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When we view an image, is the focal point of our eye on our retina? Shouldn't that hurt? Also, if that is how our eye works, then why don't lenses put the focal point their equivalent retina? I was working with a lens today that had a focal distance of only a few millimeters, however we were using it to record something half a meter away-how is this the case?

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"When we view an image, is the focal point of our eye on our retina?" Depends. How's you vision? –  dmckee May 9 '13 at 1:44
    
20/20, and I've seen pictures of the focal point being on the retina, but I feel like that would burn. When you put things at the focal point of a magnifying lens they burn! –  Anthony May 9 '13 at 1:51
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And if you let your eyes focus the sun on your retina you can do permanent damage... But you should notice the size of the magnifying glass you use to set leaves on fire and compare it's area to that of your pupil; there's a lot less light involved. –  dmckee May 9 '13 at 1:54
    
So then how does it process the now larger image when it comes to near/far sided-ness? –  Anthony May 9 '13 at 1:56
    
Badly. Things are fuzzy. You squint a lot (effectively making a partial pinhole camera to improve the image) and get headaches. You can get a taste of it by borrowing some glasses, but you won't like it. –  dmckee May 9 '13 at 1:59
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The focal point of your lens is indeed on your retina, when you look at an object far away. If you look at an object closer by, that object is also imaged onto your retina (because you are changing your eye's lens. So it's not really that important whether your lens' focal point is on the retina, but whether you are imaging a given object (sometimes the word focus is casually used for the image location, but that doesn't have to be at the focal point).

Only if the object is very bright (sun, bright small lamp) does it actually hurt. Your retina has a very high dynamic range, much higher than current camera systems.

The dynamic range is the ratio in intensity (or say number of photons) between the brightest object you can still see clearly (say the sun for a very short time) and the darkest object that you can make out against a perfectly black background. This high dynamic range makes it necessary for us to grade intensity in approximately logarithmic terms (Weber's law) when we compare brightness.

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Note that you're taking "focal point" to mean a point at a distance $s_i$ from the lens, whereas the OP may mean the point at a distance $f$ of the image from the lens. A common student misconception is that images are formed at the distance $f$. A poll of physics teachers on the PHYS-L mailing list a while back found that most of them preferred not to use the term "focal point," and that there was not a commonly understood definition of the term as referring to $f$ or $s_i$. –  Ben Crowell May 9 '13 at 11:26
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