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University of Wisconsin - Milwaukee

# A patient has a near point of 45.0 cm and far point of 85.0 cm. (a) Can a single lens correct the patient’svision? Explain the patient’s options. (b) Calculate the power lens needed to correct the near point so that the patient can see objects 25.0 cm away. Neglect the eye–lens distance. (c) Calculate the power lens needed to correct the patient’s far point, again neglecting the eye – lens distance.

## a. \text { two lenses }b. 1.78 \text { diopters }c. -1.18 \text { diopters }

#### Topics

Electromagnetic Waves

Wave Optics

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### Video Transcript

in this problem, a patient has a New York point at 45 centimeter and Farpoint at 85 centimeter. So to solve the part in this problem, the answer to whether a single corrective lens can solve his problem is yes, it can, since the patient needs corrective reason for both near vision and far reason he can use by focal lens. Okay, so the answer to that is yes. Or alternatively, he can also use two pairs of glasses, one for reading and one for distance wearing. So for part B to correct the near of reason. If the patient is looking at the object at the normal near point, then the object distance would be Peco's 25 centimeter that we write down in meters 0.25 meter. Okay, then a virtual upright image should form at his actual near point. Right? So cue, in that case would be the actual near point Negative value 45 centimeter. I'm gonna write this down in mirrors to a 0.45 negative zero points for five meter now from here. If you want to calculate the power, we can just use the list equation. So the lens equation is one of our f. The power equals one over half in meters. Your calls. One of her p plus one of Rick you. And this would be one of her 0.25 minus 1/0 0.45 And this would keep me positive. 1.7 age dieters. Not too good, too part See me, which is to create the fire of Reason. If the patient is looking at the fire object, which means the optic distances at infinity, then virtual upright image should form at the fire point. That means the aim is distance has to be the negative off the 85 centimeter. I'm gonna write this down in meters, since, well, ever 0.85 So I'm going to remind you again. The sign negative here is because the maze is virtual and in front of the list, all right. And the same case goes for Part B s. What? So in this case, if we want to find the power far reason the previous one was near vision. This one is for far vision. We're going to use the Lindsay question again. Our records one of rex, which is going to recall the one over P, which is infinity and plus one over 85. So one over infinity minus 1/85. Because you can see that the Q has negative matter there. And this one is 0.85 actually in terms of meter. So this will give me negative 1.18 Diack tres no.

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics