Question

A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?

Answer 1

(a) When the eye lens is fully relaxed, we have:
u = ∞
Distance of the retina from the eye lens, v = 2 cm = 0.02 m
The lens formula is given by

`1/v -1/u =1/f`

Putting the respective values, we get:

`1/f = 1/0.02 - 1/∞ = 1/0.02`

∴ Power of the lens =`1/f` = 50 diopters

So, in a fully relaxed condition, the power of the eye lens is 50 D.

(b) When the eye lens is most strained:
u = – 25 cm = – 0.25 m
The lens formula is given by

`1/v -1/u =1/f`

Putting the values, we get:

`1/f = 1/0.02 -1/((-0.25))`

= 50+4 =54

∴ Power of the lens = `1/f` = 54 diopters
So, in the most strained condition, the power of the eye lens is 54 D.