Question

a. What power contact lens is needed to correct the vision of a person who has...

a. What power contact lens is needed to correct the vision of a person who has a near point of 61.8 cm and allow him to see objects 25.0 cm away?

b. Is this person nearsighted or farsighted?
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Answer #1

a) Assuming the spectacle (corrective) lens is held 1.50 cm away from the eye by eyeglass frames.

When an object is held 25.0 cm from the person’s eyes, the spectacle lens must produce an image 61.8 cm away (the near point). An image 1.00 m from the eye will be 60.3 cm to the left of the spectacle lens because the spectacle lens is 1.50 cm from the eye. Therefore, di=−60.3cm. The image distance is negative, because it is on the same side of the spectacle as the object. The object is 23.5 cm to the left of the spectacle, so that do=23.5 cm.

di = -60.3 cm since the eye would be 1.5 cm from the lens

do = 23.5 cm

Since di and do are known, the power of the spectacle lens can be found using

P=1/do+1/di

P = 1/(-60.3) + 1/(23.5)

P = 2.597 D

b) Since the power is positive in this case we have a person with farsightedness

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