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

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.25.0 cm. If a person has a near point of 49.5 cm,49.5 cm, what power reading glasses should be prescribed to treat this hyperopia? Assume that the distance from the eye to the lens is negligible.

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Answer #1

For a normal eye near point is considered to be 25 cm . For a hyperopia eye image of object placed at near point will be formed behind retina as shown in figure , hence for a hyperopia eye near point shifted away from eye and it will be greater than 25 cm. To correct this defect converging lens is used as shown in figure. Focal length of converging lens is determined using the lens equation

(1/v) - (1/u) = ( 1/f ) ....................(1)

v = lens-to-image distance = -49.5 cm ( Cartesian sign convention is followed )

u = lens-to-object distance = 25 cm

f is focal length of lens

we get , (1/f) = -(1/49.5) + (1/25) = 0.0198 cm-1 or f = 50.51 cm

power of lens = 1/0.505 = +1.98 D

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