Brewster’s angle

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