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Image Distance of a concave mirror (need help with algebra!)

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A concave mirror with a focal length of 34.4 cm produces an image whose distance from the mirror is one-third the object distance. Calculate the object distance.


    2. Relevant equations
    (1/f)=(1/do) + (1/di)


    3. The attempt at a solution
    I know that i have to use the mirror equation (posted above). I substitute 1/3(do) in for Di, because the problems states that "... produces an image whose distance from the mirror is one-third the object distance" so my equation now looks like this: (1/f)=(1/do) + [1/(1/3)do].
    My problem, sad as it seems, comes during the algebra work. I can't for the life of me get the do terms by themselves to solve. Could someone please help refresh my memory? Basically what i've been doing is adding the do terms to get (1 and 1/3)do = .344m and then divide 1 by the answer that i get solving for do. Thanks for any help you can give!
     
  2. jcsd
  3. Apr 21, 2009 #2

    LowlyPion

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    Doesn't it work out from your equation that 4/3*f = do ?
     
  4. Apr 21, 2009 #3
    Yes, that's how it works out. But solving for do, i get a wrong answer. Perhaps i have my equation set up wrong?
     
  5. Apr 21, 2009 #4

    LowlyPion

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    To be clear, if you arrived at
    this is incorrect. It should be 3/4*do = .344
     
  6. Apr 21, 2009 #5
    (1/f)=(1/do)+(1/di)
    (1/f)=(1/do)+(1/((1/3)do))
    (1/f)=(1/do)+(1/(do/3))
    (1/f)=(1/do)+(3/do)
    (1/f)=(4/do)
    do=4f

    remember:
    is equal to 1/(do/3)=3/do
     
  7. Apr 22, 2009 #6
    Thanks v_bachtiar, that helped a lot!
     
  8. Apr 22, 2009 #7

    LowlyPion

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    Yikes. You're right. I took it as 1/(3*do) which is wrong. Good catch.
     
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