# Homework Help: Derivation of Fresnel equations / minus sign

1. Aug 6, 2015

### fluidistic

1. The problem statement, all variables and given/known data
For an s-polarized wave (E and B fields are orthogonal to the plane of incidence) passing from medium to medium 2, I'm not understanding a minus sign.
The matching conditions are $\hat n \times (\vec E_2 - \vec E_1)=\vec 0$ and $\hat n \times (\vec H_2 - \vec H_1 )=\vec 0$, where $\vec E_1 = \vec E_R + \vec E_I$ and $\vec E_2 = \vec E_T$. So far so good: the transmitted E field is equal to the incident one plus the reflected one, at the interface between the 2 media. The same applies for the H field.

2. Relevant equations
Matching conditions.

3. The attempt at a solution
The idea is that $\vec E_I$ is known and so the 2 matching conditions should allow us to solve for $\vec E_R$ and $\vec E_T$. We know the relationship between H and E: $Z_i\vec H_i = \hat k \times \vec E_i$, where Z is the impedance of the medium. So far so good.
Soon comes the problem: using this last relation, I obtain that $\hat k \times (\vec E_I + \vec E_R)Z_2=\hat k \times \vec E_T Z_1$. So that in terms of magnitude, $(E_I -E_R)Z_2=E_TZ_1$. That's because $\vec E_I$ and $\vec E_R$ points in opposite directions.
However if I use this same argument then the 1st matching condition becomes $(E_I-E_R)=E_T$ which is wrong. It should be $(E_I+E_R)=E_T$. I don't understand why.

2. Aug 6, 2015

### RUber

Let's say you have an incident field in the x-y plane (transverse to z) defined by:
$\vec{E_I} = \hat z E_Ie^{-i(x\cos\theta + y \sin \theta) }$
And it reflects off an interface at y = 0.
Then,
$\vec{E_R} = -\hat z E_Re^{-i(x\cos\theta - y \sin \theta) }$
Where $E_R = \Gamma E_I$ and $\Gamma>0$.
In this way, your exterior field has magnitude $E_I - E_R$ with fields defined by $\vec{E_I}+\vec{E_R}$.

When you say "in terms of magnitude" you are implying that the scalar terms $E_I, E_R$ are both positive.
When you use the vector notation, you can add opposing vectors, as in $\vec{E_I}+\vec{E_R}$.

Does that address your question, or is there something I am missing?

3. Aug 6, 2015

### fluidistic

Hmm where the time dependence? Also let's keep it simple and let's consider normal incidence so that $\theta=0$, cos theta becomes 1 and sin theta becomes 0.
As you said, in region 1 the E field is worth $E_I-E_R$ in magnitude, while from a vector point of view we have that $\vec E=\vec E_I + \vec E_R$.
So then why the matching condion $\hat n \times (\vec E_2 - \vec E_1) =\vec 0 \Rightarrow \vec E_2 = \vec E_1 \Rightarrow \vec E_T= \vec E_I + \vec E_R$ simplifies as $E_I+E_R=E_T$ from a magnitude point of view instead of $E_I-E_R=E_T$?

4. Aug 6, 2015

### RUber

I have seen some notation that keeps the negative sign inside term $E_R$, so that $|E_R| = -E_R$.
The important thing is that the energy in is equal to the energy out, so $|\vec E_I| = |\vec E_R| + |\vec E_T|$. As long as that requirement is satisfied, you can have faith in your notation.

5. Aug 6, 2015

### fluidistic

I am still completly lost. The book (and everywhere else) get $E_I+E_R=E_T$ and $Z_2(E_I-E_R)=Z_1E_T$.
I don't understand how come there is a plus in one equation but a minus in the other one.

6. Aug 6, 2015

### RUber

I did a little more digging and found a good reference:
http://ocw.mit.edu/courses/electric...ring-2011/lecture-notes/MIT6_007S11_lec29.pdf
This shows that
$E_R = rE_I, \quad E_T = t E_I$
which allows for $E_R$ to be negative, so not "magnitude" per se.
The second condition is derived from the H field, which in terms of the E field has a differential relationship, so the reflected field will have a negative sign in front of it.
In normal incidence, time-harmonic, scenario:
$\vec E_I = \hat x E_I e^{ -i k_1z}$
$\vec E_R = \hat x rE_I e^{ i k_1z}$
$\vec E_T = \hat x tE_I e^{ -i k_2z}$
And $\vec H = \frac{1}{\mu} \nabla \times \vec E = \text{sign}(i) \hat y \frac{E}{Z}e^{ -i kz}$
The continuity condition for the H field is:
$\hat n \times(\vec H_1 - \vec H_2 ) = 0$ Which leads directly to the condition you refer to with $Z_2 (E_I - E_R) = Z_1E_T.$

7. Aug 6, 2015

### fluidistic

Ok I get it. The thing is that for s-polarized waves, $\vec E_I$ and $\vec E_R$ (and $\vec E_T$ for that matter) have the same direction, while $\vec H_I$ and $\vec H_R$ don't. Hence the conditions $E_I+E_R=E_T$ and $(E_I -E_R)Z_2=E_TZ_1$ so I was wrong in my 1st post to say that $\vec E_I$ and $\vec E_R$ point in opposite directions.

8. Aug 6, 2015

### RUber

Sorry I was on the wrong track at first. I was thrown off by the discussion of magnitude. Glad to get it squared away.

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