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In the above slide, it says that "Eyepiece '2' makes the image distance $S'$ approach $-\infty$." However, What I don't understand is that if I place my eye right at the Eyepiece 2, then I would see the image as if the object (for example, moon) is right in front of me as if I'm observing a rock in my hand...

Therefore, I feel like $S'$ has to be closer to zero or some finite number! not negative infinity...

Also, eventhough $S'$ is actually negative infinity, I quite don't understand how the moon can be observed as if I am observing a rock right in my hand... If $S'=-\infty$, then my brain says that we can't really observe the cracks, rocks, or even sand in the moon since the object is so much farther away from me...

Could somebody explain why I am confused???

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  • $\begingroup$ Thanks for including the image but it's still quite hard to read, could you have a go at cutting out all the unnecessary stuff in the screenshot and giving us an enlarged image, or alternatively redraw the diagram clearly for us? $\endgroup$
    – zephyr
    Aug 7, 2014 at 17:17
  • $\begingroup$ I should clarify, it's more the diagram that's hard to read than the text... $\endgroup$
    – zephyr
    Aug 7, 2014 at 17:38
  • $\begingroup$ You can magnify the internet by pressing "ctrl + '+'" in your keyboard (if you are using window not apple). I should say, the diagram doesn't matter for you to explain this problem, if you know how the telescope works mathematically... Plus, the diagram isn't indicating where the $s'$ of the eyepiece 2 anyway. $\endgroup$
    – User
    Aug 7, 2014 at 17:45
  • $\begingroup$ That is true but it just makes working it out a bit of a strain... Oh well, I'll have a shot at answering it below $\endgroup$
    – zephyr
    Aug 7, 2014 at 17:53

2 Answers 2

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Anyway, as a brief answer to your question remember that your eye is yet another lens focusing the light to a point on your retina. Thus you don't want an image coming into your eye which is already being focused to a point.

Your eye is built to focus light coming in roughly perpendicular to the lens, i.e. (no pun intended) your eye will focus light incident on it in the form of straight parallel rays.

That's effectively what a focus point at $-\infty$ means. Remember that parallel lines meet at infinity, and therefore also at $-\infty$, i think the choice of sign just being convention (though maybe someone else knows better?)

n.b. it's true that the eye can focus light from a focal point nearer than infinity, otherwise you wouldn't be able to see anything but the stars, but for the image to have a nearby focal point would require it to be inverted, and there may be other reasons why we prefer to focus this light as coming in from infinity?

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  • $\begingroup$ You just explained what I already know... I don't want this kind of explanation.. I just want to know why we can see the moon as if I am looking at a rock in front of me in my hand. The mathematical convention in the lecture says that the moon is at infinity $s'=-\infty$. But what I see through the telescope is the moon right in front of me $s>-\infty$. Do you get it? The mathematical result is contradicting the intuition. $\endgroup$
    – User
    Aug 7, 2014 at 18:08
  • $\begingroup$ On the other hand, the negative sign in front of $\infty$ indicates that the image is virtual. Wow, I'm actually teaching you, not learning from you. $\endgroup$
    – User
    Aug 7, 2014 at 18:12
  • $\begingroup$ Even a rock in your hand is at a distance of roughly infinity compared to the lens in your eye that's doing the focusing. As for the -ve sign, you may be right, although geometrically the two infinities are equivalent here, as they simply describe where lines meet. $\endgroup$
    – zephyr
    Aug 7, 2014 at 18:22
  • $\begingroup$ Well, that is not true. a rock in my hand is not at a distance of roughly infinity compared to the lens in my eye. It is because, if that is true, than, there shouldn't be near point of the lens in my eye which makes my muscles around my lenses of my eye to change its focal length. $\endgroup$
    – User
    Aug 7, 2014 at 18:30
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The purpose of the eyepiece is to let you look at the tiny, real, inverted image formed by the objective lens.

It does this the way any magnifying glass does it. It forms a virtual, huge, still inverted image at a convenient distance away from the eyepiece. By convenient, I mean as far away as the user's eye can focus.

When you look at a rock in your hand, you can do so because the lens in your eye can take light from the arm's-length rock, and form a good image on the retina of your eye.

So, you want to see more detail in the rock (fossils, flecks of gold, whatever). So you bring it closer to your eye, so it will be bigger. But your old, inflexible eye-lenses refuse to focus that closely. So you whip out a magnifying glass, and use it. You probably don't notice, but the lenses in your eye can now relax and focus on whatever is your far point. You move the magnifying glass around until it forms an image of the rock at that distance, but larger than before.

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