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.
(a) What is the power of the (normal) eye lens, when ciliary muscles are fully relaxed?
(b) What is the power of the (normal) eye lens, when ciliary muscles are in their maximum contract position?
(c) The maximum variation in the power of the eye lens, when it adjust itself, from the normal relaxed position to the position where the eye can see the nearby object clearly?

Answer 1

Given:
When object is at infinity:u=-∞
V=2.3cm( given)
By lens formula:1/f=1/v-1/u
1/fmax=1/2.3+1/∞
=1/2.3 +0
=1/2.3
fmax=2.3cm
Therefore when object is at Infinity , the parallel rays from object  makes the ciliary muscles relaxed  and focal length has its maximum value which is equal to its distance from retina=2.3cm

When an eye is focussed on a closer object, the ciliary muscles are strained and focal length of eye lens decreases.
here u=-25cm
v=2.3cm
f=?
1\f=1/v-1/u
1/fmin=1/2.3+1/25
=10/23 +1/25
=250+23/23x25
=273/575
fmin=575/273=2.1 

If the position of an object is between Infinity and the point of least distance of distinct vision then eye lens adjust's its focal length in between 2.3cm to 2.1cm

PLs
.
.

.....
....
. FOLLOW ME