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(N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.28, what do you expect the critical angle acrit to be (in degrees)? Cladding- Core → core refracted reflected

(N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.28, what do you expect the critical angle acrit to be (in degrees)? Cladding- Core → core refracted reflected

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Understanding Fiber-Optic Cable Refraction Experiment** Consider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence \( \alpha \). When a ray strikes the end of the fiber, it is bent toward the normal. By Snell's law, the angle inside the core is \( \sin \theta_{\text{core}} = \frac{n_0}{n_1} \sin \alpha \). Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding \( i \) is the complementary angle of \( \theta_{\text{core}} \). If \( i \) is a small angle, the ray will propagate into the cladding and be lost from the fiber. If \( i \) is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is \( \sin i_{\text{crit}} = n_2/n_1 \). In the experiment, you cannot directly measure angles inside the fiber but you can measure \( \alpha \). Using the facts that: (i) for complementary angles, \( \sin \theta_{\text{core}} = \cos i \), (ii) the trigonometric identity \( \cos i = \sqrt{1 - \sin^2 i} \), and (iii) the index of refraction of air is \( n_0 = 1 \), it can be shown that \( \sin \alpha_{\text{crit}} = \sqrt{n_1^2 - n_2^2} \). This quantity is known as the numerical aperture (N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.28, what do you expect the critical angle \( \alpha_{\text{crit}} \) to be (in degrees)? --- **Diagram Explanation:** The diagram below illustrates the process: - **Cladding**: The outer layer that surrounds the core. - **Core**: The central part of the fiber-optic cable where light travels. - **Ray Path**: - The incident ray enters at angle
Transcribed Image Text:**Understanding Fiber-Optic Cable Refraction Experiment** Consider a ray incident on the end of a fiber-optic cable as illustrated in the figure below. In the experiment, you will explore the behavior of the fiber as a function of the angle of incidence \( \alpha \). When a ray strikes the end of the fiber, it is bent toward the normal. By Snell's law, the angle inside the core is \( \sin \theta_{\text{core}} = \frac{n_0}{n_1} \sin \alpha \). Once inside the core, the ray travels until it strikes the cladding of the fiber. The angle of incidence at the cladding \( i \) is the complementary angle of \( \theta_{\text{core}} \). If \( i \) is a small angle, the ray will propagate into the cladding and be lost from the fiber. If \( i \) is large, however, the ray will be internally reflected and bounce down the fiber. By Snell's law, the critical angle for the ray to be internally reflected is \( \sin i_{\text{crit}} = n_2/n_1 \). In the experiment, you cannot directly measure angles inside the fiber but you can measure \( \alpha \). Using the facts that: (i) for complementary angles, \( \sin \theta_{\text{core}} = \cos i \), (ii) the trigonometric identity \( \cos i = \sqrt{1 - \sin^2 i} \), and (iii) the index of refraction of air is \( n_0 = 1 \), it can be shown that \( \sin \alpha_{\text{crit}} = \sqrt{n_1^2 - n_2^2} \). This quantity is known as the numerical aperture (N.A.) of the fiber. For a cable with a core that has an index of refraction of 1.5 and a cladding with an index of refraction of 1.28, what do you expect the critical angle \( \alpha_{\text{crit}} \) to be (in degrees)? --- **Diagram Explanation:** The diagram below illustrates the process: - **Cladding**: The outer layer that surrounds the core. - **Core**: The central part of the fiber-optic cable where light travels. - **Ray Path**: - The incident ray enters at angle
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