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Homework Help: Optics - rotating ellipse

  1. Mar 24, 2010 #1

    fluidistic

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    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 [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].


    2. Relevant equations
    None given.


    3. 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.
     
  2. jcsd
  3. Mar 24, 2010 #2

    rl.bhat

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    Re: Optics

    I cannot visualize the problem. Can you attach the diagram?
     
  4. Mar 24, 2010 #3

    fluidistic

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    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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  5. Mar 24, 2010 #4

    rl.bhat

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    Re: Optics

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

    fluidistic

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    Re: Optics

    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!
     
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