Calculate the power of the eye lens of the normal eye, when it is focused at far point and near point, given the diameter of the eye is ` 2.5 ` cm. Find the maximum variation in the power of normal eye lens.

Explanation

The far point of a normal eye is infinity. When the object is at infinity, the image is formed at the focus, i.e., image distance v = f, where f is the focal length.
Diameter of the eye = distance between lens and the focus = ` 2.5` cm(given)
` :. f = 2.5 cm = 2/5 xx 10^(-2)` m .
Power of the lens, ` p= 1/f`
` p = 1/(2.5 xx 10^(-2)) = 10^(2)/(2.5) = 40` D.
2. The near point of a normal eye is 25 cm
= object distance
= - 25 cm = u
(form sign convention)
` v = 2.5 ` cm
= distance of the eye lens from the retina 9 i.e., the focus)
` 1/f = 1/(2.5 xx 10^(-2)) - 1/(-25 xx 10^(-2))`
` 1/f = 100/2.5 + 100/25`
` 1/f = 40 + 4`
` 1/f = 44`
Power = ` 1/f = 44` D.
Thus, the maximum variation in the power of the lens is ` 44 D - 40 D = 4 D`.