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| Re: silly inverse square law question. |
Lets do one last thought experiment and then I am out of here. I really must be doing something more productive with my time then engaging in an endless discussion on exposure
Suppose for convenience that moon is a big flat disc and neil armstrong is sitting on it looking down and taking the picture of it. Lets say his picture covers a 5 ft x 5 ft portion of the moons surface. Far far away at earth we have our camera ready and about to take the picture of moon. Unfortunately, space mice come in and eat all of the moon around that 5x5 ft portion before we could take a picture. Now we are only left with that patch to take a picture of. First of all we wouldn't even be able to see that patch. By the time the light reaches us it would be so dim that we will not be able to register it. Secondly the thing will be so small in terms of spatial resolution then neither our sensor nor the eye could capture it. Now thats all thats left of the moon, we are desperate to take a picture of the moon so we work very very hard and come up with a sensor which has resolution high enough and sensitivity high enough to capture that portion. Then we will see that the amount of light that is reaching us from that portion will be only 1.6 x 10^18 or whatever of the amount of light that Neil Armstrong's camera was getting.
Yes, because for this scenario under the conditions you list, you have made a flat object esentially a point source. Point sources suffer from spherical divergence. This happens when the viewer-object distances are big enough, relative to the size of the subject. The moon is big enough so that this won't happen.
If that were the case, then Neil Armstrong would have had to use an inexistent, super low ISO emulsion to capture the moon, since, following your logic, the moon would be so much brighter as the Armstrong to moon distance is lesser than a earth photographer shooting the moon from Kuala Lumpur.
The moon is not a point source. It does not emit light, it reflects it.