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University of Wisconsin - Milwaukee

# 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 2.80 cm. In the absence of the lens, an image of a distant object (formed by refraction at the cornea) falls 5.33 cm behind the implanted lens. 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.

## +16.9 \text { diopter } s

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

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### Video Transcript

So this is the problem off an artificial lens implantation. So the first thing is the distance between the artificial lens and Regina is actually the name is distance of the lens. So let's write that down. So that's gonna be a miss distance. It calls. Q. That is gonna be 2.8 centimeter. And now, without this lens, the cornea forms and emails at 5.33 centimeter behind Alliance and this can be considered as virtual optic distance. So that means the object distance P because negative 5.33 centimeter. All right, so now let's just use the Lindsay equation, which is one over F because one of her p class one of a cue. So remember, the definition for the power is also the reciprocal of the focal length, which is one of Raph. So what I'm gonna do here is I'm just gonna write down power because won over F equals one of her P plus one over Q. And that is gonna be 2.8. Send a reader minus 5.33 centimeter. Okay. Minus 5.33 centimeter over 2.8 times minus 5.33 and this is going to give a 16 point nine Diop tres. All right, so this is gonna be 16.9 doctors.

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics