Still Looking For Brewster

From the post "Late Bloomer Gets The Splits" dated 10 Aug 2015,

$B_1sin(\alpha_{adj\,\,1})=B_2sin(\alpha_{adj\,\,2})$

$\cfrac{B_1}{\mu_1}cos(\alpha_{adj\,\,1})=\cfrac{B_2}{\mu_2}cos(\alpha_{adj\,\,2})$

$\alpha_{adj\,\,1}=90^o+\alpha_1-\theta_1$

$\alpha_{adj\,\,2}=90^o+\alpha_2-\theta_2$

but when,

$\alpha_1=\theta_1$

$cos(\alpha_{adj\,\,1})=cos(90^o)=0$

or when,

$\alpha_2=\theta_2$

$cos(\alpha_{adj\,\,2})=cos(90^o)=0$

in which case,

$\mu_1tan(\alpha _{adj\,\, 1 })=\mu_2tan(\alpha _{adj\,\, 2 })$

is not valid,

The tangential component of the incident ray is zero and the ray enters the medium perpendicular to the inter-surface, parallel to the normal.  $\alpha_{2s}$ passes through the second medium perpendicularly.

In a similar way, when we consider the left arm,

$B_1sin(\alpha_{adj\,\,1})=B_2sin(\alpha_{adj\,\,2})$

$\cfrac{B_1}{\mu_1}cos(\alpha_{adj\,\,1})=\cfrac{B_2}{\mu_2}cos(\alpha_{adj\,\,2})$

$\alpha_{adj\,\,1}=\theta_1+\alpha_1-90^o$

in the first medium, and into the second medium,

$\alpha_{adj\,\,2}=\theta_2+\alpha_2-90^o$

If $\theta_1+\alpha_1=90^o$

$sin(\alpha_{adj\,\,1})=sin(0^o)=0$

or when,

$\theta_2+\alpha_2=90^o$

$sin(\alpha_{adj\,\,2})=sin(0^o)=0$

the perpendicular component of the left incident arm is zero.  The tangential component  $\alpha_{2p}$, glides along the surface (if it is flat) and does not enter the second medium.

Both scenario occurs when,

$\theta_1=\alpha_1$

and

$\theta_2+\alpha_2=90^o$

The incident ray is split $90^o$ one along the surface $\alpha_{2p}$, and the other enters the second medium perpendicularly $\alpha_{2s}$.  This is wrong, velocity along the ray was not accounted for.

This is still not Brewster angle.  But, from

$\theta_2+\alpha_2=90^o$

$\theta_2=90^o-\alpha_2$

$\theta_1=\alpha_1$

Since, from the post "A Bloom Crosses Over" dated 10 Aug 2015,

$\cfrac{ tan(\theta_1)}{tan(\theta_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

$\cfrac{ tan(\alpha_1)}{tan(90^o-\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

${ tan(\alpha_1)}{tan(\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

From Snell's Law,

$n_1sin(\alpha_1)=n_2sin(\alpha_2)$

$\cfrac{n_1}{n_2}\cfrac{ sin^2(\alpha_1)}{cos(\alpha_1)cos(\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

and

$\cfrac{ sin^2(\alpha_1)}{cos(\alpha_1)cos(\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}$

$\alpha_1+\alpha_2=90^o$  This is wrong!

$\cfrac{ sin^2(\alpha_1)}{cos(\alpha_1)cos(90^o-\alpha_1)}=\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}$

$\cfrac{ sin(\alpha_1)}{cos(\alpha_1)}=tan(\alpha_1)=\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}$

which is still not Brewster.