Not Snell's Law

If photon collapsing at the medium boundary is the cause of refraction,

where the torus photon collapses at the contact point $C$ and the photons stack up to give a wave front spacing of $\lambda$.  Bearing in mind that,

$2\pi a_{\psi}=\lambda$

Consider the line segment $OC$,

$\cfrac{\lambda_i}{sin(\theta_i)}=OC=\cfrac{\lambda_r}{sin(\theta_r)}$

and so,

$\cfrac{sin(\theta_i)}{sin(\theta_r)}=\cfrac{\lambda_i}{\lambda_r}$

And the beam does not change color, $f$ is a constant,

$\cfrac{sin(\theta_i)}{sin(\theta_r)}=\cfrac{f\lambda_i}{f\lambda_r}=\cfrac{v_i}{v_r}$

where $\lambda_i$ measures the thickness of the torus and $\lambda_r$ measures the diameter of the collapsed spherical photon.

Maybe...

But the details are not,

Also consider the line segment $OC$, and not considering the wave fronts of the incident and refracted waves,

$\cfrac{\lambda_i*sin(\theta_i)+a}{cos(\theta_i)}=OC=\cfrac{\lambda_r}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(\theta_i)+\cfrac{a}{\lambda_i}}{cos(\theta_i)}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

If we approximate  $\cfrac{a}{\lambda_i}\approx 1$ (ie. torus with a small hole) and let $90^o-\theta_{i\,c}=\theta_i$,

$\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(90^o-\theta_{i\,c})+1}{cos(90^o-\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}\cfrac{cos(\theta_{i\,c})+1}{sin(\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}$

$\cfrac{\lambda_i}{\lambda_r}=\cfrac{tan(\small{\cfrac{1}{2}}\theta_{i\,c})}{tan(\small{\cfrac{1}{2}\theta_r})}$

Do we have a problem here?  Yes, this is not Snell's Law.