Finding Snell’s Law as a Max-Min Problem Using Fermat’s Principle

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Image: Rhett Allain

Let me be clear. My main goal is to get to the calculus of variations and use the Euler-Lagrange equation for some cool stuff. However, I think it’s important to start off with some problems that sort of do the same thing but are less complicated.

So, here is the situation. I want to find the path of light as it travels from one medium (say air) into another medium (like water). For this, I will use Fermat’s Principle. This says that light will take a path with that will take the least amount of time. Obviously in a single material, this would also be the shortest distance (which is usually a straight line) since the velocity is constant. However, with materials like air and water the apparent speed of light changes as the following.

Where c is the speed of light and n is the index of refraction. But with two different speeds (apparent speed) of light, the light would need to have a shorter distance in the slower medium. OK, let’s say light is going from point A in medium 1 (with index of refraction n1) to point B in a different material (with an index of refraction n2). Here is a diagram.

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Yes, there’s a bunch of stuff there — but don’t worry, I’m going to go through all of this.

First, there is that point on the interface between the two materials. It’s a distance of x along the interface from the x-value for point A. The light from A travels in a straight line from A to that point, and then from another straight line from the point to B. I can move this interface point back and forth until I find the path with minimum time. Honestly, this is what is so awesome about this problem. It’s really finding a path of least time (calculus of variations), but we can set it up as a max-min problem since there is only one variable that is changing (x).

Once I find the the path lengths in the two materials (s1 and s2), then the total time would be calculated as:

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But of course, s1 and s2 both are functions of x (the distance to the interface point). OK, it’s going to get a little messy — just hang on. I’m going to say the interface is at y = 0. Also, the x position of point A is at x = 0. (why make it harder than it has to be?) That means s1 and s2 can be found as:

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This gives a total time of (notice the common denominator of c):

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Now I just need to take the derivative of this (with respect to x) and set that equal to zero to solve for x. That’s your standard max-min problem. Let me rewrite the time function to be easier to take a derivative.

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Just a reminder — the only variable here is x. Now to take the derivative (using the power rule):

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Now setting this equal to zero and doing a little bit of algebra, I get:

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This looks tricky to solve — so let’s just NOT. How about a trick instead? Let’s look at the diagram for the path of light again — but this time with some angles.

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So, there are two angle — θ1 and θ2. Let’s look at θ1 as part of the right triangle above. I know the value for s1, so I can find an expression for sin(θ1) and sin(θ2).

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Well, what do you know? Those are the same expressions in the equation for x-min above. If I substitute, I get:

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Boom. That’s Snell’s Law. It’s a relationship between the indexes of refraction and the incident and refracted angles of light.

Written by

Physics faculty, science blogger of all things geek. Technical Consultant for CBS MacGyver and MythBusters. Former WIRED blogger.

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