An artificial lens is implanted in a person\'s eye to replace a diseased lens. T

Question

An artificial lens is implanted in a person's eye to replace a diseased lens. The distance between the artificial lens and the retina is 2.78 cm. In the absence of the lens, an image of a distant object (formed by refraction at the cornea) falls 5.00 cm behind where the implanted lens would be inserted. The lens is designed to put the image of the distant object on the retina. What is the power of the implanted lens? Hint: Consider the image formed by the cornea to be a virtual object.

I keep getting 7.14 D. If anyone could provide the correct answer, I will provide lifesaver ratings. I've tried it numerous times but cannot get the correct answer!

Explanation / Answer

if we consider the image formed by cornea as object for artificial lens and applying the general lens formula

1/v - 1/u = 1/f

where u = +5 cm (since in the positive direction)
v= + 2.78 cm
1/f= 1/2.78 - 1/5

calculating from here f= 6.261 cm = 0.0626 m

power = 1/f = 1/0.0626 = 15.97 D

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