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Homework Help: Gaussian Optics / Paraxial Approximation

  1. Nov 28, 2004 #1
    Derive

    [tex] \frac{n_1}{s_o} + \frac{n_2}{s_i} = \frac{n_2-n_1}{R} [/tex]

    for Gaussian optics from the following equation

    [tex] \frac{n_1}{l_o} + \frac{n_2}{l_i} = \frac{1}{R} \left( \frac{n_2s_i}{l_i} - \frac{n_1 s_o}{l_o} \right)[/tex]

    by approximating

    [tex] l_o = \sqrt{R^2 + \left( s_o + R \right) ^2 - 2R\left( s_o + R \right) \cos \phi} [/tex]

    and

    [tex] l_i = \sqrt{R^2 + \left( s_i - R \right) ^2 - 2R\left( s_i - R \right) \cos \phi} [/tex]

    with the aid of

    [tex] \cos \phi \approx 1 [/tex]

    [tex] \hline [/tex]

    Here is what I have:

    [tex] l_o \approx \sqrt{R^2 + \left( s_o + R \right) ^2 - 2R\left( s_o + R \right) } = \sqrt{R^2 + s_o ^2 + R^2 + 2s_o R - 2s_o R - 2R^2 } = s_o [/tex]

    and

    [tex] l_i \approx \sqrt{R^2 + \left( s_i - R \right) ^2 - 2R\left( s_i - R \right) } = \sqrt{R^2 + s_i ^2 + R^2 -2s_i R - 2Rs_i + 2R^2} = 2R-s_i [/tex]

    But, I expected to find [tex] l_i \approx s_i [/tex] .... what I have doesn't seem to work out. I can't find where I made a mistake, though.

    Thanks
     
    Last edited: Nov 28, 2004
  2. jcsd
  3. Nov 28, 2004 #2
    Well keep in mind that the li and l0 are in the denominator...

    You need to use the fact that [tex]\frac{1}{2R(1-\frac{s_i}{2R})} = \frac{1}{2R}(1 + \frac{s_i}{2R} +...)[/tex]

    Because of the paraxial approximation you have that si >>> R


    regards
    marlon : ps are you sure about these formulas ??? Shouldn't there be a minus sign in the left hand side of your first given formula...
     
    Last edited: Nov 28, 2004
  4. Nov 29, 2004 #3
    I've double checked my equations. Unfortunately, they're exactly equal to those in my calc book (Calculus: Concepts and Contexts 2nd Ed. by James Stewart - page 632). Well, I did some more work, but couldn't get the desired result. Anyway, here you go:

    If

    [tex] \frac{n_1}{s_o} = \frac{n_1}{l_o} [/tex]

    and

    [tex] \frac{n_2}{s_i} = \frac{n_2}{2R-s_i} = \frac{n_2}{2R\left( 1 - \frac{s_i}{2R} \right)} = \frac{n_2}{2R} \sum _{n=0} ^{\infty} \left( \frac{s_i}{2R} \right)^n \approx \frac{n_2}{2R} [/tex]

    Then

    [tex] \frac{n_1}{s_o} + \frac{n_2}{s_i} = \frac{n_1}{l_o} + \frac{n_2}{2R} = \frac{2Rn_1 + s_o n_2}{2Rs_o} = \frac{1}{R}\left( \frac{Rn_1}{s_o} + \frac{n_2}{2} \right) \neq \frac{n_2 - n_1}{R} [/tex]

    In other words, I'm stuck! :smile:
     
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