Commit 12aa8582 by Han-Miru Kim

### move phys-iii, remove main.org, comAlg w13

parent 12946367
 % \section{Organisation} % % What is this class about? % % % \begin{enumerate} % \item Optics % \begin{itemize} % \item Geometrical optics % \item Wave optics % \item Interference % \item Polarisation % \end{itemize} % \item Statistical mechanis % \begin{itemize} % \item States and how to count them % \item Boltzmann factor % \item Ideal gas % \item Blackbody radiation % \end{itemize} % \item Segue into the quantum world % \begin{itemize} % \item Atoms % \item Photons % \item Wave-particle duality % \end{itemize} % \item Quantum mechanics % \begin{itemize} % \item Wavefunctions % \item Schrödinger equations % \end{itemize} % \end{enumerate} % % 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 $c = 3 \times 10^8 \m/\s$ 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 \begin{align*} n_{\text{glass}} := \frac{c}{v_{\text{glass}}} \simeq 1.5 \end{align*} This allows us to define the \textbf{optical path length} of a path to be the quantity \begin{align*} D := n \cdot \ell \end{align*} , where $l$ is the physical length of the path.\\ If we can parametrize the space of variations from a path with a scalar quantity $\eta$ then \textbf{Fermat's Principle} can be reformulated to say \begin{align*} \frac{d D(\eta)}{d\eta}|_{\eta = 0} = 0 \end{align*} So what path does the light take if we take the situation depicted in figure \#\#? We can easily calculate the total optical path length to be \begin{align*} D = n_1 \sqrt{d_1^2 + h_1^2} + n_2 \sqrt{d_2^2 + h_2^2} \end{align*} Because $h, d_1, d_2$ are fixed and $h_2 = h - h_1$ we can write it using only the parameter $h_1$ so in order to find the extremum of $D(h_1)$ we need \begin{align*} 0 &= \frac{dD(h_1)}{dh_1}\\ &= \frac{n_1 h_1}{\sqrt{d_1^2 + h_1^2}} - \frac{n_2h_2}{\sqrt{d_2^2 + (h-h_2)^2}}\\ &= n_1 \sin \theta_1 - n_2 \sin \theta_2 \end{align*} From this we get \textbf{Snell's Law:} \begin{empheq}[box=\bluebase]{align*} n_1 \sin \theta_1 = n_2 \sin \theta_2 \end{empheq}
 \subsubsection{Lenses} 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 \begin{align*} D_{AC'B} = n_1 \sqrt{h^2 + (d_1 + g(h))^2} + n_2 \sqrt{h^2 + (d_2 - g(h))^2} \end{align*} Using the taylor approximation of $(1 + \epsilon)^{\alpha}$ for $\abs{\epsilon} \ll 1$, we have $(1 + \epsilon)^{\alpha} \simeq 1 + \alpha \epsilon$.\\ 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 \begin{align*} D_{AC'B}(h) &= n_1(d_1 + g(h)) \sqrt{1 + \frac{h^2}{(d_1 + g(h))^{2}}} + n_2(d_2 - g(h)) \sqrt{1 + \frac{h^2}{(d_2 - g(h))^{2}}}\\ &\simeq n_1(d_1 + g(h)) \left(1 + \frac{h^2}{2(d_1 + g(h))^{2}}\right) + n_2(d_2 - g(h)) \left(1 + \frac{h^2}{2(d_2 - g(h))^{2})}\right)\\ &= n_1d_1 + n_2d_2 - (n_2-n_2)g(h) + \frac{h^2}{2} \left(\frac{n_1}{d_1} + \frac{n_2}{d_2}\right) \end{align*} where the errors are of order $\epsilon^2 \propto h^4$.\\ After taking the difference between the two pathlengths and setting it to zero, we can get the right curvature of the lens surface. \begin{align*} \Delta D = D_{AC'B} - D_{ACB} = \frac{h^2}{2} \left(\frac{n_1}{d_1} + \frac{n_2}{d_2}\right) - (n_2 - n_1) g(h) = 0\\ \implies g(h) = \frac{h^2}{2(n_2 - n_1)} \left(\frac{n_1}{d_1} + \frac{n_2}{d_2}\right) \end{align*} which is a parabola.\\ 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 \begin{empheq}[box=\bluebase]{align}\label{eq:redstar} 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 \begin{align*} \frac{1}{R_2} &= \frac{1}{1-n} \left(\frac{n}{d_1} + \frac{1}{d_2}\right)\\ &= \frac{1}{1-n} \left(- \frac{n-1}{R_1} + \frac{1}{f}\right) \\ \frac{1}{f} &= (n-1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \end{align*} 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 $h'$.\\ This would give us \begin{align*} h' &= f_1 \tan(\Delta \theta) \simeq f_1 \Delta \theta\\ \Delta \theta' &= \arctan \left(\frac{h'}{f_2}\right) \simeq \frac{f_1}{f_2} \Delta \theta \end{align*} 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!
