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Photon Emission At A Distance

  • Thread starter hy23
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Homework Statement


A light bulb 10km away from you emits visible photons at a rate of 3x1018 photons/s. Assuming this is the only source of light (the whole place is dark), and your dark-adjusted retina has a diameter of 7mm, then how many photons/s hits your retina?


Homework Equations


Not really sure if any formulas are needed, but the ones presented in the chapter containing this question are
E=hf (Plancks).....lambda =h/p (matter waves)

r=dsin(a)=m(lambda)....path length difference ( for constructive interference)
y=Lsin(a)....(position of fringes on a a screen L distance away from slit)


The Attempt at a Solution


I've taken the light bulb to be point source since it's so far away. I believe the lightbulb also emits photons in all directions (i.e. 360 degrees). Thus the portion of its photon emission that actually travels in the direction of your retina should correspond to the fraction of the angle of incident light rays divided by 360 degrees.

So, using small angle approximation, this angle is a=0.007m/10000m

my final answer was 3x1018x a/360 = 5.8333 x 109 photons/s

I doubt this is the correct answer because my initial assumption that the light bulb emits photons in all directions means that by assuming 360 degrees I have incorrectly simplified this problem to a 2D situation which is not what it is; in addition I have used none of the formulas in the chapter which worries me.


Thanks in advance for any help
 

Answers and Replies

  • #2
gneill
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The radiation is going in all directions (3D), so think in terms of surface areas. Your 3 x 1018 photons per second are going to be spread out over a surface of a shell 10km in radius...
 
  • #3
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ohhhhhhhhhhhhhhh!!!! very clever!!

so it should be the area of your retina divided by the surface area of the 10km shell, why didn't I think of that...thanks a lot

but overall is this way of approaching the problem correct? I have this fear that there may be some other formulas involved..
 
  • #4
gneill
Mentor
20,781
2,759
ohhhhhhhhhhhhhhh!!!! very clever!!

so it should be the area of your retina divided by the surface area of the 10km shell, why didn't I think of that...thanks a lot

but overall is this way of approaching the problem correct? I have this fear that there may be some other formulas involved..
I would be surprised if it were incorrect...

This is typically how things like solar irradiance at a given distance from the Sun are handled.
 

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