A nearsighted eye is corrected by placing a diverging lens in front of the eye.

Question

A nearsighted eye is corrected by placing a diverging lens in front of the eye. The lens will create a virtual image of a distant object at the far point (the farthest an object can be from the eye and still be in focus) of the myopic viewer where it will be clearly seen. In the traditional treatment of myopia, an object at infinity is focused to the far point of the eye. If an individual has a far point of 31.5 cm, prescribe the correct power of the lens that is needed. Assume that the distance from the eye to the lens is negligible.

Explanation / Answer

Solution: given:individual has a far point of 31.5 cm

to find out: the correct power of the lens that is needed

1/S +1/S' = 1/f

1/infinity -1/31.5 = 1/F

F = 31.5cm = -0.315 m

P = 1/F = -1/0.315 = -3.17 D

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