Question

The near point of a hypermetropic eye is 75cm if the person uses eye glasses having power +1.0D, calculate the distance of distinct vision for him.

Answer 1

GIVEN :

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

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