Question

The ciliary muscles can change the focal length of the eye lens. Find the ratio of focal lengths of the eye lens when it is focused on two different objects, one at a distance of 2 m and the other at a distance of 1m. The diameter of normal eye is 2.5 cm.

• A. 17:16
• B. 42:41
• C. 41:42
• D. 82:81

Answer 1

Answer: (D)

82:81

According to question image should form at a distance of 2.5cm hence
$v=2.5cm$
$u_1=-2m$ $u_2=-1m$
from lens formula
$\frac{1}{v}-\frac{1}{u_1}==\frac{1}{f_1}$ and $\frac{1}{v}-\frac{1}{u_2}=\frac{1}{f_2}$
hence
$\frac{100}{2.5}+\frac{1}{2}=\frac{1}{f_1}$ and $\frac{100}{2.5}+\frac{1}{1}=\frac{1}{f_2}$
$40.5=\frac{1}{f_1}$ and $41=\frac{1}{f_2}$
so$\frac{f_1}{f_2}=\frac{\frac{1}{f_2}}{\frac{1}{f_1}}=\frac{41}{40.5}=\frac{82}{81}$
hence $\frac{f_1}{f_2}=\frac{82}{81}$