![[Pasted image 20261018163124.png|277]] ![[Pasted image 20261018200412.png|233]] Left: Transverse Electric (TE) wave reflection Right: Transverse Magnetic (TM) wave reflection Since optical beams typically have diameters that are large compared to the free-space wavelength, plane waves are a good approximation of optical wave propagation. When a plane wave strikes an interface between two media, the tangential $\vec{E}$ and $\vec{H}$ must be continuous at the interface where $z=0$. As a result, the incident, reflected, and transmitted waves must share an identical wavenumber in $x$, i.e. $k_{x}^{(1)}=k_{x}^{(2)}=k_{x}$. This gives us **Snell's Law**, as illustrated in the figure above. $ \sqrt{ \mu_{1}\epsilon_{1} }\sin\theta_{1}=\sqrt{ \mu_{2}\epsilon_{2} }\sin\theta_{2} $ At optical frequencies, $\mu\simeq \mu_{0}$ so we define the refractive index of a material as $n\equiv \sqrt{ \frac{\epsilon}{\epsilon_{0}} }$ such that Snell's Law can also be written as $n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}$. We define the following: $Y_{0}^{(1)}\equiv \frac{k_{z}^{(1)}}{\omega \mu_{1}}=\sqrt{ \frac{\epsilon_{1}}{\mu_{1}} }\cos\theta_{1} =$ characteristic admittance of medium 1 $Z_{0}^{(1)}\equiv \frac{1}{Y_{0}^{(1)}}=$ characteristic impedance of medium 1 $Z\equiv-\frac{E_{y}}{H_{x}}=$ wave impedance at interface The wave impedance must be continuous. Superposition of the planar wave solutions $E_{\pm}e^{\pm jk_{z}z-jk_{x}x}$ gives us that at $z=0$, $ Z_{0}^{(1)} \frac{E_{+}^{(1)}+E_{-}^{(1)}}{E_{+}^{(1)}-E_{-}^{(1)}}=Z_{0}^{(2)} $ The reflection coefficient is defined as $\Gamma\equiv \frac{E_{-}}{E_{+}}$, which from the above equation can be determined to be given by the following: $ \Gamma^{(1)}= \frac{Z_{0}^{(2)}-Z_{0}^{(1)}}{Z_{0}^{(2)}+Z_{0}^{(1)}} $ It follows from the Poynting Theorem that the power flow density in the $z$ direction is $ \frac{1}{2}\mathrm{Re}(\vec{E}\times \vec{H}^{\star})=\frac{1}{2}Y_{0}^{(1)}\lvert E_{+}^{(1)} \rvert ^{2}(1-\lvert \Gamma^{(1)} \rvert^{2} ) $ Therefore, $\lvert \Gamma \rvert^{2}$ is the ratio of reflected to incident power flow. ## Key Formulas | Impedance form | Admittance form | | ------------------------------------------------------------------ | ------------------------------------------------------------------ | | $\dfrac{Z(z)}{Z_0} = \dfrac{1+\Gamma(z)}{1-\Gamma(z)}$ | $\dfrac{Y(z)}{Y_0} = \dfrac{1-\Gamma(z)}{1+\Gamma(z)}$ | | $\Gamma(z) = \dfrac{Z(z)-Z_0}{Z(z)+Z_0}$ | $\Gamma(z) = \dfrac{Y_0-Y(z)}{Y_0+Y(z)}$ | | $Z(z) = Z_0\,\dfrac{Z(0) - jZ_0\tan k_z z}{Z_0 - jZ(0)\tan k_z z}$ | $Y(z) = Y_0\,\dfrac{Y(0) - jY_0\tan k_z z}{Y_0 - jY(0)\tan k_z z}$ | ### Polarization-specific | Quantity | TE | TM | | ----------------- | -------------------------------------------------------------- | -------------------------------------------------------------- | | Reflection coeff. | $\Gamma(z) = \dfrac{E_-}{E_+}\,e^{+j2k_z z}$ | $\Gamma(z) = -\dfrac{H_-}{H_+}\,e^{+j2k_z z}$ | | Char. impedance | $Z_0 = \sqrt{\dfrac{\mu}{\varepsilon}}\,\dfrac{1}{\cos\theta}$ | $Z_0 = \sqrt{\dfrac{\mu}{\varepsilon}}\,\cos\theta$ | | Impedance | $Z(z) = -\dfrac{E_y}{H_x}$ | $Z(z) = \dfrac{E_x}{H_y}$ | | Char. admittance | $Y_0 = \sqrt{\dfrac{\varepsilon}{\mu}}\,\cos\theta$ | $Y_0 = \sqrt{\dfrac{\varepsilon}{\mu}}\,\dfrac{1}{\cos\theta}$ | | Admittance | $Y(z) = -\dfrac{H_x}{E_y}$ | $Y(z) = \dfrac{H_y}{E_x}$ | ## Special Angles If medium 1 has a larger