to the nuclear, and ``feels'' stronger bounding electrostatic potential. As a result
+the corresponding energy levels are pulled down compare to those of hydrogen, and the states with the
+same principle number $n$ but different angular momenta $l$ are split (\emph{i.e.} have different
+energies).
+
+
+To take onto account the modification of the atomic spectra while still using the same basic
+equations as for the hydrogen, it is convenient to introduce a small correction $\Delta_l$ to the
+principle quantum number $n$ to take into account the level shifts. This correction is often called a
+{\it quantum defect}, and its value % an effective nuclei charge $Z^*$ keeping the For each particular value
+%of angular momentum $l$ the energy spectrum follows the same scaling as hydrogen atom, but with an
+%effective charge $Z^*$:
+%\begin{equation}\label{heq}
+%E_n=-\frac{1}{2}\frac{Z^{*2}e^4}{(4\pi\epsilon_0)^2}\frac{mc^2}{\hbar^2c^2}
+%\frac{1}{n^2}=-Z^{*2}\frac{hcRy}{n^2}
+%\end{equation}
+%The value of the effective charge $Z^*$
+depends on the angular momentum $l$, and does not vary much between states with different principle
+quantum numbers $n$ but same $l$\footnote{The accepted notation for different electron angular
+momentum states is historical, and originates from the days when the proper quantum mechanical
+description for atomic spectra has not been developed yet. Back then spectroscopists had categorized
+atomic spectral lines corresponding to their appearend: for example any spectral lines from electron
+transitions from $s$-orbital ($l=0$) appeared always \textbf{S}harp on a photographic film, while
+those with initial electron states of $d$-orbital ( $l=2$) appeared always \textbf{D}iffuse. Also
+spectral lines of \textbf{P}rinciple  series (initial state is $p$-orbital, $l=1$) reproduced the
+hydrogen spectrum most accurately (even though at shifted frequencies), while the
+\textbf{F}undamental (initial electron state is $f$-orbital, $l=3$) series matched the absolute
+energies of the hydrogen most precisely. The orbitals with higher value of the angular momentum are
+denoted in an alphabetic order ($g$, $h$, \textit{etc}.) }:
+\begin{equation}\label{qdef}
+E_{nl}=-\frac{hcRy}{(n-\Delta_l)^2}%=-\frac{hcRy}{(n-\Delta_l)^2}
+\end{equation}
+
+%\begin{tabular}{ll}
+%States&$Z^*$\\
+%s~($l=0$)&$\approx$ 11/9.6\\
+%p~($l=1$)&$\approx$ 11/10.1\\
+%d~($l=2$)&$\approx$ 1\\
+%f~($l=3$)&$\approx $ 1\\
+%\end{tabular}
+In particular, the energies of two states with the lowest angular momentum ($s$ and $p$) are
+noticeably affected by the more complicated electron structure of Na, while the energy levels of the
+states with the higher values of angular momentum ($d$, $f$) are identical to the hydrogen energy
+spectrum.
+%
+%An alternative (but equivalent) procedure is to assign a {\it quantum defect} to the principle
+%quantum  $n$ instead of introducing an effective nuclei charge. In this case Eq.(\ref{heq}) can be
+%written as:
+%
+%where $n*=n-\Delta_l$, and $\Delta_l$  is the corresponding quantum defect. Fig. \ref{nadell} shows
+%values of quantum defects which work approximately for the alkalis. One sees that there is one value
+%for each value of the angular momentum $l$. This is not exactly true for all alkali metals, but for
+%Na there is very little variation in $\Delta_l$ with $n$ for a given $l$.
+\begin{figure}
+\includegraphics[height=\columnwidth]{./pdf_figs/nae}
+\caption{\label{nae}Energy spectrum of Na. The energy states of H are shown in far right for
+comparison.}
+\end{figure}
+%\begin{figure}
+%\includegraphics[width=0.5\columnwidth]{nadell.eps}
+%\caption{\label{nadell}Quantum Defect $\Delta_l$ versus $l$ for different alkali metals. Taken from
+%Condon and Shortley p. 143}
+%\end{figure}
+%\begin{figure}
+%\includegraphics[height=3in]{nadel.eps}
+%\caption{\label{nadel}Quantum Defect $\Delta_l$ variation with $n$. The
+%difference between the quantum defect of each term and that of the lowest term
+%of the series to which it belongs is plotted against the difference between
+%the total quantum numbers of these terms. Again from Condon and Shortley p. 144.}
+%\end{figure}
+
+The spectrum of Na is shown in Fig. \ref{nae}. One can immediately see that there are many more
+optical transitions because of the lifted degeneracy of energy states with different angular momenta.
