A farsighted eye is corrected by placing a converging lens in front of the eye.

Question

A farsighted eye is corrected by placing a converging lens in front of the eye. The lens will create a virtual image that is located at the near point (the closest an object can be and still be in focus) of the viewer when the object is held at a comfortable distance (usually taken to be 25.0 cm). If a person has a near point of 71.5 cm, what power reading glasses should be prescribed to treat this hyperopia? Assume that the distance from the eye to the lens is negligible.

Answer: _____ diopters

Please provide equations or explanations.

Explanation / Answer

1/S + 1/S' = 1/F

(1/25) - (1/71.5) = 1/F

F = 38.44 cm

P = 1/F (F in m)

P = +1/ 0.3844 = + 2.6 D

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