The problem is in the expression,
\(n=\cfrac{a_\psi}{a_\psi-d}\)
which is clearly not
\(n=\cfrac{\psi_{n\,i}}{\psi_{n\,r}}\)
But from previously,
\(h.f=E=\psi+x.F\)
in the post "Other Time Dimensions Counts" dated 21 Dec 2014, we have Planck's expression (\(h.f\)) being the sum of energy density \(\psi\), and work done on \psi (\(x.F\)). \(f\) is being referred to as the frequency of the photon, ie. the color of light. These lead us to the idea that the spread in \(f\), ie light dispersion, is the result of a spread in work done, \(x.F\), and \(\psi\) remains constant.
As \(\psi\) deforms on entering the medium, work is done on it. Such work is recovered on exit from the medium at the opposite parallel boundary. In the case where the exit boundary is not parallel or even anti-parallel as in perpendicular to the incident boundary, as in the case of a 90o prism, such work is only partially recovered. If we represent incident photons as,
\(h.f_o=\psi\)
On entering the medium, work is done on \(\psi\),
\(h.f_{r}=\psi+x.F\)
On exit from the medium, partial work remains,
\(h.f_{d}=h.f_o+x_r.F\)
where \(f_o\) is the frequency of un-dispersed incident photons, \(f_{d}\) is the dispersed photons just after the exit boundary, and \(x_r.F\) is the residue work that remains with the dispersed photons.
Formulated this way, \(x_r.F\) is the cause of dispersion. Dispersion is the result of work done on \(\psi\).
Intuitively, when \(\psi\) collide into the medium, it is compressed; around a smaller perimeter, \(\psi\) oscillates faster. A second collision on exit from the medium remove the work done on \(\psi\) wholly or partially. When partial work remains on \(\psi\), \(\psi\) is still partially compressed and oscillates at a higher frequency.
But in common literature \(f\), the frequency of light is held constant and the change in \(f\) is carried by the concept of phase velocity,
\(c\to v_p\)
\(v_p\) being the phase velocity. A change in phase velocity, is totally expressed as a change in wavelength, \(\lambda\).
The important difference here is that \(\psi\) is a standing wave around the particle oscillating at frequency \(f_o\). The particle is at light speed, \(c\) in a straight line, crossing at the incident boundary into the medium. The expression,
\(f_o\lambda_o=c\)
refers to \(\psi\) around the particle, not the particle in translational motion across the medium. \(\psi\) remains a wave on being compressed and is still at light speed \(c\). In this way photon is both a particle and a wave. More precisely, it is a wave wrapped into a particle.
We then consider the expression,
\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=\cfrac{sin\,\theta}{sin\,\alpha}=n\)
again. If we assume that such dispersion work done is only in the normal direction of the medium boundary on impact,
\(\cfrac{\psi_{n\,i}}{\psi_{n\,r}}=\cfrac{h.f_o}{h.f_{r}}=\cfrac{sin\,\theta}{sin\,\alpha}=n\)
From which we obtain,
\(\cfrac{\lambda_{r}}{\lambda_o}=\cfrac{sin\,\theta}{sin\,\alpha}=n\)
using \(f_o\lambda_o=f_r\lambda_r=c\).
These \(\lambda\)s are of the standing wave \(\psi\) around the particle, they are not wavelength normally associated with photons.
Since,
\(x_r.F\ge 0\)
and dispersion is the work done that remains with \(\psi\) as the photon exit the medium, it would seems that refraction and dispersion can only positive, ie
\(f_r\ge f_o\)
This is not so. If a photon is initially compressed,
\(h.f_o=\psi_o+x_o.F\)
but exit from the medium,
\(h.f_d=\psi_o+x_d.F\)
when
\(x_d.F\lt x_o.F\) then \(f_d\lt f_o\)
When more work is removed from the compressed photon than the work done on the photons on impact at the incident boundary, dispersion is then anomalous. This is the case of X-ray, where the photons are initially compressed. This explanation is also consistent with the fact that X-ray photons being small (compressed) is more penetrating.