 \subsection{Wave Optics} 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 \begin{enumerate} \item $\nabla \cdot \vec{E} = 0$ \item $\nabla \cdot \vec{B} = 0$ \item $\nabla \times \vec{E} = - \frac{\del \vec{B}}{\del t}$ \item $\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\del \vec{E}}{\del t}$ \end{enumerate} After taking the curl we get \begin{align*} \nabla \times (\nabla \times \vec{E}) &= \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E}\\ &= \nabla \times \left(- \frac{\del \vec{B}}{\del t}\right) = - \frac{\del}{\del t}(\nabla \times \vec{B}) = \mu_0 \epsilon_0 \frac{\del^2 \vec{E}}{\del t^2} \end{align*} And we end end up with the wave equation \begin{align*} \nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\del^2 \vec{E}}{\del t^2} = \frac{1}{c^2} \frac{\del^2 \vec{E}}{\del t} \end{align*} 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 \begin{align*} \vec{E}(\vec{r},t) = E_0 e^{i(\vec{k} \cdot \vec{r} - \omega t)} \end{align*} and the spherical wave equation \begin{align*} \vec{E}(\vec{r},t) = \frac{A}{r}e^{i(kr - \omega t)A} \end{align*} , 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 \begin{align*} U_{AC}(\vec{r}) &\propto \frac{1}{\abs{\vec{r}_C - \vec{r}_A}} e^{ik \abs{\vec{r}_C - \vec{r}_A}}\\ U_{CB}(\vec{r}) & \propto \frac{1}{\abs{\vec{r}_B - \vec{r}_C}} e^{ik \abs{\vec{r}_B - \vec{r}_C}} \end{align*} Now at $B$, we should see the sum of the two waves, so we should have \begin{align*} U_B \propto \int_{-a/2}^{a/2}dx \int_{-a/2}^{a/2} \frac{1}{\abs{\vec{r}_C - \vec{r}_A} \cdot \abs{\vec{r}_B - \vec{r}_C}}e^{ik(\abs{\vec{r}_C - \vec{r}_A} + \abs{\vec{r}_B - \vec{r}_C})} \end{align*} Since we can assume that the aperture is very small, we have $a \ll r_A,r_B$ and we get \begin{align*} \abs{\vec{r}_C - \vec{r}_A} = r_A - \frac{x x_A + yy_A}{r_A} + \frac{x^2 + y^2}{2r_A} \end{align*} 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 \begin{align*} U_B \propto \int_{-a/2}^{a/2} dx \int_{-a/2}^{a/2}dy \frac{e^{ik(r_A + r_B)}e^{-ik(Xx + Yy)}}{ \left(r_A - \frac{xx_a + yy_A}{r_A}\right) \left(r_B - \frac{xx_B - yy_B}{r_A}\right)} \end{align*} , for $X = \frac{x_A}{r_A} + \frac{x_B}{r_B}$ and $Y = \frac{y_A}{r_A} + \frac{y_B}{r_B}$\\ In our third approximation, we will drop the term $\frac{xx_a + yy_A}{r_A}$ and $\frac{xx_B - yy_B}{r_A}$ and we get\\ \begin{align*} U_B \propto \frac{e^{-k(r_A + r_B)}}{r_Ar_B}\int_{-a/2}^{a/2} e^{ik(Xx + Yy)}dxdy = \frac{4 e^{ik(r_A + r_B)}}{r_Ar_B} \cdot \text{sinc} \left(\frac{kXa}{2}\right) \cdot \text{sinc} \left(\frac{kYa}{2}\right) \end{align*} where $\text{sinc}(x) = \frac{\sin(x)}{x}$. Further, we get \begin{align*} U_B \propto \frac{a^2 e^{ik(r_A + r_B)}}{r_Ar_B}\text{sinc} \left(\frac{kXa}{2}\right) \text{sinc} \left(\frac{kYa}{2}\right) \end{align*} The intensity will then be $I = \abs{E(\vec{r},t)}^2 = \abs{U(\vec{r})}^2$ and we have \begin{align*} I_B = I_0 \text{sinc}^2 \left(\frac{kXa}{2}\right) \text{sinc} \left(\frac{kYa}{2}\right) \end{align*} so we end up with the characteristic distribution with a large peak at the center and smaller peaks periodically in the distance to the center.