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The near point of a hypermetropic person is 75 cm if the person uses eye glasses having power +1.0D, Calculate the distance of distinct vision for him?
- a)+400cm
- b)-400cm
- c)-300cm
- d)+300cm
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The near point of a hypermetropic person is 75 cm if the person uses e...
u = -75 cm
P = 1/f
f = 1 m = 100 cm
So, by lens formula ;
1/f = 1/v - 1/ u
So, 1/ v = 1/f +1/u
1/ v = 1/100 + (-1/75)
1/v = -1/300
v = -300 cm
P = 1/f
f = 1 m = 100 cm
So, by lens formula ;
1/f = 1/v - 1/ u
So, 1/ v = 1/f +1/u
1/ v = 1/100 + (-1/75)
1/v = -1/300
v = -300 cm
Most Upvoted Answer
The near point of a hypermetropic person is 75 cm if the person uses e...
Given:
Near point of hypermetropic person = 75 cm
Power of eye glasses = +1.0D
To find:
Distance of distinct vision
Formula:
Power of lens = 1/f, where f is the focal length of the lens in meters
Focal length of lens = 100cm/Power of lens, where power of lens is in dioptres
Calculation:
Power of lens = +1.0D
Focal length of lens = 100cm/1.0D = 100cm
Distance of distinct vision = Near point + Focal length of lens
Distance of distinct vision = 75cm + 100cm = 175cm
Therefore, the distance of distinct vision for the hypermetropic person is 175cm.
Option (C) is the correct answer.
Near point of hypermetropic person = 75 cm
Power of eye glasses = +1.0D
To find:
Distance of distinct vision
Formula:
Power of lens = 1/f, where f is the focal length of the lens in meters
Focal length of lens = 100cm/Power of lens, where power of lens is in dioptres
Calculation:
Power of lens = +1.0D
Focal length of lens = 100cm/1.0D = 100cm
Distance of distinct vision = Near point + Focal length of lens
Distance of distinct vision = 75cm + 100cm = 175cm
Therefore, the distance of distinct vision for the hypermetropic person is 175cm.
Option (C) is the correct answer.
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Community Answer
The near point of a hypermetropic person is 75 cm if the person uses e...
U = -75cm
P=1/f
so, f= 1m = 100cm
so, by lens formula, 1/f = 1/v - 1/u
so, 1/v= 1/f+1/u
1/v= 1/100 +(-1/75)
1/v= -1/300
v=-300cm
I hope this helps u
P=1/f
so, f= 1m = 100cm
so, by lens formula, 1/f = 1/v - 1/u
so, 1/v= 1/f+1/u
1/v= 1/100 +(-1/75)
1/v= -1/300
v=-300cm
I hope this helps u
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The near point of a hypermetropic person is 75 cm if the person uses eye glasses having power +1.0D, Calculate the distance of distinct vision for him?a)+400cmb)-400cmc)-300cmd)+300cmCorrect answer is option 'C'. Can you explain this answer?
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