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$\cdot$ The diameter of Mars is $6 / 94 \mathrm{km},$ and its minimum distance from the earth is $5.58 \times 10^{7} \mathrm{km}$ . (a) When Mars is at this distance, find the diameter of the image of Mars formed by aspherical, concave telescope mirror with a focal length of 1.75 $\mathrm{m} .$ (b) Where is the image located?

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Diameter of the image is 0.213 $\mathrm{mm}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 24

Geometric Optics

Electromagnetic Waves

Reflection and Refraction of Light

University of Michigan - Ann Arbor

Hope College

University of Sheffield

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.

08:14

The diameter of Mars is $6…

07:41

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02:38

Find the diameter of the i…

02:47

When Mars is nearest the E…

02:44

The telescope at Yerkes Ob…

02:34

03:28

The moon's diameter i…

okay. And this problem, we're trying to image Mars. And so we know that Mars is 5.58 kilometres from Earth. So our times 10 of the seven kilometers So we'll call that the image distance by 0.57 times. So times 10 cubed times 10 to the seventh. So that's times 10 to the 10 meters and the diameter reduced double Check this question. There's a little bit of a typo and the thing I'm reading. So I just want to make sure I'm using the right number six of Hey 6794 kilometers is the diameter can glad I checked so we can call that like the height, basically, or in this book you're using. Why? So I'll go ahead and use why as well. 6794 And I think that was kilometers. Yeah. Times 10 to the three meters. And our goals took up the diameter of the image formed by this telescope mirror. Um and the focal length is 1.75 meters. So, um, let's get where the image is located first, let's with the second part of the question asks using one over ass plus one over us Prime equals one over off algebraic Lee and manipulating this. Let me see if they actually do it in the book. So I could just start citing a formula. They probably don't. But let me just check. It will save me a lot of work on these problems. A lot of writing of algebra. Oh, um, kind of No, not really. Okay, I guess I don't. So algebraic Lee, manipulating this we want to solve for us prime by bringing the one over s term here. So you get one over s prime is equal to one over half my ass, one over us, and then as prime is the inverse of this. So I'm just gonna not algebraic Lee simplify it. That will make it easier on me. Um, okay, so it's just basically put all of the calculator and then do the minus one. So I'm gonna go ahead and positive video and do that, and I'll tell you what I get. Okay, so, um, I actually got 1.75 is the position when I plugged all these numbers and I guess if I stopped and thought about it before I plugged in the numbers. I could have predicted it because the distant if the distance is really far. Um, if, as it's really big compared to the focal length, then this term basically goes to zero. And so it's just the inverse of the focal length. Um S o then, too. So that would be the position that would be the answer to be. And then the magnification it's equal to in the book. It has a formula y prime over. Why is asked prime over equal to negative s crime over s. So then, um, I'll go ahead and write that out. Its primary were as equals, minus groups of minus. Should be over here. Um, why prime over? Why? So why prime is the height of the image and why is the height of the object therefore, um, why prime is equal to why multiplied by negative s prime over us and plugging that into a calculator, a guy got 0.213 um millimeters so it will make a small image

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