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If a person with hypermetropia struggles to see objects placed closer than 50 cm, what is the power of the lens needed to correct this vision?
- a)+2 D
- b)-2 D
- c)+1 D
- d)-1 D
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
If a person with hypermetropia struggles to see objects placed closer ...
For hypermetropia where the near point is significantly further than normal, a convex lens with a positive power, such as +2 D, is required to converge light rays correctly on the retina.
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Most Upvoted Answer
If a person with hypermetropia struggles to see objects placed closer ...
Given: u= -25cm, v= -50cm
P= ?
P= 1/f (in m)
We have to find focal length.
Using lens formula
1/f = 1/v - 1/u
1/f = 1/(-50) - 1/(-25)
1/f = -1/50 + 1/25
1/f = -1 + 2/50
1/f = 1/50
f = 50cm
or, f = 0.5m
Now, P= 1/f(in m)
P = 1/0.5
P = +2 D
Hence the power of the lens needed to correct this vision is +2 D.
P= ?
P= 1/f (in m)
We have to find focal length.
Using lens formula
1/f = 1/v - 1/u
1/f = 1/(-50) - 1/(-25)
1/f = -1/50 + 1/25
1/f = -1 + 2/50
1/f = 1/50
f = 50cm
or, f = 0.5m
Now, P= 1/f(in m)
P = 1/0.5
P = +2 D
Hence the power of the lens needed to correct this vision is +2 D.
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Community Answer
If a person with hypermetropia struggles to see objects placed closer ...
Correction for Hypermetropia:
To correct hypermetropia, a converging lens is needed to help focus light rays properly onto the retina. In hypermetropia, the light rays coming from near objects focus behind the retina instead of directly on it, causing difficulty in seeing close objects clearly.
Given Information:
- Distance of near point = 50 cm
- Power of lens needed to correct vision
Calculation:
The near point of a person with hypermetropia is at 50 cm. The formula to calculate the power of the lens needed is:
\[
\text{Power (D)} = \frac{1}{\text{Focal Length (m)}}
\]
The focal length can be calculated as the reciprocal of the near point distance:
\[
\text{Focal Length (m)} = \frac{1}{50 \text{ cm}} = 0.02 \text{ m}
\]
Now, plug in the focal length into the formula for power:
\[
\text{Power (D)} = \frac{1}{0.02 \text{ m}} = 50 \text{ D}
\]
Therefore, the power of the lens needed to correct the hypermetropia of the person is +2 D.
To correct hypermetropia, a converging lens is needed to help focus light rays properly onto the retina. In hypermetropia, the light rays coming from near objects focus behind the retina instead of directly on it, causing difficulty in seeing close objects clearly.
Given Information:
- Distance of near point = 50 cm
- Power of lens needed to correct vision
Calculation:
The near point of a person with hypermetropia is at 50 cm. The formula to calculate the power of the lens needed is:
\[
\text{Power (D)} = \frac{1}{\text{Focal Length (m)}}
\]
The focal length can be calculated as the reciprocal of the near point distance:
\[
\text{Focal Length (m)} = \frac{1}{50 \text{ cm}} = 0.02 \text{ m}
\]
Now, plug in the focal length into the formula for power:
\[
\text{Power (D)} = \frac{1}{0.02 \text{ m}} = 50 \text{ D}
\]
Therefore, the power of the lens needed to correct the hypermetropia of the person is +2 D.
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If a person with hypermetropia struggles to see objects placed closer than 50 cm, what is the power of the lens needed to correct this vision?a)+2 Db)-2 Dc)+1 Dd)-1 DCorrect answer is option 'A'. Can you explain this answer? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If a person with hypermetropia struggles to see objects placed closer than 50 cm, what is the power of the lens needed to correct this vision?a)+2 Db)-2 Dc)+1 Dd)-1 DCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a person with hypermetropia struggles to see objects placed closer than 50 cm, what is the power of the lens needed to correct this vision?a)+2 Db)-2 Dc)+1 Dd)-1 DCorrect answer is option 'A'. Can you explain this answer?.
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