How can we determine the focal length and power of the convex lens required to correct a hypermetropia eye ?

Explanation

Let x = distance of the near point N. from the hypermetropic eye. We know that the near point N of the normal eye is at distance D = 25 cm. The object placed at N forms its virtual image at N. due to the convex lens.
So, object distance = u = -D,Image distance = v = -x. focal length = f = ?
By lens formula,
`(1)/(f) = (1)/(v) - (1)/(u) = (1)/(-x)-(1)/(-D) = (x-D)/(xD)`
`:.` required focal length,
Required Power , `P = (1)/(f) = (x-D)/(xD)`
As `x gt D` both f and D are positive. Thus, the correcting lens must be a convex lens.