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

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

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

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

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

now the reflected wave is 

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

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

since k_{i}=k_{r}=\beta _{1} =\omega \sqrt{\mu_{1} \epsilon_{1} }

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

and the transmitted fields in the second  medium is  

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

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

where \beta _{2} = \omega \sqrt{\mu _{2}\epsilon _{2}}.

Transmission coefficient:-

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


(E_{i}+E_{r}) \cos \theta _{i} = E_{t}\cos \theta _{t}

H_{tan1}-H_{tan2} = \overrightarrow{J_{s}} , H_{tan1} = H_{tan2}  since \overrightarrow{J_{s}} =0

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

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

(1)+(2) implies 

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

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

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

Reflection coefficient:-

from EQN (1)

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

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

after simplification

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

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

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

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Author: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.