\experiment{Barrier Penetration}

\section{Introduction}


A striking consequence of quantum mechanics is the prediction that a
particle of total energy $E$ located in a potential well of depth
$V_{0} > E$ has a finite probability of escaping if the walls of the
potential well have a finite thickness. This phenomenon, known as
\emph{barrier penetration} or tunneling, is not uncommon at the atomic
or subatomic scale; for example, $\alpha$ decay occurs via tunneling
of the $\alpha$ particle through the Coulomb barrier of the
radioactive nucleus (see Section 9.3 of Townsend or Section 16-2 of
Eisberg and Resnick). While barrier penetration is hardly commonplace
on the macroscopic scale, it can be seen; in fact, barrier penetration
is a property of both classical and quantum mechanical wave motion.

An optical analog of barrier penetration, known as ``frustrated total
internal reflection,'' is described formally by the same equations
that describe quantum mechanical tunneling. In this phenomenon, a
light beam traveling through glass (or any other transparent medium
with an index of refraction $n > 1$) is incident on the glass-air
interface. For sufficiently small angles of incidence, the light is
partly reflected and partly transmitted into the air. But for angles
of incidence greater than the ``critical angle'' $\sin^{-1}(1/n)$, the
beam is totally reflected back into the glass; no light is transmitted
into the air. The oscillating electromagnetic field of the light does
not stop precisely at the interface, however; it extends some
distance into the air. Ifanother piece of glass is brought close
enough to the interface, this electromagnetic field can then
propagate away from the interface (thus the total internal reflection
is ``frustrated''). The trick is getting the second piece of glass
close enough, to within about a wavelength of the interface. Unless
the interface is very flat, the effect won't occur; in any case, the
gap is so small as to be invisible. 
One can change the scale of electromagnetic radiation to the
microwave region, where wavelengths are on the order of centimeters. Then this
phenomenon can be easily observed. For radiation with wavelengths of
a few centimeters, polyethylene becomes a good substitute for glass; it is
almost transparent to microwaves and has an index of refraction very
similar to that of glass for optical frequencies. A microwave beam
traveling through a polyethylene block and incident on the
polyethylene-air interface at an angle of \dg{45} undergoes total
internal reflection, provided the interface is isolated. Again, there
is an oscillating electromagnetic field extending into the air beyond
the interface, as you will see. If another polyethylene block is
brought close enough to the interface, it should allow a transmitted
wave to propagate away from the interface. You will study this
phenomenon. 

In the experiment there are two \dg{45}-\dg{45}-\dg{90} polyethylene
prisms arranged so that the two hypotenuse faces can be brought close
together. A microwave beam is incident on the first prism
perpendicular to one base, travels through the prism and strikes the
hypotenuse at \dg{45}. If the perpendicular separation of the two prisms
is $d$, then the fraction of the microwave radiation intensity that can
penetrate the gap between the prisms ($T$) is given by 
\begin{equation}
	\label{eq:Barrier1}
	T = \left( 1 + \alpha \sinh^{2} \beta d \right)^{-1}
\end{equation}
The form of this equation is identical to that seen for quantum
mechanical barrier penetration (see Section 4.7 of Townsend).
The coefficients $\alpha$ and $\beta$ can be obtained from
classical electromagnetic theory. For the geometry of this experiment
one obtains\footnote{See, e.g., J. Strong, Concepts of Classical
  Optics (1958), Section 6-9.}
\begin{align}
	\alpha &=
    \frac{(n^{2}-1)^{2}}{n^{2}(n^{2}-2)}  \label{eq:Barrier2} \\
    \beta &= \frac{2 \pi}{\lambda} \sqrt{(n^{2}-2)/2}  
\end{align}
where $n$ is the index of refraction of the polyethylene and
$\lambda$ is the wavelength in air. 

\section{Preliminary Measurements}

To predict the intensity of the microwave radiation transmitted across
the gap, you need to know the microwave wavelength and the index of
refraction of polyethylene. And there is one more subtle question,
namely, does the detector measure the intensity of the microwave
radiation? That is, does the detector respond linearly to the square
of the microwave electric field strength? You will perform some
preliminary measurements to determine these three parameters.

\begin{description}
\item[Transmitter and receiver] The microwave transmitter operates at
  10.5 GHz. The microwaves emitted from the horn are polarized
  parallel to the long axis of the Gunn diode (the slender shiny
  cylinder located at the base of the horn). While the transmitter can
  be rotated to change the polarization axis, for this experiment the
  Gunn diode should be kept vertical (0 on the scale). The microwave
  receiver consists of a detector diode mounted similarly at the base
  of the detector horn. The diode responds only to the component of
  the microwave signal that is parallel to the diode axis. There are
  four amplification ranges and a variable gain control on the
  receiver.  \scream{Always start at the least sensitive range
    $(30\times)$ to avoid damaging the electronics.}

\item[Wavelength] A good way to measure the wavelength is to use a
Fabry-Perot interferometer. Follow the procedures in the PASCO
Instructions and Experiments Manual, Experiment 9, p.\ 26. Compare
your result with the expected value for \hval{10.5}{GHz} radiation. 

