Calculating Magnitude Difference: Is it Right?

AI Thread Summary
The calculated magnitude difference of -0.073 using the formula -2.5log(11347/10606) raises questions about its accuracy and uncertainty calculation. A standard error of 100 is mentioned, but it is clarified that this represents a background level rather than an uncertainty. To accurately determine uncertainty, adjustments for background counts are necessary. The standard deviation for total-count measurements should be derived from a Poisson distribution. Finally, applying error propagation rules will yield the correct final result.
ChrisWM
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Homework Statement
We make two observations of the same star on different nights. In the first observation, the integrated flux value is 11,347 counts and in the second observation the integrated flux is 10,606 counts. Assume that the background level(dark & bias)is 100 counts for each observation. Also assume one count per photon(i.e. QE=100% and gain = 1).
a. Compute the difference in the star’s brightness and the uncertainty in that difference.
Relevant Equations
M=-2.5log(f1/f2) SE=𝜎/sqrt(2n-2)
I have found the difference in the magnitude from the counts to be -.073 using -2.5log(11347/10606) but I m unsure if this is right or how to calculate the uncertainty
 
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It looks like the number 100 is a given standard error(uncertainty) for a single measurement. Perhaps you could use that to find the uncertainty for the difference.
 
guv said:
It looks like the number 100 is a given standard error(uncertainty) for a single measurement. Perhaps you could use that to find the uncertainty for the difference.
I think the '100' is the background level, not an uncertainty; it is basically the reading you'd still get with no star in view (e.g. like background level when measuring ionising radiation).

@ChrisWM first needs to correct the given total-count values to allow for this background.

It's not clear where 'SE=𝜎/sqrt(2n-2)' comes from. The standard deviation for each total-count measurement can be found by using the standard deviation for aPoisson distrubution.

Then the final answer can be found using the rules for error propagation.
 
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