Finding the height of a focus point via ray tracing @ Snell's Law

[peeballs]

Homework Statement

I am given a diagram of two layers in the earth that a planar wave travels through: the wave is propagating upward at 4km/s and when it crosses into the next layer, the velocity is reduced to 2km/s. The rays encounter a cylindrical-shaped depression small enough that sin\theta = \theta. The radius of the cylinder is 1km. Find the height of the focus point.

Relevant Equations

eta = u*cos(theta), (u_1)*\(theta_1) = (u_2)*\(theta_2)

So far all I can work out is that the angle of incidence of the outer two and inner two rays is zero degrees, however, I can't work out how to get started on the problem. I feel like I need to use vertical slowness rather than the normal snell's law since I'm working with a dZ rather than a dX, but all of the equations that feature dZ also feature time, so I'm at a bit of a loss here. A nudge to get started would be appreciated.

 

[Merlin3189]

Perhaps you can use the equal time principal?
The waves travel twice as fast in the lower region than in the upper, but they must all arrive at the same time.
So the shorter distance in the lower material must equal twice the longer distance in the upper material.

 

[peeballs]

Perhaps you can use the equal time principal?
The waves travel twice as fast in the lower region than in the upper, but they must all arrive at the same time.
So the shorter distance in the lower material must equal twice the longer distance in the upper material.

In that case, I don't know the distance traveled in the lower layer either so doesn't that just introduce another unknown?

Would the equation dz/ds = (1-p^2/u^2)^1/2 where p is just theta*slowness and use U = 1/2? Although in this case theta still seems like 0 for 3/5 of the incident rays.

 

[Merlin3189]

It doesn't matter how far it has traveled before the wavefront reaches the depression (green). From then on some of the wave (the centre part) is in the slower medium and some (the outer part) still in the fast medium.
For any given part of the wave, the extra distance it travels in the fast medium must equate to a shorter distance in the slower medium after it crosses the boundary.

(Edit: Image changed.)

 

[peeballs]

It doesn't matter how far it has traveled before the wavefront reaches the depression (green). From then on some of the wave (the centre part) is in the slower medium and some (the outer part) still in the fast medium.
For any given part of the wave, the extra distance it travels in the fast medium must equate to a shorter distance in the slower medium after it crosses the boundary.

https://www.physicsforums.com/attachments/257331

I appreciate the diagram! In this case, don't I still need the refracted angle of the two outer rays? If the ray parameter has to stay constant, how can 4*the incident angle = 2*outer refracted angle while the same is true for the middle angle despite the two being different?

 

[Merlin3189]

I've added a bit to the diagram.
I hadn't intended to use Snell's law as such, though one derivation of Snell's law uses the same principle.

I thought you could simply use a single off axis ray and simple geom /trig to calculate the focal distance, using lengths rather than angles. But I forget the math manipulations that are used. You could possibly find the maths if you look up the derivation of the thin lens formula?

Like most optics, there's a bit of fudge, because the rays don't really come to a focus.

If sin(Θ)=Θ as they suggest, then i =2 r by Snells law, but I don't know if that's what was intended nor whether it helps.
===============
Edit: Now I've had a look at some other derivations, it looks as if they do use Snells law rather than the equal time principle. Using the derivation for refraction at spherical surface, it works out more easily than my route.

 

[peeballs]

I've added a bit to the diagram.
I hadn't intended to use Snell's law as such, though one derivation of Snell's law uses the same principle.

I thought you could simply use a single off axis ray and simple geom /trig to calculate the focal distance, using lengths rather than angles. But I forget the math manipulations that are used. You could possibly find the maths if you look up the derivation of the thin lens formula?

Like most optics, there's a bit of fudge, because the rays don't really come to a focus.

If sin(Θ)=Θ as they suggest, then i =2 r by Snells law, but I don't know if that's what was intended nor whether it helps.
===============
Edit: Now I've had a look at some other derivations, it looks as if they do use Snells law rather than the equal time principle. Using the derivation for refraction at spherical surface, it works out more easily than my route.

Would I be looking for an equation containing dZ to plug things into since that's the change in height? I don't know if I have ds or dX, though. I understand Snell's law in general but not the more advanced scenarios, I suppose.

 

[Merlin3189]

I'm not sure what your dZ, dX and ds are.
Also, I don't recognise your equation. Your *\ symbol and commas are confusing. Maybe you can give a reference to it?
Have you got a diagram with things like dZ,dX, ds, eta, theta, theta2, u, u1, u2 , etc marked on?

Looking at the Snells law method, it turns out to give a very simple answer when you use the sin(x)=x approximation. Just take one edge ray, mark in the angles. You can use the sine rule to make an equation to find the focal distance, but you hardly need to, because the answer is so obvious once you have all the angles.

 

[peeballs]

I'm not sure what your dZ, dX and ds are.
Also, I don't recognise your equation. Your *\ symbol and commas are confusing. Maybe you can give a reference to it?
Have you got a diagram with things like dZ,dX, ds, eta, theta, theta2, u, u1, u2 , etc marked on?

Looking at the Snells law method, it turns out to give a very simple answer when you use the sin(x)=x approximation. Just take one edge ray, mark in the angles. You can use the sine rule to make an equation to find the focal distance, but you hardly need to, because the answer is so obvious once you have all the angles.

I drew this diagram, it looks like theta is just 90 for the incident edge angles? I don't know an equation for focal distance, the lensmaker's equation has two distance variables and I don't know them

 

[Merlin3189]

I don't know if there is an equation either. I just think about what happens, draw it and work out equations to fit the diagram.
You have lots of thetas, but it's not clear what they are. I don't think any are 90 deg.

The picture you were given was not to scale. The depression would really be very shallow. It has been exaggerated to make it clear to you. Theta can not be anywhere near 90, because you are told it is near zero! Maybe it could be 5 deg, but definitely not more than 10. It is totally irrelevant what it is, so long as it is very small. If you like you could pick any small number, like 4 degrees, if you like.

Your dotted lines are the rays, I think, but you need to show the centre and the radius of the depression.
The original picture showed the focus much further away than the centre of curvature. I would make it look a bit more like that.