Question

The near point of a hypermetropic eye is 1 m . What is the power of the lens required to correct this defect

Answer 1

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...

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...