Find the minimum and maximum focal lengths of the eye lens.

Explanation

1) When the object is at infinity ,the parallel rays from the object falling on the eye lens are refracted and they form a point sized image on the retina.
2) In this situation, eye lens has a maximum focal length.
3) When the object is at infinity.
`u=-oo`, v=2.5 cm ( distance between eye lens and retina)
Using Lens formula , `(1)/(f) = (1)/(v)-(1)/(u) implies(1)/(f_(max)) = (1)/(2.5)+(1)/(oo)implies(1)/(f_(max))=(1)/(2.5)+0`
`implies(1)/(f_(max))=(1)/(2.5) therefore f_(max) = 2.5 cm`
Consider that an object is placed at a distance of 25 cm from our eye.
5) In this situation eye hs minimum focal length.
Here u=-25 cm , v=2.5 cm
Using Lens formula , `(1)/(f) = (1)/(v)-(1)/(u)`
`(1)/(f_(min)) = (1)/(2.5)+(1)/(2.5) implies (1)/(f_min) = (10)/(25)`
`(1)/(f_(min)) = (11)/(25) therefore f_(min)= (25)/(11) = 2.27 cm`
6) If the position of an object is between infinity and the point of least distance of distance vision, then the eye lens adjusts its focal length in between 2.5 cm to 2.27 cm to form a clear image on the retina.