my edits

1 files changed, 15 insertions, 10 deletions

diff --git a/blackbody_new.tex b/blackbody_new.tex
index 966020a..203b759 100644
--- a/blackbody_new.tex
+++ b/blackbody_new.tex
@@ -44,7 +44,7 @@ where $u(\lambda)(J/m^4)$ is the spectral radiance -- energy radiated per unit a
 \begin{equation} \label{weins}
  \lambda_{max}T = 2.898\times 10^{-3} m\cdot K
 \end{equation}
-and approaches zero for short  wavelengths.
+and the spectral radiance approaches zero for short  wavelengths.
 
     The breakthrough came when Planck assumed that the energy of the oscillation modes can only take on discrete values rather than a continuous distribution of values, as in classical physics. With this assumption, Planck's law was derived:
 \begin{equation}
@@ -70,7 +70,7 @@ where $c$ is the speed of light and $h=6.626076\times 10^{-34} J\cdot s$ is the
 
 \section*{Radiation sensor operation principle}
 
-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 at the joint, but the difference in material properties causes $\Delta V=V_1 - V_2 \neq 0$ between the separated ends. 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.}
@@ -79,7 +79,7 @@ Imagine a metal wire connected to a cold reservoir at one end and a hot reservoi
 \\
 \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 $2mm\times2mm$:
 \begin{displaymath}
-S[W/m^2]=\frac{S[mV]\cdot 10^{-3}[V/mV]}{22 [V/W]}\cdot \frac{1}{4\cdot
+S[W/m^2]=\frac{\Delta V[mV]\cdot 10^{-3}[V/mV]}{22 [V/W]}\cdot \frac{1}{4\cdot
 10^{-6}[m^2]}
 \end{displaymath}
 
@@ -100,7 +100,7 @@ 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 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:
+\item To indirectly measure the temperature of the filament, we will use the known dependence of its resistance on the temperature, given in table shown in Fig.~\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.
@@ -109,7 +109,7 @@ Before starting actual experiment take some time to have fun with the thermal ra
 	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: $T[K]=T[^oC]+273$.
+	\ref{w_res:fig}. Don't forget to use Kelvin scale for the temperatures (conversion equation is $T[K]=T[^oC]+273$).
 	
 \item Calculate the values of $T^4$ - these are going to be the $x$-values for the graph. Are they more or less equally distributed? If not (which is probably the case), estimate the big gaps, and measure additional points to fill them in.
 	
@@ -118,7 +118,7 @@ Before starting actual experiment take some time to have fun with the thermal ra
 
 \begin{figure}[h]
 \includegraphics[width=\columnwidth]{./pdf_figs/w_res}
-\caption{\label{w_res:fig}Table of tungsten's resistance as a function of temperature.}
+\caption{\label{w_res:fig}Table of tungsten's resistance as a function of temperature. Here, $R_{300K}$ is the resistance of tungsten at the temperature of 300~K.}
 \end{figure}
 
 
@@ -127,9 +127,9 @@ In the lab report plot the reading from the radiation sensor (convert to $W/m^2$
 \begin{equation}\label{SBl}
  S =\epsilon\sigma T^4
 \end{equation}
-where the coefficient $\epsilon$ is called \emph{emissivity} and is defined as the ratio of the energy radiated from a material's surface to that radiated a perfect blackbody at the same temperature. The values of $\epsilon$ vary from 0 to 1, with one corresponding to an ideal blackbody. All real materials have $\epsilon<1$, although some come quite close to the ideal (for example, carbon black has $\epsilon=0.95$). However, emissivity of a tungsten wire varies from $\epsilon=0.032$ (at $30^{\circ} C$) to $\epsilon=0.35$ (at $3300^{\circ}C$). Using the result of your fit, and assuming we know the Stephan-Boltzman constant $\sigma$ by some other means, what is $\epsilon$ and what is the uncertainty on it?  Is it consistent with tungsten? What else could be affecting this measurement?
+where the coefficient $\epsilon$ is called \emph{emissivity} and is defined as the ratio of the energy radiated from a material's surface to that radiated by a perfect blackbody at the same temperature. The values of $\epsilon$ vary from 0 to 1, with one corresponding to an ideal blackbody. All real materials have $\epsilon<1$, although some come quite close to the ideal (for example, carbon black has $\epsilon=0.95$). However, emissivity of a tungsten wire varies from $\epsilon=0.032$ (at $30^{\circ} C$) to $\epsilon=0.35$ (at $3300^{\circ}C$). Using the result of your fit, and assuming we know the Stephan-Boltzman constant $\sigma$ by some other means, what is $\epsilon$ and what is the uncertainty on it?  Is it consistent with tungsten? What else could be affecting this measurement?
 
-Let us examine the quality of the fit more carefully. For that it is convenient to make a separate plot of the \emph{residual} - the difference between the experimental points and the fit values. For a proper fit function we expect the experimental points to be randomly distributed around zero. Analyze your results. Do the points seem to systematically differ from the fit line in a particular region?  Can you think of a reason why that would be? 
+Let us examine the quality of the fit more carefully. For that it is convenient to make a separate plot of the \emph{residual} - the difference between the experimental points and the fit values. For a proper fit function, we expect the residuals to be randomly distributed around zero. Analyze your results. Do the points seem to systematically differ from the fit line in a particular region?  Can you think of a reason why that would be? 
 
 
 
@@ -178,9 +178,14 @@ S(r)=\frac{S_0}{2\pi r^2}
 
 \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 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} 
+\begin{figure}[h]
+	\centering
+	\includegraphics[height=2.5in]{./pdf_figs/blackbody_radn_curves} 
+	\caption{Black body radiation spectrum for objects with different temperatures.}%
+	\label{fig:bbStars}
+\end{figure}
 
-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.  
+The human bodies, of course, are much cooler than stars and emit in infrared range. This radiation is invisible to a 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}