So at the boundary, what's really happening is: the frequency of light must remain the same (wave crests/troughs cannot exit any faster than they enter) and therefore the different speed of light changes the wavelength of the light.
Of a certain intensity of light, a proportion $T$ transmits into the new medium and a proportion $R$ exits. The barrier does not absorb energy, so $T + R = 1.$ It can only absorb momentum perpendicular to its surface, so there must be conservation of momentum in the other direction. Now for a light wave, the momentum is proportional to the inverse of the wavelength $\lambda$, so if $\theta$ is measured from the line perpendicular to the surface, momentum balance in this region means:$$\frac{1}{\lambda_1} ~ \sin \theta_1 = \frac{T}{\lambda_1}~\sin\theta_1 + \frac{R}{\lambda_2}~\sin\theta_2$$Conservation of momentum and energy therefore together give us Snell's law, since you can work out that if the frequency stays the same then $\lambda_1 / \lambda_2 = n_2 / n_1.$
You need another equation to successfully pin down the exact value of $R$ and see how it varies with respect to $\theta_1.$ This other equation is harder for me to explain, and it comes from the continuity of fields, which makes it polarization-dependent.
Typically light is a wave made of two fields, an electric and magnetic field, that oscillate perpendicular to each other, and perpendicular to the direction that the wave goes, too. So if light is going "up" then possibly the electric field is oscillating east-west while the magnetic field is oscillating north-south. The energy is contained in the square of the amplitude of these oscillations, and, if the wave is moving at a speed $v$, the magnetic field's amplitude is usually related to the electric field's amplitude by $E / v$. A full writeup of how exactly all this works is available here.
Here's one example. If the electric field points alongside the surface between the two media, then electric field continuity means that $E_i + E_r = E_t$. Defining the transmission amplitude $\tau = E_t / E_i$ and the reflection amplitude $\rho = E_r / E_i$, this means that $1 + \rho = \tau$. In general because it's the same medium, $R = \rho^2$ but $T \ne \tau^2$, so we need another equation. That's given by the magnetic field, $$\frac{1}{v_1} ~ (1 - \rho) ~ \cos\theta_1 = \frac 1 {v_2} ~\tau~\cos\theta_2, $$therefore
$$ (1 - \rho) ~ n_1 ~ \cos\theta_1 = (1 + \rho) ~ n_2 ~ \cos\theta_2$$or once you work it all out, $$R = \left({n_1 \cos\theta_1 - n_2\cos\theta_2 \over n_1 \cos\theta_1 + n_2\cos\theta_2 }\right)^2.$$Again, that's only true for this one polarization, but let's look at this formula.
First off, observe that this is actually really well-defined for $\theta_2 = \pi/2$, which is when we know $\theta_1$ is critical: it limits to 1. For all $\theta_1$ less than this, we get a reflection coefficient R which smoothly increases from 0 to 1 before abruptly having a "kink" at 1 (because Snell's law can no longer be satisfied beneath that point and so it stays $R = 1$). And this means that $T$ goes to 0.
So, as the transmitted ray gets closer and closer to being "alongside the surface", it also gets weaker and weaker in intensity. By the time the transmitted ray lies purely "alongside" the surface it has smoothly transitioned into being nonexistent.
The intuitive answer uses symmetry.
For a prism (or any linear optics) you should be able to reverse the direction of light and get the same result. This means that if a given angle of incidence $\alpha$ results in a certain exit angle $\beta$, then incident angle $\beta$ will result in exit angle $\alpha$.
Now a maximum or minimum in the deviation occurs when a small change doesn't change the outcome. By symmetry this has to happen when $\alpha=\beta$.
Best Answer
The corpuscular model says that light is composed of tiny discrete particles.
It can explain reflection if we assume that the particles are so small that they're very unlikely to collide with each other, or that they don't interact with each other for some other reason.
For example, if the corpuscles in a beam of light take up only one trillionth of the volume of the beam, then in a light beam reflecting back on itself there will be hardly any collisions. This would also explain why two beams of light can pass through each other without interference. (If the theory had held up over time, we'd probably have experiments designed to demonstrate these rare collisions.)
Or we could consistently assume that the corpuscles interact with ordinary matter, but can freely pass through each other.
The wave model can also explain reflection. Both sound waves and waves on the surface of a body of water can reflect off hard surfaces; the reflected waves pass through the incoming waves without being disrupted.