+However, not all electronic transition are allowed: since the angular momentum of a photon is $1$,
+then the electron angular momentum cannot change by more than one while emitting one spontaneous
+photon. Thus, it is important to remember the following \emph{selection rule} for atomic transitions:
+\begin{equation}\label{selrules}
+\Delta l = \pm 1.
+\end{equation}
+According to that rule, only transitions between two ``adjacent'' series are possible: for example $p
+\rightarrow s$ or $d \rightarrow p$ are allowed, while $s \rightarrow s$ or $s \rightarrow d$ are
+forbidden. The strongest allowed optical transitions are shown in Fig. \ref{natrns}.
+\begin{figure}
+\includegraphics[height=\columnwidth]{./pdf_figs/natrans}
+\caption{\label{natrns}Transitions for Na. The wavelengths of selected transition are shown in {\AA}.
+Note, that $p$ state is now shown in two columns, one referred to as $P_{1/2}$ and the other as
+$P_{3/2}$. The small difference between their energy levels is the ``fine structure''.}
+\end{figure}
+%\begin{figure}
+%\includegraphics[height=4in]{series.eps}
+%\caption{\label{series}Series for Hydrogen, Alkalis are similar.}
+%\end{figure}
+Note that each level for given $n$ and $l$ is split into two because of the \emph{fine structure
+splitting}. This splitting is due to the effect of electron \emph{spin} and its coupling with the
+angular momentum. Proper treatment of spin requires knowledge of quantum electrodynamics and solving
+Dirac equation; for now spin can be treated as an additional quantum number associated with any
+particle. The spin of electron is $1/2$, and it may be oriented either along or against the non-zero
+electron's angular momentum. Because of the weak coupling between the angular momentum and spin,
+these two possible orientation results in small difference in energy for corresponding electron
+states.
+
+\section*{Experimental setup}
+\textbf{Equipment needed}: Gaertner-Peck optical spectrometer, hydrogen discharge lamp, sodium
+discharge lamp.
+
+Fig. \ref{expspec} gives a top view of the Gaertner-Peck optical spectrometer used in this lab. The
+spectrometer consists of two tubes. One tube holds a vertical collimator slit of variable width, and
+should be directed toward a discharge lamp. The light from the discharge lamp passes through the
+collimator slit and then get dispersed on the diffraction grating mounted in the mounting pedestal in
+the middle of the apparatus. The other tube holds the telescope that allows you to magnify the image
+of the collimator slit to get more accurate reading on the rotating table. This tube can be rotated
+around the central point, so that you will be able to align and measure the wavelength of all visible
+spectral lines of the lamp in all orders of diffraction grating.
+\begin{figure}
+\includegraphics[height=4in]{./pdf_figs/expspec}
+\caption{\label{expspec}Gaertner-Peck Spectrometer}
+\end{figure}
+
+It is likely that you will find a spectrometer at nearly aligned condition at the beginning on the
+lab. Nevertheless, take time making sure that all elements are in order to insure good quality of
+your data.
+
+\textit{Telescope Alignment:} Start by adjusting the telescope eyepiece in or out to bring the
+crosshairs into sharp focus.  Next aim the telescope out the window to view a distant object such as
+leaves in a tree.  If the distant object is not in focus, you will have to adjust the position of the
+telescope tube - ask your instructor for directions.
+
+\textit{Collimator Conditions:} Swing the telescope to view the collimator which is accepting light
+from the hydrogen discharge tube through a vertical slit of variable width.  The slit opening should
+be set to about 5-10 times the crosshair width to permit sufficient light to see the faint violet
+line and to be able to see the crosshairs.  If the bright column of light is not in sharp focus, you
+should align the position of the telescope tube again (with the help of the instructor).