value of $\sqrt{ \mu\epsilon }$ than medium 2, Snell's law cannot yield a real angle $\theta_{2}$ when $\theta_{1}>\theta_{c}$ for the critical angle $\theta_{c}$ which is given by: $\sin\theta_{c}=\sqrt{ \frac{\epsilon_{2}}{\epsilon_{1}} }=\frac{n_{2}}{n_{1}}$ In order to maintain continuity of $k_{x}$, we must have a negative imaginary propagation constant resulting in an **evanescent wave** where the following is true: - $k_{z}^{(2)}$ is imaginary - $\lvert \Gamma \rvert=1$ - No power is transmitted - $E_{y}=E_{t}e^{-jk_{x}x}e^{-\alpha z}$, where $\alpha=\sqrt{ k_{x}^{2}-k_{2}^{2} }$ ## Fresnel Coefficients (explicit, $\mu_1=\mu_2$) These follow from $\Gamma=\dfrac{Z_0^{(2)}-Z_0^{(1)}}{Z_0^{(2)}+Z_0^{(1)}}$ with the impedances above. $\Gamma$ is the ratio of tangential $E$ (the TM sign convention matches the table: $\Gamma=-\dfrac{H_-}{H_+}$). $ \Gamma_{TE}=\frac{n_1\cos\theta_1-n_2\cos\theta_2}{n_1\cos\theta_1+n_2\cos\theta_2},\qquad \Gamma_{TM}=\frac{n_1\cos\theta_2-n_2\cos\theta_1}{n_1\cos\theta_2+n_2\cos\theta_1} $ (Many optics books define $r_p$ with the opposite sign. Check before comparing.) - Normal incidence (both polarizations): $ \Gamma=\frac{n_1-n_2}{n_1+n_2},\qquad R=\left(\frac{n_1-n_2}{n_1+n_2}\right)^2 $ Air to glass ($n=1.5$) gives 4%. - Power: $R=|\Gamma|^2$ and $\mathcal T=1-R$ (lossless). Don't compute $\mathcal T$ as $|E_t/E_i|^2$: the impedance and beam-area ($\cos\theta$) factors matter. ## Brewster's Angle For TM with $\mu_1=\mu_2$: $\Gamma_{TM}=0$ when $Z_0^{(1)}=Z_0^{(2)}$, i.e. $\dfrac{\cos\theta_1}{n_1}=\dfrac{\cos\theta_2}{n_2}$. Combined with Snell: $ \tan\theta_B=\frac{n_2}{n_1},\qquad \theta_B+\theta_2=\frac{\pi}{2} $ - The reflected direction would be along the dipoles induced in medium 2, and a dipole can't radiate along its axis. That's the physical picture. - There is a Brewster angle from **both** sides: the internal one is the complement of the external one, e.g. 56.3° and 33.7° for glass (Haus 3.1). A slab at $\theta_B$ transmits TM perfectly at both faces. - **No TE Brewster angle** when $\mu_1=\mu_2$. By duality a TE Brewster angle needs $\mu_1\neq\mu_2$: $ \tan\theta=\sqrt{\frac{\mu_2}{\mu_1}}\quad\text{when }\epsilon_1=\epsilon_2 $ - Sensitivity near $\theta_B$ (Haus 2.1): $\Gamma$ is linear in $\theta-\theta_B$, so $R$ is quadratic. For $n=1.7$, $|\Gamma|\approx0.75\,|\Delta\theta|$ with $\Delta\theta$ in radians. - Uses: Brewster windows in gas lasers (the output is TM-polarized), and the MacNeille PBS (below). ## Total Internal Reflection: Phase and Goos–Hänchen For $\theta_1>\theta_c$, $k_z^{(2)}=-j\kappa$ with $ \kappa=\sqrt{k_x^2-k_2^2}=k_0\sqrt{n_1^2\sin^2\theta_1-n_2^2} $ so $Z_0^{(2)}$ is imaginary and $|\Gamma|=1$, but with a phase: $ \Gamma_{TE}=e^{j\phi_{TE}},\qquad \tan\frac{\phi_{TE}}{2}=\frac{\sqrt{n_1^2\sin^2\theta_1-n_2^2}}{n_1\cos\theta_1} $ $ \Gamma_{TM}=-e^{j\phi_{TM}},\qquad \tan\frac{\phi_{TM}}{2}=\frac{n_1^2}{n_2^2}\,\tan\frac{\phi_{TE}}{2} $ - The TM–TE phase difference $\phi_{TM}-\phi_{TE}$ (Kong 4.3.7), with $n=\dfrac{n_1}{n_2}$: $ \tan\frac{\phi_{TM}-\phi_{TE}}{2}=\frac{\cos\theta_1\sqrt{\sin^2\theta_1-n^{-2}}}{\sin^2\theta_1} $ Two TIR bounces giving 45° each turn linear into circular polarization (Fresnel rhomb). - The field in medium 2 decays as $e^{-\kappa z}$ into the medium (normal to the interface) and carries no time-averaged power. The power flow is reactive. - The reflection phase makes the