How will you calculate the focal length of a biconvex lens that is used to correct the defect f hypermetropia ? Explain it mathematically.

Explanation

The person who has hypermetropia cannot see near objects. He can see the objects those are beyond near point (H). For correction of this eye defect the image of the image of the object placed at "least distance of distinct vision (L)" should be at near point (H).
u =-25 cm , v=-d cm
`(1)/(f)=(1)/(v)-(1)/(u)`
`implies (1)/(f)=(1)/(-d)-(1)/(-25)`
`(1)/(f) = (1)/(25)-(1)/(d)`
`(1)/(f) = ((d-25))/(25d)`
`f = (25d)/(d-25)`
Here `d gt 25`, so 'f' gets positive value . Hence, convex lens should be used.