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. A nearsighted eye. A certain very nearsighted person cannot focus on anything farther than 36.0 $\mathrm{cm}$ from the eye. Consider the simplified model of the eye described in problem $49 .$ If the radius of curvature of the comea is 0.75 $\mathrm{cm}$ when the eye is focusing on an object 36.0 $\mathrm{cm}$ from the comea vertex and the indexes of refraction are as described in problem 49 , what is the distance from the cornea vertex to the retina? What does this tell you about the shape of the nearsighted eye?

2.77 $\mathrm{cm}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 25

Optical Instruments

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

University of Michigan - Ann Arbor

University of Washington

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

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A certain very nearsighted…

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05:55

Some myopic person is unab…

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Curvature of the Cornea. I…

01:37

The distance from the vert…

03:55

$\bullet$ Curvature of the…

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In a simplified model of t…

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Measurements on the cornea…

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Suppose that the lens syst…

Okay, So we were going to use this formula and a over image distance over object a sensory plus and be over image distances e call to their difference over radius of curvature, house ones and so on. From the previous problem, we know that on a is air So that's one Ah, and B is one point for 1.40 or 1.4 Ah and radius of curvature. And this problem is given to be 0.75 centimeters s. So all we need to do is find that's prime because this has also given. So let's write them down. S is 36 centimeters. Ah, that Then we have 1.4 over as prime is equal to 1.4 minus one over 10.75 centimetres. Ah, and therefore a CZ prime is 2.77 centimetres or 27.7 millimeters is where the image is formed

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