Solved Example of Human Eye

Solved Example of Human Eye

1. A person cannot see objects closer than 75 cm from the eye. Calculate the power of the corrected lens should he use.

Solution: Since the person cannot see objects lying closer than 75 cm, he suffers from hypermetropia. His near point has shifted from 25 cm to 75 cm. The focal length of the corrective lens can be calculated by considering u = –25 cm, v = -75 cm, f = ?

Now,  \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{{-75}}-\frac{1}{{-25}}$

or  \frac{1}{f}=\frac{2}{{75}}f=\frac{{75}}{2}cm=\frac{{0.75}}{2}m$

∴  Power = =  D = +  D = 2.66D

Example 2: The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to enable him to see very distant objects distinctly?

Solution: Since the person suffers from myopia, concave lens of focal length 80 cm = –0.80 m should be used.

∴        P = \frac{1}{{-0.80}}$  = –1.25D

 Example 3: The far point of a person suffering from myopia is 2 m from the eye. Calculate the focal length and the power of the corrective lens.

Solution: The far point lies at 2 m. Therefore, a concave lens of focal length 2 m should be used so that the objects lying at infinity can be focused at the far point.

∴ For corrective lens, focal length,f = –2 m

∴ Power,                      P = \frac{1}{{-2}}$ = –0.5D

Example 4: The near point of an elderly person lies at 50 cm from the eye. Calculate the focal length and power of the corrective lens.

Solution: The person suffers from hypermetropia. His near point lies at 50 cm. Therefore, a convex lens should be used for the correction of his vision. The focal length of the corrective lens is calculated by

\frac{1}{{-25}}-\frac{1}{{-50}}=\frac{1}{f}$

or                \frac{1}{f}=\frac{1}{{50}}$

or                           f = 50 cm = 0.5 m

\ Power of the corrective lens,

P =  \frac{1}{f}=\frac{1}{{0.5}}$= + 2D