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| -rw-r--r-- | single-photon-interference.tex | 19 |
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diff --git a/single-photon-interference.tex b/single-photon-interference.tex index 53881da..e0f21d6 100644 --- a/single-photon-interference.tex +++ b/single-photon-interference.tex @@ -124,7 +124,7 @@ The experiment consists of three steps: \subsection*{Step 1: Observing interference and finding dial settings} \begin{framed} {\center{\large \bf \textcolor{red}{Important personal safety information}\\} } -The apparatus has a 5-mW diode laser with an output wavelength of $670 \pm 5$~nm. It could potentially harm your eyes if you are not careful. Don't observe the beam directly (i.e., by staring into it) and be careful of reflections off of reflective surfaces. +The apparatus has a 5-mW diode laser with an output wavelength of $670 \pm 5$~nm. It could potentially harm your eyes if you are not careful. Never look directly onto the laser and be careful with the reflected beams. \end{framed} @@ -220,11 +220,14 @@ Now you need to change the apparatus to use the light bulb and PMT. Keep the shu \item[Bulb settings] Now set the 3-position toggle switch to the ``bulb'' position and dial its intensity to 8. As long as you haven't touched anything else the apparatus is still aligned and light from the bulb will form interference patters that you can measure with the PMT. +\item[Source slit positioning] The bulb is not a point light source. Thus, we need to follow the prescription of Young and add a single slit (\bf{S1} in Fig.~\ref{young.fig}) in front of the double slit (there is a holder for it between the bulb and the double-slit holder). + \item[Double slit mode] Use the dial to move the slit blocker into double slit mode. \item[Close it up] Now close and latch the cover. -\item[High voltage and PMT pulses] Have an instructor show you how to use the DVM to measure the voltage supplied to the PMT and how to observe the PMT output on an oscilloscope. With the instructor, gradually turn on the HV to get up to a setting of around \unit[700]{V}. You should see pulses on the oscilloscope screen. +\item[High voltage and PMT pulses] Have an instructor show you how to use the DVM to measure the voltage supplied to the PMT and how to observe the PMT output on an oscilloscope. With the instructor, gradually turn on the HV to get up to a setting of around ``5'' mark on the dial.%\unit[700]{V}. +You should see pulses on the oscilloscope screen. % instructor: going for a darkrate of ~100Hz and a signal rate of at least 1kHz at the center of the pattern. May need to play with HV and discriminator settings. @@ -350,23 +353,23 @@ plane to the screen. \begin{enumerate} %\item Begin by roughly estimating the number of photons per second arriving to the detector. -\item Begin by computing the amount of electrical power that is converted to light by the light bulb. Assume we operate the bulb at \unit[6]{VDC} and it draws \unit[0.1]{A} of current. From this you can get the power drawn by the bulb. % Ans: 1.2W +\item Begin by computing the amount of electrical power that is converted to light by the light bulb. Assume we operate the bulb at \unit[6]{VDC} and it draws \unit[0.1]{A} of current. From this you can get the power drawn by the bulb. % Ans: 0.6W \item Most of the power goes into heating the bulb. Assume 5\% of it is converted to light evenly distributed from 500-1500~nm. -\item The bulb has a green filter that blocks the light outside the range 541-551~nm. From this you can get the amount of power radiated as green light. %1.2W*0.05*10nm/1500nm = 6e-4W +\item The bulb has a green filter that blocks the light outside the range 541-551~nm. From this you can get the amount of power radiated as green light. %0.6W*0.05*10nm/1000nm = 3e-4W \item You can convert the power (energy/time) into a rate of photons using the fact that $E = \frac{hc}{\lambda}$. From this you can find the rate of green photons emitted from the bulb+filter. % One 546nm photon has E=6.626e-34 J sec * 3e8 m/sec / 546e-9m = 3.64e-19 J -% (6e-4 J/sec) / (3.64e-19 J/photon) = 1.75e15 photons/sec +% (3e-4 J/sec) / (3.64e-19 J/photon) = 8.74e14 photons/sec \item Only the small fraction of the photons emitted by the bulb get through the double slits. Let's assume that slits are about $R=\unit[50]{cm}$ from the bulb, and each of them has the width of \unit[0.1]{mm} and a height of \unit[1]{cm}. Assuming that the bulb radiates evenly onto a spherical surface of radius $R$, compute the fraction of the light that would pass through the slits. % slit area/4 pi R^2=3.18e-7 -% rate passing slit is 3.18e-7 * 1.65e15 photons/sec = 5.56e8 photons/sec +% rate passing slit is 3.18e-7 * 2* 8.74e14 photons/sec = 5.56e8 photons/sec %\item After the single slit the beam is diffracted into a single slit interference pattern and impinges on the double slit. If you had the apparatus in front of you, you could estimate the area that the interference pattern covers. A good guess is $\unit[1]{cm^2}$. Assume this and that the double slits have an area twice that of the single slit. You can compute the rate of photons passing through the double slits. % 1.31e10 photons/sec * 2 * 7.96e-6 = 2.1e5 photons/sec -\item We know that a photon travels at a speed $c=\unit[3\times 10^8]{m/sec}$. Divide that by the number of photons per second that pass through the slits. This gives you the typical space between photons in meters/photon. % 1428 m/photon +\item We know that a photon travels at a speed $c=\unit[3\times 10^8]{m/sec}$. Divide that by the number of photons per second that pass through the slits. This gives you the typical space between photons in meters/photon. % 0.54 m/photon -\item Invert the number from the previous bullet point and multiply by the 0.5~m between the double slit and detector slit to determine how many photons are in the setup on average. Is it a number less than one? If so, there is a good reason to believe that the photons must be interfering with themselves to form the light and dark fringes you see. % 3.5e-4 photons in the half meter downstream of the double slit +\item Invert the number from the previous bullet point and multiply by the 0.5~m between the double slit and detector slit to determine how many photons are in the setup on average. Is it a number less than one? If so, there is a good reason to believe that the photons must be interfering with themselves to form the light and dark fringes you see. % 0.9 photons in the half meter downstream of the double slit \end{enumerate} |