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Approximating the eye as a single thin lens 2.70 $\mathrm{cm}$ from theretina, find the focal length of the eye when it is focused on anobject at a distance of (a) 255 $\mathrm{cm}$ and (b) 25.5 $\mathrm{cm} ?$

a) 2.67 $\mathrm{cm}$b) 2.44 $\mathrm{cm}$

Physics 103

Chapter 27

Optical lnstruments

Wave Optics

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University of Michigan - Ann Arbor

University of Sheffield

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Okay, so and this problem again, we have on I producing an image of the object. So we have the object. I'm here, we have the lens of the eye and here we have the retina in here and we know that a distance from the object to the lens, it's just zero in the distance from the lands to directing a It's just e I. And this lens has a focal lens, but the problem was to discover what it is. Defoe co lens in this problem. Let's see when the eye is focused in the object at a distance. First off zero equals 255 Same team enters in second and a distance off 25.5. Same team monitors. Okay, so what? We know, we know that we can use the fill in's equation to discover the focal length. This is just going to be f equals one divided by yeah, lost one divided by zero. All of this to the hour of minus one. Okay, so let's carefully this the first item on a distance off two, about 255 centimeters away. So what's currently it's the first. I think it's just going to be. It's cool f one just going to be one divided by 2.7. That is the distance. The problem gives us off the I plus one divided by 255 all of this to the power of minus one. So this is just going to be equal to two point 67 Sancti mergers. The second all right and we have the second focal land. Using the same equation is to have one divided by 2.7 close one divided by 25.5 all just to the power of minus one. This is going to be 2.44 saying.

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