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\begin{center}
{\bf Notes on Geometric Optics and Interference effects of Electromagnetic Radiation}
\end{center}

\bigskip

We just finished discussing Maxwell's Equations and the free space solutions of
electromagnetic radiation.  Now we examine some practical applications 
for electromagnetic radiation in the visible region.  We start with geometric
optics, which is followed by interference effects.

\bigskip

\noindent{\it Ray Tracing}\\

\bigskip

Electromagnetic radiation can reflect off surfaces, transmit through materials, interact
with charge, etc.  Applying Maxwell's equations directly would involve solving for
$\vec{E}(\vec{r},t)$ and $\vec{B}(\vec{r},t)$ for the appropriate boundary conditions
of the situation.  Carrying out such calculations can be difficult, however for many
situations one can follow the "rays" of the radiation.  A ray is a line that
is perpendicular to the plane of the electric and magnetic fields of the radiation.
It is a line that is in the direction of Poynting's vector, $\vec{S}$.  To determine
what the radiation does, we will follow the rays as they reflect off of, or travel
through materials.

How accurate is ray tracing?  We have all seen a laser beam, or seen a thin light
beam after it has passed through a small hole in piece of paper.  It is as if the
light stays as a small thin line as it travels.  So one can imagine breaking up
light into thin lines, and following what each line does as it travels.  One has
to be careful with this picture.  As light travels, it's cross sectional area cannot
remain constant.  We will discuss this more later under the topic of diffraction.
I mention one point now.  The amount of spreading, that is how much the cross section
area increases, depends on the ratio:

\begin{equation}
     spreading \; angle \sim {\lambda \over d}
\end{equation}

\noindent where $\lambda$ is the wavelength of the radiation, and $d$ is the diameter
of the beam.  The wavelengths for visible light range from $400-700 \; nm$.  So if
a beam of light is $1 \; mm$ in diameter, then 
$(7 \times 10^{-7})/(10^{-3}) \approx 7 \times 10^{-4}$ and one not notice 
the spreading unless the beam travels hundreds
of meters.  So, for visible light, it is a good approximation to break up the radiation 
into thin rays and follow each ray as it travels through the environment as long
as it doesn't travel hundreds of meters.  Next we will take a "ray tracing" approach
in analyzing how light reflects off surfaces and transmits through different media.

\bigskip

\centerline{\bf Geometric Optics and Mirrors}

\bigskip

If a beam of light strikes a flat mirror, or smooth reflecting surface, the angle
that the light makes when it is reflected from the mirror equals the angle
that it stuck the mirror.  We know this from experience, and we will demonstrate
in lecture that the incident and reflected angles are equal.  When we discuss refraction,
we will understand why this is true in general for waves reflecting off of a flat
surface.

What if the mirror is not flat, but is curved.  In this case, the situation is not
difficult if the beam is very narrow.  That is, if the diameter of the beam is much
smaller than the radius of the curved surface where it strikes the mirror.  If the
beam is very narrow, then the mirror is essentially flat where the beam hits it.
So, as with the completely flat mirror, the incident and reflected angles will be 
equal.  In order to handle curved surfaces, the best angles to use are the angles
with respect to a line that is perpendicular to the surface that the beam hits.

In the figure we show how the incident and reflected beam, or ray, is measured.
Go to where the beam, or ray, stikes the surface.  Then draw a line perpendicular
(or normal) to the surface.  The incident angle, $\theta_i$, is measured from the
incoming ray to the normal.  The reflected angle, $\theta_r$, is measured from
the outgoing ray to the same normal.  For reflection:

\begin{figure}
\includegraphics[width=14cm]{fig2342a.png}
\end{figure}

\begin{equation}
   \theta_i = \theta_r
\end{equation}

\noindent This equation is true even if the surface is curved.

\bigskip

\noindent {\it Flat Mirror}

\bigskip

Consider an object that is in front of a flat mirror.  Let the distance
from the object to the mirror be $o$.  The object emits light in all
directions, which we show by rays.  Some of the light will reflect off the 
mirror.  Let's follow the rays that reflect off the mirror.  Each ray
will reflect such that $\theta_r=\theta_i$ as shown in the figure.

We can extend the reflected ray back into the mirror, as is drawn in
the figure as dashed lines.   The dashed lines intersect at a point.
{\it This point is where one would see the object if one looked in the mirror.}
That is, if you looked in the mirror, you would see the object at that 
position no matter where your eyes are.  This position is where the image
is located.  We will label the distance from the mirror to the image as
$i$.  For the flat mirror, you can see that

\begin{equation}
     i = o
\end{equation} 

\noindent by using geometry and $\theta_r = \theta_i$.   An image that
is seen by looking into a mirror (or lens) is called a {\it virtual image}.
The image cannot be projected onto a screen, but rather it can only
be seen by looking into the mirror.

\bigskip

\noindent {\it Curved Mirrors}

\bigskip

We first consider a special kind of mirror, one that has axial symmetry and
is curved in such a
way as to focus light to a point.  The mirror has the following property:
{\it rays parallel to its axis focus at a point}, the focal point. 
The distance from the mirror to the {\it focal point} is called
the {\it focal length} of the mirror.  We label this distance $f$.  To 
produce the focusing of rays at a point, the mathematical shape of the mirror
needs to be a parabola (rotated about its axis).

Note that if we put a light source exactly at the focal point, then the light
that reflects from the mirror will emerge parallel to the axis.  That is,
by putting a light bulb at the focal point, the reflected light will be a
nice beam, like a flashlight.

Now, consider an object that is placed between the focal point and the
mirror.  Let the object have a height $h_0$ and be located a distance
$o$ ($o<f$) from the mirror.  The object is represented by an arrow in the figure.
We can trace three rays from the tip of the arrow that reflect off the mirror:\\

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will reflect off the mirror and pass
through the focal point.  The dashed line extends the reflected ray back
behind the mirror.\\

\noindent {\it A ray directed away from the focal point} ($2 \rightarrow 8$)\\
The ray that is directed away from the focal point will reflect parallel
to the axis.  The dashed line extends the reflected ray back behind the
mirror.\\

\begin{figure}
\includegraphics[width=14cm]{fig2342b.png}
\end{figure}

\noindent {\it A ray that strikes the center of the mirror} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will reflect back
at the same angle that it strikes the mirror.  This part of the mirror
is perpendicular to the axis of the mirror.  The dashed line extends
the reflected ray back behind the mirror.\\

\bigskip

All three of the dashed lines intersect at a point.  Although we only
considered these three rays, the dashed lines 
for all rays from the tip would intersect there as well.  Wherever you look into
the mirror, you will see the tip of the arrow at this intersection point.  The
intersection point is the image of the arrows tip.  Since the image is
"in the mirror", it is a virtual image.  We label $i$ as the distance
from the image to the mirror, $h_i$ as the height of the image, and $h_o$ has
the height of the object.

Using geometry, we can determine the relationship between $o$, $i$, $f$,
$h_i$ and $h_o$.\\

\noindent Triangle $1-4-5$ and triangle $1-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{f+i} \over f}
\end{equation}

\noindent Triangle $2-6-7$ and triangle $4-5-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{i+f} \over f} \\
  {1 \over o} - {1 \over i} & = & {1 \over f}
\end{eqnarray*}

The ratio of $h_i/h_o$ is how much the image increased in size, and is called the
{\it magnification} $m$.  For the case just treated, we see that $m=h_i/h_o=i/o$.
Since for this case $i>o$, the magnification is greater than 1.\\

Now, lets use the same mirror, but place the object such that $o>f$.  We can follow the
same rays from the tip of the arrow as before:

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will reflect off the mirror and pass
through the focal point.  The dashed line extends the reflected ray back
behind the mirror.\\

\noindent {\it A ray directed through the focal point} ($2 \rightarrow 8$)\\
Since the $o>f$ a ray from the tip can pass through the focal point.
The ray that passes through the focal point will reflect parallel
to the axis.  The dashed line extends the reflected ray back behind the
mirror.\\

\noindent {\it A ray that strikes the center of the mirror} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will reflect back
at the same angle that it strikes the mirror.  This part of the mirror
is perpendicular to the axis of the mirror.  The dashed line extends
the reflected ray back behind the mirror.\\

\bigskip

From the figure we can see that the dashed lines do not converge.  One will
not see an image of the arrow in the mirror.  However, in this case, the
reflected rays do converge.  The rays from point "2" do converge at point
"4".  However, now point "4" is in front of the mirror.  Although we just
followed three rays, all rays from point "2" that hit the mirror will
converge at point "4".  If we put a piece of paper, or screen, where
point "4" is, we would see an image of the arrow on the screen.  An
image that can be projected on a screen is called a {\it real image}.
If $o>f$ the image is real.

