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If an eye has near point at a distance of 0.5m, What is the power of lens required to correct it?
- a)2 dioptre
- b)3 dioptre
- c)1 dioptre
- d)2.5 dioptre
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If an eye has near point at a distance of 0.5m, What is the power of l...
Since the person can't read an object at a distance of 50 cm, therefore focal length must be 0.5 m. We know that ;
P = 1/f , hence, P = 1/0.5 = 2 dioptre.
P = 1/f , hence, P = 1/0.5 = 2 dioptre.
Most Upvoted Answer
If an eye has near point at a distance of 0.5m, What is the power of l...
Explanation:
Power of lens is defined as the reciprocal of the focal length of the lens.
Formula for power of lens is:
P = 1/f
where P is power of lens and f is focal length of the lens.
Given, near point of eye = 0.5 m
For a normal eye, near point is considered to be at a distance of 25 cm or 0.25 m.
So, in this case, the near point is farther than the normal near point. This condition is known as hypermetropia or long-sightedness.
To correct this condition, a converging lens is used.
Converging lens has positive power.
Formula for focal length of a lens is:
1/f = (μ - 1) (1/R1 - 1/R2)
where μ is refractive index of lens material, R1 is radius of curvature of first surface, and R2 is radius of curvature of second surface.
For a thin lens, R1 and R2 are assumed to be much larger than the thickness of the lens.
So, the formula becomes:
1/f = (μ - 1) / R
where R is the radius of curvature of the lens surface.
For a converging lens, R is positive.
Let the power of lens required to correct the condition be P.
Then, using the formula for power of lens and the formula for focal length of a lens, we get:
P = 1/f = (μ - 1) / R
Here, μ is taken to be 1.5, which is the refractive index of the lens material used in most corrective lenses.
Substituting the values, we get:
P = (1.5 - 1) / R
Since the near point is farther than the normal near point, the focal length of the lens required is positive.
So, the radius of curvature of the lens surface is also positive.
We do not know the exact value of R, but we know that it is positive.
Therefore, the power of lens required is positive.
Option A: 2 dioptre is the correct answer.
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Community Answer
If an eye has near point at a distance of 0.5m, What is the power of l...
Since the person can't read an object at a distance of 50cm...therefore focal length must be 0.5m
we know that P=1/f... hence P=1/0.5=2 dioptre(S.I unit of power)
we know that P=1/f... hence P=1/0.5=2 dioptre(S.I unit of power)
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If an eye has near point at a distance of 0.5m, What is the power of lens required to correct it?a)2 dioptreb)3 dioptrec)1 dioptred)2.5 dioptreCorrect answer is option 'A'. Can you explain this answer?
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