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$\bullet$ A converging lens with a focal length of 12.0 $\mathrm{cm}$ formsvirtual image 8.00 $\mathrm{mm}$ tall, 17.0 $\mathrm{cm}$ to the right of the lensDetermine the position and size of the object. Is the imageerect or inverted? Are the object and image on the same side oopposite sides of the lens?

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$s=7 \mathrm{cm}$$y=3.31 \mathrm{mm}$The image is erectThe image is virtual, so the object and the image are in the same side of the lens

Physics 102 Electricity and Magnetism

Physics 103

Chapter 24

Geometric Optics

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

Rutgers, The State University of New Jersey

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.

05:29

A converging lens with a f…

04:19

A converging lens with foc…

09:32

08:18

01:00

03:46

02:51

A converging lens forms an…

06:49

A diverging lens with a fo…

06:01

06:41

02:38

All right, so we have to find ah, distance and height of the object. So we're given image distance on Were Given Focal Inc So we have that one of her focal ing is one of our an object, a senseless woman for image distance. So one of her image distance are rather one of the object distance we want here is equal to one over F minus one over X prime ad s prime will be negative, 17 centimeters. And that's because it's on the right side of the lens. Nah, as they are right side of lens. And so it's a conversion lands. And if you're on the right side of this converging lens, your value you will be a negative distance. And that's just the saint invention we use. So it's a negative there. So we put those in for going to 12 centimeters minus one over minus 17 centimeters of the first sister. Plus here of the values and so s the image. This object distances seven centimeters from the lens on then and then we can figure out magnification from that. It's negative image, distance over object distance. So that's minus minus 17. That's a plus 17 over seven. So magnification is plus 2.4 uh, plus 2.4 times. And so and so this is also equal to image height over object. Hi. So, object height is just image height of the magnification. And so that's Tio. Uh, oops. That's 0.8 centimeters. It's eight millimeters over 2.4. So its point object height as 0.36 centimeters. Whoops, that's ah, 4.34 centimeters or 3.4 millimeters. So since M is greater than zero Ah, this image is direct and not inverted on DH in this instance s prime is negative and s is a CE prime is negative and s is positive. This implies that the image an object are on opposite sides So object will be crowd mission object on opposite side so object will be on the left and image will be on the right

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