\item[Receiver Response] You can check to what extent the meter
  reading on the microwave receiver is proportional to the intensity
  by seeing how the meter reading changes as the diode axis of the
  receiver is rotated relative to the polarization direction of the
  transmitted electric field (transmitter diode axis). If the meter
  responds directly to the electric field strength, then the meter
  reading should be proportional to $\cos \theta$, where $\theta$ is
  the angle of the detector diode relative to the E field. If the
  meter responds to intensity, then the meter reading should be
  proportional $E^{2}$, hence to $\cos^{2}\theta$. In fact, the
  detector diode is a nonlinear device, so that the way the meter
  responds can vary with the strength of the field. To test the meter
  in the relevant range, you should separate the transmitter and
  receiver by about \val{2.5}{m}, roughly the same distance you will
  be using in the barrier penetration experiment. (The goniometer arm
  on which the transmitter and receiver were mounted for the
  wavelength determination should be removed from the table to
  minimize spurious reflections.) Adjust the detector position to get
  a maximum signal when both transmitter and receiver diodes are
  vertical. Set the meter reading to full scale. Then rotate the
  receiver in \dg{15} increments up to \dg{90} and record the meter
  readings. Compare with $\cos\theta$ and $\cos^{2}\theta$.  The meter
  response should be quite close to $\cos^{2}\theta$ (intensity) at
  this separation. If it is not, see your instructor.

\item[Index of refraction] Light (or microwave radiation) incident on a
prism is refracted on entering and leaving the prism. If the prism is
oriented so that the angle with which the beam leaves the prism
(where as usual the angle is measured relative to the normal to the
surface it is exiting) is the same as the angle at which it enters
(relative to the normal to the front surface)---see the figure---then
the index of refraction is given by
\begin{equation}
	\label{eq:Barrier3}
	n \equiv \frac{\sin \theta_{i}}{\sin \theta_{r}} =
    \frac{\sin(\psi + \alpha/2)}{\sin(\alpha/2)} 
\end{equation}
where $\theta_{i}$ is the angle of incidence, $\theta_{r}$ is the
angle of refraction, $\alpha$ is the apex angle of the prism, and
$\psi$ is defined as shown in \fref{Barrier1}.
\begin{figure}[h]
  \centering
  \igw[2.5in]{Barrier1}
  \caption{Symmetric propagation.}
  \label{fig:Barrier1}
\end{figure}

(\scream{Note}: As part of your writeup for this experiment, you
should derive this formula.) Two rotating goniometer arms are attached
to the platform supporting the fixed polyethylene prism. (You may have
to roll back the second polyethylene prism to see the second
goniometer arm.) The angle scales marked for each arm correspond to
the angle $\psi$ in the figure above. Position the transmitter on one
arm and the receiver on the other. Rotate the arms symmetrically
relative to the \dg{45} apex angle of the prism (i.e., both angles
must be the same) and locate the angle $\psi$ where the receiver
signal is a maximum.  Use this to determine the index of refraction of
the prism.
\end{description}

\section{Barrier penetration measurement}

Remove the transmitter and receiver from the goniometer arms and
rotate the arms so that they are parallel to the base of the prism.
Use the pegs provided to fix these arms in position. Now roll the
second prism all the way up to the first prism. Place the receiver in
the guides on the movable platform behind the second prism. (The
separation of the prism and the receiver should be about 30 cm to the
base of the receiver horn.) Place the source about \val{190}{cm} from
the front face of the fixed prism and align it carefully to give a
maximum reading on the receiver meter. Slide the receiver forward and
backward a few centimeters in the guides to maximize the signal. Set
the sensitivity for full-scale reading when the two prisms are
touching.  Now roll the prism back. You should see the meter reading
drop quickly, with essentially 0 reading for a separation of several
centimeters.

If the reading does not drop to at most 5-10\% of the initial reading,
you need to realign the source and detector and look for any causes of
extraneous reflections. With the second prism still several
centimeters away, adjust the position of the second receiver, located
to detect the beam reflected from the hypotenuse of the fixed
prism. This detector should be about 30 cm from the fixed
prism. Adjust its position for a maximum meter reading and set this to
1 when the movable prism is ``far'' away. Now roll the second prism
toward the fixed prism and note the meter readings on the two
receivers.  Describe qualitatively what you see.

Since the receiver monitoring the reflected beam can produce spurious
reflections which affect the transmitted beam receiver, it should be
removed for the rest of the experiment. \scream{Be sure to turn off
  the receivers after you are finished with them, as their batteries
  run down quickly}. Recheck the transmitter receiver for the extreme
positions of the movable prism, and if necessary readjust and reset
before taking quantitative measurements. Vary the movable prism
location, measure the perpendicular separation $d$ of the prisms, and
plot the resulting transmission coefficient $T$. On the same graph,
plot the predicted transmission coefficient. Compare the two and
comment.

