In the second medium,
\(2\theta_1-\Delta\theta-\Delta\alpha=90^o\)
where,
\(\Delta\theta=\theta_1-\theta_2\)
\(\Delta\alpha=\alpha_1-\alpha_2\)
which implies \(2\theta_1\gt90^o\) and so, \(\theta_1\gt45^o\)
So,
\(\theta_1+\theta_2-\alpha_1+\alpha_2=90^o\)
both material properties \(\varepsilon\) and \(n\) are involved. Alternatively,
\(2\theta_1-90^o=\Delta\theta+\Delta\alpha\)
the change in orientation needed, is from both a change in \(\theta\) and \(\alpha\).
When we set,
\(\alpha_1=\theta_1\) --- (1)
then,
\(\theta_2+\alpha_2=90^o\) --- (2)
the ray splits into two perpendicular directions with orthogonal polarization. Given a light source with a spread of values in \(\theta\), there can be a number of \(\theta\) for which a split of the refracted rays orthogonal to each other can occur. Expressions (1) and (2) are the same as those we obtained in the post "Still Looking For Brewster" dated 16 Aug 2015. From that post,
\({ tan(\alpha_1)}{tan(\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}\)
\(\cfrac{ sin^2(\alpha_1)}{cos(\alpha_1)cos(\alpha_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}\)
but \(\alpha_1+\alpha_2\ne90^o\) as oppose to the often quoted \(\alpha_1+\alpha_2=90^o\) in the derivation for Brewster's angle. Instead,
\(sin(\alpha_2)=\cfrac{n_1}{n_2}sin(\alpha_1)\)
\(cos(\alpha_2)=\sqrt{1-\left(\cfrac{n_1}{n_2}\right)^2sin^2(\alpha_1)}\)
So,
\(tan(\alpha_1).\cfrac{n_1}{n_2}\cfrac{sin(\alpha_1)}{\sqrt{1-\left(\cfrac{n_1}{n_2}\right)^2sin^2(\alpha_1)}}=\cfrac{\varepsilon_2}{\varepsilon_1}\)
\(tan^2(\alpha_1)sin^2(\alpha_1)=\left(\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}\right)^2\left\{1-\left(\cfrac{n_1}{n_2}\right)^2sin^2(\alpha_1)\right\}\)
\(tan^2(\alpha_1)sin^2(\alpha_1)+\left(\cfrac{\varepsilon_2}{\varepsilon_1}\right)^2sin^2(\alpha_1)=\left(\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}\right)^2\)
\(sin^4(\alpha_1)+\left(\cfrac{\varepsilon_2}{\varepsilon_1}\right)^2sin^2(\alpha_1)(1-sin^2(\alpha_1))=\left(\cfrac{\varepsilon_2}{\varepsilon_1}\cfrac{n_2}{n_1}\right)^2(1-sin^2(\alpha_1))\)
\(sin^{ 4 }(\alpha _{ 1 })+\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \right) ^{ 2 }sin^{ 2 }(\alpha _{ 1 })-\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \right) ^{ 2 }sin^{ 4 }(\alpha _{ 1 })=\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }-\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }sin^{ 2 }(\alpha _{ 1 })\)
\( sin^{ 4 }(\alpha _{ 1 })\left\{ 1-\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \right) ^{ 2 } \right\} +\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \right) ^{ 2 }\left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} sin^{ 2 }(\alpha _{ 1 })=\left( \cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } } \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }\)
\(sin^{ 4 }(\alpha _{ 1 })\left\{ \left( \cfrac { \varepsilon _{ 1 } }{ \varepsilon _{ 2 } } \right) ^{ 2 }-1 \right\} +\left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} sin^{ 2 }(\alpha _{ 1 })=\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }\)
\(sin^{ 4 }(\alpha _{ 1 })\left\{ \left( \cfrac { \varepsilon _{ 1 } }{ \varepsilon _{ 2 } } \right) ^{ 2 }-1 \right\} +\left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} sin^{ 2 }(\alpha _{ 1 })=\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }\)
which is quadratic in \(sin^2(\alpha_1)\) and can yield two positive solutions for \(\alpha_1\).
When \(\cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } }=1\),
\(\left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} sin^{ 2 }(\alpha _{ 1 })=\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }\)
\(sin^{ 2 }(\alpha _{ 1 })=\frac { \left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } }{ \left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} }\)
\(tan(\alpha_1)=\cfrac{n_2}{n_1}\)
and this Brewster!
Thank you very much.
When \(\cfrac { \varepsilon _{ 2 } }{ \varepsilon _{ 1 } }=1\),
\(\left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} sin^{ 2 }(\alpha _{ 1 })=\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 }\)
\(sin^{ 2 }(\alpha _{ 1 })=\frac { \left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } }{ \left\{ 1+\left( \cfrac { n_{ 2 } }{ n_{ 1 } } \right) ^{ 2 } \right\} }\)
and this Brewster!
Thank you very much.