+
+\textit{Diffraction Grating Conditions:} In this experiment you will be using a diffraction grating
+that has 600 lines per mm. A brief summary of diffraction grating operation is given in the Appendix
+of this manual. If the grating is not already in place, put it back to the baseclamp and fix it
+there. The table plate that holds the grating can be rotated, so try to orient the grating surface to
+be maximally perpendicular to the collimator axis. However, the accurate measurement of angle does
+not require the perfect grating alignment. Instead, for each spectral line in each diffraction order
+you will be measuring the angles on the left ($\theta_l$) and on the right ($\theta_r$), and use both
+of the measurements to figure out the optical wavelength using the following equation:
+\begin{equation}\label{nlambda}
+m\lambda=\frac{d}{2}(\sin\theta_r+\sin\theta_l),
+\end{equation}
+where $m$ is the diffraction order, and $d$ is the distance between the lines in the grating.
+
+   %  Use
+%    the AUTOCOLLIMATION procedure  to achieve a fairly accurate
+%    alignment of the grating surface.  This will determine how to
+%    adjust the three leveling screws H1, H2, and H3 and the
+%    rotation angle set screw for the grating table.
+%
+%    \textbf{AUTOCOLLIMATION} is a sensitive way to align an optical
+%    element.  First, mount a ``cross slit'' across the objective lens of
+%    the collimator, and direct a strong light source into the
+%    input end of the collimator.  Some of the light exiting through
+%    the cross slit will reflect from the grating and return to the
+%    cross slit.  The grating can then be manipulated till this
+%    reflected light retraces its path through the cross slit
+%    opening.  With this the grating surface is normal to the
+%    collimator light.
+     Then, with the hydrogen tube ON and in place at
+    the collimator slit, swing the rotating telescope slowly
+    through 90 degrees both on the left \& right sides of the forward
+    direction.  You should observe diffraction maxima for most
+    spectral wavelength, $\lambda$, in 1st, 2nd, and 3rd order.  If these
+    lines seem to climb uphill or drop downhill
+    the grating has to be adjusted in its baseclamp to
+    bring them all to the same elevation.
+
+    Also, turn on the sodium lamp as soon as you arrive, since it requires 10-15 minutes to warm up
+    and show clear Na spectrum.
+
+\subsection*{Experimental studies of Hydrogen Balmer line}
+
+Swing the rotating telescope slowly and determine which spectral lines from Balmer series you
+observe. You should be able to see three bright lines - Blue, Green and Red - in the first (m=1) and
+second (m=2) diffraction orders on both left \& right sides. In the third order (m=3) only the Blue,
+\& Green lines are visible, and you will not see the Red.
+
+One more line of the Balmer series is in the visible range - Violet, but its intensity is much lower
+than for the other three line. However, you will be able to find it in the first order if you look
+carefully with the collimator slit open a little wider.
+%
+%\emph{Lines to be measured:}
+%\begin{itemize}
+%\item \emph{Zero order} (m=0): All spectral lines merge.
+%\item \emph{First order} (m=1): Violet, Blue, Green, \& Red on both left \&
+%    right sides.
+%\item \emph{Second order} (m=2): Violet, Blue, Green, \& Red on
+%    both left \& right sides.
+%\item \emph{Third order} (m=3):  Blue, \& Green.
+%\end{itemize}
+%    You might not see the Violet line due to its low
+%    intensity. Red will not be seen in 3rd  order.
+
+After locating all the lines, measure  the angles at which each line occurs. The spectrometer reading for each line should be measured at least \emph{twice} by \textit{different} lab partners to avoid systematic errors. \textbf{Don't forget}: for every line you need to measure the angles to the right and to the left!
+
+You should be able to determine the angle with accuracy of $1$ minute, but you should know how to
+read angles with high precision in the spectrometer: first use the bottom scale to get the rough
+angle reading within a half of the degree. Then use the upper scale for more accurate reading in
+minutes. To get this reading find a tick mark of the upper scale that aligns perfectly with some tick
+mark of the bottom scale - this is your minute reading. Total angle is the sum of two readings.
+
+To measure the frequency precisely center the crosshairs on the line as accurately as possible.
