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 53.5 cm, what power reading glasses should be prescribed to treat this hyperopia? Assume that the distance from the eye to the lens is negligible.

ps: answer in diopters please

Explanation / Answer

Here ,

object distance , do = 25 cm

image distance , di = -53.5 cm

let the focal length be f

using lens formula

1/f = 1/di + 1/do

1/f = - 1/53.5 + 1/25

f = 46.93 cm

f = 0.47 m

power = 1/f

power = 1/0.47

power = 2.13 m^-1

the power of the glasses should be 2.13 m^-1

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