Phase Shift, No Normal

There's nothing wrong with Snell's Law,

This diagram is constructed by touching a particle of radius $\lambda_i$ on a medium boundary, with a ray $AO$ through its center that intersect the boundary at point $O$.  The particle shrinks to size $\lambda_r$ at the point of contact $C$.  The line through its center to point $O$ is the new wave front.

Consider $CO$,

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

$\cfrac{\lambda_i }{\lambda_r}=\cfrac{1}{tan(\theta_i)tan(\theta_r)}$

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

or

$\cfrac{\lambda_i }{\lambda_r}=\cfrac{cos(\theta_i-\theta_r)+cos(\theta_i-\theta_r)}{cos(\theta_i-\theta_r)-cos(\theta_i-\theta_r)}$

The right diagram should be,

What happens on one wave front happens on the next and the two green triangles on the right shares the same hypotenuse but are displaced.

$OC=O'C'=\cfrac{2\lambda_i}{sin(\theta_i)}=\cfrac{2\lambda_r}{sin(\theta_r)}$

and so,

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

The particle collapse to a smaller radius at the contact point.  The wave fronts are disjointed and so are the incident and refracted rays.  The is no normal that meets both incident and refracted rays, as the wave fronts are not continuous.

A discontinuous wave front suggests a phase change with refraction.

Centers of the particles are at $a$ and $a'$

$oa=\cfrac{\lambda_i}{cos(\theta_i)}$

so,

$of=\cfrac{oa-\lambda_i}{sin(\theta_i)}$

$cf=oc-of=\lambda_i tan(\theta_i)-\cfrac{\cfrac{\lambda_i}{cos(\theta_i)}-\lambda_i}{sin(\theta_i)}$

$cf=\lambda_i \left(tan(\theta_i)-\cfrac{\cfrac{1}{cos(\theta_i)}-1}{sin(\theta_i)}\right)=\lambda_i \left[tan(\theta_i)-{2}{cosec(2\theta_i)}+{cosec(\theta_i)}\right]$

Similarly the distance from the contact point $c$ to where the wave front meets the boundary at $f'$ is,

$cf'=\lambda_r tan(\theta_r)-\cfrac{\cfrac{\lambda_r}{cos(\theta_r)}-\lambda_r}{sin(\theta_r)}$

$cf'=\lambda_r \left(tan(\theta_r)-\cfrac{\cfrac{1}{cos(\theta_r)}-1}{sin(\theta_r)}\right)=\lambda_r \left[tan(\theta_r)-{2}{cosec(2\theta_r)}+{cosec(\theta_r)}\right]$

and since between two consecutive wave fronts along the medium boundary is a phase difference of $2\pi$, so the phase difference between the incident and refracted wave is,

$\angle P=\cfrac{cf+cf'}{OR}*2\pi=\cfrac{cf+cf' }{2\lambda_i}*{2\pi}{sin(\theta_i)}=\cfrac{cf+cf' }{\lambda_i}{sin(\theta_i)}*\pi$

or

$\angle P=\cfrac{cf+cf' }{\lambda_r}{sin(\theta_r)}*\pi$

And we have the phase difference after refraction!