Question

You are given that the diameter of the eyeball is about $2.3$ cm and a normal eye can adjust the focal length of its eye lens to see objects situated anywhere from $25$ cm to an infinite distance away from it.

What is the power of the (normal) eye lens, when ciliary muscles are fully relaxed?

What is the power of the (normal)eye lens, when ciliary muscles are in their maximum contract position?

The maximum variation in the power of the eye lens, when it adjusts itself, from the normal relaxed position to the position where the eye can see the nearby object clearly?

Answer 1

Ciliary muscles will be relaxed when object distance, $u=-\infty$

Image distance $v=2.3cm$

Using,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f_{max}}=\frac{1}{2.3}-\frac{1}{-\infty }$

$f_{max}=2.3cm$

$Powe{r}_{min}=43.47D$

Ciliary muscles are in their maximum contract position when object distance, $u=-25cm$

$v=2.3cm$

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f_{min}}=\frac{1}{2.3}-\frac{1}{-25}$

$\frac{1}{f_{min}}=\frac{10}{23}+\frac{1}{25}$

$f_{min}=\frac{272}{575}$

$f_{min}=2.1cm$

$Powe{r}_{max}=47.61D$

If position of object is between infinity and 25 cm, the power of eye lens will be between 43.47 D and 47.61 D.