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Experiment to determine the focal distance and magnification of a converging lens

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data

    My task was to determine the focal distance and the magnification of a converging lens. I've done everything as stated in the procedure. The problem that i have is that when i am calculating the magnification in two differents ways i.e M1=b/a and M2=h'/h where a=distance form object to mirror, b=distance from image to mirror, h=height of the objet and h' height of the image, those two magnifications are very divergent. for example in 5 trials if i choose two : M1= 2 while M2=0.95
    M1=1.57 while M2=0.85
    ...
    So how can i explain it? Can you give me a hint or just some error analysis methods (link if possible)? Cause i've never get a course about how to analyse errors.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 13, 2008 #2

    Redbelly98

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    That's strange. Just what are the 4 distances involved in your experiments? (Image distance, object distance, image height, object height)

    Possible errors are either inaccuracies in the measurements, or that the object or image distance is comparable to the lens thickness (the lens should be much thinner than either distance for the calculations to work out simply).
     
  4. Jul 13, 2008 #3
    But do you know any mathematical theory used in error analysis so i can try to interpret my results? Please give me a refence!
     
  5. Jul 13, 2008 #4

    Redbelly98

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    The simplest theory works as follows:

    When you are adding or subtracting two numbers, the error of the result is the sum of the errors.

    When multiplying or dividing two numbers, the percentage error of the result is the sum of the percentage errors.

    There are more complicated ways of calculating errors, but if you're new to error analysis I would use this simple method.

    Example 1, adding or subtracting numbers:

    [tex]
    (5.0 \ \pm \ 0.3) - (2.0 \ \pm \ 0.1) = 3.0 \ \pm \ (0.3 + 0.1)
    = 3.0 \ \pm \ 0.4
    [/tex]

    Check by adding or subtracting errors from terms, in order to get the maximum result possible:
    (5.0 + 0.3) - (2.0 - 0.1)
    = 5.3 - 1.9
    = 3.4
    This "maximal error" calculation gives an answer that is 0.4 higher than the original calculation of 3.0, so the calculated error of 0.4 is correct.

    Example 2, multiplying or dividing:

    [tex]
    \frac{6.0 \pm 0.5}{2.0 \pm 0.1} = 3.0 \ \pm \ ???
    [/tex]

    To find the error, add the percentage or fractional errors:

    [tex]
    \frac{0.5}{6.0}\ 100 \ + \ \frac{0.1}{2.0} \ 100
    [/tex]
    = 8% + 5% = 13%

    So the error is 13% of 3.0:
    0.13 x 3.0 = 0.39 or roughly 0.4

    The final answer is

    [tex]3.0 \pm 0.4[/tex]

    Check the result by adding or subtracting errors to produce the maximum result possible:

    [tex]
    \frac{6.0 + 0.5}{2.0 - 0.1} = \frac{6.5}{1.9} = 3.42...
    [/tex]

    This result is 0.4 higher than 3.0 (to the nearest 0.1), so the calculated error of 0.4 is correct.

    Hope that helps.
     
  6. Jul 13, 2008 #5
    Thank you very much!
     
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