The near point of a patient’s eye is 50.0 cm. What power (in diopters) must a corrective lens have t
Source: JAMB · 2023
The near point of a patient’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?
Explanation
The lens must make an object at 25 cm appear as a virtual image at the eye’s near point, 50 cm away.
1/f = 1/u + 1/v = 1/25 − 1/50 = 1/50, so f = 50 cm = 0.5 m.
Power = 1/f = 1 ÷ 0.5 = 2 diopters.
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