A Question about Fermat's Principle

[neelakash]

Fermat's principle is well-known to everybody. If light travels from point 1 to point 2, it will take the path along which $$\int_{1}^{2}n dl$$ is stationary where $$\ n$$ is refractive index.

When points 1 and 2 are two points on the path of the light ray, there is no problem. However, 1 or 2 are source and image point, we know that the principle works fine. For example, remember the case where we derive law of reflection or refraction from Fermat's principle. Although we do not usually mention about the refractive index in these contexts, actually it is the optical path we are interested in.

What I want to clarify is does the inclusion of point 1 and 2 included in the integral make sense? Because, if point 1 is a source point, there we cannot define $$\ n$$ or and point 2 (image point) we cannot define $$\ n$$. $$\ n(1)$$ or $$\ n(2)$$ is not defined. What is actually done in these case?

-Neel

 

[K^2]

n is a function, n(x,y,z). Since the path can be represented as a curve x(l), y(l), z(l), this function can be parametrized as n(l) for purposes of this integral. If you have just two materials n(l) is a step function with jump discontinuity at boundary. In that case, you can simply introduce point 0 between 1 and 2, and do the two integrals separately, each with its own constant index. That way, solution is a straight line, and then you can do simple one variable variation to find where point 0 is. You should get Snell's Law.

If index of refraction is actually a continuous function, such as the case with atmospheric lensing, you have to do honest variation to find the solution.

 

[neelakash]

Thank you for your answer...However, I am not sure I understood it fully. I think n(l) is undefined at 1 and 2, but how do you know they are discontinuous? Again, let's say, we introduce a point 0 between 1 and 2. Then, the integrals are (1 to 0) and (0 to 2). In both these cases, the boundary point leads to divergence, isn't it?

 

[Born2bwire]

You do not need to define the value at the interface. The interface has well defined left-hand and right-hand limits and you use these limits in the integrals. An integral is the area under a curve. How would your interpretation work if we wanted to take the integral of say the positive pulse of a square wave function?

 

[neelakash]

Then the interval must be a semi-open interval...for precisely at the boundary, the value is undefined.

 

[Born2bwire]

Then the interval must be a semi-open interval...for precisely at the boundary, the value is undefined.

The integral is taken over the limit. This is purely a mathematical question and has nothing do with the current application. I could ask the same question on how do we integrate any discontinous but finite signal like a square wave or saw tooth. The function has well defined left and right hand limits and since the interior is continuous with the exclusion of these points then the resulting integral is well defined as well.

 

[neelakash]

That's right...I think we can integrate over the limit in case where the boundary point contribution can be safely neglected...for example, if we happen to calculate the potential difference between two point charges (say one +ve and other -ve), that will definitely diverge only because of the inclusion of the source/sink points. However, $$\Phi(2)-\Phi(1)$$ will tend to a common limit for different electric field paths once we choose the points just outside the point charges. Isn't it?