+Choose the width of lines by turning the collimator slit adjustment screw. If the slit is too wide,
+it is hard to judge the center of the line accurately; if the slit is too narrow, then not enough
+light is available to see the crosshairs. For Violet the intensity is noticeably less than for the
+other three lines, so you may not see it. However, even if you find it, a little assistance is
+usually required in order to locate the crosshairs at this line. We suggest that a low intensity
+flashlight be aimed toward the Telescope input, and switched ON and OFF repeatedly to reveal the
+location of the vertical crosshair relative to the faint Violet line.
+
+%\subsubsection*{ Calibration of Diffraction Grating:} Replace the hydrogen tube with
+%    a sodium (Na) lamp and take readings for the following two
+%    lines from sodium: $568.27$~nm (green) and $589.90$~nm (yellow).  Extract from
+%    these readings the best average value for $h$ the groove
+%    spacing in the diffraction grating.  Compare to the statement
+%    that the grating has 600 lines per mm.  Try using your measured value
+%    for $h$ versus the stated value $600$ lines per mm in
+%    the diffraction formula when obtaining the measured
+%    wavelengths of hydrogen. Determine which one provide more accurate results, and discuss the conclusion.
+
+\subsubsection*{ Data analysis for Hydrogen Data}
+Calculate the wavelength $\lambda$ for each line observed in all orders, calculate the average
+wavelength value and uncertainty for each line, and then identify the initial and final electron
+states principle numbers ($n_1$ and $n_2$) for each line using Fig.~\ref{Hspec.fig}.
+
+Make a plot of $1/\lambda$  vs $1/n_1^2$ where $n_1$ = the principal quantum number of the electron's
+initial state. Put all $\lambda$ values you measure above on this plot. You data point should form a
+straight line. From Equation~(\ref{Hlines_inf}) determine the physical meaning of both slope and
+intercept, and compare the data from the fit to the expected values for each of them. The slope
+should be the Rydberg constant for hydrogen, $Ry$. The intercept is $Ry/(n_2)^2$. From this,
+determine the value for the principal quantum number $n_2$. Compare to the accepted value in the
+Balmer series.
+
+\subsection*{Experimental studies of Sodium Spectrum}
+
+Switch to Sodium lamp and make sure the lamp warms up for approximately 5-10 minutes before starting
+the measurements.
+
+Double check that you see a sharp spectrum in the spectrometer (adjust the width of the collimator
+slit if necessary). In the beginning it will be very useful for each lab partner to quickly scan the
+spectrometer telescope through all first-order lines, and then discuss which line you see corresponds
+to with transition in Table~\ref{tab:sodium} and Fig.~\ref{natrns}. Keep in mind that the color
+names are symbolic rather than descriptive!
+
+After that carefully measure the left and right angles for as many spectral lines in the first and orders
+as possible. The spectrometer reading for each line should be measured at least \emph{twice} by
+\textit{different} lab partners to avoid systematic errors.
+
+Determine  the wavelengths of all measured Na spectral lines using Eq. \ref{nlambda}. Compare these
+measured mean wavelengths to the accepted values given in Fig.~\ref{natrns} and in the table below.
+Identify  at least seven of the lines with a particular transition, e.g. $\lambda = 4494.3${\AA}
+corresponds to $8d \rightarrow 3p$ transition.
+
+\begin{table}
+
+\centering
+\begin{tabular}{l|l|l} 
+  Color&Line$_1$(\AA)&Line$_2$(\AA)\\ \hline
+Red&6154.3&6160.7\\
+Yellow & 5890.0&5895.9\\
+Green & 5682.7&5688.2\\
+&5149.1&5153.6\\
+& 4978.6&4982.9\\
+Blue&4748.0&4751.9\\
+&4664.9&4668.6\\
+Blue-Violet&4494.3&4497.7\\
+\end{tabular}
+\caption{\label{tab:sodium}Wavelength of the visible sodium lines.}
+\end{table}
+Line$_1$ and Line$_2$ correspond to transitions to two fine-spitted $3p$ states $P_{1/2}$ and
+$P_{3/2}$. These two transition frequencies  are very close to each other, and to resolve them with
+the spectrometer the width of the slit should be very narrow. However, you may not be able to see
+some weaker lines then. In this case you should open the slit wider to let more light in when
+searching for a line. If you can see a spectral line but cannot resolve the doublet, record the
+reading for the center of the spectrometer line, and use the average of two wavelengthes given above.
