- #1
yungman
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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. <br />
\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.
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.