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How many Photons per second are entering the pupil?

  1. Apr 9, 2017 #1
    1. The problem statement, all variables and given/known data
    (a)Estimate the number of photons per second emitted by a 100-W lightbulb, assuming a photon wavelength of 550nm.(b) A person can just see this bulb for a distance of 800m, with the pupil dilated to 7.5mm. How many photons per second are entering the pupil?

    2. Relevant equations
    E=hc/λ
    h=6.63*10^-34 J s
    c=3.0*10^8 m/s
    1eV=1.602*10^-19 J
    1 watt second = 100 joules
    3. The attempt at a solution
    For part (a)
    E=hc/λ=(6.63*10^-34 J s * 3.0*10^8 m/s)/(550nm) =3.616*10^-19 J

    convert to eV: (3.616*10^-19 J * (1eV/1.602*10^-19J)=2.257 eV per photon

    100 watt sec = 100 joules

    (100 joules * (1eV/1.602*10^-19J))/2.257eV) = 2.77*10^20 photons

    For part (b):

    Not excatly sure how to proceed. I made a circle and determined the percent of the circumference was the 7.5mm pupil then multiplied that by the total number of photons, but the answer I got did not mach the book.
    C=pi*d=pi*1600m=5026.548m
    (7.5mm/5026.548m)*2.77*10^20 photons = 4.13*10^14 photons.
    The answer in the book is 6.1*10^9 photons.
    How can I get to this answer?
     
  2. jcsd
  3. Apr 9, 2017 #2

    gneill

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    Staff: Mentor

    Does a lightbulb send out all its photons in a plane?
     
  4. Apr 9, 2017 #3
    Ah, I need to consider a sphere, and a conical section of that sphere with the circular base being the 7.5mm pupil. I shall attempt.
     
  5. Apr 9, 2017 #4
    Volume of the sphere = 4/3 pi r^3 = 2.14×10^9 m^3
    Volume of the cone = pi r^2 h/3 = 0.047 m^3
    ratio = 2.196 * 10^-11
    2.196 * 10^-11 * 2.77*10^20 =6.1*10^9 protons!
    Thanks!
     
  6. Apr 9, 2017 #5

    gneill

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    Staff: Mentor

    Actually, I believe that the given answer is incorrect. You want to look at the flux of photons passing through the surface area of the pupil per second, so the ratio should be between surface areas, not volumes.
     
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