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4

Disclaimer: I've never done the particular class of measurements you ask about, but I have done other low raw-rate, precision measurements (neutrino mixing and weak form-factors). The focus of experimental work for low count rates is multi-pronged: Maximize the quantity of data. For counting experiments the raw fractional statistical uncertainty goes by ...

0

The problem you encounter is that you are applying the rule that is applicable to "linear addition" of uncertainty, to non-linear applications (multiplication). Valid application .. $(4 \frac{+}{-}2)$ + $(8 \frac{+}{-}3) = 12 \frac{+}{-}5$ (7 -> 17 range) Invalid application $(4 \frac{+}{-}2)$ x $(8 \frac{+}{-}3) = 32 \frac{+}{-}?$ To obtain the correct ...

2

The general rule to determine the number of significant figures comes from the error propagation formula applied to the rounding errors : $\Delta F=\sqrt{(\frac{\partial F}{\partial a}\Delta a)^2+(\frac{\partial F}{\partial m}\Delta m)^2}$ Writing m = 2.86 means that 2.86 - 0.005 < m < 2.86 + 0.005 Writing a = 12.893 means that 12.893 - 0.0005 < a ...

0

It depends on what error you want to quantify. You can take several images of the same beam at different times (frames of a movie), then for every pixel you find the time-average and standard deviation. This will give the time average and uncertainty, related to the stability of the laser intensity.

2

Since it is a product of two quantities, the answer should have significant figures equal to that number which had the least number of significant figures of the two. Example: Say I have an object of mass $2.86\:\mathrm{kg}$, accelerated to $12.893\:\mathrm{m\: s}^{-2}$. Here the force on that object is equal to $$F=ma.$$ Here \begin{align} m & = ...

5

As a simple rule you use the number of significant figures of the least precise value. So in the example you would quote the force two the significant figures, because that is the precision with which you know the mass. You can't justify any more significant figures than that because of the implied uncertainty in the mass. As to why this is the case, ...

0

You do not have enough information. If you could relate the intensity in the picture with the number of photons detected by the CCD, you could use square root of that number. So say 49 in your graph correspond to 49 photons. Then the error bar on that point is 7. But if the same intensity correspond to 4900, the square root is 70, and your error bar is 0.7

1

ANSWER 1 You're asking about error propagation. In your case: $\sigma_{M+N} = \sqrt{\sigma_M^2 + \sigma_N^2}$ $3\dfrac{\sigma_A}{A} = \sqrt{2(\dfrac{\sigma_B}{b})^2 + (\dfrac{\sigma_{M+N}}{M+N})^2}$ The wrinkle in my calculation is that I assume that the errors in B, M and N are independent. Since M and N are functions of B that isn't strictly true. (I ...

1

Total time measurement: $25s \pm 0.2s$ Error in one oscillation: $\frac{25s \pm 0.2s}{20} = 12.5s \pm 0.01s$ Hence Percentage error: $\frac{0.01s}{12.5s} \approx 8 \cdot 10^{-4}$ Depending on how exactly your clock works you could argue that all the errors are halved, but the above is the safe bound.

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