Answer
See answers.
Work Step by Step
a. The focal length of a lens is the reciprocal of its power.
$$f=\frac{1}{P}=\frac{1}{3.50D}=0.286 m$$
b. The lens will produce a virtual image at the near point of Sam’s eye.
The object distance is 23 cm (because it is 25 cm from the eye). Solve for the image distance, which is created at Sam’s near point.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$d_i=(P-\frac{1}{d_o})^{-1}=(3.50D-\frac{1}{0.23m})^{-1}=-1.18m$$
Add the 2 cm, the distance between the lens and the eye, to find the near point.
$$N=|d_i|+0.02m=1.18m+0.02m=1.2m$$
c. Pam wears these glasses. Find the object distance that creates an image at her near point, which is at an image distance of -0.23m.
$$P=\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$
$$d_o=(P-\frac{1}{d_i})^{-1}=(3.50D-\frac{1}{-0.23m})^{-1}=0.13m$$
Pam’s near point while wearing Sam’s glasses is 13 cm from the lens, which is 15 cm from her eye.
