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Focus of the eye. The cornea of the eye has a radius of curvature of approximately $0.50 \mathrm{cm},$ and the aqueous humor bbehind it has an index of refraction of $1.35 .$ The thickness ofthe cornea itself is small enough that we shall neglect it. The depth of a typical human eye is around 25 5 $\mathrm{mm}$ . (a) What would have to be the radius of curvature of the cornea so that it alone would focus the image of a distant mountain on the retina, which is at the back of the eye opposite the cornea? (b) If the cornea focused the mountain correctly on the retina as described in part $(a),$ would it also focus the text from a computer screen on the retina if that screen were 25 $\mathrm{cm}$ in front of the eye? If not, where would it focus that text, in front of orbehind the retina? (c) Given that the cornea has a radius of curvature of about $5.0 \mathrm{mm},$ where does it actually focus the mountain? Is this in front of or behind the retina? Does thishelp you see why the eye needs help from a lens to completethe task of focusing?

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a) 6.481 $\mathrm{mm}$b) 27 $\mathrm{mm}$c) 19.3 $\mathrm{mm}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 24

Geometric Optics

Electromagnetic Waves

Reflection and Refraction of Light

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

McMaster University

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.

07:54

Focus of the eye. The corn…

05:54

The cornea of the eye has …

08:50

09:55

09:31

Focus of the Eye. The come…

06:17

The cornea behaves as a th…

04:46

05:52

Resolution of the Eye. The…

03:45

The cornea as a simple len…

05:38

The maximum resolution of …

02:16

As objects are moved close…

06:03

Figure $34-47 a$ shows the…

05:34

No, I said throughout this problem will be using this formula. Ah, but refractive index of medium a over object distance from refracting object plus for fact of index about meeting be over image distance from refracting object is equal to ah, the difference over the radius of curvature radius off the res of curvature off the refracting object. So in part and we know the object is at a very large distance. It's a distant object and distant object means we can make this approximation that distance from object to reflecting surfaces very large. So one of a very large number is approximately zero. So this first term drops out okay. Ah, so we're left with this proactive index off media be over s prime equals and B minus and a over R and R is what we want. So medium be would be their refractive index specified in the problem 1.35 image distance is 2.5 centimeters on DH. That's equal to 1.35 minus 11 being refracted the necks of air over our and so this gives us our equals 0.648 centimeters or 6.48 millimeters. Okay, Uh, part B. Wei don't make this approximation anymore. So we have all the terms of this equation and part B. We have a fraction Vanek Severa one over the distance of the text from the cornea. And so that would be 25 centimeters plus 1.35 over as prime, which is where the image will form. And that's what we don't know here, a sequel to 1.35 My Swan over R and R. We just found the last problem are in the last part, and so are is 0.64 eight centimeters, right? So this works out to 1.35 over us. Prime equals 0.5, so you can see that as prime is 22.7 centimeters, which is equal to 27 millimeters. We were given in the problem that the depth of the average human eyes about 25 millimeters and so this image is formed deeper than the than the eye. So it's it's form. So if we consider the retina as those farthest deepest point of the eye, this image it's formed behind the retina and finally heart Syrians again we go back to the states. An object approximation so that this term doesn't factory. So again, you have this equation here, SL for part B part. See, you have 1.35 over s prime. That's what you don't know. Is it cool? 2.35 over the actual radius of curvature, which is 0.5 centimeters. And so, as prime comes out to be 1.93 centimeters or 19.3 millimeters and so this is less than s O. This is within the death of the eye. And so this ah tells us that the image is in front his form in front of the retina, which means that you need additional lens to focus the image that the mix

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