Question

the near point of hypermetropic eye is 1 m what is the power of lens required to correct this effect assume that near point of normalised 25cm

Answer 1

The convex lens is required to correct hypermetropic object in the eye.
object distance,u = 25cm
image distance, v= 1m =100 cm
by applying lens formula =1/f=1/v=1/u
= 1/f= 1/-100 - 1/-25= -1/100+1/25= -1+4/100= 3/100
now p= 1/ f×m= 3/100×100= 3D
hope it hlps u

Answer 2

Answer:near point of hypermetropic eye=V= -75cm

object distance=u=?

P=1D

As we know that P=1/f

p=100/f

f=100/p

f=100cm

From lens formula :

1/f=1/v-1/u

1/u=1/v-1/f

1/u=1/-75 +1/00 [ by sign conventions]

1/u=-4+3/300

1/u=-1/300

u= - 300cm

∴ the distance of distinct vision for him is -300cm