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Optics using Bessel's method

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data
    In Bessel's method for finding the focal length f of a lens, an object and a screen are separated by distance L, where L is greater than 4f. It is then possible to place the lens at either of two locations, both between the object and the screen, so that there is an image of the object on the screen, in one case magnified and in the other case reduced. Show that if the distance between those two lens locations is D, then the focal length is given by f=(1/4)(L2-D2)/L.


    2. Relevant equations
    Thin lens equation: [tex]\frac{1}{s}[/tex]+[tex]\frac{1}{s'}[/tex]=[tex]\frac{1}{f}[/tex]
    Focal length: [tex]\frac{1}{f}[/tex]=(([tex]\frac{n}{nair}[/tex])-1)([tex]\frac{1}{r1}[/tex]-[tex]\frac{1}{r2}[/tex])

    3. The attempt at a solution
    I know that you guys are not able to help me out until I show you my attempt, but I do not even know where to start. I guess if you guys could tell me whether or not these equations are relevant, then I could work from there and get back to you with an actual attempt.
     
    Last edited: Apr 15, 2009
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  3. Apr 15, 2009 #2

    Redbelly98

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    The thin lens equation is relevant, the focal length equation isn't.

    But there is another relevant equation. What can you say about s, s', and L?


    Code (Text):



    o<-------- s -------->()<---- s' ---->|

     <---------------- L ---------------->

     

    p.s. to get subscripts in Latex, type eg. n_{air} rather than [noparse]nair[/noparse]
     
  4. Apr 15, 2009 #3

    rl.bhat

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    At one position of the lens s is the object distance and s' is the image distance. Let s<s'.When you shift the lens towards the screen, again you get a sharp image. In this position s and s' interchange. If D is the distance moved by the lens, then s + s' = L and s' - s = D. From these equations find s and s' and then f.
     
  5. Apr 16, 2009 #4
    Okay. Using the thin lens equation and the relationships between s, s', D, and L, I was able to solve for s and s' and plug them into the lens equation. However, doing this alone gets me only close to the right answer. I end up with f= (1/4)(L^2-D^2). I am missing the L term dividing into the (L^2-D^2). I think it has something to do with the relationship of L>4f. But in terms of f, this is L/4 which still isn't the right answer. Where am I going wrong?
     
  6. Apr 16, 2009 #5

    rl.bhat

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    S = (L+D)/2 and s' = (L-D)/2 from these values find f.
     
  7. Apr 16, 2009 #6
    I know that those are the values for s and s' and I used them in the lens equation. What I was saying was that when I plugged them in and solved for f, I did not have the extra L term. I ended up with (1/4)(L^2-D^2) instead of (1/4)(L^2-D^2)/L. I have a feeling the L does not come from either the s or s' equation. I was wondering if it may have come from L>4f.
     
  8. Apr 16, 2009 #7

    Redbelly98

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    Note that your equation is inconsistent in the units, in that it equates a length f with a length2 on the right-hand-side. That's a dead give-away that a mistake has been made.

    Presumably you started with equations that were consistent in the units. Find out where in your derivation the units inconsistency first appears, and that is where the error is.
     
  9. Apr 16, 2009 #8
    The unit error occurs when I combine the denominators in the lens equation. Using the equation (1/s)+(1/s')=(1/f), where s=(1/2)(L-D) and s'=(1/2)(L+D). This gives me a good equation that relates L, D, and f. If I then want to find a common denominator, I multiply the two denominators together to get (1/4)(L^2-D^2). This is where I have length squared equaling a length. For the sake of units, I know I could just divide by L to make the units work out, but I am at a loss as to how the L enters the equation. My initial thought was that it had something to do with the initial condition given: f<(L/4).
     
  10. Apr 16, 2009 #9

    Redbelly98

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    It probably would be helpful to review how to combine fraction expressions and getting a common denominator.

    You realized that we desire a common denominator when combining the expressions

    [tex]\frac{1}{\frac{L-D}{2}}+\frac{1}{\frac{L+D}{2}}
    [/tex]

    Multiply the first fraction by (L+D)/(L+D) to make it

    [tex]\frac{L+D}{\ \frac{(L-D)(L+D)}{2} \ }[/tex]

    or equivalently

    [tex]\frac{2(L+D)}{L^2-D^2}[/tex]

    If you do similarly for the 2nd fraction, you'll then be able to combine the two fractions.

    (And notice that the units are "1/length" at each step so far.)
     
  11. Apr 16, 2009 #10
    Wow. Thanks a lot. I got it now. I'm really embarrassed to have made such a simple mistake in finding the common denominator. Thanks everyone for helping out. I greatly appreciate it.
     
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