# Astrophysics: Finding flux of a star given magnitude

1. Feb 5, 2012

### Xyius

1. The problem statement, all variables and given/known data

The faintest stars a naked eye can see under the ideal conditions are of m = 6
magnitude. Diameter of a maximally dilated pupil is d = 9mm. Calculate the
magnitude of the faintest star a person can see if observing through binoculars
(d = 5cm), a large backyard telescope (d = 8 inches), and a professional
telescope (d = 2m). Compare these magnitudes to those found in part (a) -
what kind of instrument would you need to see a Sun-like star at given distances?

2. Relevant equations
The magnitude equation:
$m-M=5log\left( \frac{d}{10} \right)$
Where m and M are the apparent and absolute magnitudes of a star respectively. And the number 10 is in parsecs.

3. The attempt at a solution
My thoughts upon how to do this problem is to find the flux of the star that has a magnitude of 6, then see how much flux goes into a diameter of 9mm (pupil). From there I can relate that number to the amount of flux that goes into a backyard a professional telescope. The only problem is, it seems like I am not give enough information to find the flux! I probably have the wrong approach to this problem. Can anyone help? :\

NOTE:
This was part A, which I answered using the magnitude equation.
Given that apparent magnitude of the Sun is m = 26:7 (at 1A:U:), nd its
apparent magnitude at distances of 1pc; 10pc; 100pc and 1000pc.

2. Feb 6, 2012

### Staff: Mentor

Perhaps you should consider the variation in brightness due to increased light-collecting ability of the binoculars over the naked eye. What's the ratio? How many magnitudes does this ratio represent?

3. Feb 6, 2012

### Xyius

How can I get the brightness though? Is that the same as the flux? I cannot get a flux ratio because I do not know how far the magnitude 6 star is. :\

4. Feb 6, 2012

### Staff: Mentor

You should be able to determine the flux ratio from the details of the light-collecting apparatus. How much more light does the binocular provide compared to the naked eye?

5. Feb 6, 2012

### Xyius

I would imagine that the ratio of their surface areas, would that give the correct answer?

So for the eye: $S_e=\pi r^2= \pi (0.009)^2=2.54\times 10^{-4}m^2$
And for the binoculars: $S_b=\pi r^2= \pi (0.05)^2=0.15708m^2$
So..
$\frac{S_b}{S_e}= \frac{0.15708}{2.54\times 10^{-4}}=617.284$

But how do I relate this to a magnitude 6 star?

Last edited: Feb 6, 2012
6. Feb 6, 2012

### Staff: Mentor

You'll want to check your calculation for the binocular area. Also note that you've used the diameter values for radii -- which actually won't matter when the ratio is taken as the "error" then cancels out. You could save yourself a bit of calculator work by writing the ratio symbolically to start with and cancelling mutual constants before plugging in numbers.

You should have in our notes or text a description that relates observed brightness ratios to magnitude differences. By how many magnitudes does the increased light-collecting ability of the binocular change the magnitudes of observed objects?

7. Feb 6, 2012

### Xyius

The only equation that I have about this subject is simply the magnitude formula. I did something here and I am not sure if it is correct.

I re-did the surface area ratio and got a more reasonable 30.86. I then plugged this in for the flux ratio in the magnitude formula..

$m_1-m_2=2.5log \left(\frac{F_2}{F_1} \right)$
$m_1-6=2.5log \left(30.86 \right)$
$m_1=9.72$

Since 9.72 > 6 it is a fainter star. I do not know how sound this answer is though..

8. Feb 6, 2012

### Staff: Mentor

Well, it looks quite reasonable to me. The binoculars add nearly four magnitudes to the "depth" of star that one can see.

9. Feb 6, 2012

### Xyius

Same. I am going to do this as I feel this is correct. Thanks for your help! It is very much appreciated :]