# Plane wave normal incidence on dielectric, direction of fields

I'm having an extremely hard time to figure out the direction of the $\vec E_R$ (reflected electric field) and $\vec H_R$ (reflected magnetic field) as well as the transmitted fields.

I've been reading Griffiths's Introduction to electrodynamics, Jackson's Electrodynamics, Zangwill's modern electrodynamics as well as seeking help at #physics on IRC and I just don't get it.

Here's the problem: We consider vacuum on the left and a semi-infinite dielectric medium that extends on the right side. We consider an incident plane wave going from the left to the right, at normal incidence on the surface of the dielectric. To keep things simple, we'll assume that the incident electric field points in the x direction only (up) while the z direction is the direction of incidence (right). This settles the direction of the H field of that wave (because $\vec E \times \vec H$ must give the direction of the Poynting's vector, the direction of motion which in this case is the y direction.)

The reflected wave will be travelling in the -z direction (so its Poynting vector must have that direction) while the transmitted wave will be travelling in the z direction.

The matching conditions at the surface of the dielectric are:

(1) $\hat n \cdot (\vec D_1 - \vec D_2 ) =0$

(2) $\hat n \cdot (\vec B_1-\vec B_2)=0$

(3) $\hat n \times (\vec E_1 - \vec E_2)=\vec 0$

(4) $\hat n \times (\vec H_1 - \vec H_2)=\vec 0$

where $\hat n = \hat z$, thus the first 2 equations don't tell me anything because $\vec D$ and $\vec B$ are orthogonal to $\hat n$. However the eqs. (3) and (4) give me the information:

(3') $\vec E_I + \vec E_R = \vec E_T$

(4') $\vec H_I + \vec H_R = \vec H_T$

and since $\vec H= \hat k \times \vec E$, eq. (4') reduces to (4'') $n_1(\vec E_I - \vec E_R)- n_2 \vec E_T=\vec 0$, where $n_1=1$ and $n_2$ is the refractive index of the dielectric medium.

Now comes my problem of understanding: since we have chosen an incident electric field parallel to the plane of incidence, according to the books, the eqs. (3) and (4'') reduce to something like:

(5) $E_I-E_R-E_T=0$

(6) $n_1(E_I+E_R)- n_2E_T=0$

So that it appears that $\vec E_I$ and $\vec E_T$ have the same direction, opposite to $\vec E_R$ while the $\vec H$ fields of the 3 waves have the same direction. I don't understand how they figured this out. I see no explanation in the books.

I do know that $\vec E_R \times \vec H_R$ must yield the direction -z, but that's not enough to settle down the directions of both fields. For instance both $\hat E_R=-\hat x$, $\hat H_R=\hat y$ and $\hat E_R=\hat x$, $\hat H_R=-\hat y$ yield $\hat k_R=-\hat z$. So how did they figure out that the former is correct instead of the latter?

The rough idea that I got is: from Jackson and Zangwill: because the sketch is like that. They deduced it from the sketch. From Griffiths, it's because of the direction of the Poynting vector.

To me none is satisfying.

The short answer is that it doesn't matter. Whichever direction you choose to define the reflected E-field initially you will always gets its direction with respect to the incident E-field coming out the same.

In the case you have in your question (where you have reversed the sign of $E_R$ and defined it to be in the -x direction and opposite to the incident E-field), the reflection coefficient is given by $$\frac{E_R}{E_I} = \frac{n_2 - n_1}{n_2+n_1}$$

So if $n_2 > n_1$, then the reflection coefficient is positive and this confirms that the reflected E-field is in the -x direction (or to be more accurate, confirms that the reflected and incident E-fields are opposite, as you assumed in your eqn 6).

On the other hand, if you set up the simultaneous equations hypothesising that the reflected and incident E-fields are aligned in the +x direction and hence that $H_R$ is defined opposite to $H_I$, then the reflection coefficient becomes $$\frac{E_R}{E_I} = \frac{n_1 - n_2}{n_2+n_1}$$

In this case if $n_2 > n_1$, the reflection coefficient is negative, and this indicates that in fact the reflected E-field is opposite to the incident E-field, as before.

Thus it does not matter which way you define $E_R$ in your equation 6; the direction of the reflected E-field compared to the incident E-field is always correctly derived and is unambiguous.

When I teach this I always advise students to define all the E-fields in the same direction, not assume that the reflected E-field is reversed as you have done in your eqn 6, and then find a negative reflection coefficient that can be straightforwardly interpreted as a $\pi$ phase change.

• I can follow you up to the 1st reflection coefficient. From it I can deduce that since $n_2>n_1$, the coefficient is positive as you said, but to me this implies that there's no phase difference between $\vec E_I$ and $\vec E_R$. So I am tempted to conclude that they have the same direction (which I know is wrong). But for now I'll assume that I follow you and that this implies that $\vec E_R$ as the $-\hat x$ direction as I started with. I got lost for the comments about the 2nd case. We start with both $\vec E_I$ and $\vec E_R$ pointing in the +x direction (more to come) – thermomagnetic condensed boson Nov 15 '15 at 2:33
• and we end up with $\vec E_R$ pointing in the -x direction; I didn't get this. To add more to that, for either p or s polarization (parallel like we did here, or perpendicular to the plane of incidence), the reflection coefficients for normal incidence are the same. Yet the fields have difference directions, for instance \vec E_I and \vec E_R point in the same direction if we had picked an s-polarized wave, despite that the reflection coefficient would have been the same. So from that point of view, I really don't see how checking the coefficient can tell us about the direction of the fields. – thermomagnetic condensed boson Nov 15 '15 at 2:37
• @no_choice99 What you are not seeing is that equations 5 and 6 in your Q are not vector equations. You have changed the sign in front of $E_R$. The reflection coefficient, which acts on that amplitude, is positive and so confirms your direction choice that $E_R$ is opposite to $E_I$. In my second example the reflection coefficient is negative, indicating that $E_R$ is in the opposite direction to my hypothesis and hence opposite to $E_I$ for both cases. Perhaps if you just left everything as vectors it would be more clear to you. – Rob Jeffries Nov 15 '15 at 8:14
• @no_choice99 I don't follow what you are saying about different polarisations. There is no distinction between s and p polarisations for normal incidence. The sign of the reflection coefficient tells you by what you must multiply the amplitude of the reflected wave, according to your definition of its E-field direction. – Rob Jeffries Nov 15 '15 at 8:25
• @no_choice99 For a negative reflection coefficient to indicate a phase change, you should start by assuming all E-field vectors are defined in the same direction. Not do what you've done in eq. 6. – Rob Jeffries Nov 15 '15 at 8:29