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Help with thin lenses and uncertanty

  1. Aug 12, 2009 #1
    Heyy this is my first time on here and I'm only doing first year physics so my understanding might not be up to scratch =/

    I'm doing a practical on thin lenses where we have to do a various amount of measurements and calculations to do with convex lenses. However I'm not very confident with some of the things needed.

    If anyone could help me with how to calculate the uncertainty of 1/X (where X has an uncertainty of delta X) and how I would plot the object distance and image distance to create a straight line (It hints to examine the relationship between these distances and focal length in the thin lens equation)

    Any help would be greatly appreciated and if I confused you with really bad descriptions, sorry :confused:
  2. jcsd
  3. Aug 12, 2009 #2


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    Staff: Mentor

    The brute-force way to calculate the uncertainty is to calculate [itex]1 / (x + \Delta x)[/itex] and [itex]1 / (x - \Delta x)[/itex]. The difference between them is [itex]\Delta (1 / x)[/itex].
  4. Aug 12, 2009 #3
    Does that work with absolute uncertainty too (just read the sheet again and realized it said absolute and not the other one which I've forgotten the name of)
  5. Aug 12, 2009 #4


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    Staff: Mentor

    Correction: for [itex]\Delta(1/x)[/itex], use either

    [tex]\frac{1}{x} - \frac{1}{1 + \Delta x}[/itex]


    [tex]\frac{1}{x - \Delta x} - \frac{1}{x}[/itex]

    The formula I gave before gives you about twice the value you want.

    In general, for anything (call it y) calculated as a formula of x, to find the uncertainty:

    1. Calculate the formula using the "original" value of x, to get an "unmodified" value of y.

    2. Calculate either [itex]x + \Delta x[/itex] or [itex]x - \Delta x[/itex], and substitute it into the formula to get a "modified" value of y.

    3. Find the difference between the two values of y. The absolute value is [itex]\Delta y[/itex].

    This uses the absolute uncertainty. If you have a measurement like x = 10.3 cm +/- 0.2 cm, then in my notation [itex]\Delta x[/itex] is 0.2 cm.

    If you want the relative or percent uncertainty, to a pretty good approximation it's the same for both x and 1/x. Check this if you like, by calculating the absolute uncertainties for an example, then use them to get the relative uncertainties.
  6. Aug 12, 2009 #5
    Do you know the equation connecting object distance, image distance and focal length?If not look it up, compare this to the general equation for a straight line and you should see it.If you get stuck come back.
  7. Aug 12, 2009 #6
    Alright I'll do that, thanks a bunch, my prac is today (Yeah I kinda planned my time badly) so wish me will :uhh:

    Thanks again
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