# Equation for the boundary of rays turning to plane waves

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1. Nov 4, 2016

### Cocoleia

1. The problem statement, all variables and given/known data
I am given the following figure:

These are converging rays that appear to be going to a point F convert to a plane wave upon hitting the boundary between n2 and n1, and I am asked to find the equation for the boundary between n1 and n2 that perfectly accomplishes this conversion to a plane wave. Assume the distance between the vertex of the boundary and the point F is a known quantity.
2. Relevant equations

3. The attempt at a solution
I believe the answer is 1/f = 1/n1 + 1/n2
But I am not sure how to get to this answer. Could someone please explain the steps and help me understand the problem? Thanks.

2. Nov 4, 2016

### Charles Link

I believe that the material $n_2$ needs to be higher index than $n_1$ and the necessary shape (the boundary between $n_1$ and $n_2$) will be an ellipsoid. Showing this would take some calculus and application of Snell's law at the boundary. Derivations such as this with ellipsoids (in this case a 2-D ellipse would suffice), are non-trivial and require considerable effort. (The lensmaker's equation 1/f=1/b+1/m will not produce the necessary result.) In this case you need to take a ray that is pointed at the far focus and show with proper choice of "a" and "b" for the ellipse (for a given n1 and n2) that the ray will necessarily come out pointing horizontal.

3. Nov 5, 2016

### Cocoleia

Can you give me an example or get me started on how to do the derivations and Snell's law?

4. Nov 5, 2016

### Charles Link

I can try, but this one will take some effort. Begin with $x^2/a^2+y^2/b^2=1$ (an ellipse) and the focii are at (-c,0) and (c,0) where $a^2-b^2=c^2$. You write the equation for an arbitrary ray with slope m that is aimed to go through (c,0). The equation of that line is $m=(y-0)/(x-c)$. You need to find the point where this line intersects the ellipse. Next, you need to apply Snell's law at that point. In order to do that you need to find the slope of the ellipsoid $m_p$ at that point by evaluating dy/dx at that point. The slope of the perpendicular at that point is $m_{perp}=-1/m_p$. Snell's law is not real easy to apply, but you use $n_2 sin(\theta_i)=n_1 sin(\theta_r)$ where $\theta_i$ is the angle between the line of slope $m$ and the perpendicular of slope $m_{perp}$. $\$ $\theta_r$ is the angle between the exiting ray (which must have slope zero) and the perpendicular. With a little effort, using $m=tan(\phi)$ etc. for each of the angles (measured from the horizontal direction) and their slopes, you should be able to determine what $a$ and $b$ need to be in terms of $n_1$ and $n_2$ in order for the ray to emerge in the horizontal direction. It will take a lot of algebra, etc., but it is quite workable.

5. Nov 5, 2016

### Cocoleia

The algebra is getting me confused, I can't get an intersection point because my equation is too hard to simplify. If I use wolfram alpha it gives me that x=+a and x=-a. I also don't understand what you mean when you say find the ellipsoid mp

6. Nov 5, 2016

### Charles Link

$y=mx-mc$ intersects the ellipse at two points. Substituting into the equation for the ellipse, it is a simple quadratic expression that you can solve for the point(s) of intersection.

7. Nov 5, 2016

### Cocoleia

Where am I going wrong?

At this point I stopped because I figure I must have messed up somewhere before this

8. Nov 5, 2016

### Charles Link

I think you correctly applied the quadratic formula. These problems simply are non-trivial. I had previously ( a number of years ago) worked a similar problem where an ellipsoid (with higher index inside and air outside) is used as a focussing element and incident parallel rays converge at the second focus. The algebra is non-trivial, and you need to get every "+" and "-" sign correct in order to get the correct result. editing... because of the azimuthal symmetry it is sufficient to work the problem in 2 dimensions... Even in the simplified 2-D version, it is quite difficult.

Last edited: Nov 5, 2016
9. Nov 5, 2016

### Cocoleia

What would the 2D version be?

10. Nov 5, 2016

### Charles Link

This is the 2-D version. The rays are assumed to travel in a plane (the refracted ray and incident ray define a plane which also contains the axis of the ellipsoid) so there is no need to introduce a 3-D parametrized line or a 3-D ellipsoid. You simply use an ellipse in 2-D to prove it and it will apply to 3 dimensions.

11. Nov 5, 2016

### Cocoleia

Let's say that at some point I am able to get the point(s) where it intersects. What is the ellipsoid mp?

12. Nov 5, 2016

### Charles Link

What is called "implicit differentiation" works the best: $2x \, dx/a^2+2y \, dy/b^2=0$. Compute $dy/dx$. ...I wrote that poorly. It should say find the slope $m_p$ of the ellipsoid at the point of intersection.

