% We generally want to become confortable with uncertainty and having to deal with approximation.\\

% In Optics, we want to know how well we can measure something by looking at it. In stat mech, there are just too many objects we want to consider and in quantum mechanics our measurements directly impact the object we want to study.\\

%

% Books

% \begin{itemize}

% \item ``Optics'', E. Hecht

% \item ``Fundamentals of Photonics'', B. Saleh \& M. Teich

% \item ``Statistical Physics'', F.Mandl

% \item ``Introduction to Quantum Mechanics'', D. Griffiths

% \end{itemize}

%

\section{Optics}

Usually, we think of light as a wave. But often, we can simplify and just think of it as a ray. We can do it when the wavelength is much smaller than the size of the objects the light touches.\\

Going in the other direction, the more complete picture is electro-magnetic optics. We can go even further and look at light as a quantum phenomenon.\\

We will start out in the simplest model, Ray optics or also called Geometrical optics.

\subsection{Ray Optics}

\subsubsection{Fermat's Principle}

One of the underlying principles of Ray optics is \textbf{Fermat's principle}:\\

\emph{Light travels the path whose optical path length is extremal to variations in the path}

\\

Intuitively, this means that if we consider a path that is infinitesimally close the the actual path, the optical path lenth does not change.\\

The path is usually a minimum, but that is not always the case.

In a homogeneuous medium, where the speed of light is the same everywhere, this is just saying that light travels in a straight line.\\

Now consider two media, let's say vacuum and glass and Points $A$, $B$ in each medium\\

In vacuum, the speed of light is about $c =3\times10^8\m/\s$ and in glass, it is about $1.5$ times slower than that. We then can define the \textbf{refractive index} of glass to be the fraction

Imagine a situtation, where a point $A$ is emitting light in all directions and you want to ``send'' all the light rays towards $B$.\\ Normally, Fermat's principle would tell us that there is only one path form $A$ to $B$ that the light would reach $B$.\\

Now what a lens would do is to change the optical path lengths for some paths such that more than one of them is extremal. This results in ``more Light'' reaching $B$. (See Figure 1.6)\\

One example of this is the \textbf{single surface lens}, which an arced surface between two materials. (For example light and glass).

We can describe the surface of the surface using some function $g(h)$, which allows us the write down the optical path length for the straight line and the point going trough the point $(g(h),h)$.\\

For the straight line, we simply have $D_{ACB}= n_1d_1+ n_2d_2$, whereas for the path $AC'B$ we have

Using the taylor approximation of $(1+\epsilon)^{\alpha}$ for $\abs{\epsilon}\ll1$, we have $(1+\epsilon)^{\alpha}\simeq1+\alpha\epsilon$.\\

Using the so called \emph{Paraxial approximation}, we are considering only rays that make a small angle with the optical axis ($ACB$). This means that we consider $h$ and $g(h)$ to be small in comparison to $d_1$ and $d_2$.\\

This allows us to simplify the optical path length $D_{AC'B}$. After dividing out the $(d_i + g(h))$ term and taking the taylor approximation of this term we get

In practice, creating parabolic surfaces can be hard so we can further approximate the surfac as a sphere with Radius $R \gg h$:

\begin{align*}

g(h) = R - \sqrt{R^2 - h^2}\simeq\frac{h^2}{2R}

\end{align*}

for $\frac{1}{R}=\frac{1}{n_2- n_1}\left(\frac{n_1}{d_1}+\frac{n_2}{d_2}\right)$.\\

Next we may consider a light source living infinitely far away from you. Here, it's emitted light rays are coming in paralel to you and we want to focus all light rays into a single point using a lens.\\

Using our previous equations, we would have $d_1=\infty$ and $d_2= f$ for the distance to the focal point we want to get

R = \frac{f(n_2 - n_1)}{n_2}, \quad f = \frac{R n_2}{n_2 - n_1}, \quad\frac{n_2}{f} = \left(\frac{n_1}{d_1} + \frac{n_2}{d_2}\right)

\end{empheq}

Check the images in the slides for lecture 2.\\

When the light enters the lens, we can have multiple scenarios. If we get a positive focal distance $f > 0$, we wee that the light gets converges into a single point.\\

If however we get $f < 0$, we get a \emph{virtual focal point} at distance $f$\emph{before} the lens.\\

If we look from the other side of the lens, it would look like the light is coming from the virtual focal point.\\

If we chose a spherical surface, we see that the light rays don't intersect at exactly the same point. We call this phenomenon the \emph{spherical abberations}

\begin{empheq}[box=\bluebase]{align*}

\Delta D_{\text{err}} = \phi(n_1,n_2) h \left(\frac{h}{f}\right)^{3}

\end{empheq}

Next we consider a lens with two surfaces. The first surface with Radius $R_1$ will have focal length $f'$. Here we will look at the rays hitting the second surface as coming from a virtual source from behind the lens and use the equation \ref{eq:redstar} with $d_1=-f$ and we use $n_2=1, n_1= n$. We then get

This accurately describes our Biconvex lense $R_1 > 0, R_2 < 0\implies f > 0$

This gives rise to the \textbf{lensmaker's equation}

\begin{empheq}[box=\bluebase]{align*}

\frac{1}{f} = \frac{1}{d_1} + \frac{1}{d_2}

\end{empheq}

where we an object $A$ at distance $d_1$ before the lens and Image of the object at point $B$ at distance $d_2$ after the lens.\\

This gives us the following sign conventions

\begin{tabular}{ll}

$f > 0$& Lightrays converge\\

$f < 0$& Lightrays diverge\\

$d_2 > 0$& Real Image\\

$d_2 < 0$& virtual image\\

\end{tabular}

\subsubsection{Ray Tracing}

If we have an object which does not lie on the lens axis, we can use the three rules to find out, where the image is.\\

Rule 1 gives us a straight line through the center of the lens. Using either rule 2 or rule 3, we intersect rule 1 with the line that goes parallel until it hits the lens and go through the focal point.\\

For a convex lens, we use the focal point of the other side for rule 2, and for concave lenses, we use the focal point on the same side as the object or rule 2. For rule 3, we use the other focal point\\

To better understand how big the image (virtual or real) will be, we introduce the \textbf{Magnification factor}$M :=\frac{h'}{h}=-\frac{d_2}{d_1}$.\\

It's absolute value tell us its size and the sign tells us the orientation.\\

If we are interested in seeing things far away from us, we dont' necessarily want to increase Magnification factor as much as we want to increase the angular size of the object. One way to do this is to use multiple lenses.\\

Let's say an object with angular size $\Delta\theta$ whose rays travel trough lens 1 and form an image $h'$ at focal point $f_1$. We can then magnify the image by lens 2 whose focal point lies at the image $h'$.\\

A microscope does something even better: Instead of putting the image $h'$ at the focal point $f_2$, we can place the image $h'$ between the focal point and the lens to create a virtual image that is bigger, but further away than $h'$.\\

Some important things to consider is that the lensmaker's equation assumes that the index of refraction of the surrounding medium is air. If we place a lens in, for example water. The properties of a lens can change!

When we were talking about Light as Rays, we made quite a few appproximations and in certain cases, these approximations can lead to minor errors. The next step in accuracy comes, when we look at light as a wave\\

The starting point for this section is the wave equation, which can be derived from the Maxwell equations. In vacuum, they are

where $c =\frac{1}{\sqrt{\mu_0\epsilon_0}}$ is the speed of light in vacuum.\\

The wave equation has he \textbf{superposition principle}, where if we have two solution, $\vec{E}_1$ and $\vec{E}_2$, then any linear combination $a \vec{E_1}+ b \vec{E_2}$ is also a solution.\\

Next, consider solutions of the form

\begin{align*}

\vec{E}(\vec{r},t) = U(\vec{r}) e^{i\omega t}

\end{align*}

where the physical field is given be its real part $\text{Re}(\vec{E})$.\\

Here there are two possible solutionn. The plane wave equation

, where the wave vector $\vec{k}$ is in relation with the wavelength $\lambda$ in its absolute value: $\abs{\vec{k}}=\frac{2\pi}{\lambda}$\\

\subsubsection{Huygen's Principle}

After Fermat's Principle, where light is treated as travelling in straight lines, we will introduce \textbf{Huygen's Principle} , from which Fermat's Principle can be derived under some assumptions.

\begin{center}

\emph{Every Point on the wavefront of a wave acts as a secondary source of hemispherical waves that progate in the forward direction}

\end{center}

Now let's test this: We imagine two points $A(x_A,y_A),B(x_B,y_B)$ in space and a light ray travelling from $A$ to $B$.