\\ If we take the limit as $X = 0$ (which implies $\frac{x_A}{r_A} = - \frac{x_b}{r_B}$, then we get the delta-distribution, which mimics ray-optics!\\ If we look at the equation again \begin{align*} U_B \propto \int_{-a/2}^{a/2}dx \int_{-a/2}^{a/2}dy e^{ik(\abs{\vec{r}_C - \vec{r}_A} + \abs{\vec{r}_B - \vec{r}_C})} \end{align*} we can see the optical path length $D \simeq r_A + r_B - Xx - Yy$\\ If we try to restate Fermat's principle $\frac{d D(\eta)}{d\eta}|_{\eta = 0} = 0$ we get the condition in two variables \begin{align*} \frac{\del D(x)}{\del x} = -X = 0, \quad \frac{\del D(y)}{\del y} = -Y = 0 \end{align*} 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 \begin{align*} U_B \propto e^{ik(r_A + r_B)} \int_{-\infty}^{\infty}dx \int_{-\infty}^{\infty}dy \tau(x,y) e^{-ik(Xx - Yy)} \end{align*} 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 \begin{empheq}[box=\bluebase]{align*} X_2 := \frac{x_{A_2}}{r_{A_2}} + \frac{x_B}{r_B} = \sin\phi + \sin \theta \end{empheq} ,where $\theta$ is the angle of the outgoing lightrays from the lens.\\ The angle that satisfies Rayleigh-critera must have \begin{align*} \sin \left(\frac{ka X_2(\theta = 0)}{2}\right) = \sin \left(\frac{k\sin\phi a}{2}\right) = 0 \end{align*} And we obtain \begin{align*} \frac{ka \sin\phi}{2} = \pi \implies \sin\phi = \frac{\lambda}{a} \end{align*} This is what we would expect. In order to tell apart to different objects far away, we need a bigger aperture i.e. a bigger telescope.\\ Next we consider a plane wave entering a double slit, distance $d$ apart and then hitting a lense on the other side with Lens length $L$.\\ The transmission function of the double slit will be \begin{align*} \tau(x) \propto \delta(x - \frac{d}{2}) + \delta(x + \frac{d}{2}) \end{align*} Using the generalized result from the previous section we have that the Field at the lens will be of the form \begin{align*} U(x_B) &\propto e^{ik(x_b^2 + s^2)}\int_{-\infty}^{\infty}dx \tau(x) e^{ikx_b \frac{X}{\sqrt{x_B^2 + s^2}}}\\ &\propto e^{ik(x_B^2 + s^2)}\cos \left(\frac{kxd}{2 \sqrt{x_B^2 + s^2}}\right) \end{align*} 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}: \begin{align*} \frac{kd \frac{L}{2}}{2 \sqrt{(\frac{L}{2})^2 + s^2}} = \pi, \quad d \gg \lambda \frac{\sqrt{s^2 + (\frac{L}{2})^2}}{L} \end{align*} 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 \begin{align*} \vec{E}(\vec{r},t) &= (E_x \hat{x} + E_y\hat{y})e^{i(kz + \omega t)}\\ &= E_0 (\cos\theta \hat{x} + \sin \theta \hat{y}) e^{i(kz - \omega t)} \end{align*} , 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 \begin{align*} \vec{E}(\vec{r},t) = \left(E_x \hat{x} + E_ye^{i\phi}\hat{y}\right) e^{i(kz - \omega t)}. \end{align*} 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. \begin{align*} \lambda_0 \to \frac{\lambda_0}{n}, \quad k_0 \to \frac{n}{k_0} \end{align*} If we had a material, where the index of refraction differences in multiple directions, i.e. $n_x \neq n_y$ we get the \textbf{birefringence} effect: \begin{align*} \vec{E}(\vec{r},t) = \left[E_x \hat{x} e^{ik_xz} + E_y \hat{y}e^{i(k_yz + \phi)}\right]e^{-i\omega t} \end{align*} 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 \begin{align*} \nabla \cdot \vec{D} = 0, \quad \nabla \times \vec{E} = - \frac{\del B}{\del t}\\ \nabla \cdot \vec{B} = 0, \quad \nabla \times \vec{H} = \frac{\del D}{\del t} \end{align*} For a linear material, we will have \begin{align*} \vec{B} = \mu\mu_R \vec{H}, \quad \vec{D} = \epsilon_0\epsilon_r \vec{E} \end{align*} And for non-magnetic materials we have $\mu_r = 1$, so \begin{align*} \nabla^2 \vec{E} = \epsilon_r\epsilon_0\mu_0 \frac{\del^2 \vec{E}}{\del t^2} \end{align*} which changes the speed of light \begin{align*} v = \frac{1}{\sqrt{\epsilon_r\epsilon_0\mu_0}} = \frac{c}{n} \end{align*} where $n = \sqrt{\frac{\epsilon_r}{\mu_r}}$ is the index of refraction.\\ For plane-waves we get \begin{align*} \vec{k} \times \vec{E} = \omega \vec{B} \implies B = \frac{k}{\omega}E = \frac{nk_0}{\omega}E = \frac{n}{c}E \end{align*}
 In the integral form over the closed surface $S$ of a cylinder with area $A$ and height $d$ we have \begin{align*} \int_S \vec{B} \cdot\hat{n}dA = 0 \end{align*} As we reduce the height of the surcace: we get that \begin{align*} d \to 0: \quad AB_{\bot1} - AB_{\bot 2} = 0 \implies B_{\bot 1} = B_{\bot 2} \end{align*} 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 \begin{align*} E_i \cos \theta_i - E_r \theta_i = E_t \cos \theta_t \end{align*} where $E_i$ is the ampluditde of the \emph{incoming}, $E_r$ of the \emph{reflected} and $E_t$ of the \emph{transmitted} field.\\ From the relation \begin{align*} \vec{H} = \frac{\vec{B}}{\mu_0} = \frac{n}{\mu_0 \omega} \vec{k} \times \vec{E}_0 \end{align*} and the boundary condition, we obtain \begin{align*} n_1E_i + n_1E_r = n_2E_t \end{align*} Combining the newly found equations, we obtain the \textbf{Fresnel equations for $p$-polarisation}: \begin{empheq}[box=\bluebase]{align*} t_p := \frac{E_t}{E_i} = \frac{2n_1 \cos \theta_i}{n_2 \cos \theta_i + n_1 \cos \theta_t}\\[1em] r_p := \frac{E_r}{E_i} = \frac{n_2 \cos \theta_i - n_1 \cos \theta_t}{n_2 \cos \theta_i + n_1 \cos \theta_t} \end{empheq} From Figure 1.37 Steve, we see that the $x$ component of the reflecting $E$ field gets a inus sign, For the $s$-polarisation, where the Electric field is pointing parallel to the surface (which we will derive as homework exercise) we have \begin{empheq}[box=\bluebase]{align*} t_s := \frac{E_t}{E_i} = \frac{2n_1 \cos \theta_i}{n_1 \cos \theta_i + n_2 \cos \theta_t}\\[1em] r_s := \frac{E_r}{E_i} = \frac{n_1 \cos \theta_i - n_2 \cos \theta_t}{n_1 \cos \theta_i + n_2 \cos \theta_t} \end{empheq} 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.