Calculating percentage uncertainty of diffraction grating?

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metrize
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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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Teenie2
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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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metrize
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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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Plagioclase
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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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metrize
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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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Plagioclase
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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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marmbite
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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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metrize
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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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username2488725
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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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GarlicBread01
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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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JackSpinner1
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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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Alexion
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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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JackSpinner1
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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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Slambaa
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it is 0.05, half the smallest increment. When measuring the metre, there is uncertainty on both sides of the ruler so its 0.05 on each side hence 0.1 in total
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JNGLZ
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The reason the absolute error is 0.1*10^3 is because you are using a ruler to measure out 1m to then count how many lines lie in that range, and in doing that measurement of 1m you are actually taking to readings, a reading of where the zero mark is and a reading of where the 1m mark is so you’re absolute error of each reading is 0.5*10^3 lines so overall absolute error is 1*10^3
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Cholmes
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The value you are given is 3.5x10^-3 so in this value the uncertainty is the last decimal placeGreater than 3.45 and less than 3.55 which gives an uncertainty range of 0.1x10^-3Percentage uncertainty is (1x10^-3/3.5x10^-3) x100 = 2.9%It isn’t anything to do with measuring using a ruler to the nearest mmAnd the uncertainty isn’t /- 0.05 like it would be with a reading from a top pan balance for example It is the uncertainty in a given value which is the last decimal place of the quantity quoted0.1
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