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.. Figure 24.47 shows an object and its image formed by athin lens. (a) What is the focal length of the lens and what typeof lens (converging or diverging) is it? (b) What is the heightof the image? Is it real or virtual?

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a) 58.67 cm, converging lensb) Virtual

Physics 102 Electricity and Magnetism

Physics 103

Chapter 24

Geometric Optics

Electromagnetic Waves

Reflection and Refraction of Light

University of Michigan - Ann Arbor

Simon Fraser University

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.

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$\bullet$ Figure 24.46 sho…

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$\bullet$ Figure 24.45 sho…

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Figure 24.44 shows an obje…

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Figure 34.58 shows an obje…

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Figure 34.57 shows an obje…

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The figure below Shows an …

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Figure $\mathrm{P} 34.85$ …

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An object is a distance $2…

we do part air by using the Finland's equation says that one over object distance plus one over image distance is equal to one over focal length. Okay, so in this problem we have object distance of 16 centimeters. We have image distance, which is 16 plus six. But there is a negative before it. And this is because the images on the same side of the lens as the object. So that is why it's ah negative 22 centimeters instead of positive 22 centimeters. And so therefore you have won over 16 plus one over negative 22 equals one over f. And so one over f is just one of her 16 minus one over 16 2122 s o f. The focal length is 58.7 centimetres. These air, all centimeters s o a positive 58.7 centimeters. So this means that efforts were zero positive, which means that it is a converging lens. Ah oh, and part B. He's magnification and equals negative of image, distance over object distance and so negative of negative. 22 is just 22 over 16 and so magnification is one point 38 and notice that. And so Ama's positive meaning this this is, ah, upright image as we expected From the diagram on DSO, This remember am is also equal to image height over object hype. So image height, Why? Prime is just 1.38 times object. Heightened riches, 3.25 millimetres. And so this gives us 4.48 millimeters for the image height. So the image is magnified relative to the object. And finally, since s prime is on the same side of the mirror and is negative, uh, this is a virtual image.

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