\\ From writing \begin{align*} A_i e^{i(\vec{k}_i \cdot \vec{r} - \omega_i t)} + A_R e^{i(\vec{k}_r \cdot \vec{r} - \omega_r t)} = A_t e^{i(\vec{k}_t \cdot \vec{r} - \omega_t t)} \end{align*} we obtain Snell's law: \begin{align*} n_1 k_0 \sin \theta_i = n_1 k_0 \sin \theta_r = n_2 k_0 \sin \theta_t \\ \implies \theta_i = \theta_r \quad \text{and} \quad n_q \sin \theta_1 = n_2 \sin \theta_t \end{align*} If we want to actually solve some problems, we can use the following: If we have any arbitrary polarisation, it can be written as a superposition of $p$ and $s$ polarizations. And in order to relate $\theta_i$ and $\theta_t$, we can apply Snell's law.\\ \subsubsection{Brewster's Angle} Starting from the Fresnel we can use some trigonometric tricks'' to get \begin{align*} r_p &= \frac{n_2 \cos \theta_i - n_1 \cos \theta_t}{n_2 \cos \theta_i + n_1 \cos \theta_t}\\ &= \frac{\sin \theta_1 \cos \theta_i - \sin \theta_t \cos \theta_t}{\sin \theta_i + \sin \theta_t \cos \theta_t}\\ &= \frac{\sin(2 \theta_i) - \sin(2 \theta_t)}{\sin(2 \theta_i) + \sin(2 \theta_t)}\\ &= \frac{\tan(\theta_i - \theta_t)}{\tan(\theta_i + \theta_t)} \end{align*} 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 \begin{align*} \theta_B = \frac{\pi}{2} - \theta_t \implies \cos\theta_B = \sin(\theta_t) = \frac{n_1}{n_2} \sin \theta_B\\ \implies \theta_B = \arctan(n_2/n_1) \end{align*} 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. \end{demo}
 \subsection{Spectroscopy} 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}} \\ &= \frac{e^{i\phi (\frac{N}{2}} - e^{i\phi \frac{N}{2}})}{e^{i \frac{\phi}{2}} (e^{- \frac{\phi}{2}} - e^{i \frac{\phi}{2}})} = e^{i(N-1) \frac{\phi}{2}} \left(\frac{\sin(N \frac{\phi}{2})}{\sin (\frac{\phi}{2})}\right) \end{align*} So the Intensitiy will have \begin{align*} I = \abs{U}^2 \propto \frac{\sin^2 \left(N \frac{\phi}{2}\right)}{\sin^2 \left(\frac{\phi}{2}\right)} \end{align*} which will have its maxima at \begin{align*} \frac{\phi}{2} = m \pi \quad \text{for} \quad m \in \Z \end{align*} At these points, the Intensity will be \begin{align*} \lim_{\epsilon \to 0} \frac{\sin^2 \left(N \frac{\phi}{2} + \epsilon\right)}{\sin^2 \left(\frac{\phi}{2} + \frac{\epsilon}{N}\right)} = N^2 \end{align*} See (Steve Fig. 1.47) \begin{align*} \sin \theta_i - \sin \theta_j = \frac{2 \pi m}{dk} = \frac{m \lambda}{d} \end{align*} 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