Calculate a: Dist. b/w Lens Positions for 40 cm Object-Image Distance

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The discussion focuses on calculating the distance between two positions of a converging lens that forms a sharp image for a fixed object-image distance of 40 cm. The thin lens formula, 1/f = 1/o + 1/i, is used to derive the relationship between object distance (o) and image distance (i). The user seeks clarification on how to apply the formula correctly, noting that switching the object and image distances leads to another valid lens position. The conversation also humorously touches on mechanical advantages in unrelated contexts. Ultimately, the goal is to determine the distance 'a' between the two lens positions.
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The distance between an object and its image is fixed at 40.0 cm. A converging lens of focal length f = 3.62 cm forms a sharp image for two positions of the lens. What is the distance a between these two positions?

Determine:

The distance between these two positions a = ____cm

Could someone help me with the formula i want to be using?
 
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The thin lens formula gives
\frac{1}{f}=\frac{1}{o}+\frac{1}{i}
with i the image distance and o the object distance. Now say you find a position between the object and fixed screen (image position) such that you get a sharp image on the screen. The object image is now o and the image distance i while the sum of these two being 40 cm. But according to the formula if you make the object distance i the image distance should be o! Which gives you the other position of the lens where you should get a sharp image on the screen. In the second case the object distance will therefore be i and the image distance will be o.
 
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I would guess that with a mech advantage of 2 you would have problems lifting your lecturer! Three would give you a fighting chance!
 
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