**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

Given, Index of refraction of core n1=1.5 Index of refraction of cladding n2=1.28 we know, Numerical…

(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

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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Concept explainers

Ray Optics

Optics is the study of light in the field of physics. It refers to the study and properties of light. Optical phenomena can be classified into three categories: ray optics, wave optics, and quantum optics. Geometrical optics, also known as ray optics, is an optics model that explains light propagation using rays. In an optical device, a ray is a direction along which light energy is transmitted from one point to another. Geometric optics assumes that waves (rays) move in straight lines before they reach a surface. When a ray collides with a surface, it can bounce back (reflect) or bend (refract), but it continues in a straight line. The laws of reflection and refraction are the fundamental laws of geometrical optics. Light is an electromagnetic wave with a wavelength that falls within the visible spectrum.

Converging Lens

Converging lens, also known as a convex lens, is thinner at the upper and lower edges and thicker at the center. The edges are curved outwards. This lens can converge a beam of parallel rays of light that is coming from outside and focus it on a point on the other side of the lens.

Plano-Convex Lens

To understand the topic well we will first break down the name of the topic, ‘Plano Convex lens’ into three separate words and look at them individually.

Lateral Magnification

In very simple terms, the same object can be viewed in enlarged versions of itself, which we call magnification. To rephrase, magnification is the ability to enlarge the image of an object without physically altering its dimensions and structure. This process is mainly done to get an even more detailed view of the object by scaling up the image. A lot of daily life examples for this can be the use of magnifying glasses, projectors, and microscopes in laboratories. This plays a vital role in the fields of research and development and to some extent even our daily lives; our daily activity of magnifying images and texts on our mobile screen for a better look is nothing other than magnification.

Question

**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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