Refraction across two interfaces: Snell's laws for a light ray and the law of sines for a triangle

Question

I got intrigued by a paradoxical thought when thinking of light rays that are refracted twice.
Polite request: please use answers for answering, and comments for commenting or asking.

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A light ray travelling across a medium impinges a transparent slab at a certain angle of incidence. This slab has a finite thickness and is made of a medium having other optical properties. The ray gets refracted into this second medium. The ray is further refracted into the outer medium after traversing that intermediate medium.

The three ray angles with respect to the normal of the interface are $\theta_1$, $\theta_2$, $\theta_3$. The speeds of light in the three media are $c_1, c_2, c_3$. The Snell's laws at the two interfaces result from the application of Fermat's principle:

$ \frac{\sin \theta_1}{c_1} = \frac{\sin \theta_2}{c_2} $ and $ \frac{\sin \theta_2}{c_2} = \frac{\sin \theta_3}{c_3} $.

Trivially this implies:

$ \frac{\sin \theta_1}{c_1} = \frac{\sin \theta_2}{c_2} = \frac{\sin \theta_3}{c_3} \tag{1}\label{1} $

Typically it is observed that, when $c_3=c_1$, the last angle of refraction is the same as the first angle of incidence ($\theta_3 =\theta_1$) as if there was no slab. Only an offset occurs because of the slab in the middle.

So far, plain sailing.

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Turning to trigonometry, the last equality has a striking similarity with the law of sines (Wikipedia). Given a triangle with sides $a,b,c$ and opposite angles $\alpha, \beta, \gamma$, it holds:

$ \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} \tag{2}\label{2} $

Trivially, if I draw a triangle with side lengths $c_1, c_2, c_3$, or a scaled measure of that, the angles $\theta_1, \theta_2, \theta_3$ are set.
This would imply that the optical properties of the media completely determine the angles of the ray path, no matter what. The first angle of incidence cannot be a free parameter either.

This conclusion looks like plain nonsense from a physical viewpoint.

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Where is the fallacy/blunder in the reasoning? Which physical consideration breaks the parallel between the chained Snell's laws \eqref{1} and the law of sines \eqref{2}?

Answer 1

Not every case of the Snell's law equations will correspond to an equivalent triangle representation.

Think about it, it is necessary for the sides and angles of any triangle to satisfy the law of sines, but its nowhere necessary to have a triangle for every numerical combination of thetas and c's that satisfy the law of sines form of equations.

For example, take a random triangle with known sides and angles and write down its law of sine. Now in the equations, keep the length of sides fixed while changing one of the angles slightly, then what will happen is that to maintain the necessary ratios in the equations the other angles will also have to change slightly, but now this new configuration of angles along with the old sides won't form a triangle! Why? Because the new angles won't sum up to 180 degrees anymore.

So, to conclude, for a given set of 3 side lengths there are infinte corresponding angles possible for satisfying the law of sine relations (corresponding to the intuitively infinite cases of refraction), but only that combination where the sum of angles is 180 degrees, will we get a possible triangle.