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 \(90^o-\theta_{i\,c}=\theta_i\),
\(\cfrac{\lambda_i}{\lambda_r}\cfrac{sin(90^o-\theta_{i\,c})+1}{cos(90^o-\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}\)
\(\cfrac{\lambda_i}{\lambda_r}\cfrac{cos(\theta_{i\,c})+1}{sin(\theta_{i\,c})}=\cfrac{1}{tan(\cfrac{1}{2}\theta_r)}\)
Do we have a problem here? Yes, this is not Snell's Law.
