Introduction to Electrodynamics/Chapter 8/2/5

8.2.5 Reflection and Transmission at Oblique Incidence

What happens when an EM plane wave travels at an oblique angle towards a boundary between two different linear media?

Hi, I'm Jonathan Gardner. Last time in section 8.2.4, we considered what happens for EM plane waves striking perpendicular. This time, in section 8.2.5 of Griffith's Introduction to Electrodynamics, 2nd Edition, we will consider what happens at an angle.

Setup

The boundary is the y-z plane at x=0. On the left, we have \epsilon_1 and \mu_1, and EM waves travel at v1. The index of refraction is n_1. On the right, the same things but for 2 - \epsilon_2, \mu_2, v_1, n_1.

We have the incident wave coming in from the left side, forming some angle \theta_I with the x axis. It will transmit and reflect, those waves forming an angle \theta_R and \theta_T. Each of these waves may have different wave numbers \kappa_I, \kappa_R, \kappa_T. Thankfully, they all share the same \omega. The reasoning is simple: We have some boundary conditions to meet at the boundary, and that requires that the waves wave at the same frequency.

Fields

As before, we have E and B fields for each of the waves. However, this time we can't use the simple kappa-x. We need to use the kappa vector we discussed earlier in 8.2.1. The r vector is the vector pointing from the origin to some arbitrary point in space. The Kappa vectors have a direction matching the direction of the wave, and a magnitude reflecting the wave number rather than the speed. The speed, of course, depends on the material and not the wave.

Note that our E_0's are vectors. The direction is the polarization of the wave.

<math> \begin{align}

\tilde{\vec{E_I}}(\vec{r}, t) &= \tilde{\vec{E_{0_I}}}e^{i(\vec{\kappa_I} \cdot \vec{r} - \omega t)} & \qquad \tilde{\vec{B_I}}(\vec{r}, t) &= {1 \over v_1} (\hat{\kappa_I} \times \tilde{\vec{E_I}}) \\

\tilde{\vec{E_R}}(\vec{r}, t) &= \tilde{\vec{E_{0_R}}}e^{i(\vec{\kappa_R} \cdot \vec{r} - \omega t)} & \tilde{\vec{B_R}}(\vec{r}, t) &= {1 \over v_1} (\hat{\kappa_R} \times \tilde{\vec{E_R}}) \\

\tilde{\vec{E_T}}(\vec{r}, t) &= \tilde{\vec{E_{0_T}}}e^{i(\vec{\kappa_T} \cdot \vec{r} - \omega t)} & \tilde{\vec{B_T}}(\vec{r}, t) &= {1 \over v_2} (\hat{\kappa_T} \times \tilde{\vec{E_T}}) \\

\end{align} </math>

Of course, the total E and B fields are given by the sums of the incident and reflected waves on the left side, and the transmitted wave on the right.

Remember that these are plane waves. Perpendicular to the direction, the E and B fields are the same. (Draw dotted perpendicular lines)

Angular Frequency

The angular frequency \omega is the same for all three waves. The reasoning is actually quite simple, exactly the same reasoning we had for the string and the normal incident wave. At the boundary condition, at a particular point, as we vary the time, the incident, reflected, and transmitted waves must all move together. If it were not so, the boundary conditions would be violated.

From this, we have the relationship:

<math>\kappa_I v_1 = \kappa_R v_1 = \kappa_T v_2</math>

Which says that:

<math>\kappa_I = \kappa_R = {v_2 \over v_1} \kappa_T = {n_1 \over n_2} \kappa_T</math>

We already know a lot about these waves. What we're really interested in next is how big they are and what direction they travel.

Applying Boundary Conditions

I won't even bother listing the boundary conditions here --- we've covered them in the last two sections and previously before that.

Suffice it to say, you're going to get conditions that state something like:

<math>C_1 F_1(\vec{r}, t) = C_2 F_2(\vec{r}, t)\;</math>

Here, F is the E or B field, and C is some constant. Maybe it's 1, or maybe it's the permittivities or permeabilities or whatever. The vector r, of course, must lie in the yz plane at x=0 -- that's where the two sides meet.

