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Find the Focal Length Of a lens

  1. Apr 30, 2013 #1
    1. The problem statement, all variables and given/known data
    When an object is placed 69.0cm from a certain converging lens, it forms a real image. When the object is moved to 39.0cm from the lens, the image moves 19.0cm farther from the lens.

    Find the focal length of this lens.

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


    3. The attempt at a solution
    1/69.0cm + 1/di = 1/f

    1/39.0cm + 1/di + 19cm = 1/f

    1/69.0cm + 1/di = 1/39.0cm + 1/di + 19cm

    which gives di = 20cm

    1/69.0 cm + 1/20.0 cm = 1/f

    f = 15.5??? this is not the right answer though, where am i going wrong?
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2

    TSny

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    Parentheses need to be used here: 1/(di + 19cm)

    Likewise here.

    The algebra is a little tedious, so you will need to be careful soving for di.
     
  4. Apr 30, 2013 #3
    Even with the parenthesis in the equations wouldnt di still be 20, 39-19 is 20 either way

     
  5. Apr 30, 2013 #4

    TSny

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    The equation is

    ##\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}##

    Do you see why you can't simply subtract the 39 and 19?
     
  6. Apr 30, 2013 #5
    not really im having a little trouble understanding this problem

     
  7. Apr 30, 2013 #6

    TSny

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    ##\frac{1}{69} + \frac{1}{d_i} = \frac{1}{39} + \frac{1}{d_i+19}##

    There are sevaral approaches to solving this equation for ##d_i##.

    You might start by rearrainging the equation so that the two terms containing ##d_i## are on the left side and the two fractions without the unknown are on the right.
     
  8. Apr 30, 2013 #7
    okay so i have

    1/di - 1/di + 19 = 1/39 - 1/69

    di + 19/di = 69/39

    39di + 741 = 69di

    741 = 30di

    24.7 = di....is this correct?

     
  9. Apr 30, 2013 #8

    SammyS

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    You continue to ignore the required parentheses !

    Your equation:
    1/di - 1/di + 19 = 1/39 - 1/69​

    should be:
    1/di - 1/(di + 19) = 1/39 - 1/69​

    The next step is wrong.
    So what follows that is pointless.


    What is the common denominator for ##\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}\ ? ##

    What is the smallest common denominator for ##\displaystyle\ \frac{1}{39}-\frac{1}{69}\ ? ##
     
  10. May 1, 2013 #9
    my apologies for ignoring the parenthesis again, i believe the common denominators are

    di2+19di and 897

    so then it would be 19/di2+19di = 36/897

    which would give you di equal .0401337793

    1/.0401337793 = 24.91666667

     
  11. May 1, 2013 #10

    TSny

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    Can you combine ##\displaystyle\ \frac{1}{d_i}-\frac{1}{d_i+19}## to make one fraction?
     
  12. May 1, 2013 #11
    i believe i did that in the post above it gives you 19/di2 + 19di. is this wrong?

     
  13. May 1, 2013 #12

    TSny

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    Oh, I didn't see it! Yes, that's right if you include the parentheses!:

    19/(di2 + 19di)

    [EDIT:The 36/897 is incorrect.]
     
  14. May 1, 2013 #13
    okay how is the 36/897 wrong? The first number that 39 and 69 go into evenly is 897. 13*23 and 69*13 is 897. which would igve you 23/897 and 13/897

     
  15. May 1, 2013 #14

    TSny

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    Yes, but remember that you are subtracting the fractions.
     
  16. May 1, 2013 #15
    Ohh okay so it would be 19/(di2 + 19di) = 10/897



     
  17. May 1, 2013 #16

    TSny

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    Yes, that's correct.
     
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