Can someone please help with this question? I do not know what to do. Thanks. An

Question

Can someone please help with this question? I do not know what to do. Thanks.

An office administrator realizes that she is having trouble doing her work due to eyesight problems. Her optometrist determines that without glasses she can focus clearly only on objects that lie at distances between 0.580 m and 4.40 m. The optometrist decides that the best way to handle her eyesight problems is to prescribe bifocals that will give her normal visual acuity from 25.0 cm out to infinity. After trying on several pairs of glasses frames, they agree that the glasses will be a distance of 2.20 cm from her eyes. (a) Determine the refractive power needed for the bifocal part that will correct for her nearsightedness (include the sign). D What type of lens is needed? converging diverging (b) Determine the refractive power needed for the bifocal part that will correct for her farsightedness (include the sign). What type of lens is needed? converging diverging

Explanation / Answer

Part A)

For nearsightedness, we need to take objects from infinity and focus them at 4.40 m in front of the eye.

Since the lenses will sit 2.2 cm in front of the eye, the image will be formed 440 - 2.2 = 437.8 cm in front of the lens

Apply the lens equation...

1/f = 1/(infinity) + 1/-437.8

f = -437.8 cm

D = 1/-4.378

D = -.228 Diopters

This lens is diverging (D is negative)

Part B)

For the farsightedness, we will take objects at 25 cm from the eye and refocus them at 58 cm from the eye.

That is 25 - 2.2 = 22.8 cm from the lens and 58 - 2.2 = 55.8 cm from the lens.

1/f = 1/p + 1/q

1/f = 1/22.8 + 1/-55.8

f = 38.55

D = 1/.3855

D = 2.59 Diopters

This lens is converging (D is positive)

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