# Understanding some implications of Fermat principle

I am modelling the propagation of light rays from one point to another, with some object in between, in this case a disk to simplify matters, and I compared the travel times of a refracted ray and a hypothetical, unphysical "direct ray" unaffected by refraction. Let's say the ray starts at position $$A = (-0.18, 0.668)$$ (all in meters) and is received at position $$B = (-0.02, -0.668)$$. A disk of radius $$r = 0.1$$ is located in the centre at $$(0,0)$$, with an index of refraction of $$n_d = 1.45$$. A refracted ray that hits position $$B$$ can be determined to intersect the circle first at $$C_1 = (-0.0081, 0.0997)$$ and then leave the circle at $$C_2 = (-0.0209, -0.0978)$$. On the other hand, a straight ray that ignores refraction will intersect the circle at $$E_1 = (-0.1, 0.0)$$ and $$E_2 = (-0.0972, -0.0236)$$. From these values, we can calculate the travel times:

$$t_{\text{refr}} = (d_1 + d_3) / c * n_a + d_2 / c * n_d$$ $$t_{\text{dir}} = (d_1' + d_3') / c * n_a + d_2' / c * n_d$$ with $$d_1 = |\overline{AC_1}|$$, $$d_2 = |\overline{C_1C_2}|$$, $$d_3 = |\overline{C_2B}|$$ and $$d_i'$$ the same but replacing $$C_j$$ with $$E_j$$. $$n_a$$ also denotes the index of refraction of air. The resulting travel time is then $$t_{\text{refr}} = 4.84 \, \text{ns}$$ and $$t_{\text{dir}} = 4.53 \, \text{ns}$$, i.e. the direct ray actually has a shorter travel time because its travel length in the object is far shorter than for the refracted ray. Usually, however, if people talk about Snell's law, it is said that the straight line has a longer travel time (and is therefore not taken) because of its longer way through the object, i.e. refraction causes the travel length through the medium of higher refractive index to shorten, which is the opposite of what is happening here.

Where, exactly, does this discrepancy come from? The literature examples usually include only one medium transition, is that it? Or is it the shape of my object? Is there something I'm missing?

• It would be very helpful if you could provide a sketch showing the geometry and labeling points. It would also be helpful if you could use simpler locations ... they seem to be arbitrary and perhaps unnecessarily "not nice". Perhaps to make your point it has to be the way you describe, but it's hard to visualize. Mar 14, 2022 at 17:14
• You could remove your concern about having only one medium transition and simplify the problem by placing point $B$ be inside the disk. Mar 14, 2022 at 20:21
• @garyp the points are like that because they resemble our set-up, unfortunately... but I added a sketch, so hopefully that clears things up a bit. Mar 15, 2022 at 8:57
• @Andrew that is not our objective, unfortunately, we need to have two medium transitions. we cannot receive a signal by placing an antenna inside an object (without destroying the object in some way) Mar 15, 2022 at 8:57
• If you're doing a simulation, you could still use this as a way to check your code is working correctly. Anyway, it's just a thought. Mar 15, 2022 at 12:36

EDIT: At second glance, from your diagram it doesn't look like the first refraction of your ray obeys Snell's law, so you may have made a mistake in determining the points $$C_1$$ and $$C_2$$.