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$\cdot$ The rod of the previous problem is immersed in a liquid. Anobject 90.0 $\mathrm{cm}$ from the vertex of the left end of the rod and onits axis is imaged at a point 1.60 $\mathrm{m}$ inside the rod. What is therefractive index of the liquid?

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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

Simon Fraser University

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.

03:00

The glass rod of Exercise …

03:38

01:59

One end of a cylindrical g…

01:30

Find the mass of a rod of …

are from the previous problem we have that the radius of this classroom is three centimeters and it's a convex lens. So that's a positive three. Um, and and then on down the index of refraction Ah, it's college and subi of this. Now this class is 1.60 Come. Ah, here we have this problem. We have ah object. Distance is 90 centimeters Image distance as prime That's 1.6 meters and remember, 1.6 meter one meter is 100 centimetres so that's 160 centimetres. OK, so we use Equation 24.11 from the book which is that n a over image distance ah index of refraction of medium and medium, from which like this coming plus and B that Cenex of perfection to which light is going over image distance as his object distance That's prima's image distance, eyes a quote too and B minus Anna over the radius of curvature are okay, so any is the only unknown in our and in this equation equation here we don't know and s so we have to solve for a name. So here it is, and a over 90 centimeters plus and B, which we know is one point 1.6 zero over 1 60 centimetres. Eyes equal to 1.60 minus and a over three centimeters. Because that's our and and we take in a common and we pull out this end over native any of a three centimeter term to the left we get is one over 90 plus one over three times and there is equal to way notice at 1.6 is common for the remaining two terms. So we take that 1.6 times. One over three Ah, a minus one over 1 60 No. And so you work out these the two different sides and you get that any is 1.5 to see index of refraction off of this liquid medium.

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