\(B_1sin(\alpha_{adj\,\,1})=B_2sin(\alpha_{adj\,\,2})\)
\(\cfrac{B_1}{\mu_1}cos(\alpha_{adj\,\,1})=\cfrac{B_2}{\mu_2}cos(\alpha_{adj\,\,2})\)
\(\alpha_{adj\,\,1}=90^o+\alpha_1-\theta_1\)
\(\alpha_{adj\,\,2}=90^o+\alpha_2-\theta_2\)
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,
\(B_1sin(\alpha_{adj\,\,1})=B_2sin(\alpha_{adj\,\,2})\)
\(\cfrac{B_1}{\mu_1}cos(\alpha_{adj\,\,1})=\cfrac{B_2}{\mu_2}cos(\alpha_{adj\,\,2})\)
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,
\(\theta_2+\alpha_2=90^o\)
\(sin(\alpha_{adj\,\,2})=sin(0^o)=0\)
\(sin(\alpha_{adj\,\,2})=sin(0^o)=0\)
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.
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.