We then add an aperture centered at $C$ inbetween the points $A$ and $B$. Using Fermat's principle, we would expect the aperture to ``see'' the light as it should pass through it and then emit another wave towards $B$

The spatial part of the electromagnetic wave will be

Our next approximation is called the \textbf{Fraunhofer approximation}, where we drop the last term $\frac{x^2+ y^2}{2r_A}$. We can think of this as saying that the (hemi-)spherical wave will appear flat in a small region. We then get

We see that the intensity is proportional to $a^4$, which we wouldn't expect with ray-optics as the size of the aperture shouldn't matter.\\

The conditions to recover ray optics from this is that we have large $\gamma :=\frac{a}{\sqrt{\lambda s}}$, for $s = r_A = r_B$ or equivalently $a \gg\lambda$\\

We also must have, that the distance to the source and the detecter is $\geq\lambda$\\

To generelize, we will define our square aperture in terms of a transmission function $\tau(x,y)$, which in our case was

\begin{align*}

\tau(x,y) = \left\{\begin{array}{rcl}

1, &\text{if}& x \in (-a/2,a/2) \text{ and } y \in (-a/2,a/2), \\

0, &\text{elsewhere}&

\end{array}\right.

\end{align*}

In general for any transmission function we will have

We saw that if we the intensity is proportional to $a^4$, so the wider our slit is, the weaker the peak intensity is.\\

The width of the diffration pattern is also proportional to the wavelength.\\

So, the \textbf{ray-optics limit} occurs when the aperture size is much bigger than the wavelength, which implies that the width of the diffraction pattern becomes very small and we recover Ray optics.

Now let's talk about imaging resolution. When we have Fraunhofer Diffraction, we assumed that the wave hitting the aperture were planar. This assumes that the wave is coming from infinitely far away.\\

In reality, we can mimic a wave coming from infinitely far away by putting lenses between the source and the aperture.\\

If we put the source at the focal point of a convex lens, the lightwaves will look like a plane and we can then use the Fraunhofer approximation.\\

Looking at a couple situations where we have a lightsource and a lense that focuses the light into a source, but there is an aperture is blocking some of the light. We can mimic this by imagining two lenses, a diverging and a converging one in series.

The first lens cancels the convergent behaviour just before the aperture and the second lens redos the cancellation. If we assume that the two lenses get infintely close together, we obtain the original path again, all the while the wave hits the aparatus perpendicularly.

So even if it doesn't look like Fraunhofer approximation applies, by imagining extra lenses, we can assume it does.\\

\subsubsection{Spatial resolution of telescopes}

In this section, we want to observe objects that emit light by letting their light go trough an aperture thats far away and looking at the resulting refraction pattern.

Remember that as the aperture gets smaller, the sinc function intensity peaks get broader and broader.\\

This was a direct consequence of the effect of the fourier transformation, where if you have a shorter signal, you can make less accurate measurements of the frequency spectrum.\\

So if imagine two stars $A_1, A_2$ and a lens perpendicular to the optical path from $A_1$ to the lens and the angle from the second star being $\phi$.\\

We will get a refraction pattern which is the sum of the refraction patters of each star, with the angle determining how much the patters are shifted.\\

But in order to even tell that we have two refraction patterns in the firs place, we must have that the intensities of the two patters are shifted in a way that their peak intensities don't overlap.\\

This can be stated as the \textbf{Rayleigh-criteria} which can be formulated in the equation

This means that if the Lens length is small, (i.e. $L \ll\frac{d}{\lambda s}$), the cosine will be just $1$, so it will look like there is only one slit.\\

In order to find out that there are two slits, we must have that the distance is above the \textbf{Abbé limit}:

Which can be interpreted as asking: How far must two objects be apart, in order for a lens to detect that we have two objects.\\

If we define the \emph{Numerical aperture}$NA =\frac{\frac{L}{2}}{\sqrt{s^2+(\frac{L}{2})^2}}$, we can restate the formula above as

\begin{align*}

d > \frac{\lambda}{2 NA}

\end{align*}

\begin{demo}[Abbé limit]

We have a light source pointing through a circular hole and a wire grid, through a lens and an adjustable slit in horizontal direction.

We notice that even if we rotate the slit, the image stays the same, as the slit width stays the same.\\

However, if we tighted the slit, the vertical lines get washed out, whereas the horizontal lies stay sharp.\\

If we rotat the slit again, we see that now the horizontal lines are sharp and the vertical lines are blurred.

\end{demo}

\subsubsection{Polarisation}

We first look at a couple different types of polarisatiion, for a light travelling along the $z-axis$, i.e. $\vec{k}\parallel\vec{z}$\\

We will first look at \textbf{Linear polarisation}, where the field always points at a fixed direction. So the field is oscillating along a straight line. The field can be expressed as

, where $E_0=\sqrt{E_x^2+ E_y^2}$ and $\tan\theta=\frac{E_y}{E_x}$ and $\hat{\epsilon}=(\cos\theta\hat{x}+\sin\theta\hat{y})$ is called the \textbf{polarisation unit vector}.\\

In a more general form, we can have \textbf{Elliptical polarisation}, where the field will be of the form

