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.
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Answered by
16
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
Answered by
5
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
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