A turnip sits before a thin converging lens,o | vedclass

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(A) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u$ is the object distance (negative),$v$ is the image distance,and $

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decrease the squeeze of the lens

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(A) The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u$ is the object distance (negative),$v$ is the image distance,and $f$ is the focal length.As the turnip (object) moves away from the lens,the magnitude of $u$ increases.For a fixed image distance $v$ (since the card position is fixed),the term $\frac{1}{v}$ remains constant.As $|u|$ increases,$\frac{1}{|u|}$ decreases.To maintain the equality $\frac{1}{f} = \frac{1}{v} - \frac{1}{|u|}$,as $\frac{1}{|u|}$ decreases,$\frac{1}{f}$ must decrease,which means the focal length $f$ must increase.Increasing the focal length $f$ of a lens requires decreasing its curvature.Since squeezing the lens increases its curvature (making it more convex and decreasing $f$),to increase $f$,we must decrease the squeeze of the lens.

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