The near point of a hypermetropic eye is 1 m. What is the nature and power of the lens required to correct this defect ? (Assume that the near point of the normal eye is 25 cm).

Explanation

The eye defect called hypermetropia is corrected by using a convex lens. So, the person requires convex lens spectacles. We will first calculate the focal length of the convex lens required in this case. This hypermetropic eye can see the nearby object kept at 25 cm (at near point of normal eye ) clearly if the image of this object is formed at its own near point which is 1 metre here. So, in this case :
Object distance, u = -25 cm `" "` (Normal near point)
Image distance, v = -1m `" "` (Near point of this defective eye)
= - 100 cm
And, Focal length, `f = ?" "` (To be calculated)
Putting these value in this lens formula,
`(1)/(v)-(1)/(u) = (1)/(f)`
We get : `(1)/(-100) - (1)/(-25) = (1)/(f)`
or `- (1)/(100) + (1)/(25) = (1)/(f)`
`(-1 + 4)/(100) = (1)/(f)`
`(3)/(100) = (1)/(f)`
`f = (100)/(3)`
`f = 33.3 cm`
Thus, the focal length of the convex lens required is + 33.3 cm. We will now calculate the power. Please note that 33.3 cm is equal to `(33.3)/(100)` m or 0.33 m. Now,
Power, `P = (1)/(f ("in metres"))`
`= (1)/(+ 0.33)`
`= + (100)/(33)`
`= + 3.0 D`
So, the power of convex lens required is + 3.0 dioptres.
Myopia and hypermetropia are the two most common defects of vision (or defects of eye). We will now study another defect of vision which occurs in old age. It is called presbyopia.