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The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?​
  • a)
    diverging lens of power +1.25 D
  • b)
    converging lens of power +1.25 D
  • c)
    diverging lens of power -1.25 D
  • d)
    converging lens of power -1.25 D
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The far point of a myopic person is 80 cm in front of the eye. What is...
Far point of the defective eye, v = -80 cm

Object distance, u = -∞ (-infinity)

To find :

Nature and power of the corrective lens.

Solution :

1/v - 1/u = 1/f

1/f = 1/(-80) - 1/(-∞) 

1/f = - 1/80 + 0          [Since, 1/-∞ = 0]

1/f = - 1/80

f = -80 cm

Therefore, the corrective lens should be of the focal length 80 cm.

Power, P = 1 / focal length

As focal length is in centimetres, 1 m = 100 cm.

P = 100 / -80

P = -1.25 D

Therefore, the corrective lens is diverging or concave lens of power -1.5 D.
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Community Answer
The far point of a myopic person is 80 cm in front of the eye. What is...
The corrective lens is diverging or concave lens of power -1.25 D.
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