# Transmission and Reflection

## Normal Incidence

Consider if we have to dielectric media, \(n,n'\) with the xy plane being the interface between them. Let a wave propogate in the z-direction. What is the transmission and reflection of the wave? If \(E = E\exp(i(kz-\omega t))\hat{x}\) then \(B = \frac{nE}{c}\exp(i(kz-\omega t))\hat{y}\). The transmitted wave vector is \(k' = n'\frac{\omega}{c}\hat{z}\) and reflected is \(k'' = -k\hat{z} = -n\frac{\omega}{c}\hat{z}\). Then, \(E' = E'\exp(i(kz-\omega t))\hat{x}\) and \(E' = E'\exp(i(kz-\omega t))\hat{y}\). For the reflected wave, \(E'' = E''\exp(i(-kz-\omega t))(-\hat{x})\) and \(B'' = B''\exp(i(-kz-\omega t))\hat{y}\).

From our boundary conditions, \(E_{t1} = E_{t2}\). Since our wave is already only tangential, the electric field is continuous. Similarly, \(H_{t}\) is continuous so \(H=B/\mu_0\) is continuous. From these we get,

\begin{align} E - E'' &= E' \\ E + E'' &= \frac{n'}{n}E' \end{align}So,

\begin{align} E' &= \frac{2n}{n'+n}E = tE. \end{align}This gives us the transmission coefficient of the interface for a normal wave propogation. \(t = \frac{2n}{n'+n}\). \(r = 1-T = \frac{n'+n-2n}{n'+n} = \frac{n'-n}{n'+n}\).

When \(n'\) is a denser medium, we get a phase flip in the reflected wave and a smaller transmitted wave. If it is not as dense, the phase is unchanged in the reflected wave and we get a larger transmitted wave.

If we consider the full expression for intensity, \(I = |\vec{S}| = \frac{1}{2}nc\ve_0|\vec{E}|^2\). \(S_i = S_r + S_t = n|E|^2 + n|E''|^2 + n'|E'|^2\).

\(1 = \frac{S_r}{S_i} + \frac{S_t}{S_i} = R + T\), the reflectance and transmittance. So, \(1 = \frac{|E''|^2}{|E|^2} + \frac{n'|E'|^2}{n|E|^2} = |r|^2 + \frac{n'}{n}|t|^2\).

## Oblique Incidence

Now consider if we have an oblique angle of incident light, \(\theta_i\) from \(-z-axis\). So, \(\vec{E} = \vec{E}_0\exp(i(\vec{k}\cdot\vec{x} - \omega t))\). Then we have reflected \(\vec{k}''\) and \(\vec{k}'\) be the transmitted. Starting out with the true,

\begin{align} c\vec{B} = n\hat{k}\times\vec{E} \end{align}We know the tangential components must be continuous, \(E\exp(i(\vec{k}\cdot\vec{x}-\omega t)) + E''\exp(i(\vec{k}''\cdot\vec{x}-\omega t)) = E'\exp(i(\vec{k}'\cdot\vec{x}-\omega t))\). So, \(E\exp(i(\vec{k}\cdot\vec{x})) + E''\exp(i(\vec{k}''\cdot\vec{x})) = E'\exp(i(\vec{k}'\cdot\vec{x}))\)

Note, \(\vec{k} = k_x\hat{x} + k_z\hat{z}\). Then, consider the phases (remembering continuity causes z-phases to cancel), so \(f(x) = g(x,y) = h(x,y)\), thus it must only depend on \(x\). So, at \(z=0\), \(\hat{k}\cdot\vec{x} = \hat{k}'\cdot\vec{x} = \hat{k}''\cdot\vec{x}\). Hence, \(k_x = k_x'x + k_y'y = k_x''x + k_y''y\). Then, \(k_y' = k_y'' = 0\). Thus, they are confined to the same plane. \(k_x = k_x'' = k\sin\theta_i\). \(k_x'' = k''\sin\theta_r = k\sin\theta_r\). Therefore, \(\theta_i = \theta_r\). \(k_x' = k_x \Rightarrow k\sin\theta_i = k'\sin\theta_t\). Thus, \(n\sin\theta_i = n'\sin\theta_t\).

Considering the pattern the incident light has on the boundary, we see that the reflected and refracted light must have the same pattern at the boundary, a period of \(\frac{\lambda}{\sin\theta_i} = \frac{2\pi}{k_x} = \frac{\lambda}{\sin\theta_t} = \frac{2\pi}{k_x''}\).

Consequences:

- \(\vec{k}\) confines \(\vec{k}'\) and \(\vec{k}''\) to the plane of incidence (i.e. xz)
- \(\theta_i = \theta_r\)
- \(n\sin\theta_i = n'\sin\theta_t\)