Brewster’s angle

In parallel polarization the incident angle at which there is no reflection is called Brewster’s angle.

\rho _{parallel} = 0.

As   \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})}.

\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})}=0.

(\eta _{2}\cos \theta _{t}-\eta _{1}\cos \theta _{i})=0.

\eta _{2}\cos \theta _{t}=\eta _{1}\cos \theta _{i}.

by squaring on both sides  \eta^{2} _{2}\cos^{2} \theta _{t}=\eta^{2} _{1}\cos^{2} \theta _{i}.

\eta^{2}_{2}(1-\sin^{2} \theta _{t})=\eta^{2} _{1}(1-\sin^{2} \theta _{i})....EQN(I)

By using Snell’s law  \frac{\sin \theta _{i}}{\sin \theta _{t}} =\sqrt{\frac{\mu _{2}\epsilon _{2}}{\mu _{1}\epsilon _{1}}} =\frac{r_{2}}{r_{1}}.

using the above equation  \sin^{2} \theta _{t} =\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i}. substituting this in EQN (I).

\eta^{2}_{2}(1-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=\eta^{2} _{1}(1-\sin^{2} \theta _{i}).

(\eta^{2}_{2}-\eta^{2}_{2}\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}} \sin^{2} \theta _{i})=(\eta^{2} _{1}- \eta^{2} _{1}\sin^{2} \theta _{i}).

after simplification    \sin^{2} \theta _{i}(\eta^{2} _{1}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\eta^{2} _{2} )=(\eta^{2} _{1}-\eta^{2} _{2 }).

as \eta _{1} = \sqrt{\frac{\mu _{1}}{\epsilon _{1}}}  and  \eta _{2} = \sqrt{\frac{\mu _{2}}{\epsilon _{2}}}.

\sin^{2} \theta _{i}(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{1}\epsilon _{1}}{\mu _{2}\epsilon _{2}}\frac{\mu _{2}}{\epsilon _{2}} )=(\frac{\mu _{1}}{\epsilon _{1}}-\frac{\mu _{2}}{\epsilon _{2}})..

by simplification  \sin^{2} \theta _{i}= \frac{(1-\frac{\mu _{2}\epsilon _{1}}{\mu _{1}\epsilon _{2}} )}{(1-\frac{\epsilon^{2} _{1}}{\epsilon^{2} _{2}})}.

here \theta _{i} is called as Brewster’s angle.

Let us assume two mediums are lossless dielectrics and are non-magnetic  then

\mu _{1} =\mu _{2}=\mu _{0}.

\sin^{2} \theta _{Brewster}= \frac{1}{(1+\frac{\epsilon _{1}}{\epsilon _{2}})}.

\sin \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{2}+\epsilon _{1}}}.

\tan \theta _{Brewster}= \sqrt{\frac{\epsilon _{2}}{\epsilon _{1}}} = \frac{r_{2}}{r_{1}}.

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Author: Lakshmi Prasanna Ponnala

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

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