Ask your homework questions to teachers and professors, meet other students, and be entered to win $600 or an Xbox Series X 🎉Join our Discord! University of Wisconsin - Milwaukee ### Problem 12 Medium Difficulty # A certain child’s near point is 10.0 cm; her far point (with eyes relaxed) is 125 cm. Each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens–cornea combination vary? (b) Calculate the power of the eyeglass lens the child should use for relaxed distance vision. Is the lens converging or diverging? ### Answer ## a.$+60 \mathrm{diopters}$b.$-0.8 \mathrm{diopters}\$

#### Topics

Electromagnetic Waves

Wave Optics

### Discussion

You must be signed in to discuss.

### Video Transcript

in this problem. A child has a new point at 10 centimeter and she also has a fire point, which is that 125 centimeter. Also, it is given that each eye is two centimeter from Latina. So to solve the problem, eh of this, When the child is able to see the object at the Farpoint without using the corrective lis, she will have the objective object distance at 125 centimeter. So her object distance is going to be 1 25 centimeter. And that's gonna be 1.25 meter. Okay. And remember, in this case, he's going to be able to focus the Emmys. Andre, Tina, which is apt two centimeter from the the lens. Right. So that means Q is going to recall 22 centimeter, which is sirrah points here too. Meter. So now what we need to do here is to find the power for the Farpoint for the Farpoint, which is also one of her focal length for the Farpoint. It's basically one over p plus whatever Q and P is here going to BP Max and this is going to be won over 1.25 plus one of her 0.2 All right, so from here. So this is zero here, by the way. So whatever zubrin's over to from here, the power of we get is going to be positive. 50.8 die Attar's. So this is going to be power for the Farpoint. Now, to find the power for the near point, we're going to use, uh, near point as theon object distance. Okay, so for that p mean which is the object distance is going to be 10 centimeter and the images distance is the same, which is key calls to Senator. So from here, p near which is the power is going to be one over 0.1 plus one over 0.0 or two. Let me read this stone as 0.1 meter. Okay, so from here we will get positive. 60 diameters defer. The range of the power of the lens cornea combination is between 50.8 to 16 chapters. So this solved part B in part B. If the child wants to see a distant objects clearly, then we know that the distant object means P equals infinity and her corrective Lis would have to be what it would have to form a virtual and upright images at her far points, which means Q would have to be a far point. Right? So the Q in this case would be at negative 1 25 centimeter. We also missed negative 1.25 meter. Okay, so now, in this case, the power would be one of her ass. Plus so whatever ethical sze whatever f because one of her p plus one of her cute and one of her p is one of the infinity, which is zero. So it's gonna be zero plus one. Virtue is going to be won over negative 1.25 And this is going to give us the power to be negative. 0.8 dieters. All right, so this is the power of the leads. Okay, Now, based on this, you can see the power is the negative. So that means the less is the diversion glass. So the lenses divers ing

University of Wisconsin - Milwaukee

#### Topics

Electromagnetic Waves

Wave Optics