Once again, we can use geometry to find the relationship between $o$, $i$,
$f$, $h_0$ and $h_i$:

\noindent Triangle $1-4-5$ and triangle $1-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{i-f} \over f}
\end{equation}

\noindent Triangle $2-6-7$ and triangle $4-5-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Once again, equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{i-f} \over f} \\
  {1 \over o} + {1 \over i} & = & {1 \over f}
\end{eqnarray*}

\noindent As before, the magnification $m$ equals $h_i/h_o = i/o$.  For the real
image, the magnification can be less than one ($o>2f$) or greater than one 
($f<o<2f$).

Note that the equation for $o<f$ is similar to the equation for $o>f$.  The only
difference is the minus sign in front of $i$.  We can combine both equations into
one, which will be valid for any value for $o$ if we assign a minus sign to
a virtual image.  That is, $i>0$ means the image is in front of the mirror,
and $i<0$ means the image is "behind" or "in" the mirror.  The universal equation
for all values of $o$, $i$, and $f$ is

\begin{equation}
  {1 \over o} + {1 \over i} = {1 \over f}
\end{equation}

\noindent where $i>0$ for a real image in front of the mirror, and $i<0$ for a
virtual image behind the mirror.  This equation also works for a flat mirror.
For a flat mirror, $f = \infty$, and we have $i=-o$.

\bigskip

\noindent {\it Non-Focusing (diverging) curved mirror}

\bigskip

Now we consider a curved mirror that is curved in the opposite way as the converging
mirror we just analyzed.  The property that the diverging mirror will have is
{\it rays parallel to the axis diverge from a point behind the mirror}.  As before, the 
"divergence" point is called the focal point.   Although parallel rays do not
"focus" at this point, they diverge from the point, it is still useful to use the same 
name.  The distance from this focal point to the mirror is the focal length, "$f$".
Note that if rays are directed at the focal point, after they reflect from the mirror
they will travel parallel to the axis.  This property is the same as the converging
mirror, but now the physical rays cannot go through the focal point.


Lets examine that the image will be for an object of height $h_o$ placed a distance
$o$ from the mirror.  We will follow the same three rays from the tip of an arror
as we did with the converging mirror, and label the diagram the same way:\\

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will reflect off the mirror and travel 
away from the focal point.  The dashed line extends the reflected ray back
behind the mirror.\\

\noindent {\it A ray directed at the focal point} ($2 \rightarrow 8$)\\
A ray from the tip of the arrow directed towards the focal point
will reflect parallel to the axis.  The dashed line extends the reflected ray back behind the
mirror.\\

\noindent {\it A ray that strikes the center of the mirror} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will reflect back
at the same angle that it strikes the mirror.  This part of the mirror
is perpendicular to the axis of the mirror.  The dashed line extends
the reflected ray back behind the mirror.\\

All three of the dashed lines intersect at a point.  Although we only
considered these three rays, the dashed lines 
for all rays from the tip would intersect there as well.  Wherever you look into
the mirror, you will see the tip of the arrow at this intersection point.  The
intersection point is the image of the arrows tip.  Since the image is
"in the mirror", it is a virtual image.  As before, we label $i$ as the distance
from the image to the mirror, and $h_i$ as the height of the image.

Using geometry, we can determine the relationship between $o$, $i$, $f$,
$h_i$ and $h_o$.\\

\noindent Triangle $1-4-5$ and triangle $1-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{f-i} \over f}
\end{equation}

\noindent Triangle $2-6-7$ and triangle $4-5-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{f-i} \over f} \\
  {1 \over o} - {1 \over i} & = & -{1 \over f}
\end{eqnarray*}

\begin{figure}
\includegraphics[width=14cm]{fig2342c.png}
\end{figure}

\noindent Wow, this equation is very similar to the equation for the converging
mirror.  If we assign a negative value to the focal length of a diverging mirror,
then we can use {\it one equation} for both converging and diverging mirrors,
and the cases which have real or virtual images.  Thus, we can produce a
{\it Universal Equation for converging and diverging mirrors}:

\begin{equation}
   {1 \over o} + {1 \over i} = {1 \over f}
\end{equation}

\noindent {\it where  $o$ and $i$ are greater than zero for real objects
and images, and less than zero for virtual objects and images.
The focal length $f$ is positive for converging mirrors and 
negative for diverging mirrors.}\\

There are a few final points that need mentioning:

\begin{enumerate}
\item The magnification is defined as $m=h_i/h_o$, which is equal to
$i/o$.  Nowever, we are defining the signs of $o$ and $i$ to represent
real or virtual quantities.  When a real object produces a real image,
the image is inverted.  So we are can define the magnification as $m=-o/i$,
to include the inverted (non-inverted) nature of the image.  A negative
$m$ means the object and image have opposite orientations.  A positive
$m$ means that the object and image have the same orientation.
\item It is much easier to make a spherical mirror than a parabolic mirror.
If the light only reflects near the center of a spherical mirror then the
rays will focus at a point fairly well.  The focusing will not be perfect,
as with a parabolic surface.  If the radius of curvature of the spherical
surface is $r$, then the focal length will be $f \approx r/2$.  This accuracy
of this approximation can be determined as follows.  From the figure we
can see that

\begin{equation}
  {{r/2} \over {r-f}} = cos(\alpha )
\end{equation}

\noindent We can solve this equation for $f$ to obtain

\begin{equation}
   f = r(1 - {1 \over {2 cos(\alpha )}})
\end{equation}

\noindent So far this equation is exact.  Now, if $\alpha << 1$ we can
use the Taylor expansion for $1/cos(\alpha)=sec(\alpha )$:

\begin{eqnarray*}
  f & = & r(1 - {1 \over 2}(1 + {\alpha^2 \over 2} - \dots )) \\
    & = & {r \over 2}(1 - {\alpha^2 \over 2} + \dots ) \\
    & \approx & {r \over 2} (1 - {1 \over 2}({h \over r})^2 + \dots )
\end{eqnarray*}

\noindent where we have substituted $\alpha \approx h/r$.  So the first
correction to $f=r/2$ is of order $(1/2)(h/r)^2$.  So parallel rays will focus
on the axis of the mirror in a region that is a distance of $r/2$ to 
($r/2 - (1/2)(h_{max}/r)^2$) from the mirror.  Thus, the "bluryness" of the focal point
will be approximately $(1/2)(h_{max}/r)^2$.  If $h_{max}<<r$ then the
focusing is pretty good at $f=r/2$.
\item A converging mirror is also called a concave mirror.  A diverging mirror
is also called a convex mirror.  The diverging mirror only has virtual images.
Real images (from a real object) can only be produced with a converging 
mirror, and only when $o>f$.
\item Our derivation of the relationship between $o$, $i$, $f$, $h_o$, $h_i$,
and $m$ was based on geometry.  Hence, the analysis is termed "Geometric Optics".
\end{enumerate}

We now consider images formed by lenses.  First we need to understand what happens
when light passes from one material (medium) to another.  We start with Snell's
law, then derive the relationships for thin lenses.

\bigskip

\centerline{\bf Snell's Law}

Snell's law deals with how light travels from one material (or medium) to another.
Light travels at the speed $c = 3 \times 10^8 \; m/s$ in vacuum.  In a transparent
material it travels slower.  If light travels at the speed $v$ in a medium,
we define the {\it index of refraction} $n$ as $v=c/n$.  In vacuum, $n=1$, and
for transparent materials $n>1$ since light travels slower in materials than in
vacuum.  We should mention that the index of refraction $n$ will depend on
the material and in general on the frequency of the radiation.

Suppose there are two materials (media) which we label as "1" and "2", and that
the interface where they are in contact is a flat plane.  Let the speed of
waves in the respective media be $v_1$ and $v_2$.  From experiment, we observe
that light "bends" as it travels from one material into another.  This phenomena
is called refraction.  We label the angle that the incident wave makes with
the interface as $\theta_i$, and the angle that the refracted wave makes as $\theta_R$. 
{\it Both angles are measured from the direction of the wave with respect to a 
line perpendicular (normal) to the interface}. Both angles are always positive and
between $0^\circ$ and $90^\circ$.  We want to come up with a 
relationship between $\theta_i$ and $\theta_R$.  

Suppose a plane wave travels from "1" into "2" as shown in the figure.  The figure is a snapshot
of the wave as it travels from "1" into "2".  The lines in the
figure are points which have the same phase in the snapshot.  If you want, one of the lines
can be the crest of the wave when the snapshot is taken.  The points $A$ and $A'$ have
the same phase.  The points $B$ and $B'$ also have the same phase, although different
than $A$ and $A'$.  
Let $x \equiv (distance \; A \rightarrow B)/\lambda_2$.  With this assignment,
the distance from $A$ to $B$ equals $x \lambda_2$.  Since $A$ and $A'$ have the same
phase, and $B$ and $B'$ have the same phase, the distance from $A'$ to $B'$ must
be $x\lambda_1$.  Let $l$ be the distance on the interface between $A$ and $B'$.
From the figure we see that

\begin{equation}
   sin(\theta_1) = {{x\lambda_1} \over l}
\end{equation}  

\noindent and 

\begin{equation}
   sin(\theta_2) = {{x\lambda_2} \over l}
\end{equation}  

\begin{figure}
\includegraphics[width=14cm]{fig2342d.png}
\end{figure}

\noindent equating $x/l$ in each equation we have

\begin{eqnarray*}
  {{sin(\theta_1)} \over \lambda_1} & = & {x \over l} = {{sin(\theta_2)} \over \lambda_2} \\
  {{sin(\theta_1)} \over \lambda_1} & = & {{sin(\theta_2)} \over \lambda_2} \\
\end{eqnarray*}

\noindent which is true for any value of $x$.  $x$ does not need to be $1.0$.  The
final line should also hold in the limit as $l \rightarrow 0$, and consequently
$x \rightarrow 0$.  

There is one more bit of physics to refraction: the frequency of the radiation in
medium "2" will be the same as the frequency in medium "1".  Why is this true? 
At the interface, medium "1" and medium "2" share the same location. So
the time dependence of the fields are the same at the interface and
consequently in the different media.  That is $f_1 = f_2 \equiv f$.  
Dividing both sides of the equation by $f$ yields

\begin{eqnarray*}
  {{sin(\theta_1)} \over {f \lambda_1}} & = & {{sin(\theta_2)} \over {f \lambda_2}} \\
  {{sin(\theta_1)} \over {v_1}} & = & {{sin(\theta_2)} \over {v_2}} \\   
\end{eqnarray*}

\noindent This refraction equation is true for waves in general.  For light, in terms of
the indicies of refraction, we have

\begin{equation}
    n_1 sin(\theta_1 ) = n_2 sin(\theta_2 )
\end{equation}

\noindent This equation is known as Snell's Law.

Is refraction the only thing that can occur when light travels from one medium 
into another?  No, light can (and usually does) reflect as well.  For the
reflected light, the angle of reflection equals the angle of incidence as with
a mirror.  Actually, the derivation of Snell's law will work for reflection.
In this case $n_1=n_2$, since the medium is the same.  If $n_1=n_2$ then
$\theta_1 = \theta_2$.  So, in general when light passes from one medium into another
there can be reflection as well as refraction.  Usually, both occur.

Some points regarding reflection and refraction:
\begin{enumerate}
\item The angles in Snell's law are measured from the ray to a line normal
(or perpendicular) to the interface.
\item If light travels from a faster medium into a slower medium, rays bend
towards the normal.  If light travels from a slower medium into a faster
medium, rays bend away from the normal.
\item Suppose light travels from medium "1" into medium "2", and $n_1>n_2$.  That
is, light travels from a slower medium into a faster one.  Since

\begin{equation}
            sin(\theta_2 ) = {n_1 \over n_2} sin(\theta_1 )
\end{equation}

\noindent the angle $\theta_2$ is greater than $\theta_1$.  In this case, the
refracted light bends away from the normal.  If $(n_1/n_2)sin(\theta_1)$ is
larger than one, there is no real solution for $\theta_2$.  The light cannot
refract into medium "2".  All the light is reflected!  We call this phenomenon
{\it total internal reflection}.  The critical angle $\theta_c$ when this happens
is when

\begin{eqnarray*}
  {n_1 \over n_2} sin(\theta_c) & = & 1 \\
       sin(\theta_c) & = & {n_2 \over n_1}
\end{eqnarray*}
  
\noindent For angles $\theta > \theta_c$ from the normal, all the incident light
is reflected.  
\end{enumerate}

Snell's law is the "physics" of refraction.  Now we can use this understanding
in practical applications.

\bigskip

\centerline{\bf Thin Lenses}

\bigskip

Glass, or other transparent materials, can be formed such that parallel rays focus
at a point on the other side of the glass.  The glass is called a {\it converging
lens}.  Alternatively, glass (or other transparent material) can be formed such
that parallel rays diverge from a point.  In this case we have a {\it diverging
lens}.  In lecture we will show that a converging lens is thicker in the middle
than on the edges, and a diverging lens is thinner in the middle than the
edges.  Let's first consider thin converging lenses, then thin diverging lenses.
The geometry will be very similar to the mirror analysis, and the final thin lens
formula will be the same as the one for mirrors.

\bigskip

\noindent {\it Converging Lenses}\\

\bigskip

As with mirrors, we will follow (the same) three rays to examine where the image is formed
and if the image is real or virtual.
First, consider an object that is placed between the focal point and the
lens, $o<f$.  Let the object have a height $h_0$ and be located a distance
$o$ ($o<f$) from the lens.  The object is represented by an arrow in the figure.
We can trace three rays from the tip of the arrow that travel through the lens:\\

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will go through the lens and pass
through the focal point.  The dashed line extends this ray back
behind the lens.\\

\noindent {\it A ray directed away from the focal point} ($2 \rightarrow 8$)\\
The ray that is directed away from the focal point will emerge parallel
to the axis.  The dashed line extends the refracted ray back behind the
lens.\\

\noindent {\it A ray that strikes the center of the lens} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will pass straight through
at the same angle that it strikes the lens.  This part of the lens
is line a plane window.  The dashed line extends
the transmitted ray back behind the lens.\\

\bigskip

All three of the dashed lines intersect at a point.  Although we only
considered these three rays, the dashed lines 
for all rays from the tip would intersect there as well.  Wherever you look into
the lens from the left side, you will see the tip of the arrow at this intersection point.  The
intersection point is the image of the arrows tip.  Since the image is
on the same side as the object, it is a virtual image.  We label $i$ as the distance
from the image to the mirror, $h_i$ as the height of the image, and $h_o$ has
the height of the object as before.

Using geometry, we can determine the relationship between $o$, $i$, $f$,
$h_i$ and $h_o$.\\

\noindent Triangle $7-4-5$ and triangle $7-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{f+i} \over f}
\end{equation}

\noindent Triangle $6-1-2$ and triangle $6-4-5$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{i+f} \over f} \\
  {1 \over o} - {1 \over i} & = & {1 \over f}
\end{eqnarray*}

The ratio of $h_i/h_o$ is how much the image increased in size, and is called the
{\it magnification} $m$ as in the case with the mirror.  For the case just treated, 
we see that $m=h_i/h_o=i/o$. \\

Now, lets use the same lens, but place the object such that $o>f$.  We can follow the
same rays from the tip of the arrow as before:\\

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will go through the lens and pass
through the focal point.  The dashed line extends the refracted ray back
behind the lens.\\

\noindent {\it A ray directed towards the focal point} ($2 \rightarrow 8$)\\
The ray that is directed towards the focal point will emerge parallel
to the axis.  The dashed line extends the refracted ray back behind the
lens.\\

\noindent {\it A ray that strikes the center of the lens} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will pass straight through
at the same angle that it strikes the lens.  The dashed line extends
the transmitted ray back behind the lens.\\

\bigskip

From the figure we can see that the dashed lines do not converge.  One will
not see an image of the arrow in the lens.  However, in this case, the
transmitted rays do converge.  The rays from point "2" do converge at point
"4".  However, now point "4" is on the other side of the lens.  Although we just
followed three rays, all rays from point "2" that hit the lens will
converge at point "4".  If we put a piece of paper, or screen, where
point "4" is, we would see an image of the arrow on the screen.  An
image that can be projected on a screen is called a {\it real image}.
If $o>f$ the image is real.

Once again, we can use geometry to find the relationship between $o$, $i$,
$f$, $h_0$ and $h_i$:

\noindent Triangle $7-4-5$ and triangle $7-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{i-f} \over f}
\end{equation}

\noindent Triangle $2-6-1$ and triangle $4-5-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Once again, equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{i-f} \over f} \\
  {1 \over o} + {1 \over i} & = & {1 \over f}
\end{eqnarray*}

\noindent As before, the magnification $m$ equals $h_i/h_o = i/o$.  For the real
image, the magnification can be less than one ($o>2f$) or greater than one 
($f<o<2f$).

\begin{figure}
\includegraphics[width=14cm]{fig2342e.png}
\end{figure}

As with the mirrors, the equation for $o<f$ is similar to the equation for $o>f$.  The only
difference is the minus sign in front of $i$.  We can combine both equations into
one, which will be valid for any value for $o$ if we assign a minus sign to
a virtual image.  That is, $i>0$ means the image is on the other side of the lens,
and $i<0$ means the image is on the same side of the lens as the object.  The universal equation
for all values of $o$, $i$, and $f$ for a converging lens is

\begin{equation}
  {1 \over o} + {1 \over i} = {1 \over f}
\end{equation}

\noindent where $i>0$ for a real image, and $i<0$ for a
virtual image.  This equation also works for a flat window.
For a flat window, $f = \infty$, and we have $i=-o$.

\bigskip

\noindent {\it Diverging Lens}\\

\bigskip

Finally, we consider a diverging lens.  This is a lens that is thinner in the center
than on the edges.  If the lens is formed properly, parallel light rays will diverge
on the other side of the lens as if they are coming from a single point.  Let's examine
the image that if formed with this kind of lens.  Place an arrow a distance $o$ from
the lens.  We trace the same three rays as before:\\

\noindent {\it A ray parallel to the axis} ($2 \rightarrow 3$)\\
The ray that is parallel to the axis will go through the lens and diverge
from the focal point.  The dashed line extends the refracted ray back
behind the lens.\\

\noindent {\it A ray directed towards the focal point} ($2 \rightarrow 8$)\\
The ray that is directed towards the focal point will emerge parallel
to the axis.  The dashed line extends the refracted ray back behind the
lens.\\

\noindent {\it A ray that strikes the center of the lens} ($2 \rightarrow 6$)\\
The ray that strikes the center of the mirror will pass straight through
at the same angle that it strikes the lens.  The dashed line extends
the transmitted ray back behind the lens.\\

\bigskip

From the figure we can see that the dashed lines do converge.  One will
see an image of the arrow in the lens.  The rays from point "2" do converge at point
"4".  However, point "4" is on the same side of the lens as the object.  Although we just
followed three rays, all rays from point "2" that hit the lens will
diverge from point "4".  If we look into the lens, we will see the
tip of the arrow at the point "4", and the arrow will be a virtual 
image. 

Once again, we can use geometry to find the relationship between $o$, $i$,
$f$, $h_0$ and $h_i$:

\noindent Triangle $7-4-5$ and triangle $7-3-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {{f-i} \over f}
\end{equation}

\noindent Triangle $2-6-1$ and triangle $4-5-6$ are similar triangles:

\begin{equation}
  {h_i \over h_0} = {i \over o}
\end{equation}

\noindent Once again, equating the $h_i/h_o$ we have

\begin{eqnarray*}
  {i \over o} & = & {{f-i} \over f} \\
  {1 \over o} - {1 \over i} & = & - {1 \over f}
\end{eqnarray*}

\noindent Wow, this equation is very similar to the equation for the converging
lens.  If we assign a negative value to the focal length of a diverging lens,
then we can use {\it one equation} for both converging and diverging lenses,
and the cases which have real or virtual images.  Thus, we can produce a
{\it Universal Equation for converging and diverging lenses}:

\begin{equation}
   {1 \over o} + {1 \over i} = {1 \over f}
\end{equation}

\noindent {\it where  $o$ and $i$ are greater than zero for real objects
and images, and less than zero for virtual objects and images.
The focal length $f$ is positive for converging lenses and 
negative for diverging lenses.}  This same equation applies universally
for mirrors and lenses because the properties of these elements are
the same:  {\it Parallel rays converge (focus) or diverge from the focal points
of both mirrors and lenses}. \\

\noindent A few final points on lenses and Snell's law:

\begin{enumerate}
\item Converging lenses are also called convex lenses.  Diverging lenses are also
called concave lenses.
\item Real images are only formed with converging lenses, and only when $o>f$.
\item The magnification $m=|h_i/h_o|$ as with mirrors.  If we adopt the convention
that + means upright and - means inverted, then $m=-i/o$ as with mirrors.
\item The index of refraction depends on the frequency (or color) of the light.
For water, a wavelength of $\lambda = 400 \; nm$, has $n = 1.35$.  For water, a 
wavelength of $\lambda = 700 \; nm$, has $n=1.33$.  Thus, for water, red light travels slightly faster
than blue light.  Blue light will therefore refract more than red in going from 
air into water, giving us a nice rainbow.
\end{enumerate}

\bigskip

\centerline{\bf Other topics in Optics}

\bigskip

We have covered the main idea used in geometric optics, Snell's law.  Considering
the different polarizations of light gives interesting results.  If there is time
in lecture we might discuss:\\

\begin{enumerate}
\item Images using two lenses: In this case, the image of the first lens is the object
of the second lens.  Applications include telescopes and microscopes.
\item Natural polarization:  If light scatters at a special angle, Brewster's angle,
it will be polarized.  The special incident angle is when the reflected ray makes an
angle of $90^\circ$ with the refracted ray.  Then the reflected light will be
polarized in a direction parallel to the surface of the interface.  Polaroid lenses
are designed to greatly reduce this "glare".
\item Polarization in the sky: Light scattered at $90^\circ$ in the sky will
also be polarized.  If you have polarized glasses, you can check this out, as
well as Brewster's angle.
\item The lens maker's formula.  The lens maker's formula relates the radii
of curvature of the lens to the focal length. 
\end{enumerate}

\bigskip

There are so many topics we can cover in optics, we have two upper division optics courses.
If you are interested in optics, I recommend you take one (or both) of
these classes: Phy417/Phy417L and/or Phy344(Applied Optics).  As a final important topic 
in optics, we cover interference effects.

\bigskip

\centerline{\bf Interference Effects}

\bigskip

We will have demonstrations in lecture that show similar interference effects
with visible light that we saw and discussed in Phy132 with sound.  We will
shine light at screens with slits, etc. and look at what happens when light
goes through the slits or reflects off thin films.  We will see that at some
positions the transmitted (or reflected) light will be bright, and at other
positions dark.  We can explain these phenomena using the principle of superposition.
In this course we will use the classical approach of treating the electromagnetic
radiation as an electric-magnetic field.  The net electric field at any point
in space will be the superposition of the electric fields from all the sources
of the radiation.  

We will consider light of one frequency (i.e. one color).
Thus, the net electric field will consist of the sum of sinusoidal functions that have
the same frequency.  We first go over the mathematics of adding sinusoidal functions that have
the same frequency, which is the basis for understanding these interference effects.

\bigskip

\centerline{\bf Adding Sinusoidal Functions that have the same Frequency}

\bigskip

The general problem is to add terms of the following form:

\begin{equation}
 \sum_i a_i cos(\omega t + \phi_i) = a_1 cos(\omega t) + a_2 cos(\omega t + \phi_2) + \cdots
\end{equation}

\noindent where we have measured all phase angles $\phi_i$ relative to the first term for
simplicity.  There is no generality lost in taking $\phi_1=0$, we have just started our time $t$
when the first sinusoidal function is at a maximum $+a_1$.  Before we go into the details,
we can guess what this sum will look like.  First, after a time equal to the period,
$T=2\pi /\omega$ each term has gone through a complete cycle and the sum is the same
as it was.  Thus, the sum of the sinusoidals will be periodic with the same period,
or frequency, as the terms.  We will also show that the sum will be a sinusoidal function
as well.  The sum will be therefore equal to $A cos(\omega t + \phi)$:

\begin{equation}
 \sum_i a_i cos(\omega t + \phi_i) = A cos(\omega t + \phi)  
\end{equation}

\noindent Our goal is to find a mathematical way to obtain $A$ and $\phi$ from the
$a_i$ and $\phi_i$.  I'll present three different ways to do this: using trig identities,
"phasors", and complex numbers.  I'll also show the special case of two functions
with $a_1 = a_2$.\\

\bigskip

\noindent {\it Using Trigonometric Identities}

\bigskip

Let's start with the simple case of two sinusoidals where $a_1 = a_2$.  We can use
the trig identity:

\begin{equation}
  cos(\alpha ) + cos(\beta ) = 2 \; cos({{\alpha + \beta} \over 2})cos({{\alpha - \beta} \over 2})
\end{equation}

\noindent Now, if $a_1=a_2$, $\alpha = \omega t$, and $\beta = \omega t + \phi_2$,  we have

\begin{eqnarray*}
 a_1 cos(\omega t) + a_1 cos(\omega t + \phi_2) & = & 2a_1 cos(\omega t + \phi_2/2)cos(-\phi_2/2) \\
                   & = & 2 a_1 cos(\phi_2/2) \; cos(\omega t + \phi_2/2)
\end{eqnarray*}

\noindent where we have used $cos(-\theta) = cos(\theta)$.  So we see that the sum of these
two cosine functions is a cosine function with the same frequency $\omega$.  The resulting phase is
$\phi_2/2$ from the first term and the Net Amplitude $A$ is $A = 2 a_1 cos(\phi_2/2)$.

From the expression for the net amplitude $A=2 a_1 cos(\phi_2/2)$ we can see how the interference
effects can be understood.  If $\phi_2 = 0, \pm 2\pi, \pm 4\pi, \cdots$, then $cos(\phi/2) = 1$.  The
net amplitude will be a maximum, the light will be bright, constructive interference will occur.
If $\phi_2/2 = \pm \pi, \pm (3/2) \pi, \cdots$, then $\cos(\phi/2) = 0$.  The net amplitude
will be zero, the light will be gone, destructive interference will occur.  For other values of
$\phi_2$ the amplitude will be $2 a_1 cos(\phi_2/2)$ and the intensity will be proportional
to the square of the amplitude $A^2 = 4 a_1^2 cos^2(\phi_2/2)$.

If $a_1 \ne a_2$ we could perform a similar analysis, but it will be easier to use the phasor
method described next.

\bigskip

\noindent {\it Using Phasors}

\bigskip

Since all the sinusoidal functions have the same frequency, we can add them just like
we add vectors.  Let's demonstrate this.  At time $t=0$ the sum is

\begin{equation}
  a_1 + a_2 cos(\phi_2)
\end{equation}

\noindent This sum is the projection on the x-axis of the vector sum of the two vectors
$\vec{a_1} = a_1 \hat{i}$ and $\vec{a_2}$, where $|\vec{a_2}|=a_2$ and $\vec{a_2}$
makes an angle of $\phi_2$ with respect to the x-axis, as shown in the figure.

\begin{figure}
\includegraphics[width=14cm]{fig2342f.png}
\end{figure}

At time $t$, both vectors in the figure have rotated by the same angle $\omega t$.  The
sum of the sinusoidal functions are, at time $t$, equal to

\begin{equation}
  a_1 cos(\omega t) + a_2 cos(\omega t + \phi_2)
\end{equation}

\noindent Once again, this sum is the projection on the x-axis of the vector sum
of the two vectors as shown in the figure.  The result of the above equation is
true for all times.  

That is, we can define the vector $\vec{a_1}(t)$ to have
a magnitude $|\vec{a_1}|=a_1$ and make an angle of $\omega t$ with respect to the
x-axis, and define the vector $\vec{a_2}(t)$ to have a magnitude or $|\vec{a_2}|=a_2$
and make an angle of $\omega t + \phi_2$ with respect to the x-axis.  At any
time $t$, we can add these two vectors.  The projection of the resulting
vector sum ($\vec{a_1}(t) + \vec{a_2}(t)$) on the x-axis will be the sum of the two
sinusoidal functions at time $t$.

Note the important result:  {\bf The magnitude of $\vec{a_1}(t) + \vec{a_2}(t)$ does not
change in time!}  The vector sum $\vec{a_1}(t)+\vec{a_2}(t)$ just rotates around with
angular velocity $\omega$!  {\it Thus the net amplitude of the sum of the sinusoidal
functions is just the magnitude of the vector sum: $|\vec{a_1}(t)+\vec{a_2}(t)|$.}
We can evaluate this sum at any time.  It easiest to evaluate the vector sum at
$t=0$, which we have chosen such that $\vec{a_1}(0)$ lies along the x-axis.

In the figure, we evaluate the vector sum for the special case of $a_2=a_1$.
Here we see that the magnitude $|\vec{a_1}(0)+\vec{a_2}(0)|$ is equal to

\begin{equation}
  |\vec{a_1}(0)+\vec{a_2}(0)| = 2a_1 cos(\phi_2/2)
\end{equation}

\noindent which is the same result we obtained using trig identities.  The angle that
the vector $\vec{a_1}(0) + \vec{a_2}(0)$ makes with respect to the x-axis is $\phi_2/2$.
This will be the phase relative of the sum to the first sinusoidal.

We can use this procedure of adding sinusoidal functions of the same frequency to any number
of functions.  One simply associates a vector with each term in the sum.  Then one adds
the vectors by the rules of vector addition to obtain a net vector.  The magnitude of the
net vector is the magnitude of the sum of sinusoidals.  The angle of the net vector
is the relative phase of the sum of sinusoidals.  One can say the same thing mathematically
as follows.  For a sinusoidal function with frequency $\omega$, amplitude $a_k$, and
phase $\phi_k$,  associate the vector $\vec{a}_k$:

\begin{eqnarray*}
  a_k cos(\omega t + \phi_k) & \leftrightarrow & \vec{a}_k \\
           & \leftrightarrow & a_k cos(\phi_k) \hat{i} + a_k sin(\phi_k) \hat{j}
\end{eqnarray*}

\bigskip

\noindent Note that $|\vec{a}_k| = a_k$ and $\phi_k$ is the angle between $\vec{a}_k$
and the x-axis.  Here I am using $k$ as the index for the sinusoidal function, since
we are using $\hat{i}$ as the unit vector along the x-axis.  The addition of the
sinusoidal functions is found by adding the vectors:

\begin{eqnarray*}
  \sum_i a_i \; cos(\omega t + \phi_i) & = & ?  \\
           & \leftrightarrow & \sum_i \vec{a}_i \; \; \; (Assign \; the \; Vectors) \\
           & \leftrightarrow & \vec{A}_{net} \; \; \; (add \; the \; vectors)\\
           & \leftrightarrow & A \; cos(\omega t + \phi) \; \; \; (Map \; back \; to \; sinusoidals) \\
  \sum_i a_i \; cos(\omega t + \phi_i) & = & A \; cos(\omega t + \phi)  \\           
\end{eqnarray*}

\noindent The addition of sinusoidal functions of the same frequency is isomorphic to
vector addition.  Instead of using the word vector, phasor is often used.  This is because
these vectors are all rotating around the z-axis with the same frequency $\omega$.  
A phasor is a vector that is rotating around an axis with a constant angular velocity.

\bigskip

\noindent {\it Using Complex Numbers}

\bigskip

Adding complex numbers obeys the same mathematics as adding two dimensional vectors.
The real parts add independently to the imaginary parts.  For every sinusoidal
function we associate a complex number:

\begin{eqnarray*}
    a_k \; cos(\omega t + \phi_k) &  \leftrightarrow & z_k \\
                                  & \leftrightarrow & a_k e^{i\phi_k}
\end{eqnarray*}

\noindent I use the index $k$, since $i$ is assigned $\sqrt{-1}$ when using
complex numbers.  As with the vector (or phasor) method one justs adds the
complex numbers:

\begin{equation}
  \sum_k a_ke^{i\phi_k} = A = |A| e^{i\phi}
\end{equation}

\noindent One just needs to express the complex number $A$ in terms of its magnitude $|A|$
and its phase $\phi$.  The reason this works for complex numbers is as follows:

\begin{eqnarray*}
  \sum_k a_k cos(\omega t + \phi_k) & = & \sum_k Re(a_ke^{i(\omega t + \phi_k)}) \\
                                    & = & Re (\sum_k a_ke^{i(\omega t + \phi_k)}) \\
                                    & = & Re (e^{i\omega t}(\sum_k a_ke^{\phi_k})) \\
                                    & = & Re (e^{i\omega t} Ae^{i \phi}\\
                                    & = & Re (Ae^{i(\omega t+\phi)})\\
                                    & = & A cos(\omega t + \phi)
\end{eqnarray*}

\noindent where Re means take the real part of the complex number.  Note that it was
critical that each function have the same frequency.  Otherwise, we could not have removed
the common term $e^{i\omega t}$ outside the summation.  
The addition of sinusoidal functions of the same frequency is isomorphic to the addition
of complex numbers.

Now we are ready to see if the "interference" experiments with monochromatic light can be 
understood using the mathematics of adding sinusoidal functions.  We will investigate what
happens when light interacts with two slits, many slits, a single slit, thin films, etc.

\bigskip

\centerline{\bf "Interference" Experiments with light}

\bigskip

\noindent {\it Double Slit Interference}

\bigskip

In the double slit experiment, light is emitted that has a {\it single frequency} and 
is from a {\it single source}.  The light hits a screen that has two thin slits,
slit "1" and slit "2".  The
slits are separated by a distance $d$.  After going through the two slits the
light hits a second screen.  On the second screen, one sees bright and dark 
regions.  

\begin{figure}
\includegraphics[width=14cm]{fig2342g.png}
\end{figure}

This phenomena of bright and dark regions can be understood as follows.  When the light
hits the second screen, the net amplitude is the superposition of the light from 
slit "1" plus the light from slit "2".  At the position on the second screen between
the two slits, the distance to each slit is the same.  The light from slit "1" has the same
phase as the light from slit "2".  A bright spot will be formed at this middle position.
At positions on the second screen away from the
middle, the distance to slit "1" is different than from slit "2".  Maxwell's picture
of monochromatic light is that it is an oscillating electric field vector.  If the distance
to slit "1" is different than to slit "2", then the oscillating electric field from
slit "1" will have a different phase with respect to the oscillating electric field from slit "2".
The net electric field will be the sum of two sinusoidal functions that have a relative
phase between them.  

Let's calculate the net electric field.  If the distance to slit "1" is not too much different than that
to slit "2", then the amplitude from each slit is approximately the same, say
$E_0$.  So the net electric field on the second screen is the sum of the electric
field from slit "1" plus that from slit "2":

\begin{eqnarray*}
   \vec{E}_{net} & = & \vec{E}_1 + \vec{E}_2 \\
                 & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega t + \phi) )  
\end{eqnarray*}

\noindent What is the phase difference $\phi$?  It is due to the extra distance that the light
must travel from one slit compared to the other.  Let the extra distance be $l$.  Then the
extra time to reach the screen is $l/c$.  So one "wave" arrives at time $t$, and the other
"wave" arrives at a time of $t + l/c$.  Substituting into the equation we have

\begin{eqnarray*}
   \vec{E}_{net} & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega (t+l/c)) ) \\  
                 & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega t+ \omega l/c)) ) \\  
                 & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega t+ 2 \pi f l/c)) ) \\  
                 & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega t+ 2 \pi l/\lambda)) ) \\  
\end{eqnarray*}

\noindent where we have used $c = f \lambda$ to get the last line.  Thus, the
difference in the sinusoidal phase of the light from the two slits is 

\begin{equation}
    \phi = 2 \pi {l \over \lambda}
\end{equation}

\noindent  We know how to add sinusoidal functions from the last section.  For the
case at hand, two slits, we have

\begin{eqnarray*}
  \vec{E}_{net}  & \approx & \vec{E}_0 (cos(\omega t) + cos(\omega t+ \phi)) ) \\
                 & \approx & \vec{E}_0 2 cos(\phi /2)cos(\omega t + \phi /2) \\
                 & \approx & \vec{E}_0 2 cos(\pi l /\lambda)cos(\omega t + \phi /2) \\  
\end{eqnarray*}

\noindent Here we see that the amplitude of the light when it stikes the second screen
is 

\begin{equation}
  |\vec{E}_{net}| = 2 E_0 cos(\pi {l \over \lambda} )
\end{equation}

\noindent This equation for $\vec{E}_{net}$ enables us to understand the two slit
experiment.  The intensity of the light is proportional to the square of the electric 
field strength:

\begin{equation}
  I = 4 I_0 cos^2(\pi {l \over \lambda} )
\end{equation}

\noindent At the middle position, $l=0$, and the intensity is $4 I_0$.  If the
extra distance $l = \pm \lambda /2$, $cos(\pi /2) = 0$, then there is a dark spot.
Dark spots will also occur whenever $l=\pm 3 \lambda /2, 5 \lambda /2, \cdots$.
Bright spots will occur whenever $l=0, \lambda, 2\lambda, \cdots$.  The intensity,
or brightness, varies as $cos^2(\pi l/\lambda )$.

We can express the extra distance $l$ in terms of the "scattering" angle $\theta$
shown in the figure.  If $\theta$ is small, then 

\begin{equation}
  l \approx d sin(\theta)
\end{equation}  

\noindent So, in terms of the scattering angle $\theta$ the intensity on the second
screen is

\begin{equation}
  I = 4 I_0 cos^2({{\pi d \; sin(\theta )}  \over \lambda} )
\end{equation}

\noindent Some final comments on the two slit experiment:

\begin{enumerate}
\item It is important that the light coming from slit "1" is in sych with the light
coming from slit "2" at the slit opening at all times.  For the case of a single
monochromatic source impinging directly on the two slits, the light eminating
from each slit is the same (at the slit opening).  If the single monochromatic
source were closer to one slit than the other, the light from "1" would still 
be in sych with the light from slit "2".  The relative phase would not change
in time.  The radiation from sources that are in sych is called {\it coherent}.
\item For interference to occur, the two "sources", slit "1" and slit "2", must
be coherent longer than a distance $l$.  The maximum distance that the
sources remain coherent is called the {\it coherence length}.  In our case,
the coherence length must be longer than $l$ for interference to exist.
\item In a real experiment, the intensity does not decrease exactly as
$cos^2(\pi d \; sin(\theta )/\lambda)$.  The slits will have a finite opening
size, and this will affect the pattern on the second screen.  We will consider
this effect later.
\item The middle point is often called the central maximum.
\item The two slit experiment was first performed by Young.  The experiment is
sometimes refered to as "Young's double slit experiment".  The experiment was one
of the first to show the interference effects of light.  It lead scientists to
the incorrect conclusion that light is a classical wave.
\end{enumerate}

\bigskip

\noindent {\it Multiple Slit Interference}

\bigskip

  Now let's consider will happen if light is incident on a screen with three
equally spaced thin slits, located on the screen.  The light will travel
though the slits and strike a wall.  On the wall, we will see an
interference pattern.  We can carry out the same analysis that we did with
two slits.  Let the spacing between each slit be $d$.  Label the
slits "1", "2", and "3".  If $E_0$
is the amplitude on the wall from each slit, then the amplitude at the central maximum
on the wall is $3E_0$.  On the wall away from the central maximum (upward in the figure), 
slit "1" will be closer than slit "2" by a length $l$, and slit "2" will be closer than 
slit "3" by a length $l$.  The phase difference between adjacent slits will be 
$\phi = 2 \pi l/\lambda$.  The net amplitude will be the sum of the amplitudes from
each of the three slits:

\begin{figure}
\includegraphics[width=14cm]{fig2342h.png}
\end{figure}

\begin{equation}
  E_{net} = E_0 cos(\omega t) + E_0 cos(\omega t + \phi) + E_0 cos(\omega t + 2\phi)
\end{equation}

\noindent We can add the sinusoidals using phasors.  The first term is associated with
a vector of length $E_0$ along the x-axis.  The second sinusoidal is associated with
a vector of length $E_0$ at an angle of $\phi$ with respect to the x-axis.  The third
sinusoidal is associated with a vector of length $E_0$ at an angle of $2\phi$
with respect to the x-axis.  The magnitude of $E_{net}$ is magnitude of the sum of
the three vectors.

One of your homework problems is to solve for $E_{net}$ in terms of $\phi$.  Here we will
determine the values of $\phi$ that result in minima and maxima.  Maxima occur
when $\phi =0^\circ$ and when $\phi = 360^\circ$.  For a minima to occur,
the three vectors must add to zero.  The first value of $\phi$ for the sum to 
be zero is when $\phi = 120^\circ$.  The next value of $\phi$ for the sum to add to
zero is when $\phi = 240^\circ$.  These are the only values for $0^\circ < \phi < 360^\circ$
where there are minima.  In the middle of the minima at $\phi = 120^\circ$ and $240^\circ$ there
must be a maximum, a "local maximum".  This will occur at $\phi=180^\circ$.  Substituting in the
equation for $E_{net}$, we have $E_{net} = E_0 - E_0 + E_0 = E_0$.  The pattern on the wall
will be the following.  $E_{net}$ has maxima of $3E_0$ when $\phi = 0, \pm 2\pi, \pm 4\pi, \cdots$.
$E_0$ has smaller local maxima at $\phi = \pm \pi, \pm 3\pi, \cdots$.

The intensity of the light is proportional to $E^2_{net}$.  So the intensity 
at the maxima when $\phi = 0^\circ$ and $\phi = 360^\circ$ will be $9$ times the 
intensity at the local maximum when $\phi = 180^\circ$.  As we move across the
wall from the central maximum, the intensity will vary as: $I_0$, $0$, $I_0/9$,
$0$, $I_0$, $0$, $I_0/9$, $0$, $I_0$, $\cdots$.

We can perform the same analysis with $4$ slits as we did with $3$.  As before,
let $\phi$ be the relative phase between light from one slit and its adjacent
slit.  The net amplitude will be the sum of four sinusoidals:

\begin{equation}
  E_{net} = E_0 cos(\omega t) + E_0 cos(\omega t + \phi ) + E_0 cos(\omega t + 2\phi ) 
              + E_0 cos(\omega t + 3\phi )
\end{equation}
 
\noindent As before we can add the sinusoidals using phasors.  The first term is associated with
a vector of length $E_0$ along the x-axis.  The second sinusoidal is associated with
a vector of length $E_0$ at an angle of $\phi$ with respect to the x-axis.  The third
sinusoidal is associated with a vector of length $E_0$ at an angle of $2\phi$
with respect to the x-axis.  The forth sinusoidal is associated with a vector of length
$E_0$ at an angle of $3\phi$ from the x-axis.  The magnitude of 
$E_{net}$ is magnitude of the sum of the four vectors.

