# Question on derivation of Fresnel's equation in Griffiths

I have a few question on the derivation of the Fresnel's equation in Griffiths' book P389. This is the diagrams representing the parallel polarization. The plane of incidence is the xz plane at y=0. boundary is the xy plane. $\vec{k_I},\; \vec{k_R},\;\vec{k_T}$ are direction of incidence, reflected and transmitted wave resp. Unit direction vector of the three waves are:

$$\hat k_I =\hat x sin \theta _I \;+\; \hat z cos \theta _I, \;\;\; \hat k_R =\hat x sin \theta _R \;-\; \hat z cos \theta _R, \;\;\; \hat k_T =\hat x sin \theta _T \;+\; \hat z cos \theta _T$$

I denote $E_{0_I} , \; E_{0_R} ,\; E_{0_T} \;$ be incidence, reflected and transmitted E with parallel polarization at z=0 resp. This convention is also used for H field also.

For the normal components of the E ( in z direction):

$$\epsilon_1 ( -E_{0_I} sin \theta_I \;+\; E_{0_R} sin \theta_R ) \; \epsilon_2 ( -E_{0_T} sin \theta_T)$$ (1)

For tangential components of E and H ( in x direction):

$$E_{0_I} cos \theta_I \;+\; E_{0_R} cos \theta_R \;=\; E_{0_T} cos \theta_T$$

$$\frac {1}{\mu_1}( H_{0_I} \;+\; H_{0_R} ) \;=\;\frac {1}{ \mu_2 } H_{0_T} \;\Rightarrow\; \frac {1}{\mu_1 v_1}( E_{0_I} \;-\; E_{0_R} ) \;=\; \frac {1}{ \mu_2 } E_{0_T}$$ (2)

These will give:

$$\tilde E_{0_I} \;-\; \tilde E_{0_R}\;=\; \beta \tilde E_{0_T} \; \hbox { where } \beta \;=\; \frac{\mu_1 v_1}{\mu_2 v_2}$$

$$\tilde E_{0_I} \;+\; \tilde E_{0_R} \;=\; \alpha \tilde E_{0_T} \;\hbox { where } \;\alpha \;=\; \frac{ cos \theta_T}{ cos \theta_I}$$

For reflection coef :

$$\tilde E_{0_T} \;=\; \frac { \alpha - \beta } { \alpha +\beta}$$

Where the reflected $\tilde E_{0_R}$ is either in phase or 180 deg out of phase depend on whether $\alpha - \beta$ is +ve or -ve respectively.

Here is my problem:

If you look at (1) the $\tilde E_R$ is assumed direction on the plane of indicence while the resulting reflection again defined whether the $\tilde E_R$ is in or out of phase depend on the polarity of $\alpha - \beta$. How can you define the direction of $\tilde E_R \;$ first and use it to derive the fresnel's equation which can change the direction? I see the same thing on Cheng's and Ulaby's books like this also.

I have been going through other books while waiting for someone to give me some clarification. I found from the book by Hyatt and Buck that they even have a different representation of the the Reflected wave!!!

In parallel polarization, the H is tangential to the boundary surface and is continuous cross boundary:

$$\tilde H_I \;+\; \tilde H_R \;=\; \tilde H_T$$

And the book set the $\; \tilde H_R = \hat y H_R \;$ which is the same direction as the incidence H. This will put the $\;\tilde E_R \;$ is 180 shift from the diagram of what I drew.

In it true that you can just take however which direction and consider that is the +ve. Then you work out the reflection coef. If it is +ve, then you follow the stated direction of your initial position. If the coef turn out to be negative, then you switch the wave to the other side ( I mean rotate the entire EM 180 degree)?

Is that how you look at this?

Meir Achuz
Homework Helper
Gold Member
The number E_R can be a negative number.

The number E_R can be a negative number.

Thanks, I understand that. I know that is a starting point and the E_R can be +ve or -ve from the reflection coef later.

But why most start out like I drew in #1. Is there a particular reason? I have no issue understanding the Fresnel's reflection and transmission coef, just how they get started. Is this the normal way how the mirror reflect?

Meir Achuz
Homework Helper
Gold Member
The original picture has no effect on the results.
Some books have E_R and E_T in the opposite direction from your picture. This has the virtue that all the B vectors are out of the page.

The original picture has no effect on the results.
Some books have E_R and E_T in the opposite direction from your picture. This has the virtue that all the B vectors are out of the page.

So you agree with me that you just start out with a "standard" and call it possitive with the drawing. Then when you use Fresnel reflection and transmittion coef, then the polarity will sort out the final absolute polarity of the signal. So no matter how you start out, the end result will all agree?

That is what I think, I just want to verify this so I can move on.

Anyone can confirm this?