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Question on aperture and magnitude.

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    A telescope with aperture 10 times the human eye could see stars down to magnitude?

    2. Relevant equations

    I was not given an equation with the question but I have found the following but I do not know if it is correct.

    M=Log(2.51188)*(A2/A1)^2

    3. The attempt at a solution

    A1= aperture of human eye-7mm
    A2= 10* 7 (10>human eye)

    Log(2.51188)*(70/7)= 40 (to 2 s.f.)

    I am finding the concept of stellar magnitudes difficult to grasp so any help would be appreciated.
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2

    cepheid

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    Yeah, the magnitude system is pretty stupid and annoying :biggrin:. You have the right concept for this question, just the wrong equation. Since you said you found magnitudes confusing, I'll try to explain:

    What you actually measure with a detector (e.g. retina, photographic plate, CCD chip, or some other kind of sensor) is what astronomers call flux, F. The flux of a source is the amount of light energy arriving from that source per second, and per square centimetre of area at the observer. In other words, it's the power per unit area, measured in watts/metre2.

    Magnitude is basically a logarithmic scale for flux, because human vision tends to work logarithmically. I.e. what appear to be "steps in brightness" by the same amount to the human eye are actually increases in flux by factors of ~10. The only tricky part is that the scale is reversed (smaller magnitudes correspond to larger fluxes). The equation for the difference in apparent magnitude in terms of the flux ratio is

    m1 - m2 = -2.5log10(F1/F2)

    As you can see, you don't need to know the individual fluxes in order to find the magnitude difference. All you need to know is their ratio. So what you have to figure out is the ratio of the flux seen by the eye to the flux seen by the telescope. As you correctly stated, the flux scales with the square of the aperture diameter. This is because the flux increases linearly with the collecting area of the aperture, and area depends on the square of the diameter. This makes sense. Flux multiplied by area gives you the total power detected. Increase the number of square metres, and you increase the power. You can get the flux ratio from the ratio of the squares of the aperture diameters, just as you were doing. Note: it would probably be better to use the letter D for aperture diameter, since A usually means area in this context.

    EDIT: Another need to know (just assumed to be common knowledge) is that humans can see down to a visual magnitude of around 6, under ideal conditions (clear, dark skies and no light pollution). If the magnitude is any larger, the star is too faint to see.
     
    Last edited: Feb 11, 2013
  4. Feb 12, 2013 #3
    Thank you for your help, I have been finding this topic to be very counterintuitive (at least for me anyway). Ok, lets see if I have made any progress!

    Ratio of the square of aperture diameters= (70/7)^2= 100

    So difference in magnitude=-2.5Log(base10)(100)=-5

    Human eye can see magnitude 6.

    Would that mean that the answer to my question is- Magnitude 11, or am I still as confused as ever? Again, thanks for your help.
     
  5. Feb 12, 2013 #4

    cepheid

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    Yeah, exactly. The equation tells you that the object will always look 5 visual magnitudes brighter when seen through a telescope, so the object with the faintest flux you can see is going to be an object that is 5 visual magnitudes fainter than the faintest one you can see with the naked eye. Hence you go from 6 to 11.
     
  6. Feb 12, 2013 #5

    cepheid

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    I had the ratio backwards for *this particular application*, and I sort of "explained away" a negative sign in the previous post. Let's get the math to match what I said in the previous post. The flux ratio you're considering should be 1/100. The proper way to do this is like so:

    Let F2 be the faintest flux you can see with the naked eye. It corresponds to an apparent visual magnitude of m2 = 6.

    Since the telescope has 100 times the collecting area as your eye, a star with a flux of F2/100 will produce the same total power as the star with flux F2 did in your eye. Therefore, the faintest source for the telescope has flux F1 = F2/100, or F1/F2 = 1/100.

    So we have:

    m1 - m2 = -2.5log(0.01) = -2.5*(-2) = 5

    m1 - m2 = 5

    m1 = 5 + m2 = 5 + 6 = 11.
     
    Last edited: Feb 12, 2013
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