Oblique Incidences Continued

Brewster and Critical Angle

Brewster Angle

Is there such an angle $\theta_i$ such that $\Gamma_\parallel(\theta_i) = 0$ and $\Gamma_\perp(\theta_i) = 0$?

• We can simplify by using $\cos\theta_t = \sqrt{1-\sin^2\theta_t} = \sqrt{1-\frac{{n_1}^2}{{n_2}^2}\sin^2(\theta_i)}$
• $\Gamma_\parallel = 0$ for $\theta _i = \theta _B$, the Brewster angle
• $\sin^2\theta_B = \infty$, no Brewster angle
• We can make polarizers where you can turn a circular/ellipsoid polarized incident wave into a linearly polarized reflection.
• Actually, in real life, reflected light tends to get TE polarized.

Critical Angle

Looking at Snell’s Law $\frac{n_1}{n_2}\sin\theta_i = \sin\theta_t$, we can see that if $n_1 > n_2$, there is a critical angle $\theta _c$ in which $\frac{n_1}{n_2}\sin\theta_c = 1$. In this case, we have $\sin\theta_t = 1 \ \rightarrow \ \theta_t = 90^\circ$

• At incident angles greater than the critical angle, we have $\frac{n_1}{n_2}\sin\theta_i > 1$

$$ e^{-jk_0n_1\sin\theta_ix}e^{-k_0n_2\underbrace{\sqrt{(\frac{n_1}{n_2}\sin\theta_i)^2-1}}_\text{attenuation rate} \ z} $$

• We can see that it is propagating along $+x$ but exponentially attenuating along $+z$.
  • This is a inhomogeneous plane wave.
  • This is in contrast to the lossy media case, where it was attenuating in the same direction as propagation.
• This is also called total internal reflection (TIR).

Important: this means exponential attenuation does not imply lossy media!

• This is how optical fibre works, as we want the wave to be confined as much as possible to the fibre, and we do not want energy lost into the medium surrounding that fibre

Poynting Vector