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Compound lens separated by a distance

  1. Feb 17, 2017 #1
    1. The problem statement, all variables and given/known data
    Capture.jpg

    2. Relevant equations

    We will call ##s## the distance of the object from the first lens, ##s'## the distance of the image from the first lens and ##s''## the distance of the image from the second lens.

    The Gauss's lens equation:
    $$\frac{1}{s} +\frac{1}{s'}=\frac{1}{f_1}$$

    3. The attempt at a solution

    Using Gauss's lens equation we get:

    $$s'=\frac{sf_1}{s-f_1} $$


    we can use the image from the first lens as the object for the second lens.

    $$\frac{-1}{s'-d}+\frac{1}{s''}=\frac{1}{f_2} $$
    Note - I feel my mistake is here, I think my algebra is correct after this.

    This gets us:

    $$ \frac{1}{s''}=\frac{1}{f_2}+\frac{1}{s'-d}=\frac{s-f_1}{sf_1-sd+f_1d}+\frac{1}{f_2}$$
    The total focal length is given by:

    $$\frac{1}{f}=\frac{1}{s}+\frac{1}{s''}=\frac{1}{s}+\frac{1}{f_2}+\frac{s-f_1}{sf_1-sd+f_1d}$$

    Which doesn't get the desired result.


    Thank you in advance
     
  2. jcsd
  3. Feb 17, 2017 #2

    BvU

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    Gold Member

    Hello Toby,
    Nope. Just fine.
    Are ##s## and ##s'' ## wrt the same zero point ? No.
    And: there is another unknown: the position ##x## of the equivalent lens wrt e.g. lens 1.... So that$$ {1\over f} = {1\over s + x} + {1\over d + s - x} \ ...$$
    Meaning you'll need another equation.

    I made a drawing with f1 = f2 = 5 cm, s= 10 cm and d = 3 cm. It comes out pretty neatly (but the ## {3\over 100}## is a rather small term).
     
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