Please solve the physics numerical related to light and human eye.?

Question

Near point of a person suffering from hypermetropia is 60 cm from his eye. What is the nature and power of the lens needed to correct this defect?

Please specify what would be the object distance and image distance and focal length and why???
Please solve the numerical in details with each and every step along with explanation.

Thanks...

P.S. The formula for lens is 1/v-1/u=1/f
here v= image distance
u=object distance
f=focal length
Power, P = 1/f

Answer 1

In optometry, the least distance of distinct vision (LDDV) or the reference seeing distance (RSD) is the closest someone with "normal" vision (20/20 vision) can comfortably look at something. In other words LDDV is the minimum comfortable distance between the naked human eye and a visible object.
The LDDV is widely accepted as being 250mm (10 inches).
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In the given problem the person can view objects at a distance of 60 cm. Below this distance the objects will not be clear.

He has to use a lens such that when the object is placed at a distance of 60 cm an imgae is formed at a distance of LDDV i.e at 25 cm.

Hence u = 60 cm
v = 25 cm .

find f.

1/60 +1/25 = 1/f

P = 1/f = 0.0566 [cm]^(-1)

f = 17.65 cm.

A dioptre, or diopter, is a unit of measurement of the optical power of a lens or curved mirror, which is equal to the reciprocal of the focal length measured in metres (that is, 1/metres).

Therefore

P = 0.0566 x 100 diptre or diopter.

P = 5.6 dioptre.

Answer 2

The answer depends on the desired near point. A 60 cm near point means the eye is applying 1/0.6 = 1.67 diopters of additional power relative to vision at an infinite distance. Suppose the desired near point is 20 cm (good for removing splinters!). Then you would want a total of 1/0.2 = 5 diopters of power relative to infinity, so you'd need 5 - 1.67 = 3.33 diopters (converging lens) of correction.
EDIT (response to pearlsawme's answer):
"He has to use a lens such that when the object is placed at a distance of 60 cm an imgae is formed at a distance of LDDV i.e at 25 cm." But the whole idea is that he wants to see an object at 25 cm, not 60 cm! And the formula is set up wrong, as shown by the corrective lens focal length of ~18 cm being less than the desired object distance of 25 cm. If f is the (unknown) focal length of the corrective lens, you have 1/25 = 1/60 + 1/f. The answer for the 25 cm object distance is 2.33 diopters or f = 42.9 cm.