For $\phi=0$, we recover linear polarisation, and for $\phi=\pm\frac{\pi}{2}$ we call that right(left) \textbf{circular polarisation}.\\

Warning, the right-left convention is not universally agreed upon, but in this convention, if we place our right thumb along $\hat{k}$, the other fingers follow the polarisation.\\

\subsubsection{Birefringence}

Recall from the previous sections, that we called denoted the Index of refraction as $n$. This had the effect that the effective wave-length changed depending on the medium or the wave vector.

where we call the axis with higher index of refraction the \emph{slow axis}.\\

If the light travels through a such a material with Length $L$, then the \emph{accumulated phase difference} will be equal to $(n_x - n_y)k_0L$.\\

For the special case where $(n_x - n_y)k_0L =\pi$ we call that a \emph{half-wave plate}. And if it equals $\frac{\pi}{2}$, we call that a \emph{quarter-wave plate}.\\

Next we will consider light reflection/transmission through a change in material. In the case where there are no free charges or currents we have

Similarly, we have $D_{\bot1}= D_{\bot2}$. In the Integral version of the fourth macroscopic maxwell equation, we have

\begin{align*}

\int_C \vec{H}\cdot d \vec{l} = \int_S \frac{\del D}{\del t}\cdot\hat{n} dA

\end{align*}

, where $C$ is the boundary of the surface and $S$ is its area with height $d$ and Length $L$. So as the height goes to zero, only the parallel components remain and we get

\begin{align*}

d \to 0: L H_{\parallel 1} - LH_{\parallel2}\implies H_{\parallel1} = H_{\parallel2}

\end{align*}

And similarly, $E_{\parallel1}= E_{\parallel2}$.\\

(Figure 1.40, 1.41) Steve

\subsubsection{Fresnel equations}

So now we can find out the amplitudes of the transmitting and reflecting waves bouncing off an interface. In the case of the p-polariszed light ($E$-ield is pointing in the plane of incidence) we have

where the indices of $n_1$ and $n_2$ are switched compared to the p-polarisation. Similar as in the $p$-polarisation case, we see that the $x$-component of the reflected $B$-field gets the minus sign.\\

Notice that the denominator diverges for $\theta_1+\theta_t =\frac{\pi}{2}$. In that case we have $r_p =0$. The special angle $\theta_i$ that causes this is called \textbf{Brewster's angle}$\theta_B$. Using Snells law again we have

From Figure 1.42 Steve, we can see that for $n_2:$ index of refractin of Water, the Brewster angle is at abou $55^{\circ}$, where $p$-polarised Light is not reflected. We also see that the reflected amplitude is always higher for $s$ polarisation than for $p$-polarisation.

\begin{demo}[]

We have a laser going through two linear polarizers we can rotate to change the relative angles of the polarizers. The light is then pointed to a photodetecter and the intensity is plotted on a screen.\\

If both are aligned in the same angle, we see that the intensity is high. As we rotate one polarizer, the intensity decreases until we hit the 90 dgree angle, at which we measure no light.\\

We also see that for angles bigger than 90 degrees, we obtain a periodic pattern, with peak intensity reached every 180 degrees.

Electro-magnetic waves can come in a huge spectrum of wavelengths. When we have a mixture of light, we might be interested in finding out what wavelengths are present.\\

Prisms are able to separate some frequencies, but they aren't really good to make accurate measurements, as they the index of refraction doesn't differ alot for waves, whose wavelengths only differ by a small amount. We could work the difference in angles using Snell's Law.\\

A more effective way to seperate by wavelength is using \textbf{Gratings}.\\

Consider an array of small mirrors with centers distance $d$ apart with some gaps inbetween. (Steve Fig. 1.44)\\

Now if we were assuming that there would be no gaps, we could argue with symmetry and find out that the angle of incoming and outgoing light is the same. But because we have gaps here, we can't do that.\\

When looking at the light reflected from neighboring mirrors, we can calculate the their phase difference as

\begin{align*}

\phi(\lambda) = k \cdot\left( d \sin\theta_i - d \sin\theta_j\right), \quad\text{where}\quad k = \frac{2\pi}{\lambda}

\end{align*}

Now look at the contribution of all $N$ strips and we can see that the Intensitiy will be

\begin{align*}

U &\propto\sum_{n=0}^{N-1} e^{i\phi n} = \frac{1 - e^{iN\phi}}{1 - e^{i\phi}}\\

Now we can ask when we will be able to resolve the difference between two wavelenegths. Looking at the (Fig. 1.47), we can frame the condition (similar to the Rayleigh criterium) as follows:\\

The conditions for the peak of wavelength $\lambda_1$ sitting at the zero of wavelength $\lambda_2$ will be