What is the boundary surface of a collimator?

[kubenun]

Hi everybody,

I’m trying to calculate the shape of a boundary line f(x) between two mediums that collimates rays from a point light source. This requires the rays to hit the boundary line under a certain angle, so I calculated the slope m(φ) of the boundary line for a ray with polar angle φ (φ is less than a known critical angle φ_c). Now in order to get from m(φ) to m(x), I need φ(x). But how do I get it?

Specifying the boundary conditions:
i) Since I will calculate the boundary line by integrating m(x), specifying f(x=0) – i.e. the minimum distance between source and boundary line – determines the constant of integration.
ii) From a physical point of view, I can further specify the radius R of the “lens.” This means x ≤ R and phi(x=R) = φ_c.

So what is φ(x)? Or, in general terms, how can I construct a function f(x) if I know the slope as a function of the polar angle φ (with given f(x=0) and R)?

It sounds easy but I’m stuck nonetheless. Any help is appreciated!

 

[mfb]

A sketch would help to understand how your variables are defined.

 

[kubenun]

Here's a sketch to clear things up:
https://ibin.co/33cEQIZIkc2S.jpg

 

[mfb]

r(φ) could be a useful intermediate function. You get it from the refraction condition and an integral, afterwards there are standard ways to convert that to a function f(x) (not always in an analytical way, however).

 

[kubenun]

Interesting idea! I'm assuming r denotes the distance the ray has traveled when hitting the boundary, right? Getting from r(φ) to f(φ) is easy once I've got r(φ), which is somehow smoothly connected to f(x=0), but it is the same story as before: Snell's law tells me the slopes of the tangents only, not the intercepts.

Geometrically speaking, all tangent lines envelope f, but all I know is the slopes, not the intercepts. If I knew those, too, I could just Legendre transform the tangents to get f(x).

By the way, I don't need an analytic solution to this, since there is no analytic expression for m(φ) either. :)

 

[mfb]

I'm assuming r denotes the distance the ray has traveled when hitting the boundary, right?

Sure.

Snell's law tells me the slopes of the tangents only, not the intercepts.

It gives you a differential equation, but this time as r'(φ) = r f(φ) not as something much more complicated. Easy to solve numerically.

If you have to use numerical methods only, where would be the problem with calculating f(x) in small steps of x, calculating f'(x) at every point?