Thin Lenses: Understanding (3/2)f in Image Formation

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An object located three times the focal length from a lens leads to an image distance of (3/2)f. The relevant equation for this problem is (1/object distance) + (1/image distance) = (1/focal length). By substituting the object distance as 3f, the calculation reveals the image distance. The discussion emphasizes using algebra to solve for the image distance based on the given conditions. Understanding this relationship clarifies the concept of image formation in thin lenses.
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Homework Statement



An object is 3 times further from a a lens than its focal length. Let f rep focal length. How man times the focal length of the lens is the image from the lense? The answer is (3/2)f but I don't understand why. Could someone please explain it to me?
 
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welcome to pf!

hi libra10489! welcome to pf! :wink:

what equations do you know relating the object distance the image distance and the focal length? :smile:
 
The only one I know is (1/object distance) + (1/image distance) = (1/focal length).
 
that's the one! :wink:

ok, now put the numbers in …

what do you get? :smile:​
 
Um...there are no numbers...if there were numbers I'd be able to plug and chug. I guess the confusing part is "An object is 3 times further from a lens than its focal length."
 
use algebra …

call the focal length "f" …

then the object distance is 3f, so the image distance is … ? :smile:
 
Oh..., now I feel stupid... OK I got it. Thanks.
 

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