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$\cdot$ When an object is 16.0 $\mathrm{cm}$ from a lens, an image is formed 12.0 $\mathrm{cm}$ from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 $\mathrm{mm}$ tall, how tall is the image? Isit erect or inverted? (c) Draw a principal-ray diagram.

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a) -48cmb) 6.4 mmc) AB is objectA'B' is imageAnd $F$ is focal length

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

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.

05:48

When an object is $16.0 \m…

04:30

A lens forms an image of a…

08:34

07:17

An object is 16.0 cm to th…

03:07

04:39

An object is $16.0 \mathrm…

03:43

An object is 16.0 $\mathrm…

06:49

A diverging lens with a fo…

06:14

A converging lens forms an…

06:41

10:47

A converging lens has a fo…

04:08

04:40

You plan to project an inv…

All right. So in this problem, you're given for part a the object distance and the image distance. So we use his equation to find the focal length. Right? So the object distance 16 centimeters. The image distance, um, is 12 centimeters on the same side of the lens as the object. Ah, so it's a negative 12 there. Ah, because that's what we do in a CZ prime hears the same assess. That's this prime. Their turn. So half turns out to be negative. 48 centimeters. And so it is a negative focal length, meaning this is a diversion lens. All right, Part B, we want you first. Want to find the magnification, and that's a simple case of regular rest. Prior over s. So that's negative. Negative. 12. So that's a positive 12 over 16. So magnification is 160.75 Okay, so this is ahh, um, direct or upright image. And so why prime over Why is Emma's we know so image height. White prime is just 0.75 Time's object high of eight. 8.5 of a kind of 8.5 millimetres. Um, and so image height comes out to be a 6.38 millimeters. All right, and this is on direct image. Is he just mentioned and put? See, I've drawn the ray diagram here. Um, so you have the focal length 48 centimetres. That's way out there. And your object is within the focal ink. And you have you have, ah, diverging rents. And that's what this looks like. Ah, and your image is images and d magnified. So it is. It comes before the object. Ah, so there's your image that there's your object. And so I've drawn with the with the different colors of three ways that that travels from the object. So in blue, you have this one way that's traveling parallel to the access of this is a terrible drawing. But the picture ideas it's traveling parallel to the access and it gets refracted at the surface. And the reflected part of that the fact that Ray comes back to a diver Teo, to converge of the image point on. Then you have this red line that sells just going the surveillance going through the Vertex to the center, I should say of the lens, and so that one also converges there and finally, of this green ray, which upon refraction travels parallel to the access and the reflective part of that travels backwards on DH forms this image here.

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