Wave Front and Wave Back

The wave front theory assumes that light travels in parallel wave fronts,

Wave Transverse Vector

The parallel lines here are not wave fronts but vector lines of the velocity component that is perpendicular to the direction of travel with the same phase on the circular path.  This velocity is the circular velocity.  Such velocity are always perpendicular to the direction of travel.   There is no artificial wave front, but an assumption that in all mediums the photon travels in similar helical paths.  That means parallel lines remains parallel lines, which is the same geometrically as wave front theory but for different reason.

What is interesting is the radius of the circular path changing from medium to medium.

If the initial radius is ${r}_{c1}$ then

${r}_{c2} ={r}_{c1}.\cfrac{cos(\phi)}{cos(\theta)}=\cfrac{{r}_{c1}}{cos(\theta)}.\sqrt{1-(\cfrac{{n}_{1}}{{n}_{2}})^2.sin^2{(\theta)}}$

since, $sin(\phi)=\cfrac{{n}_{1}}{{n}_{2}}sin(\theta)$.

If the photon enters from a less dense to a denser medium, ${n}_{2}$ > ${n}_{1}$

$\cfrac{{n}_{1}}{{n}_{2}}$ < $1$  $\Longrightarrow$  $\phi$ < $\theta$ for 0 < $\phi$, $\theta$ < $\cfrac {\pi}{2}$

so, $cos(\phi)$ > $cos(\theta)$  $\Longrightarrow$  $\cfrac{cos(\phi)}{cos(\theta)}$ > 1

that is,

${r}_{c2}$ > ${r}_{c1}$ for all 0 < $\theta$ < $\cfrac {\pi}{2}$.  The circle expanded which could mean light dispersed.

In the case of the photon entering from a denser to less dense medium, ${n}_{2}$ < ${n}_{1}$,
it is possible that,

$1-(\cfrac{{n}_{1}}{{n}_{2}})^2.sin^2{(\theta)}$ = 0

$\theta = sin^{-1}{(\cfrac{{n}_{2}}{{n}_{1}})}$

Which is just the case for critical angle ${\theta}_{c}$, where

$sin(\phi)=\cfrac{{n}_{1}}{{n}_{2}}sin(\theta_c) = 1$

 $\phi$ = $\cfrac{\pi}{2}$  the refracted beam is along the inter-surface of the two mediums and has no circular component.  If energy is still conserved inside the medium then,  all energy of the photon beam is now in the direction of travel of the beam.  However,  photon speed cannot be greater than light speed.

Beyond ${\theta}_{c}$,  ${r}_{c2}$  is complex,

${r}_{c2} =i.\cfrac{{r}_{c1}}{cos(\theta)}.\sqrt{(\cfrac{{n}_{1}}{{n}_{2}})^2.sin^2{(\theta)}-1}$

We know that in this case, the photon is reflected back into the denser medium,  a phenomenon called total internal reflection.

From the diagram on the right we see that,

$sin{(\phi+\cfrac{\pi}{2})}=cos{(\phi)}=sin{(\theta^{'})}$

This suggests that the effect of $i$, the complex imaginary unit, rotated the frame of reference by $\cfrac{\pi}{2}$ anti-clockwise,

 $\phi$  $\rightarrow$ $\phi+\cfrac{\pi}{2}$  on the new reference

and the relationship

 $sin(\phi)=\cfrac{{n}_{1}}{{n}_{2}}sin(\theta)$

still holds, except, since both paths are in the same medium ${n}_{1}$ = ${n}_{2}$, and in the new reference frame

 $sin{(\phi+\cfrac{\pi}{2})} =sin{(\theta^{'})}= sin{(\theta)}$,

and so,

$\theta^{'} = \theta$

The photon path is reflected about the surface normal with the angle of incidence equals the angle of reflection.