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Optics - focal lengths

  1. Feb 15, 2007 #1
    1. The problem statement, all variables and given/known data
    A converging lens of focal length f forms a real image of an object. Show that the smallest separation of the object and the image is 4f.

    2. Relevant equations
    I think:
    [tex]\frac{n_1}{s_0} + \frac{n_2}{s_1} = \frac{n_2 - n_1}{R}[/tex]
    small angle approximations.

    3. The attempt at a solution
    I've not gotten too far with this.

    As I understand it (I missed a couple of lectures at the end of last week and the start of this week due to illness, and am working from un-annotated handouts), s1 = f?

    So in that case I assume that I need to rearrange the equation above to give me s0 in terms of f (which I would assume to be 3f as the total separation is s0 + s1).

    I've tried taking this approach, and I can't seem to get rid of R or the n's within the equation, and I'm not sure what substitutions to make (they dont seem obvious from the diagrams - I would assume that they'd be trigonometrical ones).

    Any help with this would be greatly appreciated.
  2. jcsd
  3. Feb 15, 2007 #2


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    Staff: Mentor

    I don't think that's the equation you would use. You just have one lens, so you don't need to worry about different n values. Just start with the Lens Maker's Equation, like from wikipedia (or your text):


    [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

    And write an expression for p+q. Then use differentiation to figure out what the minimum value of p+q could be.
  4. Feb 15, 2007 #3
    Do you mean get to:
    [tex]p + q = \frac{pq}{f}[/tex].

    If so what will I differentiate with respect to? (Or even if not, what will I differentiate WRT?)

    Or do you mean something else?
  5. Feb 15, 2007 #4


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    Staff: Mentor

    Well to be honest, I didn't think much about how to set it up. I just thought about the endpoints. If the object is at infinity, then the image is focused at f. And if the object comes in to f, the image is at infinity, right? So with the two extremes of position, the total of p + q is infinity, and there will be some value of p and q where the sum is minimum.

    By the symmetry of the problem and the motion of the object and image, I'd expect the minimum to be where p = q (as an initial guess, and oh by the way that seems to agree with the answer they want us to derive...). One way to approach it instead of setting up a differentiation would be to assume that as the answer, and then show that if you add an infinitesimal delta in the position of the object closer to or farther from the lens, you get a bigger total for p+q. Can you try it that way?
  6. Feb 15, 2007 #5
    I see what you mean about doing it that way, but I'm not sure I fully understand what you mean and how to implement it.

    I've also got some problems with the terminology in the question. Is the focal length thats mentioned in the question the f that you used in the equation that you gave me?
  7. Feb 15, 2007 #6


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    Staff: Mentor

    Yeah, f is the normal abbreviation for focul length. p and q are often used for the object and image distances (although I can never remember which is which), or you can see that the wikipedia.org article uses S1 for the object distance and S2 for the image distance. What does your textbook list for the basic lens (or "Lensmaker") equation?

    Can you derive an equation for the sum of S1 + S2 in terms of only f and S1? If so, can you use differentiation to minimize it with respect to S1?
  8. Feb 15, 2007 #7
    I can't off the top of my head, and the textbook/lecture notes combo doesn't seem to be helping me though.

    However it seems to me that the shortest distance is going to be when s1 = s2, is it fair to say this? Would it be alright to put this in the equation?
  9. Feb 15, 2007 #8


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    Staff: Mentor

    Not without the mathematical proof, either with the differentiation/minimization method or with the small delta proof from your assumption.

    I'll get you started with the differentiation route, but you're going to have to do the work. It may get a bit messy, but be careful with your terms and you should be able to use it to find the minimum.

    [tex]\frac{1}{S_1} = \frac{1}{f} - \frac{1}{S_2} = \frac{S_2 - f}{f S_2}[/tex]

    [tex]S_1 = \frac{f S_2}{S_2 - f}[/tex]

    Do the same thing to get an expression for S2, then add them for an expression for S1 + S2. Then go back and substitute the equation for S2 in terms of S1 into this equation to get an equation for S1 + S2 all in terms of f and S1 only. Differentiate that equation with respect to S1 and set that result equal to zero to solve for the value of S1 that minimizes (or maximizes...) the sum of S1 + S2.

    Let's see some work :biggrin:
  10. Feb 15, 2007 #9
    I see what you mean. I thought you meant derive from other equations. I haven't worked it right through to the end yet, but after a few cockups along the way, I've just worked out that s1=0 or 2f at the minimum.

    Thank you very much for your help. Its much appreciated.
  11. Feb 15, 2007 #10


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    Staff: Mentor

    S1 shouldn't be 0 at the minimum of S1 + S2 (because S2 is infinite then, right?). But your 2f answer matches fine. Good job.
  12. Feb 17, 2007 #11
    That was the reasoning that I applied to the system.

    Thank you.
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