The near point of a hypermetropic eye is 1m. What is the power of lens required to correct this defect? Assume that near point of the normal eye is 25cm.
Answers
Answered by
28
By question we came to know that v=-100cm and u=-25
ok now
by applying lens formula
we have 1/f=1/v-1/I
1/f=1/-100-1/-25
1/f=1/-100+1/25
1/f=1/25-1/100
1/f=4-1/100
1/f=3/100
f=100/3=33.33 cm or in m=0.33m
now P=1/f
P=1/0.33
P=100/33
P=+3.03 or +3D ANS
HOPE THIS ANSWER HELPS PLS MARK IT AS THE BRAINLIEST.Goodnight
ok now
by applying lens formula
we have 1/f=1/v-1/I
1/f=1/-100-1/-25
1/f=1/-100+1/25
1/f=1/25-1/100
1/f=4-1/100
1/f=3/100
f=100/3=33.33 cm or in m=0.33m
now P=1/f
P=1/0.33
P=100/33
P=+3.03 or +3D ANS
HOPE THIS ANSWER HELPS PLS MARK IT AS THE BRAINLIEST.Goodnight
L12345:
it the ANSWER correct
do you know the answer
pls let me know
Sorry bro.
My answer was 5D but in my book it is 3D.
is it correct now
KRISHNA 725 IS IT CORRECT
it is correct i checked. but i dont understand why u took v= -10 cm
I had done something different.
Answered by
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
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