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Calculating percentage uncertainty of diffraction grating? watch

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    If you were given there are 3.5\times10^3 lines per meter, how would you calculate percentage uncertainty
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    I think that the error would be 0.1x10^3 divided by 3.5x10^3 which would give you 0.029 i.e 3%
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    (Original post by Teenie2)
    I think that the error would be 0.1x10^3 divided by 3.5x10^3 which would give you 0.029 i.e 3%
    How did you know it is 0.1\times10^-^3
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    (Original post by metrize)
    If you were given there are 3.5\times10^3 lines per meter, how would you calculate percentage uncertainty
    What exactly are you trying to find the percentage uncertainty of?
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    (Original post by Plagioclase)
    What exactly are you trying to find the percentage uncertainty of?
    Diffraction grating, the value i said is lines per meter
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    (Original post by metrize)
    Diffraction grating, the value i said is lines per meter
    You're trying to find the uncertainty of the number of lines per metre? Surely you've been given more information on the precision of your instruments? You've not given us enough information to work out the percentage uncertainty.
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    (Original post by Plagioclase)
    You're trying to find the uncertainty of the number of lines per metre? Surely you've been given more information on the precision of your instruments? You've not given us enough information to work out the percentage uncertainty.
    That's it.
    The question says: A student has a diffraction grating that is marked 3.5 × 103 lines per m.Calculate the percentage uncertainty in the number of lines per metre suggestedby this marking.
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    (Original post by Plagioclase)
    You're trying to find the uncertainty of the number of lines per metre? Surely you've been given more information on the precision of your instruments? You've not given us enough information to work out the percentage uncertainty.
    Thats all we got
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    0.01 x 10^-3 is the smallest value possible based on the reading given. So I guess that acts as the resolution?
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    I was doing this question earlier, wow are we supposed to know the resolution or whatever with only that as information?
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    The value is 3.5 (it's given to 1dp) - so you know that the smallest increment on the instrument that measured it was 0.1 and thus that is the uncertainty. so it's 0.1*10^3/3.5*10*3 *100.

    Percentage uncertainty = uncertainty/value * 100
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    (Original post by JackSpinner1)
    The value is 3.5 (it's given to 1dp) - so you know that the smallest increment on the instrument that measured it was 0.1 and thus that is the uncertainty. so it's 0.1*10^3/3.5*10*3 *100. If the value was 3.456 then it would be 0.001/3.456 *100 for the same reasons

    Percentage uncertainty = uncertainty/value * 100
    Would it not be 0.05 ~ half the smallest increment? As an error of +0.1 would bring it up to the next value...

    I'm guessing it's 0.05/3.5 = 0.0143 so ±1.43% uncertainty.
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    (Original post by Alexion)
    Would it not be 0.05 ~ half the smallest increment? As an error of +0.1 would bring it up to the next value...

    I'm guessing it's 0.05/3.5 = 0.0143 so ±1.43% uncertainty.
    It can't be 0.05 as you'll be saying that there is less uncertainty in your result than in the data used to calculate it. By it being 0.1 it means that the value could have true ranged between 3.4-3.6, if it was 0.05 then the value could have ranged between 3.45-3.55, however in the question it data is given to 2sf, so that can't be the uncertainty. As it's 3.5, then 0.05 can't be justified as the uncertainty would be more precise than the data itself.

    That being said the general rule of thumb is the smallest increment then half this value, but in this scenario it is not the case
 
 
 
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