# Homework Help: Optics - rotating ellipse

1. Mar 24, 2010

### fluidistic

1. The problem statement, all variables and given/known data
Assume that the surface S which delimits the 2 mediums is a revolution surface around the z-axis. Light rays start at point $$F_1$$ and all the rays going through the surface reach the plane $$\Sigma$$ in a same amount of time.
Show that S is the result of rotating an ellipse with eccentricity $$\frac{n_2}{n_1}$$.

2. Relevant equations
None given.

3. The attempt at a solution
$$t=\frac{d}{v}$$.
$$t_0=\frac{l_0 n_2}{c}$$, $$t_1=\frac{l_1 n_1}{c}$$.
Hence the time taken for any ray to go from $$F_1$$ to $$\Sigma$$ is $$t=\frac{1}{c} (l_0 n_2 +l_1n_1)=K$$.
Therefore $$\frac{l_0}{n_1}+\frac{l_1}{n_2}=\frac{Kc}{n_1n_2}$$.
I know that the eccentricity is defined as $$e=\sqrt {1-\frac{b^2}{a^2}$$. 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.

2. Mar 24, 2010

### rl.bhat

Re: Optics

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

3. Mar 24, 2010

### fluidistic

Re: Optics

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

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4. Mar 24, 2010

### rl.bhat

Re: Optics

you have already derived the equation
lo*n2 + l1*n1 = K*c.
Two properties of ellipse are
ii) AF1 + AF2 = 2a = constant.
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.

Last edited: Mar 25, 2010
5. Mar 25, 2010

### fluidistic

Re: Optics

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

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