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Homework Help: Large Uncertainty Became Unsignificant After Calculations?

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculating focal lengths of lenses.

    s = 144.7 ± 5.52 mm
    s' = 86.0 + 0.71 mm

    2. Relevant equations


    Focal Length
    f = (s)(s') / ( s + s' )

    Adding/Subtracting two numbers with uncertainties
    A + B ± root [ (±A)^2 + (±B)^2 ]

    Multiplying/Dividing

    AB ± root [ (±A/A)^2 + (±B/B)^2) ]

    3. The attempt at a solution

    I have:

    s = 144.7 ± 5.52 mm
    s' = 86.0 + 0.71 mm

    and am getting:

    f = 53.94 ± 0.02412 mm

    My question is, does it make sense that the uncertainty that was obviously significant at the start (5.52mm uncertainty in a 144.7mm magnitude) became so miniscule at the end? The way I am calculating this is

    1) Calculating numerator and denominator of focal length equation separately
    -Top = ( s * s' ) and get uncertainty of 0.039mm using the uncertainty multiplication rule
    -Bot = ( s + s' ) and get uncertainty of 5.57mm using the uncertainty addition rule.

    2) Then I divide the top by the bottom apply the uncertainty division rule.

    Am I doing everything correctly?
     
  2. jcsd
  3. Jan 15, 2012 #2

    Simon Bridge

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    It can be smaller in absolute terms but I'd still expect it to be of the order of 1%.
    You need to check the calculation ...
     
  4. Jan 15, 2012 #3

    Simon Bridge

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    You have a relation of the form [tex]\frac{1}{z} = \frac{1}{x} + \frac{1}{y} \Rightarrow z = \frac{xy}{x+y}[/tex]... and you want to find [itex]z \pm \sigma_z[/itex] given [itex]x \pm \sigma_x[/itex] and [itex]y \pm \sigma_y[/itex]

    rule: proportional errors do not change with inverting the fractions

    hence[tex]\frac{\sigma_{1/x}}{1/x} = \frac{\sigma_x}{x} \Rightarrow \sigma_{1/x} = \frac{\sigma_x}{x^2}[/tex]... and the same for [itex]\sigma_{1/y}[/itex]

    rule: the variance of the sum is the sum of the variances
    [tex]\Rightarrow \sigma_{1/z}^2 = { \sigma_{1/x}^2 + \sigma_{1/y}^2}[/tex]
    rule: proportional errors do not change with inverting the fractions
    [tex] \frac{\sigma_z}{z} = \frac{\sigma_{1/z}}{1/z} \Rightarrow \sigma_z = z^2\sigma_{1/z}[/tex]

    ... you should be able to take it from there.
     
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