Then Reflected

From the post "Photon Emission After Absorption" dated 25 Jul 2015, the photon is ejected perpendicular to the direction of travel of the particle; in the post "A Pump!" dated 25 Jul 2015,

$E_{p}=h.\left\{1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\right\}.f_{cir}$

when

$1-\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\lt0$

$\sqrt{\cfrac{sin(\theta)}{cos(\theta)}}\gt1$

a photon is absorbed first then emitted, this introduces a $\small{\pi}$ phase in the emission.

The refracted ray we observe in the second medium is made up of emitted photons in the direction perpendicular to $\alpha_2$ (out of the paper).  This emitted ray will be absorbed and be emitted a second time.  In the first instance of emission, the photons are emitted perpendicular to the direction of travel of the particle ($P^{'}O$ and $PO$), and is along $E\,\,PE^{'}$ and $E\,\,PE$.  The direction of incident then changes to from $PO$ to $EO$, by a difference of $90^o-\theta$.  Along $EO$ particles interact with the passing photons, as if along $PO$.

The ray $\alpha_2$ corresponds to the line $EO$.  $EO$ is absorbed and emitted into the direction it was first absorbed.  It is diverted by a difference of $-(90^o-\theta)$ to be along $PO$ again.  Reciprocity.

If we confine photon emission/absorption to be at the point of incident, this emitted ray, $PO$ will result in a reflected ray,

And the possibility of a phase lag between the absorption of a photon first, then an emission, accounts for the possible $\pi$ phase shift in the reflected ray.

This is not the reflected ray off a mirrored surface, but reflection from an interface of two mediums with different optical properties.  The reflected ray is the result of a second absorption/emission of the photons in the medium, the first absorption/emission allows us to see $\alpha_2$.  The second absorption/emission is due to the interaction of $\alpha_2$ with the medium.  $\alpha_2$ interacts with the medium only once.  The first absorption/emission of the photons in the medium results in the cone $PE^{'}\,\,E\,\,PE$ from which we started to derive the geometry of $\alpha_2$ (post "It's All Fluorescence Outside, Inside" dated 29 Jul 2015).

The ray corresponding to $90^o-\theta_2-\alpha_2$ is probably totally internally reflected.  The presence of this ray is important to vindicate the results here.  It is possible that not all of this ray is absorbed and emitted to result in a reflected ray.

When would absorption and emission stop?  Absorb along $AB$ $\to$ emit perpendicularly $XY$ $\to$ absorb along $XY$ $\to$ emit perpendicularly along $AB$, return.  The photons are emitted in the direction perpendicular to the particle's travel; this is not in the direction perpendicular to the ray on the plane containing the axis of the cone.  This emitted ray is rotated about the axis of the cone and reaches an observer perpendicular to the ray out of the paper.  It is a cone in 3D in the first place.

And this accounts for a reflected ray, even when $\alpha_2$ does not penetrate into the second medium.

Note:  The cone $PE^{'}\,\,E\,\,PE$  was not initially conceived from considering absorption and emission.  There can be a problem here;  two birds with one stone is real luck!  The cone accounts for both the direction along the ray and, emitted photons that reach the eyes of the observer, perpendicular to the ray.

Both ray $-\alpha_2$ and  $90^o-\theta_2-\alpha_2$ are at the point of incident, very small.