# How Do You Calculate Astronomical Magnitudes and Photon Count Rates?

• big man
In summary, the light from a B star behind the Coal Sack was detected and it's V and Mv were measured. The V was 10.0 and Mv was -4.0. The absorption due to the Coal Sack alone was found to be around 5*log(d/10pc)+A. The colour excess was found to be around 3.2.
big man
Hey I've been practicing these example questions that our lecturer gave us as part of the preparation for the upcoming exam. I'd appreciate any help. I find it really difficult to do his work because his lecture notes barely explain anything. It would have been nice to have a textbook for the unit but we don't...
Anyway sorry if this is too simple and is in the wrong section.

1. The Keck I telescope was used to detect an object with V~30. Estimate the photon count rate for this detection.

Thoughts: I don't really know how to do this question exactly. I think you'd need the area of the telecope to find the collecting area. The equations of magnitude show that you need the ratio of the flux from the object to a reference flux. So I guess if I knew a reference flux then I could easily find the flux of the object, but I don't so I'm not too sure.

Edit: Actually I was thinking that maybe I could use the apparent magnitude of the sun along with the apparent magnitude of the object. You know the amount of energy per square metre from the sun and you know the area of the telescope so you can find the energy/sec incident on the telescope. Then to find the number of counts per second you'd just divide that value by the energy of a photon...although the visible light has a range of wavelengths so do I just choose a wavelength??

2. A millimagnitude is 1/1000 of a magnitude. Using Excel, plot how many millimagnitudes correspond to an n percent increase in brightness, where n=1,2...20. Ignore the sign. Give a rule-of-thumb concerning the percentage brightness and millimagnitudes.

Thoughts: For this question I used the formula $$m=2.5*log(L/L_0)$$. Since it said it wanted thenumber of millimagnitudes per percentage increase in brightness, I determined m for each percentage increase. That is for 1% $$L/L_0=1.01$$. Then to find the number of millimagnitudes I just multiplied the calculated m value for the percentage increase by 1000. I get a logarithmic function so it kind of looked semi-ok...I'm really not sure if what I've done is right though.

3. A B star is known to be located behind the Coal Sack (a 20pc thick dark cloud of dust and gas near the southern cross) 1.5 kpc from the sun. Its V was measured to be 10.0 and Mv is known to be -4.0. What is the absorption due to the coal sack alone?? (b) What is the absorption in magnitudes/pc? What is the colour excess??

Thoughts:With this question I was wondering if you could just use the distances of 1.5kpc and 1.5kpc - 20pc to find the absorption using the equation: $$m-M=5*log(d/10pc) + A$$
Where A is your extinction. Then obviously you subtract the value from 1.5kpc - 20pc from the 1.5kpc value. However, I don't think that's right because you aren't really taking into account a greater amount of absorption from the cloud...or don't you bother? For part (b) I can find the colour excess given that you have the absorption based on the relationship $$A_v/C_e_x_t=3$$, but yeah I don't really understand the absorption thing. I mean I'm sure you'd have to take into account the absorption by a more dense gathering of dust (Coal Sack) and add that to the general absorption through the distance of 1.5kpc

Thanks again

Last edited:
1) How can you know the Area of some particular mirror,
but don't know the base of the measurement system?!
Calibrate to the Sun ... good idea ...use yellow photons.
2) was there a question here?
3) the 20pc thickness is novelty information, normally
not used in calculating extinction along an optical path.
(you *would* use it to find density, of course)

I don't recall Av/C = 3 ... is this an approximation?

the second one I was just asking if you thought it seemed right.

For the A/C thing, yes it is an approximation. It's meant to be around that and in his notes it says it is a constant...I think it is more accurately about 3.2.

Cheers for the feedback

## 1. What is the difference between apparent and absolute magnitude?

Apparent magnitude is a measure of how bright an object appears to an observer on Earth, while absolute magnitude is a measure of its intrinsic brightness. Apparent magnitude takes into account the distance between the object and the observer, while absolute magnitude does not.

## 2. How is the magnitude scale calculated?

The magnitude scale is a logarithmic scale that measures the brightness of objects in the sky. Each whole number increase in magnitude represents a decrease in brightness by a factor of 2.5. So, for example, a magnitude 1 star is 2.5 times brighter than a magnitude 2 star.

## 3. Can negative magnitudes exist?

Yes, negative magnitudes can exist. In fact, the brightest object in the sky, the Sun, has an apparent magnitude of -26.74. Negative magnitudes are used to represent objects that are exceptionally bright.

## 4. How does the human eye perceive different magnitudes?

The human eye can perceive a wide range of magnitudes, from the brightest objects in the sky (magnitude -26.74) to the faintest visible with the naked eye (magnitude 6). However, with the help of telescopes and other instruments, we can observe even fainter objects with magnitudes as high as 30.

## 5. Why is the magnitude scale important in astronomy?

The magnitude scale is important in astronomy because it allows us to compare the brightness of objects in the sky. By measuring the magnitude of an object, we can gain insight into its distance, size, and other properties. It also helps us track changes in the brightness of objects over time, which can be useful in studying celestial phenomena such as supernovae and variable stars.

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