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diff --git a/interferometry_new.tex b/interferometry_new.tex
index 82b5863..5cacbe5 100644
--- a/interferometry_new.tex
+++ b/interferometry_new.tex
@@ -4,8 +4,9 @@
 \chapter*{Optical Interferometry}
 
 \noindent
- \textbf{Experiment objectives}: Assemble and align Michelson and Fabry-Perot
-interferometer, check its calibration it using a laser of known wavelength, and then determine the refractive index of the air. 
+ \textbf{Experiment objectives}: Assemble and align Michelson 
+ %and Fabry-Perot
+interferometer, check its calibration by using a laser of known wavelength, and then determine the refractive index of the air. 
 
 \section*{Introduction}
 
@@ -18,7 +19,10 @@ The \textbf{Michelson interferometer}, shown in Fig.~\ref{fig1mich.fig}, is base
 \includegraphics[width=0.65\linewidth]{./pdf_figs/michelson.png} 
 \caption{\label{fig1mich.fig}A Michelson interferometer setup.}
 \end{figure}
-Such an interferometer was first used by Michelson and Morley~\cite{mmoriginal} in 1887 to determine that electromagnetic waves propagate in vacuum, giving the first strong evidence against the theory of a \textit{luminiferous aether} (a fictitious medium for light wave propagation) and providing insight into the true nature of electromagnetic radiation. Michelson interferometers are widely used in many areas of physics and engineering, including the recent discovery of gravitational waves at the LIGO facility (see Sect.~\ref{LIGO} at the end of this lab for more information).
+Such an interferometer was first used by Michelson and Morley~\cite{mmoriginal} in 1887 to determine that electromagnetic waves propagate in vacuum, giving the first strong evidence against the theory of a \textit{luminiferous aether} (a fictitious medium for light wave propagation) and providing insight into the true nature of electromagnetic radiation. Michelson interferometers are widely used in many areas of physics and engineering, including the recent discovery of gravitational waves at the LIGO facility (see 
+%Sect.~\ref{LIGO}
+materials 
+at the end of this lab manual for more information).
 
 Figure~\ref{fig1mich.fig} shows the traditional setting for a Michelson
 interferometer. A beam splitter (a glass plate which is partially
@@ -33,12 +37,13 @@ and back. Since this beam travels through the glass beamsplitter
 plate 3 times, its optical path length is longer than the first beam which only passes through the beam splitter one time.  To
 compensate for that, the first beam passes twice through a clear glass plate called the compensator plate, that has the same thickness.  At the beam splitter one-half of this light is transmitted to an observer, overlapping with the beam reflected by $M_2$, and the total amplitude of the light at the screen is a combination of amplitude of  the two beams. 
 Assuming that the initial optical field is described by $E_0\cos(kz-\omega t)$, where $E_0$ is its amplitude, $k=2\pi \lambda$ is the wave vector,  and $\lambda$ and $\omega$ are correspondingly the wavelength and the frequency of the laser . After splitting and recombining at the beam splitter, the total field at the viewing screen beam splitter is a combination of the electric fields from the two arms of the interferometer :
-\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.
+\begin{align*}
+	E_{total} &=  \frac12 E_0 \cos(kz-\omega t + kL_1) + \frac12 E_0 \cos(kz-\omega t + kL_2) \\
+ &= E_0 \cos(k\Delta L/2) \cos(kz-\omega t + k(L_1+L_2)/2),
+\end{align*}
+where $L_{1,2}$ are the round-trip optical paths 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).
+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).
 
 It is easy to see that constructive interference happens when the difference between path lengths in the two interferometer arms is equal to the integer number of wavelengths $\Delta L = m\lambda$, and destructive interference corresponds to a half-integer number of wavelengths $\Delta L = (m + 1/2) \lambda$ (here $m$ is an integer number). Since the wavelength of light is small ($600-700$~nm for a red laser), Michelson interferometers are able to measure distance variation with very good precision.
 
@@ -47,9 +52,9 @@ To simplify the alignment of a Michelson interferometer, it is convenient to wor
 \Delta L=\frac{d}{\cos \theta}+\frac{d}{\cos \theta}\cos 2\theta
 \end{equation}
 Recalling that  $\cos 2\theta = 2(\cos \theta)^2-1$, we obtain $ \Delta L=2d\cos\theta$. The two rays interfere
-constructively for any angle $\theta_c$ for which $ \Delta L=2d\cos\theta = m\lambda$ ($m$=integer); at the same
+constructively for any angle $\theta_c$ for which $ \Delta L=2d\cos\theta = m\lambda$ ($m$ is an integer number); at the same
 time, two beams traveling at the angle $\theta_d$ interfere destructively when $ \Delta L=2d\cos\theta =
-(m+1/2)\lambda$ ($m$=integer). Because of the symmetry about the normal direction to the mirrors, this means
+(m+1/2)\lambda$. Because of the symmetry about the normal direction to the mirrors, this means
 that interference ( bright and dark fringes) appears in a circular shape.
 If the  fringes are not circular, the mirrors are not parallel, and additional alignment of the interferometer is required.