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Looking at one of the faintest stars observable, I wondered how many photons per second would be striking my eyeball from a star hundreds or thousands of light years away? Would it be a few hundred, thousands, or are we into millions?
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A first magnitude star delivers a couple of million photons per square centimeter per second of visible light to an observer on Earth. If your iris is, let's say, a tenth of a square centimeter in area, your eye is admitting a couple of hundred thousand photons per second. The number of photons per second from that tsource falls off as the square of the distance, and the ultimate sensitivity of the eye is about 10 photons per second. You can do the math.
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OK, I did the math. Assuming:
1) The photon flux from a 1st magnitude star is (as you state) 2x10^6 photons/second. 2) Maximum dilation of the human eye is 7mm (area = 0.49 cm*2). 3) Faintest star visible to the unaided eye is 6th magnitude. Given: Five magnitudes difference in brightness equals 100:1 flux ratio (by definition). Therefore: "Ultimate sensitivity of the human eye" 10,000 photons/second.
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Any day you wake up on "the right side of the dirt" is a good day. T. Anderson |
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2) I assumed an iris area of 0.1 cm**2. You assumed an area of .38 cm**2 (not .49 if you work it out. I'll go with 0.1. 3) I'm calling it 7 magnitudes difference. A 6th magnitude star is still visible, so I'll add another one for good luck. Seven magnitudes difference is a 631:1 flux ratio. 2,000,000 x 0.1 / 631 = 317 photons/sec Since the eye's sensitivity in the visual range is highly nonlinear, I'm calling this close enough. 2 |
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Oops! [img]/phpBB/images/smiles/icon_redface.gif[/img] My hypothetical eye had a square pupil. [img]/phpBB/images/smiles/icon_redface.gif[/img]
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Any day you wake up on "the right side of the dirt" is a good day. T. Anderson |
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I don't mean to quibble, Bob, but 0.1cm^3 seems a bit small to me, considering this would be in the dark with the iris fully dilated. But it's not a biggie.
Also, I assume these numbers apply to the "night vision" portion of the retina, rather than the day-vision, high-resolution fovea. (I can never remember whether it's rods or cones that handle low-light situations). So to see the 6th-magnitude star, we're using averted vision, right? If the above is correct, anyone know how much more light is demanded by our central vision? Obviously the brighter stars and planets are adequately bright to be seen there. What's the cutoff? Also, how much individual variation is there? Among healthy individuals, I mean; obviously disease or defect can reduce vision to nil. I have no particular reason to ask these questions, except curiosity -- which my local PBS station urges me to maintain... [img]/phpBB/images/smiles/icon_biggrin.gif[/img] |
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heck, let's quibble. 7mm is pretty much a maximum, and that gets harder to attain with age. If you assume 5mm, then the area is (.5)^2 times pi/4, which is .196 cm^2 (note "2").Quote:
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By the way, it's cones for day/color/high res/central vision, rods for night/peripheral vision. I've managed to (re)learn something today... [img]/phpBB/images/smiles/icon_biggrin.gif[/img] |
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Around here, we're sometimes lucky to see third magnitude stars. [img]/phpBB/images/smiles/icon_frown.gif[/img] |
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Yeah, I know what you mean... there are parts of my night sky where there are seldom any stars visible (city light pollution to my north, neighbor with bright yard light to the northeast).
Maybe I should come out with a brand of binocs or telescopes called "Fovea Helpers"... [img]/phpBB/images/smiles/icon_biggrin.gif[/img] |