Cont'd from the post "In Colors", dated 11 Aug 2015,
Color dispersion is not birefringence, the spread of color is due to the spread of values of \(\theta\) in the light source that results in a spread of values in \(\Delta \theta\).
However, when the source is monochromatic, the ray will split as a result of \(\Delta\theta\). A single mono color dot will appear as two dots looking into the direction of refraction. This is birefringence. \(\Delta \theta\) is due to applying boundary conditions on \(B\) as in the post "A Bloom Crosses Over" dated 10 Aug 2015, where,
\(\cfrac{ tan(\theta_1)}{tan(\theta_2)}=\cfrac{\varepsilon_2}{\varepsilon_1}\)
\(\varepsilon_1\), \(\varepsilon_2\) are electric permittivity.
Embedded charges in the material that changes \(\varepsilon\) will change \(\Delta \theta\) locally and possibly result in color patterns when illuminated with white light.
Birefringence and color dispersion may have the same underlying mechanism, but the factor leading to their manifestation is different. Color dispersion further requires a spread of \(\theta\).
\(B\) does not appear in the expressions for \(\alpha_{2s}\) and \(\alpha_{2p}\), as long as \(B\ne0\) both expressions are valid. Since in color dispersion, both \(\alpha_{2s}\) and \(\alpha_{2p}\) depended on color, \(B\) is not color.
The problem is, does turning a polarizer in white light let different color light through at different polarizing angle?
No, a polarizer is in fluorescence. Photons at different \(\theta\) are absorbed and emitted according to an emission distribution. This emission distribution has two dominant modes about \(\small{90^o}\) apart. (post "Not Exactly A Fluorescence Polarizer" dated 28 Jul 2015) . The color of the absorbed photons are destroyed. Colors are redistributed according to the emission distribution of the polarizer convoluted with the source color \(\theta\) distribution.
Unless photon absorption is selective based on \(\theta\), ie not a uniform distribution of \(\theta\) as assumed in the post "Not Exactly A Fluorescence Polarizer" dated 28 Jul 2015, then the emission distribution will be the convolution of the source distribution with the absorption distribution, and then the result convoluted with the emission distribution based on a uniformly distributed \(\theta\).
Colors are destroyed in a polarizer.
How then do we reconcile frequency/wavelength and \(\theta\)?
Tomorrow then...