+
+\textbf{Measurements of the fine structure splitting}. Once you measured all visible spectral lines,
+go back to some bright line (yellow should work well), and close the collimator slit such that you
+can barely see any light going through. In this case you should be able to see the splitting of the
+line because of the \emph{fine structure splitting} of states $P_{1/2}$ and $P_{3/2}$ . For the Na D
+doublet the splitting between two lines $\Delta\lambda=\lambda_{3/2}-\lambda_{1/2}$.
+
+Measure the splitting between two lines in the first and the second order. Which one works better?
+Discuss this issue in your lab report. Compare to the accepted value: $\Delta\lambda=5.9$~\AA.
+Compare this approach to the use of the Fabry-Perot interferometer.
+
+
+\subsection*{Data analysis for Sodium}
+
+\textbf{Calculation of a quantum defect for $n=3, p$ state. }
+
+Once you identified all of your measured spectral lines, choose only those that correspond to optical
+transitions from any $d$ to $n=3,p$ states. Since the energy states of $d$ series follows the
+hydrogen spectra almost exactly, the wavelength of emitted light $\lambda$ is given by:
+\begin{equation}
+\frac{hc}{\lambda}=E_{nd}-E_{3p}=-\frac{hcRy}{n_d^2}+\frac{hcRy}{(3-\Delta_p)^2},
+\end{equation}
+or
+\begin{equation}
+\frac{1}{\lambda}=\frac{Ry}{(3-\Delta_p)^2}-\frac{Ry}{n_d^2},
+\end{equation}
+ where $n_d$ is the principle number of the initial $d$ state. To verify this
+expression by plotting $1/\lambda$ versus $1/n_d^2$ for the $n_d$= 4,5, and 6. From the slope of this
+curve determine the value of the Rydberg constant $Ry$. The value of the intercept in this case is
+$\frac{Ry}{(3-\Delta_p)^2}$, so use it to find the quantum defect $\Delta_p$.
+
+Compare the results of your calculations for the quantum defect with the accepted value
+$\Delta_p=0.86$.
+
+%\subsection*{Calculation of a quantum defect for $s$ states}
+%Now consider the transition from the $s$-states ($n=5,6,7$) to to the $n=3, p$ state. Using
+%$hc/\lambda=E_{ns}-E_{3p}$ and the results of your previous calculations, determine the energies
+%$E_{sn}$ for different $s$ states with $n=5,6,7$  and calculate $\Delta_s$. Does the value of the
+%quantum defect depends on $n$?
+%
+
+%\textbf{Example data table for writing the results of the measurements}:
+%
+%\noindent
+%\begin{tabular}{|p{1.in}|p{1.in}|p{1.in}|p{1.in}|}
+%\hline
+% Line &$\theta_L$&$\theta_R$&Calculated $\lambda$  \\ \hline
+% m=1 Violet&&&\\ \hline
+% m=1 Blue&&&\\ \hline
+% m=1 Green&&&\\ \hline
+% m=1 Red&&&\\ \hline
+% m=2 Violet&&&\\ \hline
+% \dots&&&\\ \hline
+% m=3 Blue&&&\\ \hline
+% \dots&&&\\\hline
+%\end{tabular}
+
+\section*{Appendix: Operation of a diffraction grating-based optical spectrometer}
+
+%\subsection*{Fraunhofer Diffraction at a Single Slit}
+%Let's consider a plane electromagnetic wave incident on a vertical slit of
+%width $D$ as shown in Fig. \ref{frn}. \emph{Fraunhofer} diffraction is
+%calculated in the far-field limit, i.e. the screen is assume to be far away
+%from the slit; in this case the light beams passed through different parts of
+%the slit are nearly parallel, and one needs  a lens to bring them together and
+%see interference.
