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Calculating Photon-Flux

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data
    The faintest objects that have been detected at optical wavelengths with the Hubble
    Space Telescope have apparent magnitudes m  31. Calculate the flux from
    an object of this magnitude, and, assuming that each photon has a typical optical
    wavelength, convert your result into the number of photons per unit time per unit
    area at Earth (this is the so-called photon flux).
    [The Sun has apparent magnitude m = -26.8, the flux from the Sun is 1350Wm^-2]


    2. Relevant equations
    m_2 - m_1 = 2.5log( f_1/f_2) log with base 10 here

    E = hf, λ = c/f


    3. The attempt at a solution

    well I worked out the flux of the object to be 1.02*10^-20 Wm^-2. (I'm pretty sure this is right) however I am not sure what the typical optical wavelength is or how to go about working out the "Photon Flux".

    I thought if I could work out the energy of a photon with typical wavelength, I can then divide the calculated flux by energy of photon which will give me the number of photons...
    I am not sure how I work out the number of photons per unit time though =/
     
  2. jcsd
  3. Sep 27, 2012 #2
    Typical optical wavelength would be those of visible light.
     
  4. Sep 27, 2012 #3
    typical optical wavelength just means a wavelength somewhere in the visible spectrum

    since the visible spectrum goes from 390nm to 750nm, looks like 570nm would be a good choice, it's right in the middle

    and so using that, you should be able to figure out the typical energy per photon

    if you have

    [itex]\frac{energy}{photon}[/itex]

    and

    [itex]\frac{energy}{seconds*area}[/itex]

    then how would you get

    [itex]\frac{photon}{seconds*area}[/itex]
     
  5. Sep 27, 2012 #4
    guess I just divide one by the other. How do I work out the energy given the flux? (The total energy would be luminosity right?) if so the formula f = L / 4(pi)d^2 is the diameter the object receiving or the one emitting?
     
  6. Sep 27, 2012 #5
    Energy of what, and, more importantly, what for?
     
  7. Sep 27, 2012 #6
    flux *is* energy

    watts per unit area is energy per unit time per unit area

    the flux tells you how much energy is moving through a surface per unit time
     
  8. Sep 27, 2012 #7
    Oh perhaps... I calculated the flux wrong then. Because using a wavelength of 570nm I get the energy per photon to be higher than the flux. so surely I can't have almost 0 photons reaching earth?
     
  9. Sep 27, 2012 #8
    can you check that I have calculated the flux right?

    m_1 = 31, m_2 = -26.8, f_1 = ? and f_2 = 1350.

    31+26.8 = -2.5log(f_1 / 1350) so

    10^(-57.8/2.5) = f_1/ 1350 so f_1 = 1350*10^(-57.8/2.5)
     
  10. Sep 27, 2012 #9
    Almost is subjective. Your flux seems correct.
     
  11. Sep 27, 2012 #10
    okay E = hc/λ. so using a wavelength of 570nm I get that E = 3.49*10^-19 J.

    so the number of photons = (1.02*10^-20)/3.49*10^-19 which is 0.03...

    I think somewhere I went wrong but can't figure out where?
     
  12. Sep 27, 2012 #11
    It is indeed a very small number, that's why the source is considered very faint.
     
  13. Sep 27, 2012 #12
    Ah okay, thank you both very much for your help!
     
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