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Lens equation

  1. Apr 30, 2005 #1
    Hi Peaple :!!)

    How do you determine the minimum distance p+q between the object and image for given focal length given

    1/p + 1/q = 1/f

    this may seem trivial, but i cant figure it out.
    any tips please

    Callisto
     
  2. jcsd
  3. Apr 30, 2005 #2
    Arent p and q distances....? Just add them.
     
  4. Apr 30, 2005 #3
    If your saying given a fixed focal length, find the relationship between p and q, that case

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

    Solve it for either p or q (the result is the same) and graph it. I believe its a logarithmic relation, but I'm not sure.
     
  5. Apr 30, 2005 #4

    Integral

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    What math tools do you have at your disposal?

    Actually the thin lens relationship is a hyperbola.
     
  6. Apr 30, 2005 #5
    I should know that, my brother did a presentation on it a week ago.
     
  7. Apr 30, 2005 #6
    solving 1/p+1/p=1/f for p
    i get
    -qf/(f-q)
    now i substitute this into p+q, then i get -q^2/(f-q).
    now if i use calculus to find the minimum value i differentiate -q^2/(f-q) and let it equal zero so
    d/dq= -q(2f-q)/(f-q)^2 =0 when q = 0 or 2f

    Repeating this process for p then p = 0 or 2f
    so the minimum distance is
    p+q=4f. why is it not zero

    am i completly wrong?any comments.

    Callisto
     
  8. May 1, 2005 #7
    Looks fine to me. You might want to check that 2f is a minimum (not a maximum or point of inflection), to be thorough. Looking at the lens equation itself do you see why the 0 solutions are inadmissible?
     
    Last edited: May 1, 2005
  9. May 1, 2005 #8
    1/0+1/0=0 , 0 doesn't = 1/f
    therefore 0's are excluded from the problem, correct?
    Callisto
     
  10. May 1, 2005 #9
    1/0 + 1/0 is 0? Are you sure?
     
  11. May 1, 2005 #10
    1/0+1/0=1/0, 1/0 doesn't = 1/f
    is this why 0's are excluded?
     
  12. May 1, 2005 #11
    1/0 is undefined, as is 1/0+1/0, and no other arithmetic with these quantities is defined either, so if p or q were 0, you could not rearrange the equation the way that you did (without first defining a new arthmetic). In order for the lens equation to make sense with usual definitions, the assumption p, q not zero must be made.
     
  13. May 1, 2005 #12
    Thanks
    2f must be minimum since if the object p is at infinity then 1/p=0 so q = f for an object at infinity.
     
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