+%\begin{figure}[h]
+%\includegraphics[width=0.7\linewidth]{frnhfr.eps}
+%\caption{\label{frn}Single Slit Fraunhofer Diffraction}
+%\end{figure}
+%To calculate the total intensity on the screen we need to sum the contributions
+%from different parts of the slit, taking into account phase difference acquired
+%by light waves that traveled different distance to the lens. If this phase
+%difference is large enough we will see an interference pattern. Let's break the
+%total hight of the slit by very large number of point-like radiators with
+%length $dx$, and we denote $x$ the hight of each radiator above the center of
+%the slit (see Fig.~\ref{frn}). If we assume that the original incident wave is
+%the real part of $E(z,t)=E_0e^{ikz-i2\pi\nu t}$, where $k=2\pi/\lambda$ is the
+%wave number. Then the amplitude of each point radiator on a slit is
+%$dE(z,t)=E_0e^{ikz-i2\pi\nu t}dx$. If the beam originates at a hight $x$ above
+%the center of the slit then the beam must travel an extra distance $x\sin
+%\theta$ to reach the plane of the lens. Then we may write a contributions at
+%$P$ from a point radiator $dx$ as the real part of:
+%\begin{equation}
+%dE_P(z,t,x)=E_0e^{ikz-i2\pi\nu t}e^{ikx\sin\theta}dx.
+%\end{equation}
+%To find the overall amplitude one sums along the slit we need to add up the
+%contributions from all point sources:
+%\begin{equation}
+%E_P=\int_{-D/2}^{D/2}dE(z,t)=E_0e^{ikz-i2\pi\nu
+%t}\int_{-D/2}^{D/2}e^{ikx\sin\theta}dx = A_P e^{ikz-i2\pi\nu t}.
+%\end{equation}
+%Here $A_P$ is the overall amplitude of the electromagnetic field at the point
+%$P$. After evaluating the integral we find that
+%\begin{equation}
+%A_P=\frac{1}{ik\sin\theta}\cdot
+%\left(e^{ik\frac{D}{2}\sin\theta}-e^{-ik\frac{D}{2}\sin\theta}\right)
+%\end{equation}
+%After taking real part and choosing appropriate overall constant multiplying
+%factors the amplitude of the electromagnetic field at the point $P$ is:
+%\begin{equation}
+%A=\frac{\sin (\frac{\pi D}{\lambda}\sin\theta)}{\frac{\pi
+%D}{\lambda}\sin\theta}
+%\end{equation}
+%The intensity is proportional to the square of the amplitude and thus
+%\begin{equation}
+%I_P=\frac{(\sin (\frac{\pi D}{\lambda}\sin\theta))^2}{(\frac{\pi
+%D}{\lambda}\sin\theta)^2}
+%\end{equation}
+%The minima of the intensity occur at the zeros of the argument of the sin. The
+%maxima are near, but not exactly equal to the solution of:
+%\begin{equation}
+%  (\frac{\pi D}{\lambda}\sin\theta)=(m+\frac{1}{2})\pi \end{equation}
+%for integer $m$.
+%
+%The overall pattern looks like that shown in Fig. \ref{sinxox}.
+%\begin{figure}
+%\includegraphics[width=\linewidth]{sinxox.eps}
+%\caption{\label{sinxox}Intensity Pattern for Fraunhofer Diffraction}
+%\end{figure}
+
+%\subsection*{The Diffraction Grating}
+A diffraction grating is a common optical element, which consists of a pattern
+with many equidistant slits or grooves. Interference of multiple beams passing
+through the slits (or reflecting off the grooves) produces sharp intensity
+maxima in the output intensity distribution, which can be used to separate
+different spectral components on the incoming light. In this sense the name
+``diffraction grating'' is somewhat misleading, since we are used to talk about
+diffraction with regard to the modification of light intensity distribution  to
+finite size of a single aperture.
+\begin{figure}[h]
+\includegraphics[width=\linewidth]{./pdf_figs/grating}
+\caption{\label{grating}Intensity Pattern for Fraunhofer Diffraction}
+\end{figure}
+
+To describe the properties of a light wave after passing through the grating,
+let us first consider the case of 2 identical slits separated by the distance
+$h$, as shown in Fig.~\ref{grating}a. We will assume that the size of the slits
+is much smaller than the distance between them, so that the effect of
+Fraunhofer diffraction on each individual slit is negligible. Then the
+resulting intensity distribution on the screen is given my familiar Young
+formula:
+\begin{equation} \label{2slit_noDif}
+I(\theta)=\left|E_0 +E_0e^{ikh\sin\theta} \right|^2 = 4I_0\cos^2\left(\frac{\pi
+h}{\lambda}\sin\theta \right),
+\end{equation}
+where $k=2\pi/\lambda$, $I_0$ = $|E_0|^2$, and the angle $\theta$ is measured
+with respect to the normal to the plane containing the slits.
