A Bloom Crosses Over

When a photon, modeled as a cone shaped spread of $E$ field, is projected perpendicular from a medium of refractive index $n_1$ into a medium of refractive index $n_2$,

The $E$ field being resolved into three components as shown, and applying the boundary conditions for an $E$ field line,

$E_{1\,3}=\cfrac{\varepsilon_2}{\varepsilon_1}E_{2\,3}$

where the subscript $3$ denotes the perpendicular direction, parallel to the surface normal.

$E_{1}cos(\beta_1)=\cfrac{\varepsilon_2}{\varepsilon_1}E_{2}cos(\beta_2)$

The tangential component remains the same,

$E_1 sin(\beta_1)=E_2sin(\beta_2)$

Dividing the two expressions above, we have,

$\cfrac{ tan(\beta_2)}{tan(\beta_1)}=\cfrac{\varepsilon_2}{\varepsilon_1}$

Since, $\beta=90^o-\theta$

$tan(\beta)=tan(90^o-\theta)=cot(\theta)$

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

When the dielectric constants,

$\varepsilon_1\lt\varepsilon_2$

The $E$ field cone spreads and polarization $\small{\theta}$ decreases.  This is however, not dispersion.

When the photon is projected at an incline to the normal onto the interface $\small{\varepsilon_1|\varepsilon_2}$,

The particle maintains its circular motion in both medium and the cone passes through the medium completely when its apex is on the interface, at the foot of the surface normal, we have,

$\cfrac { AB }{ v_{ 1 } } =\cfrac { A^{ ' }B^{ ' } }{ v_{ 2 } }$

Since $BA^{'}$ is common,

$\cfrac {BA^{'} sin(\alpha _{adj\,\, 1 }) }{ v_{ 1 } } =\cfrac {BA^{'} sin(\alpha _{adj\,\, 2 }) }{ v_{ 2 } }$

where,

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

and

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

where $\theta$ is polarization; the angle at which the circular path of the particle makes with the line of incidence.

Since,

$n.v=c$

$\cfrac{1}{v}\propto n$

and we have,

$n_{ 1 }sin(\alpha _{adj\,\, 1 })=n_{ 2 }sin(\alpha _{ adj\,\,2 })$

which is just Snell's Law with $\alpha$ substituted by $\alpha_{adj}$.  Furthermore, this particle in circular motion that generates a $E$ field is itself a $B$ field.  When we apply the boundary conditions appropriate for a $B$ field upon the circulating particle at the point through the incident normal,

the normal components of $B$ across the mediums are equal,

$B_{ 1 }sin(\alpha _{ adj\,\,1 })=B_{ 2 }sin(\alpha _{ adj\,\,2 })$

where,

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

and

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

where $\theta$ is polarization; the angle at which the circular path of the particle makes with the line of incidence.

$\cfrac { B_{ 1 } }{ B_{ 2 } } =\cfrac { sin(\alpha _{ adj\,\,2 }) }{ sin(\alpha _{ adj\,\,1 }) }$

and the tangential component of $H=\cfrac{B}{\mu}$ across the medium are equal,

$\cfrac { B_{ 1 }cos(\alpha _{adj\,\, 1 }) }{ \mu _{ 1 } } =\cfrac { B_{ 2 }cos(\alpha _{adj\,\, 2 }) }{ \mu _{ 2 } }$

So,

$\cfrac { sin(\alpha _{ adj\,\,2 }) }{ sin(\alpha _{ adj\,\,1 }) } =\cfrac { \mu _{ 1 } }{ \mu _{ 2 } } \cfrac { cos(\alpha _{ adj\,\,2 }) }{ cos(\alpha _{ adj\,\,1 }) }$

$\cfrac { tan(\alpha _{ adj\,\,2 }) }{ tan(\alpha _{adj\,\, 1 }) } =\cfrac { \mu _{ 1 } }{ \mu _{ 2 } }$

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

Snell's law is due to the change in velocity along the ray with the particle still in circular motion in the perpendicular direction, at $AA^{'}$ and $BB^{'}$.  The last expression (*) is the result of applying boundary conditions on the analogous $B$ field.  Applying boundary conditions to the $E$ field indicates a change in polarization $\theta$ as the photon crosses the interface $\small{\varepsilon_1|\varepsilon_2}$.

In the case of an EMW, the $E$ field is reversed, but all expressions developed above are still true.