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

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(A) The focal length $f$ of a thin lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When

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moves towards the lens

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(A) The focal length $f$ of a thin lens is given by the lens maker's formula: $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When you squeeze the lens,the curvature of the surfaces increases,meaning the radii of curvature $R_1$ and $R_2$ decrease.As $R_1$ and $R_2$ decrease,the term $\left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ increases,which means the power of the lens increases and the focal length $f$ decreases.The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $u$ is the object distance (negative) and $v$ is the image distance.Rearranging for $v$,we get $v = \frac{uf}{u+f}$.Since the object is outside the focal point,$|u| > f$. As $f$ decreases,the image distance $v$ also decreases.Therefore,the image moves towards the lens.

Similar Questions

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