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

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

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decreases

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(B) 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 the lens is squeezed,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 lateral magnification $m$ of a lens is given by $m = \frac{f}{f - u}$,where $u$ is the object distance (which is negative,so let $u = -|u|$). Thus,$m = \frac{f}{f + |u|}$.As $f$ decreases,the denominator $f + |u|$ decreases more rapidly than the numerator $f$ in relative terms,or more simply,as the lens becomes more converging (smaller $f$),the image moves closer to the focal point and its size decreases.Therefore,the lateral height of the image decreases.

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