Question

you are given that 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 thier 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 eye can see nearby objects 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