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$\cdot$ A converging lens with a focal length of 7.00 $\mathrm{cm}$ forms animage of a 4.00 -mm-tall real object that is to the left of the lens. The image is 1.30 $\mathrm{cm}$ tall and erect. Where are the object and image located? Is the image real or virtual?

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$S^{\prime}=-15.8 \mathrm{cm}$

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

University of Washington

University of Sheffield

University of Winnipeg

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.

05:52

A converging lens with a f…

06:42

06:54

03:51

02:00

01:41

03:49

02:28

05:23

03:12

03:05

02:33

A 1.00 - cm-high object is…

06:38

A $1.00-\mathrm{cm}$ -high…

02:57

An object that is $6.00 \m…

and this problem. We're told that we have a converging lens the focal length of seven centimeters. Go ahead and write. F is equal toe positive. Seven centimeters because it's a converging lens and, ah, it forms an image. And then the object is four millimeters tall and help convert that to centimeters. So we can say, why is point for 00 centimeters and the images to the left of the lens? So that means that the images has a negative length. And so we can say, um, we'll just kind of remember that real object. Oh, yeah, and then the images 1.37 So we can say why it prime is 1.3 oh, centimeters. It's tall and erect, so we know it's positive. I guess it doesn't have to be tall. Um, and then we want to know where the objects and images and is it really a virtual? And if it's to the usually the object is to the left of the lens. Um, so I'm just gonna Yeah, I'll answer the image. Is that image really virtual cause? I'm not sure if I can assume that the object is to the left of the lens, which is sort of standard for these problems. So I'm just gonna kind of write down the equations and, um, kind of hope that it pops out there. So here's our givens and let's write down the relevant equation. The first one's gonna be the thin lens one. So that's one over ass Plus one over X Prime equals one over f. And keep in mind, we don't know what those are. So I guess that I should put a question like, um and the other thing we know is that the magnification waken say magnification is equal to y prime over why? And then that's equal to negative s prime over us. And so we can say that, um, so we basically have two unknowns here again. Asked prime and ask but two equations to unknowns. And if they're not secretly the same equation we can solve for either. So I'm gonna go ahead and do this algebraic lee first. But if you don't feel comfortable with that, you can just do it numeric like hug in numbers. So, um, I'm just solve for us prime, and I'm gonna say ask. Prime is equal to minus us. Why Prime divided by why? Just by cross multiplying. And then I'll take that, implying that into this equation here. And so, um, doing that I get one over us plus one over s prime is gonna be negative. Why over us? Why? Prime equals one over f We can take out s so one over us is equal toe one minus. Why? Over why Prime equals one over f And then you can sort of cross multiply. So bring us to this side and in the after this side and you got the ass is equal thio f times one minus. Why? Over? Why prime? And I'm gonna go ahead and calculate that is kind of these in my head. Uh, I wish I could I don't think I can, though. Um, someone a pause, the video, I do this calculation. Okay, I got 4.85 centimeters and this is positive in Let's go ahead now. Insult for asked. Prime. So I got us prime. It's just gonna be using this equation. It's minus 4.85 times 1.3 over four. So plugging that into a calculator, I got that. It's just gonna do this well, positing 4.85 times 1.3 over 0.4, I got 15 point it centimeters minus 15.8 centimeters. Okay, so if the object is has comes out negative, the object position comes out negative. That means it's on the same side has the object. We're told it's on the left. Therefore, the object is also on the left. So this is gonna be to the left of the lens. And then this will also be by the left of the lens, because that's what they told you. And what else do they want to know? And then I wasn't really our virtual kind of say suspected. Since the image has a negative, um, length, it's virtual.

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