13. Nov 5, 2016

### Cocoleia

and how does the intersection point come into this equation, or does it not?

14. Nov 5, 2016

### Charles Link

Please see my edited post #12.

15. Nov 5, 2016

### Cocoleia

This is the best I could do to simplify it:

Can I continue and try to use this or will I have to simplify it more for it to be possible to solve?

16. Nov 5, 2016

### Charles Link

$c^2=a^2-b^2$ but I would wait before you try to simplify. Just an input on the degree of difficulty of the algebra: Even though my algebra is quite good, I seem to recall being on the other ellipsoid problem for approximately one week before I managed to get the algebra to work out correctly. $\\$ I can give you one shortcut that will get you the relation between the $a$ and $b$ and $n_1$ and $n_2$ without requiring nearly as much work: If you choose the point of intersection at the top of the ellipse at $(0,b)$, the slope $m_p =0$ (by inspection, without any calculus), and you can readily apply Snell's law for the incident angle that aims at the second focus. The refracted ray comes out horizontal and the incident ray is at the critical angle for the $n_1$ and $n_2$. You can write $n_2sin(\theta_{critical})=n_1 sin(90)$ so that $(n_2/n_1)sin(\theta_{critical})=1$ and also $tan(\theta_{critical})=c/b$ (using the geometry of the ellipse) with $c=\sqrt{a^2-b^2}$. (Solve the first equation for $cos(\theta_{critical})=\sqrt{1-sin^2(\theta_{critical})}$ and then plug in $tan(\theta_{critical})=sin(\theta_{critical})/cos(\theta_{critical})$.) This shortcut will tell you the relation between $a$ and $b$ and $n_1$ and $n_2$ without a tremendous amount of algebra. A little algebra gives: (I think I computed it correctly) $\\$ $a/b=\sqrt{n_2^2/(n_2^2-n_1^2)}$. $\\$ $\\$ $\\$ editing... Perhaps they are not asking you to prove it=if all they want is the answer, the ellipsoid of $x^2/a^2+(y^2+z^2)/b^2=1$ is the answer (using the x-axis for the optic axis) with $c^2=a^2-b^2$ along with the other relation of $a/b$ that we just determined. (If $c$ is given, along with $n_1$ and $n_2$, we can solve the two equations for $a$ and $b$.) $\\$ $\\$ (Incidentally, for the problem of using an ellipsoid as a focusing element for parallel rays, a similar shortcut can be used for the ray that is incident at the very top of the ellipse, that gets refracted to the second focus along with all of the other parallel rays. I happened to spot this shortcut a few years ago=many years after having worked the problem in its full algebraic detail.)

Last edited: Nov 5, 2016
17. Nov 5, 2016

### Staff: Mentor

I'm not about to attempt this problem myself, but I can imagine that starting with a polar form for the ellipse might be beneficial. In particular,

$r(θ) = \frac{p}{1 + e~cos(θ)}$

where $p$ is a constant (also known as the "parameter" in geometry or the "semi-latus rectum" if you're studying orbits), and $e$ is the eccentricity of the ellipse.

The angle θ is measured from the positive x-axis, and the focus of interest is the origin. The immediate benefit is that the radius vector will lie along the ray that intersects the ellipse. So no hunting for the intersection. All the angle magic occurs at the end of the radius vector.

18. Nov 6, 2016

### Charles Link

My best answer is contained in the shortcut method of post #16, but I did work this one in another manner by considering the curvature of the ellipse near $(-a,0)$ and the effective radius near that point is $R_{eff}=b^2/a$. Application of the formula $1/f=(n-1)(1/R_1+1/R_2)$ is a little tricky because you need to recognize that $n$ gets replaced by $n_1/n_2$ and that $R_1=R_{eff}$ and $R_2=+ \infty$. Then you can use $f=-(c+a)$ along with $c^2=a^2-b^2$. The results agree with the shortcut of post #16, but again this is not a proof, but simply another way of arriving at the correct answer. (Note: I edited this last post slightly=I now have the signs correct on $f$ and $R_1$ etc.) ... $\\$ Additional editing and explanation: To make the thin lens formula applicable, a second boundary can be made on the right side where a plane interface is made with $n_2$ to the far right. The lens of material $n_1$ will have the parallel horizontal rays going from right to left pass through the plane (vertical) interface on the right without any change in direction.

Last edited: Nov 6, 2016
19. Nov 6, 2016

### Charles Link

@Cocoleia Please read my additional edited post #18. For this one, I think it is sufficient to leave the complete algebraic details to the mathematicians and work to the result with either the shortcut of post #16 or that of #18.

20. Nov 6, 2016

### Cocoleia

Ok. I will try to work on this. Thanks a lot!

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