Light at an Interface

Start from the macroscopic Maxwell equations for a linear, isotropic, nonmagnetic medium with no free charge. Fields vary as $\exp(-i\omega t)$, units are SI, $\varepsilon(\omega)$ is the relative permittivity, $k_0=\omega/c$ is the vacuum wave number and $Y_0=\sqrt{\varepsilon_0/\mu_0}$ is the vacuum admittance.

In a uniform region $\varepsilon$ is constant, so $\nabla\cdot\mathbf E=0$. Taking the curl of Faraday's law, substituting the Ampère–Maxwell law and using $\nabla\times\nabla\times\mathbf E=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$ gives the Helmholtz equation, with $n^2=\varepsilon$:

Geometry. The boundary is the plane $z=0$. Medium 1 fills $z<0$, medium 2 fills $z>0$, and $x$–$z$ is the plane of incidence, so nothing depends on $y$. "Forward" means travelling toward $+z$. Medium 2 has only the transmitted, forward wave, because nothing sends light back from $z>0$. Both media are lossless here, so $\varepsilon$ is real.

Plane waves and phase matching. In medium $j$ try $\exp[ik_0(k_x x\pm q_j z)]$, where $k_x$ and $q_j$ are the $x$ and $z$ components of the wave vector divided by $k_0$. The Helmholtz equation requires $k_x^2+q_j^2=n_j^2$. Integrate Faraday's law around a thin loop straddling the boundary: as the loop's width shrinks, the magnetic flux through it goes to zero, so tangential $\mathbf E$ must agree on the two sides. The Ampère–Maxwell law does the same for tangential $\mathbf H$ when there is no surface current. These conditions must hold at every $x$, so the incident, reflected and transmitted waves share the same $k_x$. For the incident wave at angle $\theta$ (the $\theta_i$ of the lecture), $k_x=n_1\sin\theta$:

The reflected wave has the same $k_x$ and $-q_1$, which is the law of reflection.

Admittance. For TE, $\mathbf E=E_y\hat{\mathbf y}$, and the $x$ component of Faraday's law gives $H_x=-\tfrac{1}{i\omega\mu_0}\partial_z E_y$. For TM, $\mathbf H=H_y\hat{\mathbf y}$, and the Ampère–Maxwell law gives $E_x=-\tfrac{i}{\omega\varepsilon_0 n_j^2}\partial_z H_y$. Use $\partial_z=\pm ik_0q_j$ for the forward (+) and backward (−) waves. The constants that appear are $k_0/(\omega\mu_0)=1/(c\mu_0)$ in TE and $\omega\varepsilon_0/k_0=c\varepsilon_0$ in TM; both equal $Y_0$:

The backward wave has the opposite sign, because its $q$ is reversed. The overall minus in TE is common to every term and cancels below.

Amplitudes. Take unit incident tangential $E$, reflected $r$ and transmitted $t$. Continuity of tangential $E$ and $H$ at $z=0$ gives

These are amplitudes of $E_y$ or $E_x$, not of the full field vector. For TM, sources differ in how they orient the reflected field vector; relative to Byrnes's full-field convention, $r$ here has the opposite sign. Powers are unaffected.

Power. The time-averaged normal Poynting flux is $\langle S_z\rangle=\tfrac12\operatorname{Re}(E_xH_y^*-E_yH_x^*)$. For a single forward wave in either polarization this is $\tfrac12Y_0\operatorname{Re}(Y_j)\,|E_t|^2$, where $E_t$ is the tangential amplitude. At $z=0$ in medium 1 the total tangential field is $1+r$ and the total tangential $H$ is proportional to $Y_1(1-r)$, with $Y_1$ real because medium 1 is lossless and $\theta<90^\circ$. Then $\operatorname{Re}[(1+r)(1-r^*)]=1-|r|^2$, because $r-r^*$ is imaginary. Dividing by the incident flux $\tfrac12Y_0Y_1$ gives

Tangential $E$ and $H$ are continuous, so $\langle S_z\rangle$ is the same on both sides, which is $R+T=1$.

Total internal reflection. When $n_2<n_1\sin\theta$, $q_2^2<0$ and $q_2$ is imaginary. Choosing $\operatorname{Im}q_2>0$ makes $\exp(ik_0q_2z)$ decay into medium 2 instead of growing. Then $Y_2$ is imaginary in both polarizations, so $\operatorname{Re}Y_2=0$ and $T=0$. With $Y_1$ real and $Y_2$ imaginary, the numerator of $r=(Y_1-Y_2)/(Y_1+Y_2)$ is the complex conjugate of the denominator, so $|r|=1$.