The change in direction as the ray enters the medium is due to a change in \(\psi_n\), the normal component of \(\psi\) at the point of incidence. \(\psi_n\) decreases in a denser medium. It does so by an increase in physical volume. No energy is loss as \(\psi_n\) adjusts, like a particle with velocity \(c\), that enters into a gravitational field, energy is still conserved. In this case, the conservative field is extended by the medium material lattice. (There may be a dominant field that is electrical, gravitational or temperature.)
After the point of impact, \(\psi_r\) is spherical with a new radius \(a_{\psi\,r}\).
This is why \(\psi_n\) just after the point of incidence is replaced with \(\psi\) when we discuss \(n\) in relation to \(a_{\psi}\) to obtain an expression for \(n(\lambda)\).
An increase in \(a_{\psi\,r}\) results in a decrease in \(m_{a_{\psi}}\) from the expression,
\(\psi_r=h.f_r=2\pi a_{\psi\,r}m_{a_{\psi}}c\)
when \(f_r\) is held constant. In general when the kinetic energy of the photon along \(t_c\) is held constant (\(E=mc^2\)), this energy is split between \(\psi\) and \(m_{\rho\,a_{\psi}}\); its energy density and mass density. A decrease in energy density, increases the mass density, with its mass-energy equivalent held constant.
This reduction in \(\psi\) does not mean a decrease in energy of the photon. Work is in fact being performed on the photon.
\(h.f_r=\psi_o+x_rF\)
where \(x.F\) denotes the work done, added to \(\psi_o\). Part of this work is recovered on the photons' exit from the medium. Partial work that remains results in dispersion,
\(h.f_d=\psi_o+x_dF\)
No dispersion occurs when all work on the photon is recovered by the medium,
\(h.f_d=\psi_o+0=h.f_o\)
Fortunately, this can be checked by observing the color changes along a ray as it passes through the the refractive medium. Unfortunately, we may not be sensitive to different shades of the same color and come to the wrongful conclusion that no color changes occurs. And so we "invented" monochromatic light.
If a dispersed photon receive positive work, this means the prism/medium loses energy during dispersion and will cool. What happens when we cool a prism actively, does it lose its dispersion property?
On closer look, if the incidence surface does positive work and the exit surface recovers such work done, then the incidence surface should be cooler then the exit surface.
All these can be checked experimentally. If a cooled prism is less dispersive then optics is also temperature and the predominant conservative field that decreases \(\psi\) at the point of incidence is temperature/heat. In a similar way, electrical and gravitational potential of the prism can be changed to see if such changes effects its dispersive property.
The reduction in \(\psi_n\) on entering a denser medium further suggests that in an opaque medium \(\psi_n\to0\) on impact at the point of incidence. In this case, the photon fully manifest itself as a mass, \(m\) and the full momentum of the photon at light speed \(c\) is exerted on the medium.
A transparent medium is then, when \(\psi_n\) retains some values on impact,
\(\psi_{n\,r}\ne0\)
Transparency is then achieved by making sure that \(\psi\nrightarrow0\).
We have postulated that photon is a wave with light speed in space and with oscillating energies between a time dimension, \(t_c\), \(t_g\) or \(t_T\) and one other space dimension. So, \(\psi\) is either electrical, gravitational or temperature. To keep \(\psi\) from zero, the surface of incidence should readily have all such energies. Since protons also provide a gravitational potential (post "Sunshine and Proton Beam" dated 23 May 14) in addition to a charge potential, a flow of proton charges along the incidence surface may ensure that \(\psi\ne0\) as photons impact the surface.
Transparency may be achieved by a flow of protons over the surface of incidence of the intercepting medium.
Invisibility is a different issue.
