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Thin lens formula

  1. Jan 29, 2012 #1
    Hi

    The thin lens equation we know is given by

    [tex]
    \frac{1}{S_O} + \frac{1}{S_I} = (n-1)(\frac{1}{R_1}-\frac{1}{R_2})
    [/tex]

    assuming light is incident from air. The "usual" way to derive the thin lens formula (aka the Gaussian Lens Formula) is to say that when looking at the system when either the image of object is at infinity, then we introduce the focal length by

    [tex]
    \frac{1}{f} = (n-1)(\frac{1}{R_1}-\frac{1}{R_2})
    [/tex]

    From this it is usually stated that then 1/f = 1/SO + 1/SI. This I don't agree with, since the thin lens equation at the top is general, however the second equation stated is when either the object or image is at infinity. From this one can't state 1/f = 1/SO + 1/SI, as done in e.g. Hecht.

    What is the correct argument?

    Best,
    Niles.
     
  2. jcsd
  3. Jan 29, 2012 #2

    Doc Al

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    Staff: Mentor

    But the right hand side of that equation is a constant. If you put parallel rays in, the image distance defines the focal length.
     
  4. Jan 30, 2012 #3
    Yes, that is a proof showing that f = SO = SI. However, it doesn't prove that 1/f = 1/SO + 1/SI, but that is what most authors state that it does.
     
  5. Jan 30, 2012 #4

    Doc Al

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    Sure it proves it. (It's really just the definition of focal length.) 1/f = 1/SI when SO = ∞, which equals the constant expression on the right hand side of the thin lens equation. Thus f is fixed.
     
    Last edited: Jan 30, 2012
  6. Jan 30, 2012 #5
    It proves that 1/f = 1/SO + 1/SI assuming that either SO or SI is at infinity. However, many authors say that 1/f = 1/SO + 1/SI is also valid when SO and SI are not at infinity, in which case we can use it to find the image distance given some arbitrary object distance. This latter case I don't agree with.
     
  7. Jan 30, 2012 #6

    Doc Al

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    No. First you start with the thin lens formula, which it seems you accept. Then you define f as the image distance when SO is at infinity; f is a constant for the lens, and 1/f is equal to the right hand side of your original thin lens equation. Thus you can rewrite the thin lens equation in terms of f.
    If you still don't agree, start by defining f.
     
  8. Jan 30, 2012 #7
    I added something to your quote (the bolded part), which is where we disagree. The focal length is defined as the image distance when the object distance is at infinity, so 1/f is only equal to the LHS of the thin lens equation, when SO is at infinity.
     
    Last edited: Jan 30, 2012
  9. Jan 30, 2012 #8

    sophiecentaur

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    That is just the definition of f, surely. What is the problem with that? f is a very convenient parameter for a lens, which why we specify lenses in terms of focal length. Can you propose a better way?

    Have you tried doing what Doc Al suggested?
     
  10. Jan 30, 2012 #9
    Ahhh, now I see it. Great! Thanks for being patient with me.

    Best wishes to both of you,
    Niles.
     
  11. Jan 30, 2012 #10

    Doc Al

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    Careful about adding something to a quote box, as it makes it look like I said it.
    Let me ask you this: If I give you n, R1, and R2 for a lens, can you calculate the focal length? How would you do that?
     
  12. Jan 30, 2012 #11
    Yeah, sorry about that. I changed it.


    Of course I can. I set SO to be at infinity, and since SI = f in this case (by definition), I just use the RHS of the thin lens equation to find 1/f. But the RHS *is* SO + SI, which is the part I missed. But I see it now.

    Thanks.

    Best wishes,
    Niles.
     
  13. Jan 30, 2012 #12

    Doc Al

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    Staff: Mentor

    Cool. :smile:
     
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