When we apply this general form to our fields, we get:

<math>

 (\quad)\tilde{\vec{E_{0_I}}}e^{i(\vec{\kappa_I}\cdot\vec{r} + \omega t)}

+ (\quad)\tilde{\vec{E_{0_R}}}e^{i(\vec{\kappa_R}\cdot\vec{r} + \omega t)} = (\quad)\tilde{\vec{E_{0_T}}}e^{i(\vec{\kappa_T}\cdot\vec{r} + \omega t)} </math>

From this alone, we can derive a whole lot of results.

Remember, we're holding x=0. But we can vary in the y or z direction, and the condition must be consistent. This tells us that when one side is zero due to the exponential term, the other side must be as well, and when one side is at its maximum, so the other side must be as well, and so on.

So we can derive that the positional term in the exponent must be equal at x=0:

<math>\vec{\kappa_I}\cdot\vec{r} = \vec{\kappa_R}\cdot\vec{r} = \vec{\kappa_T}\cdot\vec{r}</math>

Let's write this out for r = (x,y,z) and keep x=0.

<math> (\kappa_I)_y y + (\kappa_I)_z z = (\kappa_R)_y y + (\kappa_R)_z z = (\kappa_T)_y y + (\kappa_T)_z z </math>

Now, let's suppose we look at the line where z and x are both zero - the y axis. Along that line:

<math> (\kappa_I)_y y = (\kappa_R)_y y = (\kappa_T)_y y </math>

Well, that implies that the y component of the kappas are all equal! Similar reasoning can be had for the z components.

Apparently, the only difference between these kappa vectors is in the x direction -- the direction normal to the boundary.

Let's make our lives easy and orient the axis so that the kappa vectors have no z component, constraining the problem to 2 dimensions and saving us some ink.

Second Conclusions

There are 3 conclusions Griffiths draws:

(A) There is always a plane of incidence which is the plane formed by the direction of the incident wave and the normal to the boundary. All the interesting stuff happens in this plane.

With this plane, we can break any of the Kappa vectors into their x and y components, or rather, the component perpendicular to the plane and the component parallel to the plane. We have said nothing about the perpendicular component (yet) but we know that the parallel components must be equal.

Another way to write the parallel components is simply:

<math>\kappa_I \sin \theta_I = \kappa_R \sin \theta_R = \kappa_T \sin \theta_T</math>.

This leads us to conclusion B.

(B) Since the length of \kappa_I and \kappa_R are equal (else the omegas would be different), then the angle of incidence must equal the angle of reflection.

<math>\theta_I = \theta_R</math>

(C) We can calculate the angle of refraction because we know how the magnitudes of kappa I and kapp T relate.

<math>{\kappa_I \over \kappa_T} = {n_1 \over n_2} = {\sin \theta_T \over \sin \theta_I}\;</math>

This is Snell's Law.

We have concluded a number of useful and extraordinarily simple rules from this complex system. What's surprising, as Griffith's notes, is how little electrodynamics went into this derivation. Let me trace back for you what we needed to draw this conclusion.

(1) That we have waves at all. Maxwell's Laws dictate this in the absence of charge or current, free or otherwise.

(2) That the boundary conditions dictate that at the boundaries the waves must share maxima and minima at the same points in time and along the entire y-z plane.

That's pretty much it.

For any wave system, you can derive a similar set of results as long as these kinds of conditions can be met. This is why waves are so exciting in Physics. Once you've found them, you can open any book on optics and start reading off the results and conclusions.

Boundary Conditions (for reals)

Now we have concluded that the exponentials, which vary from 1 and -1, must equal each other at all times along the boundary. Let's drop those terms and write out the boundary conditions just with the E naughts.

<math> \begin{align} \epsilon_1 \left[\tilde{\vec{E_{0_I}}} + \tilde{\vec{E_{0_R}}}\right]_x &= \epsilon_2 \left[\tilde{\vec{E_{0_T}}}\right]_x \\ \left[\tilde{\vec{B_{0_I}}} + \tilde{\vec{B_{0_R}}}\right]_x &= \left[\tilde{\vec{B_{0_T}}}\right]_x \\ \left[\tilde{\vec{E_{0_I}}} + \tilde{\vec{E_{0_R}}}\right]_{yz} &= \left[\tilde{\vec{E_{0_T}}}\right]_{yz} \\ {1 \over \mu_1} \left[\tilde{\vec{B_{0_I}}} + \tilde{\vec{B_{0_R}}}\right]_{yz} &= {1 \over \mu_2} \left[\tilde{\vec{B_{0_T}}}\right]_{yz} \\ \end{align} </math>

Keep in mind that <math>\tilde{\vec{B_0}} = (\hat{\kappa} \times \tilde{\vec{E_0}} / v\;</math>

From this, we have all we pretty much need to find the magnitudes and directions of the respective waves.

