Introduction to Electrodynamics/Chapter 8/2/4

8.2.4 Reflection and Transmission at Normal Incidence

Calculating what happens to a EM plane wave at normal incidence to a boundary between two different linear media is not hard.

Hi, I'm Jonathan Gardner, and this is section 8.2.4 of Introduction to Electrodynamics by Griffiths. 2nd edition, of course. We've recently covered the nature of EM plane waves. We've just covered what happens inside of linear media, and what conditions must be satisfied at the boundary.

This video is rather long. There really isn't a good place to break it up in the middle. So grab a notepad and pencil, and get ready for a major brain dump.

Setup

On the left is linear media 1. The wave is coming towards the boundary and is reflected back to the left, and transmitted to the right side, which is linear media 2.

Each side has its own values for epsilon and mu, epsilon_1 and mu_1 on the left, and epsilon_2 and mu_2 on the right. Of course they have their own velocities, v1 and v2, and indexes of refraction, n1 and n2.

There are three waves we are calculating: the incident (I), the reflected (R) and the transmitted (T). The incident wave is coming in from the left towards the boundary. Both the reflection and transmission are moving away from the boundary --- the transmission through the right side, and the reflection through the left.

For simplicity, we'll align the x direction perpendicular to the boundary, the y direction to the polarization, and the z direction is perpendicular to both. No matter how the problem is stated, thanks to symmetry we can always orient things this way.

The B fields for the waves traveling to the right are coming out of the page. The B field for the reflection is going into the page.

Three Waves

Since these are plane waves, the electric and magnetic fields do not vary at all with y or z. The only two numbers that can change the field are the position along the x axis and the time.

The electric fields are given by the equations:

<math> \begin{align} \tilde{\vec{E_I}}(x,t) &= \tilde{E_{0I}}e^{i( \kappa_I x - \omega_I t)}\hat{j} \\ \tilde{\vec{E_R}}(x,t) &= \tilde{E_{0R}}e^{i(-\kappa_R x - \omega_R t)}\hat{j} \\ \tilde{\vec{E_T}}(x,t) &= \tilde{E_{0T}}e^{i( \kappa_T x - \omega_T t)}\hat{j} \\ \end{align} </math>

The reflected wave is traveling backwards, hence the minus kappa.

I'm not going to assume that the omegas or the kappas are the same. I'll show you why they must be so shortly.

The magnetic fields are simply the same, but divided by the respective velocities and pointing along the z axis.

<math> \begin{align} \tilde{\vec{B_I}}(x,t) &= {1 \over v_1} \tilde{\vec{E_I}}\hat{k} \\ \tilde{\vec{B_R}}(x,t) &= -{1 \over v_1} \tilde{\vec{E_R}}\hat{k} \\ \tilde{\vec{B_T}}(x,t) &= { 1 \over v_2} \tilde{\vec{E_T}}\hat{k} \\ \end{align} </math>

Why is the reflected wave's B field backward? Because it is traveling backward, and v cross E must be B.

The Total Wave Function

Putting it all together, before we start simplifying our kappas and omegas and E_0s, we have:

<math> \begin{align}

\tilde{\vec{E}} &= \begin{cases}

   \tilde{\vec{E_I}} + \tilde{\vec{E_R}} & \text{when }x \lt 0 \\
   \tilde{\vec{E_T}} & \text{when }x \gt 0 \\
 \end{cases}\\

\tilde{\vec{B}} &= \begin{cases}

   \tilde{\vec{B_I}} + \tilde{\vec{B_R}} & \text{when }x \lt 0 \\
   \tilde{\vec{B_T}} & \text{when }x \gt 0 \\
 \end{cases}\\

\end{align} </math>

What happens at x=0 is determined by the boundary conditions. It just so happens that the E fields are equal at x=0, but the B fields are not.

Boundary Conditions

I won't justify the boundary conditions here. I've done that twice now in sections X.Y.Z and just recently in 8.2.3.

The electric fields do not have to be equal across the boundary. The perpendicular component jumps by a factor of the permittivities.

<math>\epsilon_1 E_{1 \perp} = \epsilon_2 E_{2 \perp}\;</math>

The parallel components of the E fields do not vary at all.

<math> \vec{E_{1 \parallel}} = \vec{E_{2 \parallel}}\;</math>

For the magnetic fields, it's the opposite. The perpendicular component does not vary, but the parallel components vary in proportion to the permeabilities.

<math>B_{1 \perp} = B_{2 \perp}\;</math>

<math>{ \vec{B_{1 \parallel}} \over \mu_1} = { \vec{B_{2 \parallel}} \over \mu_2}\;</math>

Doing the Algebra

Let's plug in our E and B fields to the boundary conditions and see how the various E_0s and kappas and omegas relate to each other.