Let's try and determine where the minima and maxima are.  The central maximum is when
$\phi = 0$, and $E_{net} = 4E_0$.  Equally large maxima will occur when
$\phi = \pm 2\pi, \pm 4\pi, \cdots$.  Since $E_{net}(\phi \pm 2\pi) = E_{net}(\phi )$,
we can limit our analysis for $0 \le \phi \le 2\pi $. 

To find the minima for $0 \le \phi \le 2\pi$, we can slowly increase $\phi$ and see
when the four vectors add up to zero.  The first minima will occur at $\phi = 90^\circ$.
The vectors form a square going counter-clockwise.  The next minima will be 
when $\phi = 180^\circ$.  The vectors
cancel each other out.  The next minima will be when $\phi = 270^\circ$.  The vectors
form a square going clockwise.  So there are three minima (dark spots) for
$0 \le \phi \le 2\pi$.  In between the minima, there will be local (smaller) maxima.  One for
$90^\circ \le \phi \le 180^\circ$, and another one for $180^\circ \le \phi \le 270^\circ$.
At the local maxima, the net amplitude will be approximately $E_0$ by examining how
the four vectors add.   Since
intensity is proportional to the square of $E_{max}$, the intensity at $\phi = 0$
and $\phi = 2\pi$ will be much larger, around $4^2 = 16$ times larger than at the 
two local maxima between the bright peaks.

Suppose there are $N$ equally spaced narrow slits that are separated
from each other by a distance $d$.  Extending the
above analysis, the interference pattern on the wall will have a very bright (and narrow)
central maximum.  The next very bright (and narrow) maximum will be when $\phi = 2\pi$,
since all the vectors will add in a straight line.  Between the two
bright maxima, there will be $N-1$ dark spots where $E_{net}=0$.  Between each dark
spot there will be a local maxima.  The amplitude of the $N-2$ local maxima will be much smaller than at
$\phi = 0$ or $\phi = 2\pi$.  Since the intensity is proportional to $|E_{max}|^2$,
the intensity at $\phi = 0$ or $\phi = 2\pi$ will be $N^2$ greater than the local
maxima in the middle of the pattern.

The features of the three and four slit interference patterns are enhanced when $N$ is
large.  If $N$ is large, the interference pattern on the wall will consists of very 
bright lines that are very narrow, with little light between them.  We can
determine the narrowness of the lines as follows.  The central
maximum has its first minimum (i.e. dark spot) when $\phi = 2\pi /N$.  At this
value of $\phi$ all the vectors sum to zero.  $\phi$ is related to the scattering
angle $\theta$ as $\phi = 2 \pi l/\lambda = (2 \pi d sin(\theta )/\lambda$.  So
the first minima occurs for a value of $\theta$ given by

\begin{eqnarray*}
  {{2 \pi d \; sin(\theta )} \over \lambda} & = & {{2 \pi } \over N} \\
                  sin(\theta ) & = & {\lambda \over {d N}} \\
                         \theta & \approx & {\lambda \over {d N}}\\ 
\end{eqnarray*}

\noindent where we have used the approximation $sin(\theta) \approx \theta$ for $\theta << 1$.
The expression $\lambda /(dN)$ is called the {\it line width} of the diffraction line.  This
simple expression is applicable for the central maximum.

Screens with many slits are called {\it diffraction gratings}.  Diffraction gratings can
have thousand of slits and produce very narrow interference lines.  They are used
in spectrometers to measure wavelength to great accuracy.  In Phy417, you will use
complex numbers to solve for $N$ slit interference.  Briefly, the calculation goes
as

\begin{eqnarray*}
  I & \propto & |\sum_{n=0}^N e^{in\phi }|^2 \\
    & \propto & |{{1 - (e^{i\phi})^N} \over {1-e^{i\phi}}}|^2 \\
  I  & = & I_0 {{sin^2(N \phi /2)} \over {sin^2(\phi /2)}}
\end{eqnarray*}

\noindent At $\phi \rightarrow 0$, $I \rightarrow N^2 I_0$.  The intensity of the
local maxima are around $I_0$, much smaller than the central maximum.  

\noindent Summarizing equally spaced multiple slit interference:

\begin{enumerate}
\item The main property of radiation we used was that the net amplitude on the wall
is equal to the sum of the amplitudes from each of the slits.
\item The process of adding the amplitudes from each of the slits involves adding
sinusoidal functions that have a common frequency.  Each sinusoidal function has
approximately the same amplitude, but with different phase shifts.  The difference
in phase from one slit to its neighbor is the same value, $\phi$, since the slits
are equally spaced.   
\item In all our discussions, the phase shift $\phi$ has always been the relative phase from
one slit compared to an adjacent slit.  In terms of the extra distance $l$, 
$\phi = 2\pi l/\lambda$.  In terms of the scattering angle $\theta$, $l=d \; sin(\theta )$.
So, $\phi = 2\pi \; d \; sin(\theta )/\lambda$. 
\end{enumerate}

\bigskip

\noindent {\it Single Slit Interference}

\bigskip

As a final slit interference experiment, we analyze the interference effects when light
passes through a single slit.  Consider a coherent light beam that hits a screen that
has a slit of length $d$.  After the light passes through the slit, it "hits" a
wall.  In lecture, we will demonstrate this, and notice the interference pattern
on the wall.  Let the magnitude of the electric field on the wall at the central 
maximum be $E_0$.  $E_{net}$ will have its largest value here.  

\begin{figure}
\includegraphics[width=14cm]{fig2342i.png}
\end{figure}

For the case of a thin slit, each slit was a source of light.  We can carry this
model over to the finite sized slit by dividing the opening of length $d$ 
into $N$ thin slits separated by a distance of $d/N$, and letting $N \rightarrow \infty$.

Consider light scattered at an angle $\theta$.  The path difference between light
emitted at the top of the opening to that at the bottom of the opening is
$d \; sin(\theta )$, which we label as $L$, $L = d \; sin(\theta )$.  The difference
in phase between light emitted at the top compared to the bottom of the slit is
$2\pi L/\lambda $, which we label as $\Phi = 2\pi L/\lambda $.

If there are $N$ equally spaced sources (of thin slits) in the slit opening, each
source will produce an electric field of amplitude $E_0/N$ at the wall.  The relative
phase between adjacent sources will be $\phi = \Phi /N$.  Adding up these $N$ 
coherent sources at the wall yields a net amplitude $E_{net}$ of

\begin{equation}
  E_{net} = \sum_{n=0}^N {E_0 \over N} cos(\omega t + {{n \Phi} \over N})
\end{equation}

\noindent We can use phasors (vectors) to add these sinusoidal functions as we did
with $N$ slits.  If $N$ is large, each vector will be very short and be rotated
a small angle, $\Phi /N$, from the adjacent vector.  An important property
of the sum is that {\it no matter what $N$ is,
the last vector will be rotated by an angle $\Phi$ relative to the x-axis}.
To find $E_{net}$, we take the limit as $N \rightarrow \infty$:

\begin{equation}
  E_{net} = lim_{N \rightarrow \infty} \sum_{n=0}^N {E_0 \over N} cos(\omega t + {{n \Phi} \over N})
\end{equation}

\noindent Adding the vectors in the limit that $N \rightarrow \infty$ results in a circular
arc of length $E_0$ with the end of the arc rotated by an angle of $\Phi$, as shown in
the figure.  $|E_{net}|$ is the length of the chord that starts at the origin and ends
at the end of the circular arc.  From the figure, it is seen that $\Phi = E_0/r$,
where $r$ is the radius of the circle.  Using geometry, the length of the chord is
$2r \; sin(\Phi /2)$.  Since $r=E_0/\Phi$, we have

\begin{eqnarray*}
  |E_{net}| & = & 2 r \; sin(\Phi /2) \\
            & = & {{2 E_0 \; sin(\Phi /2)} \over \Phi} \\
            & = & E_0 {{sin(\Phi /2)} \over {\Phi /2}}
\end{eqnarray*}

\noindent The intensity of the light on the wall is therefore

\begin{equation}
       I =  I_0 ({{sin(\Phi /2)} \over {\Phi /2}})^2  
\end{equation}

\noindent where $I_0$ is the intensity at the central maximum.

