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Repeat Exercise 34.24 for the case in which the end of the rod is ground to a $concave$ hemispherical surface with radius 4.00 cm.
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Physics 103
Chapter 34
Geometric Optics
Section 3
Refraction at a Spherical Surface
Reflection and Refraction of Light
Wave Optics
Hope College
University of Sheffield
University of Winnipeg
McMaster University
Lectures
02:51
In physics, wave optics is the study of the behavior of light, or other electromagnetic waves, as they interact with matter. It is a branch of classical optics. While the term "optics" usually refers to the study of light propagation in free space, "wave optics" is used when it is important to take into account the effects of the material the light is traveling through. Wave optics is mostly concerned with the behavior of waves near the boundary between different media. When waves from a single source interact with each other their behavior can be difficult to predict. For example, two waves of the same frequency and wavelength can interfere, leading to a phenomenon known as interference fringes.
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.
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Okay, so we're getting Chapter 34. Problem 25 here. So this problem wants us to repeat exercise 34 24 for the case, which the end of the rod is Kong cave instead of convex. And the radius of curvature being four centimeters. So for a con cave spirit, Kal lands like this. The radius of curvature for per our user. Usual sign convention is given that negative. So this will be negative. Four centimeters. Um, some other things they gave us in the last problem was the object here is 1.5 millimeters tall. They tell us the object distance here is s equals 24 centimeters in a is just one for air. And envy is 1.6 for this glass rod. So that should be everything we need to know. And what we want to find is the image distance as prime in the image. Hyde White prime. Okay, so again, we're going to use our Lindsay equation in the medium of year. So we have an A over s plus and be over s prime equals and B minus in a over our cool pluck. That end and I should see one over 24 plus 1.6 over us. Crime equals 0.6 over negative four. We re arrange that sulfur s crime and we shouldn't see that this is negative. 38 4 over 4.6. And this equals negative 8.35 centimeters and its negative because it's on the same side of the objects. That means 8.35 centimeters to the left of the rod. So this is our image distance. This is where it is. So now we want to figure out the height of it. So it's bring our magnification equation in him, which is in equals. Um, let's see, this is negative in A s prime over B s self replaying this and we should see this becomes positive cause negative in the negative. Cancel 8.35 over 1.6 times 24 school again plugging this in our calculator we see this comes out to a positive 0.217 is our magnification. So from the magnification, we can get the height because why prime is in times why with absolute values. So like prime is 0.217 times 1.5 millimeters. Cool. Getting plugging that in our calculator should see this comes out to being zero for 3 to 6 millimeters is why prime? That's the image height. So our image is virtual is close on the same size of the object, 8.8 point 35 centimeters to the left of the rod, and that is the height of 350.3 to 6 millimeters.
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