standing-wave maximum sit off the interface. For a finite beam this becomes a lateral shift of the reflected beam, the **Goos–Hänchen** shift (Haus 2.4). - Frustrated TIR: a second dense medium within a few $\dfrac{1}{\kappa}$ lets power tunnel through. Model it as a line section with imaginary $Z_0$ ([[transmission lines]]). ## Multilayers: the Impedance Recursion Algorithm (works for TE or TM; use that polarization's $Z_0^{(l)}$ and $k_z^{(l)}$; $k_x$ is the same in every layer): 1. Start in the substrate. Nothing comes back, so $Z=Z_0^{(s)}$. 2. At each interface $Z$ is **continuous**, so carry it across unchanged. 3. Through each layer of thickness $d_l$, transform it: $ Z_\text{front}=Z_0^{(l)}\,\frac{Z_\text{back}+jZ_0^{(l)}\tan k_z^{(l)}d_l}{Z_0^{(l)}+jZ_\text{back}\tan k_z^{(l)}d_l} $ 4. In the incident medium: $ \Gamma=\frac{Z_\text{in}-Z_0^{(1)}}{Z_\text{in}+Z_0^{(1)}} $ Matrix versions (Kong propagation matrix, ABCD) are in [[optical operators]]. Use them when you also need $T$ or the fields in each layer. ### Quarter-wave layers ($k_z d=\dfrac{\pi}{2}$, i.e. $d=\dfrac{\lambda}{4n\cos\theta}$) $ Z_\text{front}=\frac{\left(Z_0^{(l)}\right)^2}{Z_\text{back}},\qquad\text{half-wave: } Z_\text{front}=Z_\text{back} $ **Single-layer AR coating** at normal incidence: $ R=\left(\frac{n_1n_s-n_c^2}{n_1n_s+n_c^2}\right)^2=0\quad\text{when}\quad n_c=\sqrt{n_1n_s},\qquad d=\frac{\lambda_0}{4n_c} $ - MgF₂ ($1.38$) on glass ($1.5$) gives $R=1.4\%$, versus 4% bare. - Off the design wavelength the layer is no longer $\dfrac{\lambda}{4}$, so $R$ rises (it is periodic in $\dfrac{1}{\lambda}$). - At oblique incidence the "index" becomes $n\cos\theta$ for TE or $\dfrac{n}{\cos\theta}$ for TM (as an admittance), and the thickness condition uses $\cos\theta$. **High-reflector stack** $H(LH)^N$ on substrate $n_s$, from medium $n_0$, normal incidence. Each LH pair multiplies the admittance by $\left(\dfrac{n_H}{n_L}\right)^2$: $ Y_\text{in}=\left(\frac{n_H}{n_L}\right)^{2N}\frac{n_H^2}{n_s},\qquad R=\left(\frac{n_0-Y_\text{in}}{n_0+Y_\text{in}}\right)^2\to1 $ - With H facing the incident light, $Y_\text{in}\to\infty$, i.e. $Z_\text{in}\to0$ and $\Gamma\to-1$. With L facing it, $\Gamma\to+1$. - Example (Haus 2.3): ZnS/ThF₄ (2.5/1.5) on glass, $R>99.5\%$, needs 13 layers. - The high-reflection band has finite width; its width grows with $\dfrac{n_H}{n_L}$. ### MacNeille polarizing beam splitter (pset 3 P3) Make **every** H/L interface Brewster for TM. In layer H the angle must satisfy $\tan\theta_H=\dfrac{n_L}{n_H}$, so the Snell invariant is $ n\sin\theta=\frac{n_Hn_L}{\sqrt{n_H^2+n_L^2}} $ - This exceeds 1 for real coatings, so it is impossible from air. Inside a glass prism at 45°: $ n_g=\sqrt2\,\frac{n_Hn_L}{\sqrt{n_H^2+n_L^2}} $ This is 1.735 for TiO₂ / SiO₂ (2.30 / 1.45). - Normalized impedances: $ \frac{Z_{TE}}{\eta_0}=\frac{1}{n\cos\theta},\qquad \frac{Z_{TM}}{\eta_0}=\frac{\cos\theta}{n} $ - **TE (s):** the H and L impedances alternate, so the stack is a quarter-wave HR stack and $R_s\to1$ as $N$ grows. The TE contrast at 45° is larger than at normal incidence. - **TM (p):** H and L have **equal** impedance (that's the Brewster condition), so the $2N+1$ layers act as **one** uniform layer that is an odd number of quarter-waves thick. It is a quarter-wave transformer between glass and glass: $ R_p=\left(\frac{Z_g^2-Z_{HL}^2}{Z_g^2+Z_{HL}^2}\right)^2\quad\text{(independent of } N\text{)} $