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Confused about thin-lens equation |
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| Apr25-12, 06:37 AM | #1 |
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Confused about thin-lens equation
As you probably know, the thin-lens equation is given by:
1/s + 1/s' = 1/f where s = distance from object to lens, s' = distance from lens to image and f = focal length of lens Now, in a particular question, I needed to find s' given that s = f. From the equation, that means s' would become undefined! Is that even possible? If so, can you please explain what that means in terms of the image (i.e. real/virtual; upright/inverted)? Many thanks, |
| Apr25-12, 06:57 AM | #2 |
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hi jayadds!
![]() the image is "at infinity" … in other words, the light rays come out parallel
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| Apr25-12, 08:38 AM | #3 |
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If the lens is a converging lens then the object at the focal point is a magnifying glass with the image formed at infinity. Light rays passing into the eye (if it is used as a magnifying glass) are parallel and the eye is in its relaxed position. The image is upright and there is angular magnification.
If the lens is diverging then the image is formed at a distance of f/2 from the lens, it is virtual and diminished. |
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