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Homework Help: 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




    then how would you get

  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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