Calculating Quasar Variability Using CCD Camera and Magnitude Comparison

[StillLearningToronto]

Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable.
With your telescope/camera combination, you know that a star with a magnitude of 0 would deliver 1 × 109 photons/second to one pixel, so this allows you to work out the photons/second from the quasar delivered to one pixel.

SO- taking out the numbers i need: My quasar is the 16th magnitude.
The star I am comparing it to has a magnitude of 0.
The 0 mag star gives out 1*10^9 photons/second (1 pixel)

Homework Equations

m2-m1=-2.5log(F2/F1)

The Attempt at a Solution

Im not sure if "flux" and "photons/second" are the same thing, BUT if they are, then:
m2-m1=-2.5log(F2/F1)
16-0=-2.5log(F2/1*10^9photons/second)
F2=3.98*10^2 photons/second.**Im not looking for the answer because I really want to figure this out, I just need to know if I am going in the right direction, thank you in advance ! (:

 

[collinsmark]

Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable.
With your telescope/camera combination, you know that a star with a magnitude of 0 would deliver 1 × 109 photons/second to one pixel, so this allows you to work out the photons/second from the quasar delivered to one pixel.

SO- taking out the numbers i need: My quasar is the 16th magnitude.
The star I am comparing it to has a magnitude of 0.
The 0 mag star gives out 1*10^9 photons/second (1 pixel)

Homework Equations

m2-m1=-2.5log(F2/F1)

The Attempt at a Solution

Im not sure if "flux" and "photons/second" are the same thing, BUT if they are, then:
m2-m1=-2.5log(F2/F1)
16-0=-2.5log(F2/1*10^9photons/second)
F2=3.98*10^2 photons/second.**Im not looking for the answer because I really want to figure this out, I just need to know if I am going in the right direction, thank you in advance ! (:

You're approach looks good to me.

By the way, power "flux" here is power per unit area. And the area in question is the area of a single pixel. So in the end, you're really just comparing powers. Power is energy per unit time.

And another "by the way," this problem seems to have an inherent assumption that the photons from the reference star and the photons from the quasar have approximately equal energies, on average. Or to put it another way, it assumes that the color of the quasar is the same color as your magnitude 0 reference star, as seen from the telescope (including the effects of red-shifts -- most quasars are significantly red-shifted due to expanding space). This might be a perfectly good assumption depending on the star and quasar in question, but I thought I'd mention it.

 

[StillLearningToronto]

You're approach looks good to me.

By the way, power "flux" here is power per unit area. And the area in question is the area of a single pixel. So in the end, you're really just comparing powers. Power is energy per unit time.

And another "by the way," this problem seems to have an inherent assumption that the photons from the reference star and the photons from the quazar have approximately equal energies, on average. Or to put it another way, it assumes that the color of the quasar is the same color as your magnitude 0 reference star, as seen from the telescope (including the effects of red-shifts -- most quasars are significantly red-shifted due to expanding space). This might be a perfectly good assumption depending on the star and quasar in question, but I thought I'd mention it.

Thank you so much ! Its a part of a bigger question I am working on, but i just wanted to make sure i was off to a good start (:

 

[collinsmark]

Thank you so much ! Its a part of a bigger question I am working on, but i just wanted to make sure i was off to a good start (:

Yes, you are off to a good start.

 

[collinsmark]

By the way, earlier in Post #2 I said,

"power 'flux' here is power per unit area."​

But I meant to say power "flux density" is power per unit area. To get the "flux," you multiply the flux density by the given area. Other than that, what I said was OK.