Polarization in the Plane of Incidence

Let's consider what happens when the incident wave has its E vectors aligned with the plane of incidence. There is a problem where you can find out what happens when the polarization is perpendicular to the plane of incidence which I won't spoil for you.

In this case, the z component of the E field is zero everywhere, as is the y component of the B field.

The x component of the fields is given by the sin of their respective angles. Obviously, the y component of the E field is given by sin.

We can reduce the E field equations to: (Remember that theta_I = theta_R.)

<math>\epsilon_1 [-\tilde{E_{0_I}}\sin \theta_I + \tilde{E_{0_R}} \sin \theta_I] = -\epsilon_2[\tilde{E_{0_T}}\sin\theta_T]</math>

<math>\tilde{E_{0_I}}\cos \theta_I + \tilde{E_{0_R}} \cos \theta_I = \tilde{E_{0_T}} \cos \theta_T</math>

The B field equation for x reduces to 0 = 0 since there is no activity in the x direction (it's perpendicular to the plane of incidence.)

The B field equation for z states:

<math>{1 \over \mu_1 v_1}[\tilde{E_{0_I}} - \tilde{E_{0_R}}] = {1 \over \mu_2 v_2} \tilde{E_{0_T}</math>

If we use the same Beta we used for the last section, <math>\mu_1 v_1 / \mu_2 v_2\;</math> the equation becomes the familiar:

<math>\tilde{E_{0_I}} - \tilde{E_{0_R}} = \beta \tilde{E_{0_T}</math>

which isn't new. But the other equation reduces to:

<math>\tilde{E_{0_I}} + \tilde{E_{0_R}} = \alpha \tilde{E_{0_T}</math>

where <math>\alpha = \cos \theta_T / \cos \theta_I\;</math>

Now we have our equations that can give us how the amplitudes and phases of the reflected and transmitted waves compare to the incident.

<math>\tilde{E_{0_R}} = \left({\alpha - \beta \over \alpha + \beta}\right) \tilde{E_{0_I}}</math>

<math>\tilde{E_{0_T}} = \left({2 \over \alpha + \beta}\right) \tilde{E_{0_I}}</math>

This looks pretty familiar, huh?

These are Fresnel's Equations. (Fre'-nel) Or rather, part of them. There are equations for waves polarized perpendicular to the plane of incidence that you should find for yourself.

Notice, that like before, the transmitted wave is always in phase with the incident. Also, if alpha is bigger than beta, then you get a reflected wave that is in phase. If alpha is smaller, then it is out of phase 180 degrees.

Unlike before, the factor that determines the amplitude of the resulting waves depends on the angle of incidence and the relative indexes of refraction. You can rewrite it solely in terms of the angle of incidence by the relation that 1 = cos^2 + sin^2.

<math> \alpha = {\sqrt{1 - \left({n_1 \over n_2} \sin \theta_I\right)^2} \over \cos \theta_I}\;</math>

Angle of Incidence

Let's look at what happens at various angles of incidence.

Obviously, if the angle of incidence is 0 degrees, then we should get the equations we found in the last section.

Also noteworthy, when the angle is almost 90, then we have a case where alpha rapidly grows larger and larger. In this case, there is no transmitted waves. That's why, almost regardless of the material, everything acts like a mirror if you look at if almost parallel to the plane.

There is a special angle where alpha = beta called Brewster's angle. Here, there is no reflected wave at all. Everything is transmitted. You can plug this in, and then take the approximation that all mus are alike, and get back:

<math>\tan \theta_B = n_2 / n_1\;</math>

Of cautionary note: This only works when the polarization is in the plane of incidence! Try it with light that isn't polarized like that, and notice it is not completely absorbed. Pull out your polaroid glasses, and you'll see that only a certain polarization gets through.

Intensity

The intensity is given by S * n hat. The following equations describe that blah blah blah.

Conclusion

This is one of the hardest problems to solve, because there are so many shortcuts and so much bookkeeping involved. I hope you've been able to follow along. If not, you'll have to walk through the process several times yourself before you see how it all fits together.

Blah ablah blah

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