# Parallel Polarization

Parallel Polarizaton:-

Parallel polarization means $\overrightarrow{E}$ field lies in the XZ-plane (y=0) that is the plane of incidence, the figure illustrates the case of parallel polarization

then the incident and reflected fields is given by

$\overrightarrow{E_{is}}&space;=&space;E_{i}(-\sin&space;\theta_{i&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{i&space;}\overrightarrow{a_{x}}&space;)e^{-j(\overrightarrow{k}.\overrightarrow{r})}$

$\overrightarrow{E_{is}}&space;=&space;E_{i}(-\sin&space;\theta_{i&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{i&space;}\overrightarrow{a_{x}}&space;)e^{-j(x\overrightarrow{a_{x}}+y\overrightarrow{a_{y}}+z\overrightarrow{a_{z}}).(k_{i}\cos&space;\theta&space;_{i}\overrightarrow{a_{z}}+k_{i}\sin&space;\theta&space;_{i}\overrightarrow{a_{x}})}$

$\overrightarrow{E_{is}}&space;=&space;E_{i}(-\sin&space;\theta_{i&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{i&space;}\overrightarrow{a_{x}}&space;)e^{-j(k_{i}\cos&space;\theta&space;_{i}z+xk_{i}\sin&space;\theta&space;_{i})}$

$\overrightarrow{E_{is}}&space;=&space;E_{i}(-\sin&space;\theta_{i&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{i&space;}\overrightarrow{a_{x}}&space;)e^{-j\beta&space;_{1}(z\cos&space;\theta&space;_{i}+x\sin&space;\theta&space;_{i})}$

$\overrightarrow{H_{is}}&space;=&space;\frac{E_{i}}{\eta&space;_{1}}e^{-j\beta&space;_{1}(z\cos&space;\theta&space;_{i}+x\sin&space;\theta&space;_{i})}\overrightarrow{a_{y}}$

now the reflected wave is

$\overrightarrow{E_{rs}}&space;=&space;E_{r}(\sin&space;\theta_{r&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{r&space;}\overrightarrow{a_{x}}&space;)e^{-j\beta&space;_{1}(-z\cos&space;\theta&space;_{r}+x\sin&space;\theta&space;_{r})}$

$\overrightarrow{H_{rs}}&space;=\frac{&space;H_{r}}{\eta&space;_{1}}e^{-j\beta&space;_{1}(-z\cos&space;\theta&space;_{r}+x\sin&space;\theta&space;_{r})}\overrightarrow{a_{y}}$

since $k_{i}=k_{r}=\beta&space;_{1}&space;=\omega&space;\sqrt{\mu_{1}&space;\epsilon_{1}&space;}$

let’s find out  $\overrightarrow{k}$  then by using the equation $\overrightarrow{k}.\overrightarrow{E_{s}}&space;=0$$\overrightarrow{H_{s}}&space;=&space;\frac{k}{\omega&space;\mu&space;}X&space;\overrightarrow{E_{s}}&space;=&space;\overrightarrow{a_{k}}X&space;\frac{E_{s}}{\eta&space;}$ .

and the transmitted fields in the second  medium is

$\overrightarrow{E_{ts}}&space;=&space;E_{t}(-\sin&space;\theta_{t&space;}\overrightarrow{a_{z}}+&space;\cos&space;\theta_{t}\overrightarrow{a_{x}}&space;)e^{-j\beta&space;_{2}(z\cos&space;\theta&space;_{t}+x\sin&space;\theta&space;_{t})}$

$\overrightarrow{H_{ts}}&space;=&space;\frac{H_{t}}{\eta&space;_{2}}e^{-j\beta&space;_{2}(z\cos&space;\theta&space;_{t}+x\sin&space;\theta&space;_{t})}\overrightarrow{a_{y}}$

where $\beta&space;_{2}&space;=&space;\omega&space;\sqrt{\mu&space;_{2}\epsilon&space;_{2}}$.

Transmission coefficient:-

as $\theta&space;_{r}&space;=&space;\theta&space;_{i}$ and also that the tangential components of $\overrightarrow{E}$ and $\overrightarrow{H}$ are continuous at the boundary z=0.

$E_{tan1}=E_{tan2}$

$(E_{i}+E_{r})&space;\cos&space;\theta&space;_{i}&space;=&space;E_{t}\cos&space;\theta&space;_{t}$

$H_{tan1}-H_{tan2}&space;=&space;\overrightarrow{J_{s}}$ , $H_{tan1}&space;=&space;H_{tan2}$  since $\overrightarrow{J_{s}}&space;=0$

$\frac{E_{i}}{\eta&space;_{1}}-\frac{E_{r}}{\eta&space;_{1}}=\frac{E_{t}}{\eta&space;_{2}}------EQN(1)$

$E_{i}+E_{r}=E_{t}\frac{\cos&space;\theta&space;_{t}}{\cos&space;\theta&space;_{i}}------EQN(2)$

(1)+(2) implies

$2E_{i}=E_{t}(\frac{\eta&space;_{1}}{\eta&space;_{2}}+\frac{\cos&space;\theta&space;_{t}}{\cos&space;\theta&space;_{i}})$

$2E_{i}=E_{t}\frac{(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}{\eta&space;_{2}\cos&space;\theta&space;_{i}}$

$\tau&space;_{parallel}&space;=&space;\frac{E_{t}}{E_{i}}&space;=&space;\frac{2\eta&space;_{2}\cos&space;\theta&space;_{i}}{(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}$.

Reflection coefficient:-

from EQN (1)

$\frac{E_{i}}{\eta&space;_{1}}-\frac{E_{r}}{\eta&space;_{1}}=\frac{E_{t}}{\eta&space;_{2}}------EQN(1)$

$\frac{E_{i}}{\eta&space;_{1}}-\frac{E_{r}}{\eta&space;_{1}}=&space;\frac{2E_{i}\eta&space;_{2}\cos&space;\theta&space;_{i}}{\eta&space;_{2}(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}$

after simplification

$E_{r}\frac{(\eta&space;_{2}\cos&space;\theta&space;_{t}-\eta&space;_{1}\cos&space;\theta&space;_{i})}{(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}=E_{i}$

$\rho&space;_{parallel}&space;=&space;\frac{E_{r}}{E_{i}}=\frac{(\eta&space;_{2}\cos&space;\theta&space;_{t}-\eta&space;_{1}\cos&space;\theta&space;_{i})}{(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}$

$1+\rho&space;_{parallel}&space;=&space;\tau&space;_{parallel}(\frac{\cos&space;\theta&space;_{t}}{\cos&space;\theta&space;_{i}})$.

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