A person with normal eyesight uses a converging lens of focal length $10 \mathrm {~cm}$ as a magnifying glass to finish an engraving job. To see detail, he holds the lens very close to his eyes. At what minimum distance from the job must he hold the lens?
A) $30 \mathrm {~cm}$
B) $20 \mathrm {~cm}$
C) $7.1 \mathrm {~cm}$
D) $9.4 \mathrm {~cm}$
E) $14 \mathrm {~cm}$

Question

Quizplus · Accepted Answer

For a person with normal eyesight, the nearest point of distinct vision (the least distance of distinct vision) is typically taken as $25 \mathrm{~cm}$. When using a magnifying glass (a converging lens) for magnification, the image is formed at the least distance of distinct vision for maximum magnification. The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, where $f$ is the focal length of the lens, $v$ is the image distance, and $u$ is the object distance. For maximum magnification, $v = 25 \mathrm{~cm}$ (the image is formed at the nearest point of distinct vision), and the focal length $f = 10 \mathrm{~cm}$. Substituting these values into the lens formula gives $\frac{1}{10} = \frac{1}{25} - \frac{1}{u}$, solving for $u$ gives $u \approx -7.1 \mathrm{~cm}$. The negative sign indicates that the object is on the same side of the lens as the incoming light, which is the usual convention in optics. Therefore, the minimum distance from the job to hold the lens is $7.1 \mathrm{~cm}$.

A Person with Normal Eyesight Uses a Converging Lens of Focal