+%If we now include the Fraunhofer diffraction on each slit
+%same way as we did it in the previous section, Eq.(\ref{2slit_noDif}) becomes:
+%\begin{equation} \label{2slit_wDif}
+%I(\theta)=4I_0\cos^2\left(\frac{\pi h}{\lambda}\sin\theta
+%\right)\left[\frac{\sin (\frac{\pi D}{\lambda}\sin\theta)}{\frac{\pi
+%D}{\lambda}\sin\theta} \right]^2.
+%\end{equation}
+
+An interference on $N$ equidistant slits illuminated by a plane wave
+(Fig.~\ref{grating}b) produces much sharper maxima. To find light intensity on
+a screen, the contributions from all N slits must be summarized taking into
+account their acquired phase difference, so that the optical field intensity
+distribution becomes:
+\begin{equation} \label{Nslit_wDif}
+I(\theta)=\left|E_0
++E_0e^{ikh\sin\theta}+E_0e^{2ikh\sin\theta}+\dots+E_0e^{(N-1)ikh\sin\theta}
+\right|^2 = I_0\left[\frac{sin\left(N\frac{\pi
+h}{\lambda}\sin\theta\right)}{sin\left(\frac{\pi h}{\lambda}\sin\theta\right)}
+\right]^2.
+\end{equation}
+ Here we again neglect the diffraction form each individual slit, assuming that the
+ size of the slit is much smaller than the separation $h$ between the slits.
+
+The intensity distributions from a diffraction grating with illuminated
+ $N=2,5$ and $10$ slits are shown in Fig.~\ref{grating}c. The tallest (\emph{principle}) maxima occur when the denominator
+ of Eq.(~\ref{Nslit_wDif}) becomes zero: $h\sin\theta=\pm m\lambda$ where
+ $m=1,2,3,\dots$ is the diffraction order. The heights of the principle maxima are
+ $I_{\mathrm{max}}=N^2I_0$, and their widths are $\Delta \theta =
+ 2\lambda/(Nh)$.
+ Notice that the more slits are illuminated, the narrower diffraction peaks
+ are, and the better the resolution of the system is:
+ \begin{equation}
+\frac{ \Delta\lambda}{\lambda}=\frac{\Delta\theta}{\theta} \simeq \frac{1}{Nm}
+\end{equation}
+For that reason in any spectroscopic equipment a light beam is usually expanded
+to cover the maximum surface of a diffraction grating.
+
+\subsection*{Diffraction Grating Equation when the Incident Rays are
+not Normal}
+
+Up to now we assumed that the incident optical wavefront is normal to the pane of a grating. Let's
+now consider more general case when the angle of incidence $\theta_i$ of the incoming wave is
+different from the normal to the grating, as shown in Fig. \ref{DGnotnormal}(a). Rather then
+calculating the whole intensity distribution, we will determine the positions of principle maxima.
+The path length difference between two rays 1 and 2 passing through the consequential slits us $a+b$,
+where:
+\begin{equation}
+a=h\sin \theta_i;\,\, b=h\sin \theta_R
+\end{equation}
+Constructive interference occurs for order $m$ when $a+b=m\lambda$, or:
+\begin{equation}
+h\sin \theta_i + h\sin\theta_R=m\lambda
+\end{equation}
+
+\begin{figure}
+\includegraphics[width=\columnwidth]{./pdf_figs/pic4i}
+%\includegraphics[height=3in]{dn.eps}
+\caption{\label{DGnotnormal}Diagram of the light beams diffracted to the right
+(a) and to the left (b).}
+\end{figure}
+
+Now consider the case shown in Fig. \ref{DGnotnormal}(b). The path length between two beams is now
+$b-a$ where $b=h\sin\theta_L$. Combining both cases we have:
+\begin{eqnarray} \label{angles}
+h\sin\theta_L-h\sin\theta_i&=&m\lambda\\
+h\sin\theta_R+h\sin\theta_i&=&m\lambda \nonumber
+\end{eqnarray}
+Adding these equations and dividing the result by 2 yields the following expression connecting the
+right and left diffraction angles:
+\begin{equation}m\lambda=\frac{h}{2}\left(\sin\theta_L+\sin\theta_R\right)
+\end{equation}
+
+\end{document}
+
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