Some final points on single slit interference:\\

\begin{enumerate}
\item The first minimum occurs when $\Phi = \pm 2\pi$.  The extra distance $L$ traveled from
the bottom of the slit is $L=\lambda$.  So the first minimum occurs when the radiation emitted
from the bottom of the slit is $360^\circ$ different than that from the top of the slit.
This seem strange that cancellation happens, since $360^\circ$ or $2 \pi$ radians means that the
two sources are in phase.  The reason for the total cancellation is that each source in the upper
half of the slit cancels with a source in the lower half of the slit.  The central maximum region
is therefore rather wide.
\item The value of $\Phi$ at the first maxima is when $\Phi = \pm 3\pi $.  The intensity at these
first maxima are

\begin{equation}
   I = ({2 \over {3\pi}})^2 I_0 = {4 \over {9\pi^2}}I_0
\end{equation}
\item In general, there will be minima when $\Phi = \pm 2\pi , \pm 4\pi , \cdots$.  There
will be maxima when $\Phi = \pm 3\pi , \pm 5\pi , \cdots$. 
\item Another way to obtain the single slit formula is to use the $N$ slit result derived
by the complex number way:

\begin{equation}
  I   =  I_0 {{sin^2(N \phi /2)} \over {sin^2(\phi /2)}}
\end{equation}

\noindent Letting $\phi = \Phi /N$ we have

\begin{equation}
  I   =  I_0 {{sin^2(\Phi /2)} \over {sin^2(\Phi /(2N))}}
\end{equation}

\noindent Taking the limit $N \rightarrow \infty$ yields

\begin{eqnarray*}
  I & = & (N^2 I_0) {{sin^2(\Phi /2)} \over {(\Phi /2)^2}} \\ 
   & = & I_C {{sin^2(\Phi /2)} \over {(\Phi /2)^2}} \\ 
\end{eqnarray*}

\begin{figure}
\includegraphics[width=14cm]{fig2342j.png}
\end{figure}

\noindent where $I_C$ is the intensity of the central maximum, since $I_0$ is
the intensity from one slit.  We have used the approximation that
$lim_{N \rightarrow \infty} sin(\Phi /N) \rightarrow \Phi /N$, since $\Phi /N$
is very small as $N \rightarrow \infty$.
\end{enumerate}

\bigskip

\noindent {\it Two slits with finite size}

\bigskip

Using the results from single slit interference, we can understand a real two slit
interference pattern.  By real, we mean two slits that have a finite size.  Suppose
the screen consists of two slits, whose separation is $d$.  Let $w$ be the size of
the slit opening of each slit.  Consider a scattering angle $\theta$.  
As we just derived, the net amplitude for single slit interference at the angle theta is
$E_0 sin(\Phi /2)/(\Phi /2)$, where $\Phi = 2 \pi w \; sin(\theta )/\lambda$.
The net amplitude for the superposition of both finite sized slits is thus

\begin{equation}
  E_{net} = E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} cos(\omega t) +
                  E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} cos(\omega t + \phi) 
\end{equation}

\noindent where $\phi = 2\pi d \; sin(\theta )/\lambda$.  Note that $\Phi$ is
the phase difference between sources at the top and bottom of the slit opening.
$\phi$ is the phase difference between the first slit and the second slit.  
$\phi$ is larger than $\Phi$.  In fact:

\begin{equation}
    {\phi \over \Phi} = {d \over w}
\end{equation}

\noindent The equation for $E_{net}$ can be simplified by factoring out the term
$sin(\Phi /2)/(\Phi /2)$:

\begin{eqnarray*}
  E_{net} & = & E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} cos(\omega t) +
                  E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} cos(\omega t + \phi) \\
      & = & E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} (cos(\omega t) + cos(\omega t + \phi)) \\
      & = & 2 E_0 {{sin(\Phi /2)} \over {(\Phi /2)}} cos(\phi /2) \; cos(\omega t + \phi /2) 
\end{eqnarray*}

\noindent So we see that the net interference amplitude is the product of the single slit
times the double slit amplitudes.  Since the single slit amplitude varies slower than the double
slit amplitude, the single slit amplitude squared is an envelope for the double slit pattern.
The single slit has its first minimum when $\Phi = 2 \pi$.  As $\Phi \rightarrow 2 \pi$,
$\phi$ passes through $2 \pi$ $d/w$ times.  Thus, the number of bright spots in the
central region is $2d/w - 1$.

\bigskip

\noindent {\it Thin Film Interference}

\bigskip

Another way interference can occur is when light is reflected off a thin film
of transparent material.  We show this in the figure.  Suppose light is
incident near the normal of a thin film.  Light can reflect from the
near surface of the film, and it can reflect from the far surface of
the film.  The reflected radiation will be a linear superposition of
the radiation reflected from the near plus the far surfaces.  If we let
$t$ be the thickness of the thin film, then the extra distance that
the light reflected from the far surface travels is $2t$:

\begin{equation}
  l = 2t
\end{equation}

\noindent So, the difference in phase of the light that reflects from the near
versus the far surface is $\phi = l/\lambda_{film} = 2t/\lambda_{film}$.  There
is one other property of reflection to consider: when light reflects off of
a "slower" medium, the reflected radiation is shifted $180^\circ$ in phase.  That is,
if light reflects from a medium with $n=n_1$ off a medium with $n=n_2$, then
the phase of the reflected radiation is shifted $180^\circ$ if $n_2>n_1$.  If $n_2<n_1$,
then there is no phase change in the reflected light.  We will demonstrate
this phenomena in lecture.  Let $n_1$, $n_2$, and $n_3$ be the indices of refraction
for the media in the diagram.  Then zero relative phase due to the surfaces
will occur if $n_1 < n_2 < n_3$ or $n_1 > n_2 > n_3$.  A relative phase change of
$\pi$ will occur is $n_1 < n_2 > n_3$ or $n_1 > n_2 < n_3$.

To summarize: upon reflection from a thin transparent film, the relative phase
$\phi$ between the radiation reflected between the near and far surfaces is

\begin{equation}
  \phi = {{2t} \over \lambda_{film}} + (any \; additional \; phase \; from \; surfaces)
\end{equation}

\noindent The wavelength $\lambda_{film}$ is the wavelength of the light in the
medium of the film.  The wavelength in the film is $\lambda_0/n_{film}$, where
$\lambda_0$ is the wavelength in vacuum and $n_{film}$ is the index of refraction
of the film.  

\bigskip

\noindent {\it Other Interference Phenomena} 

\bigskip

An important apparatus that makes use of interference effects is the Michelson
interferometer.  A monochromatic light source (usually a laser) is incident upon
a beam splitter.  Half of the light travels through the beam splitter, and
half is reflected $90^\circ$.  Both halves reflect off of mirrors,and return to
the beam splitter.  One beam travels straight through and the other reflects
$90^\circ$ and they both emerge together.  Interference can occur between the two
split beams when they "recombine".  The phase difference $\phi$ between the two beams
will be 

\begin{equation}
    \phi = 2 \pi {l \over \lambda}
\end{equation}

\noindent where $l$ is the difference in path length of the two beams.  If one
mirror is a distance $d_1$ and the other mirror a distance $d_2$ from the
beam splitter, then $l = 2 |d_1 - d_2|$.  By changing one of the distances
$d_1$ or $d_2$, one observes the interference pattern shift.  Everytime a
new interference fringe is produced, $l$ has changed by one wavelength.

\bigskip

\noindent {\it Final thoughts on interference effects}

\bigskip

Interference is a remarkable phenomena.  Water waves and sound waves (mechanical
waves) experience interference when the wave amplitude from different sources
combine.  Guided by mechanical wave interference, we have explained the double slit, 
N-slit, single slit interference patterns on the wall by summing an
amplitude from each of the individual slits, then squaring the magnitude of the
summed (net) amplitude.  In the Maxwell or classical picture of electromagnetic
radiation, the amplitude represents an electric field vector.  The radiation is
a sinusoidal wave, with the electric vector field doing the "waving".  This classical
picture explains many things, but as you will see next quarter one of its failures
is the description of what happens when the light is very dim.

As the intensity is reduced, at first the brightness is also reduced.  However,
at very low intensity the radiation arrives as particles.  The particles, or photons,
arrive more often in the "bright" regions and never in the dark regions.  The
modern view is the following.  Nothing physical is waving.  An amplitude is assigned
to each path that the photon can take to get to the wall.  After summing up the
amplitudes for all possible paths, we take the absolute square of the sum.  The interference
properties come from {\it summing the amplitudes}.  The
absolute square of this sum is the probability (or probability density) that the photon
will arrive at the particular region on the wall.  Instead of corresponding to
an electric field magnitude, the sinusoidal functions represent the probability amplitude
(density) for a "free" photon (or particle) to be in a particular location.  

Interference effects can be understood without anything waving.  The pattern on the wall
is also determined from the Fourier transform of the slit openings.  The Fourier transform
employs the term $e^{ikx}$, which does not require anything to wave.   It took 
physicists many years to sort these ideas out, and they form the basis of Quantum Mechanics.  
Next quarter you will cover the experiments and reasoning that occured from 1900-1930 
and lead to our modern formalism.

Electromagnetism and its radiation has taught us many lessons.  So far it has taught
us about the interaction of charged particles and light.   Next, it will teach us about the properties
of space and time.  We conclude this course with Einstein's theory of special relativity.

\end{document}