Of note, there are no perpendicular components since we are at normal incidence. When we cover oblique incidence, that will change. These first two boundary conditions can be safely ignored.

So we only have two equations to apply the parallel conditions.

The parallel E fields have to equal each other at the boundary.

<math> \begin{align} \tilde{\vec{E_I}}(0,t) + \tilde{\vec{E_R}}(0,t) &= \tilde{\vec{E_T}}(0,t) \\ \tilde{E_{0 I}}e^{i \omega t}\hat{j} + \tilde{E_{0R}}e^{i \omega t}\hat{j} &= \tilde{E_{0T}}e^{i \omega t}\hat{j} & & \text{Note, omegas equal} \\ \tilde{E_{0 I}} + \tilde{E_{0R}} &= \tilde{E_{0T}} \\ \end{align} </math>

Note: We don't have the requirement that the derivative of the E field be equal, like we did for the strings. But we do have a requirement for the B fields. The parallel component of the magnetic fields must be proportion to their respective permeabilities.

<math> \begin{align} {1 \over \mu_1} \left(\tilde{\vec{B_I}}(0,t) + \tilde{\vec{B_R}}(0,t)\right)

   &= {1 \over \mu_2} \tilde{\vec{B_T}}(0,t) \\

{1 \over \mu_1 v_1} \left(\tilde{E_{0_I}}e^{i \omega t}\hat{k} - \tilde{E_{0_R}}e^{i \omega t}\hat{k} \right)

   &= {1 \over \mu_2 v_2}\tilde{E_{_T}}e^{i \omega t}\hat{k} \\

\tilde{E_{0_I}} - \tilde{E_{0_R}} &= { \mu_1 v_1 \over \mu_2 v_2} \tilde{E_{0_T}} \\ \end{align} </math>

We'll just use \beta for that mess on the right there.

<math>\beta = { \mu_1 v_1 \over \mu_2 v_2}\;</math>

Now we have equations that should look familiar. Let's use the same method to solve them as we did back when we looked at strings with different velocity.

<math> \begin{align} \tilde{E_{0_R}} &= {1 - \beta \over 1 + \beta} \tilde{E_{0_I}} \\ \tilde{E_{0_T}} &= {2 \over 1 + \beta} \tilde{E_{0_I}} \\ \end{align} </math>

Now, for a bit of a simplification. For most materials, the permeabilities are very similar to the permeability of free space. The ratio \mu_1 \over \mu_2 is practically 1. So we can rewrite \beta for these materials as v_1 \over v_2. Now we have the exact same equations as we did for the string.

<math> \begin{align} \tilde{E_{0_R}} &= \left|{v_2 - v_1 \over v_2 + v_1}\right| \tilde{E_{0_I}} \\ \tilde{E_{0_T}} &= {2 v_2 \over v_2 + v_1} \tilde{E_{0_I}} \\ \end{align} </math>

with the caveat that when v2 < v1, the reflected wave is out of phase from the incident by 180 degrees.

We can rewrite this rather easily in terms of the index of refraction of the material, a property much easier to measure:

<math> \begin{align} \tilde{E_{0_R}} &= \left|{n_1 - n_2 \over n_1 + n_2}\right| \tilde{E_{0_I}} \\ \tilde{E_{0_T}} &= {2 n_1 \over n_1 + n_2} \tilde{E_{0_I}} \\ \end{align} </math>

Intensities

What fraction of the energy is transmitted or reflected?

I is simply 1/2 \epsilon v E_0^2.

So, assuming the mus are similar, the ratio of reflected to incident intensities is given by:

<math>R = {I_R \over I_I} = \left({E_{0_R} \over E_{0_I}}\right)^2 = \left({n_1- n_2 \over n_1 + n_2}\right)^2\;</math>

And the same for transmission:

<math>T = {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left({E_{0_T} \over E_{0_I}}\right)^2 = {n_2 \over n_1} \left({2n_1 \over n_1 + n_2}\right)^2\;</math>

We call R and T the reflection and transmission coefficients at the surface.

Note that they add up to 1: <math>R + T = 1\;</math>. If you need to practice your algebra, you can work that out easily. But it should be obvious since the energy has to go somewhere.

Conclusion

Next, we'll cover reflection and transmission at oblique, non-normal incidence. That's a bit more difficult, but not as terribly difficult as you may imagine.

If you have any questions or comments, share them in the comments or a video response.

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Thanks for your time!

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