# Brewster’s angle

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

$\rho&space;_{parallel}&space;=&space;0$.

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

$\frac{E_{r}}{E_{i}}=\frac{(\eta&space;_{2}\cos&space;\theta&space;_{t}-\eta&space;_{1}\cos&space;\theta&space;_{i})}{(\eta&space;_{1}\cos&space;\theta&space;_{i}+\eta&space;_{2}\cos&space;\theta&space;_{t})}=0$.

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

$\eta&space;_{2}\cos&space;\theta&space;_{t}=\eta&space;_{1}\cos&space;\theta&space;_{i}.$

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

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

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

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

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

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

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

as $\eta&space;_{1}&space;=&space;\sqrt{\frac{\mu&space;_{1}}{\epsilon&space;_{1}}}$  and  $\eta&space;_{2}&space;=&space;\sqrt{\frac{\mu&space;_{2}}{\epsilon&space;_{2}}}$.

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

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

here $\theta&space;_{i}$ is called as Brewster’s angle.

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

$\mu&space;_{1}&space;=\mu&space;_{2}=\mu&space;_{0}$.

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

$\sin&space;\theta&space;_{Brewster}=&space;\sqrt{\frac{\epsilon&space;_{2}}{\epsilon&space;_{2}+\epsilon&space;_{1}}}$.

$\tan&space;\theta&space;_{Brewster}=&space;\sqrt{\frac{\epsilon&space;_{2}}{\epsilon&space;_{1}}}&space;=&space;\frac{r_{2}}{r_{1}}.$

(2 votes, average: 5.00 out of 5)