Question

The near point of a hypermetropic eye is 50 cm what is the power of the lens required to correct this defect

Answer 1

we use convex lens to correct this defect

Answer 2

Answer:far point of a hypermetropic eye( v)= -1m

near point (u)=-25cm

using lens formula

1/v - 1/u = 1/f

1/-100 - 1/-25 = 1/f

5 1

------------------ = -----------'

100 20

f=20 cm

1m=100cm

=0.2m

D =1/0.2

=5

hope it helps u

thanks...