When a photon beam enters a medium from free space, at an angle \(\alpha\) to the normal of the surface, at the point of entry,
\({v}_{sp} = v.sin{(\alpha})\), perpendicular to the normal
\({v}_{sa} = v.cos{(\alpha})\), along the normal
We know from previously,
\({v}^{+}_{sa} = v.cos{(\alpha)}.sin{(\theta})\) where \(\theta = sin^{-1}{(\cfrac{1}{n}})\)
and if both direction are equally effected by the medium,
\({v}^{+}_{sp} = v.sin{(\alpha)}.sin{(\theta})\)
The resultant direction is,
\(tan{(\beta)} = \cfrac{{v}^{+}_{sp}}{{v}^{+}_{sa} }=\cfrac{sin{(\alpha)}.sin{(\theta)}}{cos{(\alpha)}.sin{(\theta)}}=tan{(\theta)}\)
But this is not true, the path of the beam should bent inside the medium.
If \({v}_{sp}\) is not effected by the medium
\({v}^{+}_{sp} = v.sin{(\alpha)}\)
Then the new resultant direction is,
\(tan{(\beta)} = \cfrac{sin{(\alpha)}}{cos{(\alpha)}.sin{(\theta)}} > tan{(\alpha)}\), always
which is also not true. In fact, from the observation that defraction is towards the normal inside a denser medium, the perpendicular velocity component is reduced more by the medium compared to the vertical component. Since, we can rotate the beam or the medium and obtain the same results, the medium is uniform yet not. This suggests that there is asymmetry in the plane perpendicular to the direction of the beam. It suggests that the beam has an extend perpendicular to the direction of the beam and since the beam can be rotated, such an extend is circular in the plane perpendicular to the direction of the beam.
Either a beam of photons is not a beam of point particles, or it is a beam of point particles but the path of such point particles is not straight lines but helical; circular in the plane perpendicular to the direction of travel but straight along the length of the beam. The latter view is consistent with the three parameters associated with light; a frequency, \(f\), where every turn of the circle perpendicular to the direction of travel move the particle forward by a wavelength, \(\lambda\) distance and so the speed \(c\) of such a beam is \(c=f\lambda\). It might also suggests that the phase of light wave refers to the relative position of the particle in that circle. But there are others parameter associated with a helical path: circular velocity \({v}_{c}\), radius of the circle \({r}_{c}\) projected onto a plane perpendicular to the direction of travel and the centripetal force \({f}_{c}\) giving rise to the circular motion. The existence of a circular velocity also suggests that a beam of light has more energy than its linear speed can account for.
