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Whenever we take the case of an object at infinity, we say that the image formed ends up being a point of light on the focal point if we had put a screen right at the focal point.

Now, for my lab, what we did was that we pointed the lens toward the far away (about 100 m away) trees outside the windows. Then, I assumed the objects were at infinity.

So, in the thin lens equation : 1/f=1/s=1/s', 1/s tends to 0.

Then, calculate the focal length, I just moved the screen forward and backward. When the image was clearest, I assumed the distance from the screen to the lens was the focal length.

However, I find this weird when I came back to study for my lab test. How come I did not see a single point on the screen? I actually saw an inverted, very small (about 5 cm tall) image of the trees.

Similarly, if we assume the sun is at infinity, should our eyes not focus it to the focal point and see it as a mere point rather than a large circle in the sky?

This sounds stupid, but I think I am missing something conceptually important.

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  • $\begingroup$ The focal length is determined by the lens characteristics, not by the objects being imaged. $\endgroup$ – paisanco Nov 23 '14 at 16:07
  • $\begingroup$ This is an experiment. We are trying to find the focal length. I think you don't understand what I was asking. Maybe I was not clear. $\endgroup$ – yolo123 Nov 23 '14 at 16:32
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    $\begingroup$ I understood perfectly well what you were asking. @Andrew has written an answer, so I won't bother. $\endgroup$ – paisanco Nov 23 '14 at 16:44
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An on-axis pointlike object, infinitely far away will produce rays that focus at the focal point of your converging lens. The sun and the trees are far enough away that "infinity" is not such a bad approximation, but they sun and the trees are neither pointlike, nor on-axis.

An off-axis pointlike object, infinitely far away will produce rays that (nearly) focus to a point one focal length away from your converging lens, but this point will be slightly off-axis. You can regard the light coming from the sun and the trees as coming from an array of pointlike objects, each of which produces a pointlike image on your screen. How far off-axis each source is will be (nearly) proportional to how far off-axis each image is, so the shape of the array of pointlike objects is (nearly) preserved, and you see an image of sun and trees.

Make sense?

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  • $\begingroup$ Andrew, thanks. Still have a question: How would one determine the position of the off-axis pointlike object on the focal "plane" that is formed using ray tracing? Is it possible? $\endgroup$ – yolo123 Nov 23 '14 at 16:31
  • $\begingroup$ I'm surprised I can't find a better link than this, but look here: thorlabs.us/images/TabImages/FTH100-1064_dwg3_1200.gif $\endgroup$ – Andrew Nov 23 '14 at 16:43
  • $\begingroup$ From here: thorlabs.us/newgrouppage9.cfm?objectgroup_id=6430 $\endgroup$ – Andrew Nov 23 '14 at 16:43
  • $\begingroup$ Hmm... @Andrew I do not understand why it is ftan($\theta$). Can you explain? $\endgroup$ – yolo123 Nov 23 '14 at 16:48
  • $\begingroup$ For an "ideal" lens, I think the relationship is d = f$\theta$. Thorlabs doesn't sell ideal lenses, so I think they're trying to give a more accurate equation for that particular lens. $\endgroup$ – Andrew Nov 23 '14 at 16:52

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