Confused about thin-lens equation

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When the object distance (s) equals the focal length (f) in the thin-lens equation, the image distance (s') becomes undefined, indicating that the image is formed at infinity. In this scenario, for a converging lens, the light rays exit parallel, creating an upright image that allows for angular magnification, similar to using a magnifying glass. Conversely, with a diverging lens, the image is virtual, diminished, and formed at half the focal length (f/2) from the lens. This situation illustrates the unique behavior of lenses at their focal points. Understanding these concepts is crucial for applying the thin-lens equation effectively.
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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,
 
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hi jayadds! :smile:
jayadds said:
Now, in a particular question, I needed to find s' given that s = f. From the equation, that means s' would become undefined!

this is one of those rare cases when we're allowed to say that 1/0 is infinity

the image is "at infinity" …

in other words, the light rays come out parallel :wink:
 
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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