Trouble understanding Fermat's principle of least time

[FeynmanFtw]

I've been reading about Fermats last principle in Feynmans lectures on physics, and he sort of goes through the derivation of Snell's law by considering a simple refraction and applying a bit of trigonometry (see photo).

What I'm having trouble with is this. He states that if you take a two paths separated by the distance XC when hitting the 2nd medium, and introduce the assumption that XC is very small, then AC - EC ≈ AX. But what if the distance of XC is not small enough to make this assumption valid? Surely then the trigonometry would not work and you would end up with a different equation for Snell's law?

 

[Orodruin]

It is quite easy to show that the path described by Snell's law actually gives the global minimum in this case. Simply write the total optical path as a function of where on the surface the light hits and put the derivative of this equal to zero. You really do not need to consider only small perturbations (although that will tell you that it is a possible local minimum, which is enough for light to take that path).

 

[Philip Wood]

But what if the distance of XC is not small enough to make this assumption valid?

Then you make it smaller! You have control over XC. Strictly the relationship between sines only emerges as XC tends to zero. It is a limiting process.
But it's well worth doing what Orodruin recommended and find the minimum of the total optical path (or total time of transit) in the conventional manner. Then compare this with Feynman's approach. I know which impressed me more!

 

[FeynmanFtw]

Honestly, I've had a go at what you guys suggested but it seems my maths is a tad rusty (or it's just one of "those" days). I've drawn a diagram similar to the one I posted but without the infinitesimally shorter path; I've simply sketched out one path and put in the vector locations as well as the angles where the angles required for Snell's law as well. I did write out the equation for the total time of the optical path in terms of distance, and assumed that the light has two different speeds in each medium (above the boundary, it was ν₁, and below the boundary it was ν₂, with the latter being a factor n slower than the former (basically n.ν₁=ν₂). Now I also used the distances involved in each trajectory, as this was a nice starting point, and I hoped that they at some point would disappear (either replaced with angles or left as a ratio of the velocities of light in each medium), leaving me with the required equation. But I get stuck!

Honestly I'd really appreciate it if someone could start me off in the right direction so I can have another crack at this. Maybe tomorrow I'll have better luck.

 

[Orodruin]

Honestly, I've had a go at what you guys suggested but it seems my maths is a tad rusty (or it's just one of "those" days). I've drawn a diagram similar to the one I posted but without the infinitesimally shorter path; I've simply sketched out one path and put in the vector locations as well as the angles where the angles required for Snell's law as well. I did write out the equation for the total time of the optical path in terms of distance, and assumed that the light has two different speeds in each medium (above the boundary, it was ν₁, and below the boundary it was ν₂, with the latter being a factor n slower than the former (basically n.ν₁=ν₂). Now I also used the distances involved in each trajectory, as this was a nice starting point, and I hoped that they at some point would disappear (either replaced with angles or left as a ratio of the velocities of light in each medium), leaving me with the required equation. But I get stuck!

Honestly I'd really appreciate it if someone could start me off in the right direction so I can have another crack at this. Maybe tomorrow I'll have better luck.

Why don't you show us exactly what you got by doing this? It will be much easier for us to identify your mistake.

 

[Philip Wood]

You might like to start like this… To make things simple, consider only light that travels in the plane of the paper (z=0, say). Then the boundary between the media can be treated as a line. Let it be the x-axis. Consider a route consisting of two straight lines that takes us from the point (0, a) to the point (b, -a) via the point (x, 0). Use Pythagoras's theorem to express the journey time in terms of [itex]a, b, x, v_1, v_2[/itex]. Determine the value for x which makes the time a minimum. You should find that it's easy to express the equation for this x in terms of angles.

 

[FeynmanFtw]

Ah, got it guys. Thanks for the help everyone, I would have shown you my working had I failed again but I started from scratch again this morning and got it :). It's amazing how we develop tunnel vision when solving things sometimes and stepping away from the problem for a moment can help.

I have one issue though, that I would like clarified. I never assumed (like some other people do) that the first medium is air and the latter is glass. I just assumed that both mediums have different speeds of light (labelled ν₁ and v₂ respectively) and then assumed that one is simply changed by a factor k when entering a different medium, i.e. ν₁ = kv₂. I still got the final relationship regarding the angles and velocities of light in different mediums, but my FINAL formula for Snell's Law is

Sinθ₁ = kSinθ₂

My question now is, how would one infer anything about the refractive index, assuming we had no idea about it?

 

[Orodruin]

My question now is, how would one infer anything about the refractive index, assuming we had no idea about it?

You cannot, not unless you have a reference of which you know the refractive index. Snell's law with arbitrary media reads
$$
n_1 \sin \theta_1 = n_2 \sin \theta_2
$$
so your ##k## is simply the ratio between the refractive indices.