Answer
a. 10 cm
b. 7.1 cm
Work Step by Step
a. Find the focal length from equation 25–2b with the normal eye focused on the near point.
$$M=1+\frac{N}{f}$$
$$f=\frac{N}{M-1}=\frac{25cm}{3.5-1}=10cm$$
b. Find the focal length from equation 25–2a for the relaxed normal eye.
$$M=\frac{N}{f}$$
$$f=\frac{N}{M}=\frac{25cm}{3.5}=7.1cm$$
