\(|{r}_{c2}| =\cfrac{{r}_{c1}}{cos(\theta)}.\sqrt{(\cfrac{{n}_{1}}{{n}_{2}})^2.sin^2{(\theta)}-1}\)
it is possible to concentrate all energy along the direction of travel by total internal reflection off a less dense medium. In which case, the photon is travelling along the helix in a very tight circular motion.
The equation does not suggest whether the circular motion re-emerge in the same direction or opposite. But it is certain that the circular path of the helix is small just when the photon beam is in total internal reflection, and as the incidence angle increases the circular path radius increases,
A plot of the above equation where the y-axis has been scaled by 10\({r}_{c1}\) units with \({n}_{1}\) = 1.5 and \({n}_{2}\) = 1, and incident angle \(\theta\) in the x-axis in radians. The actual equation plotted is 10/cos(x)*((1.5)^2*(sin(x))^2-1)^(0.5). We see that the reflected radius is many times the incident radius. This could be the explanation for dispersion of light after total internal reflection. And when the light first emerges from total internal reflection its radius is close to zero, this is when it is observed that light is polarized. This might suggest that circular polarization refers to the photon in circular motion. That photon not in circular motion is polarized light.
More importantly, this discussion leads to the conclusion that photon carries a electric field along the direction of travel in the circular path, that photons are electric dipoles. And since the photons are moving, a magnetic field develops in the perpendicular direction.
Notice that the E-field (in red) starts at a phase of \(\frac{\pi}{2}\) and a corresponding H-field (in blue). At \(\lambda\), the wave advances by one cycle and the E-field is back to its original position. There is no indication here how large the radius is.
