Looking for Brewster

When we consider,

$tan(\theta_2-\alpha^{'}_{2s})=\cfrac{\mu_2}{\mu_1}tan(\theta_1-\alpha)$

when $\theta_1$ is large, such that

$\theta_1-\alpha\gt90^o$

Let $x+90^o=\theta_1-\alpha$ then

$tan(\theta_1-\alpha)=tan(x+90^o)=-cot(x)$

So,

$tan(\theta_2-\alpha^{'}_{2s})=-\cfrac{\mu_2}{\mu_1}cot(\theta_1-\alpha-90^o)$

$cot(\theta_2-\alpha^{'}_{2s}+90^o)=\cfrac{\mu_2}{\mu_1}cot(\theta_1-\alpha-90^o)$

When $\mu _{ 2 }\gt\mu _{ 1 }$, since $cot(x)$ is a decreasing function,

$\theta_2-\alpha^{'}_{2s}+90^o\lt\theta_1-\alpha-90^o$

$\alpha^{'}_{2s}\gt180^o+\alpha+\Delta \theta$

Since the $\alpha^{'}_{2s}$ has been reflected back into medium 1, $\Delta \theta=0$,

$\alpha^{'}_{2s}\gt180^o+\alpha$

When $\mu _{ 2 }\lt\mu _{ 1 }$,

$\theta_2-\alpha^{'}_{2s}+90^o\gt\theta_1-\alpha-90^o$

$\alpha^{'}_{2s}\lt180^o+\alpha$

We can also have,

$tan(\theta_2+\alpha^{'}_{2p})=\cfrac{\mu_2}{\mu_1}tan(\theta_1+\alpha)$

when $\theta_1$ is large, such that

$\theta_1+\alpha\gt90^o$

Let $x+90^o=\theta_1+\alpha$ then

$tan(\theta_1+\alpha)=tan(x+90^o)=-cot(x)$

So,

$tan(\theta_2+\alpha^{'}_{2p})=-\cfrac{\mu_2}{\mu_1}cot(\theta_1+\alpha-90^o)$

$cot(\theta_2+\alpha^{'}_{2p}+90^o)=\cfrac{\mu_2}{\mu_1}cot(\theta_1+\alpha-90^o)$

When $\mu _{ 2 }\gt\mu _{ 1 }$,

$\theta_2+\alpha^{'}_{2p}+90^o\lt\theta_1+\alpha-90^o$

$\alpha^{'}_{2p}\lt\alpha-\Delta\theta-180^o$

Since,  $\Delta\theta=0$

$\alpha^{'}_{2p}\lt\alpha-180^o$

$\alpha^{'}_{2p}$ is reflected back along $\alpha$

When $\mu _{ 2 }\lt\mu _{ 1 }$,

$\theta_2+\alpha^{'}_{2p}+90^o\gt\theta_1+\alpha-90^o$

$\alpha^{'}_{2p}\gt\alpha-\Delta\theta-180^o$

Since,  $\Delta\theta=0$

$\alpha^{'}_{2p}\gt\alpha-180^o$

This might seem to be the same results as the post "More Bending Of Light" dated 13 Aug 2015, but the swing of $\alpha_s$ and $\alpha_p$ are different,

When $\mu _{ 2 }\gt\mu _{ 1 }$,  $\alpha^{'}_{2s}\gt180^o+\alpha$

When $\mu _{ 2 }\lt\mu _{ 1 }$,  $\alpha^{'}_{2s}\lt180^o+\alpha$

When $\mu _{ 2 }\gt\mu _{ 1 }$,  $\alpha^{'}_{2p}\lt\alpha-180^o$

When $\mu _{ 2 }\lt\mu _{ 1 }$,  $\alpha^{'}_{2p}\gt\alpha-180^o$

Previously,

When $\mu_2\gt\mu_1$,  $\alpha_{2s}\lt-180^o+\alpha$

When $\mu_2\lt\mu_1$, $\alpha_{2s}\gt-180^o+\alpha$

This cases show the swing of $\alpha_{2s}$ and $\alpha_{2p}$ as $\theta_1$ changed.  There is no Brewster angle here.