## The near point of a patientâ€™s eye is 50.0 cm. What power (in diopters) must a corrective lens have to enable the eye to see clearly an object 25.0 cm away?

The near point of a patientâ€™s eye is 50.0 cm. What power (in diopters) must a corrective lens have to enable the eye to see clearly an object 25.0 cm away?
A. 2 diopters
B. 2.5 diopters
C. 0.5 diopters
D. 3 diopters
Correct Option: Answer is A
Hereâ€™s how we can find the corrective lens power:
1. Focal length: We first need to determine the focal length of the corrective lens required to bring the object at 25.0 cm into focus for the patientâ€™s eye with a near point of 50.0 cm. Use the lens formula:
1/f = 1/p â€“ 1/q
where:
â€¢ f is the focal length of the lens
â€¢ p is the object distance (25.0 cm)
â€¢ q is the image distance (50.0 cm, as the object will be focused at the near point)
Solving for f, we get:
f = 1 / (1/25 â€“ 1/50) = 50 cm
2. Diopter conversion: Convert the focal length to diopters by multiplying by 100:
Power (in diopters) = 100 / f = 100 / 50 cm = 2 diopters
Therefore, a corrective lens with a power of 2 diopters will enable the patient to see clearly an object 25.0 cm away.
Checking the other options:
â€¢ B. 2.5 diopters and C. 0.5 diopters are too weak and wouldnâ€™t bring the object into focus at the near point.
â€¢ D. 3 diopters is too strong and would overcorrect the vision, making objects closer than 25.0 cm blurry.
So, the answer is A. 2 diopters.

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