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Nitric
Feb13-09, 06:27 PM
1. The night-adapted eye can detect as few as several photons per second -- say five to be definite. Assuming that the pupil of the eye is 0.6 cm in diameter, from what distance would it be possible for the naked eye to detect a 60-W lightbulb?



2. I'm not sure which equation to use. In the section where the problem is from, I get teh follow equations (which don't help me)
E=pc, E=hf, h=6.63x10^-34, p=h/lamda



3. Not sure which formula to use

Delphi51
Feb13-09, 08:50 PM
Well, you will need to know how many photons the light emits per second. You can do this by picking a frequency near the low end of the visible range (to end up with the minimum), and using the energy of the bulb and the energy of the photon. Guess or look up the energy efficiency of a light bulb.

The next thing is how the photons from the bulb spread out in space. They go out in all directions equally, so light up an imaginary sphere at the distance where the eye is. The question is what portion of that lit up area is caught by the eye.

Nitric
Feb14-09, 01:22 AM
It says 5 photons/sec in the question.

Delphi51
Feb14-09, 12:35 PM
Yes, you know the 5 per sec answer so you can use the expression for the # of photons hitting the eye - which has a d in it - to solve for the distance.

Physics often works that way. You pretend you know something like the distance to the source, work out other quantities based on that unknown, and when you find one you know the answer for, you can solve for the unknown. Very useful trick!