How Does Distance Affect Image Formation in Convex Lenses?

AI Thread Summary
As the distance between a convex lens and an object increases, the image initially moves closer to the object to maintain focus, until a certain point is reached. Beyond this point, further increasing the lens's distance requires the image to move farther away from the object to stay in focus. This behavior is explained by the lens formula, which relates object distance, image distance, and focal length. The relationship between these distances indicates that the object and image are positioned on opposite sides of the lens, with their distances affecting focus. Understanding this interaction is crucial for analyzing image formation in convex lenses.
fromthepast
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Homework Statement



I did an experiment with a convex lens, object, and image. As the lens moved farther and farther away from the object, the image decreased its distance (to remain in focus) from the object as well up to a certain point. After that point, as the lens increased its distance from the object, the image had to increase its distance from the object as well to keep in focus.

Why is this?

Homework Equations



N/A

The Attempt at a Solution



It's a theoretical question with no exact solution.
 
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fromthepast said:

Homework Statement



I did an experiment with a convex lens, object, and image. As the lens moved farther and farther away from the object, the image decreased its distance (to remain in focus) from the object as well up to a certain point. After that point, as the lens increased its distance from the object, the image had to increase its distance from the object as well to keep in focus.

Why is this?

Homework Equations



N/A

The Attempt at a Solution



It's a theoretical question with no exact solution.

One could say that it happens that way because that is the way a lens behaves.

The mathematical summaries of the situations with a lens [the lens formulae] indicate this behaviour - especially if you look at the right formula.

What lens formulae do you know?
 
1/f = 1/s + 1/s'

m = h'/h = -s'/s

Does it have something to do with the focal point?
 
fromthepast said:
1/f = 1/s + 1/s'

m = h'/h = -s'/s

Does it have something to do with the focal point?

I would like to say yes to that.
Do you know the formula using the distances from the focal points rather than from the lens?
 
Both the object distance and the image distance are measured from the lens. The object and the real image are at opposite sides of the lens, so the object and the image are d=s+s' distance from each other. The image distance is s'=d-s. Plug in for s' into the lens equation, and find d in terms of the object distance s. Investigate the function, plot it. You will see, it has a minimum.

ehild
 

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