What is the result of rotating an ellipse with eccentricity n2/n1 in optics?

[fluidistic]

Homework Statement

Assume that the surface S which delimits the 2 mediums is a revolution surface around the z-axis. Light rays start at point [tex]F_1[/tex] and all the rays going through the surface reach the plane [tex]\Sigma[/tex] in a same amount of time.
Show that S is the result of rotating an ellipse with eccentricity [tex]\frac{n_2}{n_1}[/tex].

Homework Equations

None given.

The Attempt at a Solution

[tex]t=\frac{d}{v}[/tex].
[tex]t_0=\frac{l_0 n_2}{c}[/tex], [tex]t_1=\frac{l_1 n_1}{c}[/tex].
Hence the time taken for any ray to go from [tex]F_1[/tex] to [tex]\Sigma[/tex] is [tex]t=\frac{1}{c} (l_0 n_2 +l_1n_1)=K[/tex].
Therefore [tex]\frac{l_0}{n_1}+\frac{l_1}{n_2}=\frac{Kc}{n_1n_2}[/tex].
I know that the eccentricity is defined as [tex]e=\sqrt {1-\frac{b^2}{a^2}[/tex]. The problem I'm facing is that I don't have the equation of an ellipse yet.
Have I to find K?
I'll try something.

 

[rl.bhat]

I cannot visualize the problem. Can you attach the diagram?

 

[fluidistic]

I cannot visualize the problem. Can you attach the diagram?

Oops, I forgot to attach it. By the way I've been thinking about it, but I'm still stuck.

 

[rl.bhat]

you have already derived the equation
lo*n2 + l1*n1 = K*c.
Two properties of ellipse are
i) AF2/AD = e (eccentricity)
AF2 = e*AD.
ii) AF1 + AF2 = 2a = constant.
AF1 + e*AD = 2a.
lo + e*l1 = 2a.
Compare this equation with
lo*n2 + l1*n1 = K*c.
In the ellipse e<1.
In the diagram, ray is moving away from the normal after refraction. So n1<n2.
Hence eccentricity cannot be n2/n1 as you have expected in the problem.

 

[fluidistic]

you have already derived the equation
lo*n2 + l1*n1 = K*c.
Two properties of ellipse are
i) AF2/AD = e (eccentricity)
AF2 = e*AD.
ii) AF1 + AF2 = 2a = constant.
AF1 + e*AD = 2a.
lo + e*l1 = 2a.
Compare this equation with
lo*n2 + l1*n1 = K*c.
In the ellipse e<1.
In the diagram, ray is moving away from the normal after refraction. So n1<n2.
Hence eccentricity cannot be n2/n1 as you have expected in the problem.

Oh nice... I wasn't aware of many properties.
So I'm at the point of [tex]l_0+\frac{n_1 l_1}{n_2}=\frac{Kc}{n_2}[/tex].
Now if I can show that [tex]2a=\frac{Kc}{n_2}[/tex] then [tex]e=\frac{n_1}{n_2}[/tex].

I have that [tex]n_2 l_0 +l_1n_2e=2an_2[/tex]. Now if [tex]e=\frac{n_1}{n_2}[/tex] I have that all works and [tex]K=\frac{2an_2}{c}[/tex].
Thanks a lot once again. Problem solved!