# 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:

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

Created: 2023-06-25 Sun 02:29

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