/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 An optical fiber consists of an ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An optical fiber consists of an outer "cladding" layer and an inner core with a slightly higher index of refraction. Light rays entering the core are trapped inside by total internal reflection and forced to travel along the fiber (see Figure 23.56 ). Suppose the cladding has an index of refraction of \(n_{b}=1.44\) and the core has an index of refraction of \(n_{a}=1.46 .\) Calculate the largest angle \(\theta\) between a light ray and the longitudinal axis of the fiber (see the figure) for which the ray will be totally internally reflected at the core/cladding boundary.

Short Answer

Expert verified
The largest angle is approximately \(9.86^\circ\).

Step by step solution

01

Understand Total Internal Reflection

Total internal reflection occurs when light travels from a medium with a higher index of refraction to a medium with a lower index and the angle of incidence exceeds the critical angle. The light is then completely reflected back into the original medium.
02

Identify Necessary Indices

For total internal reflection to occur at the core/cladding interface, light must enter at a critical angle or greater from the core (index of refraction \(n_a = 1.46\)) into the cladding (index of refraction \(n_b = 1.44\)).
03

Calculate the Critical Angle

The critical angle \(\theta_c\) can be calculated using the formula: \(\sin(\theta_c) = \frac{n_b}{n_a}\). Plugging in the indices, we get \(\sin(\theta_c) = \frac{1.44}{1.46}\).
04

Compute the Sine of the Critical Angle

Calculate \(\sin(\theta_c) = 0.9863\) using the values from the previous step.
05

Find the Critical Angle

The critical angle \(\theta_c\) is found by taking the inverse sine of \(0.9863\): \(\theta_c = \sin^{-1}(0.9863) \approx 80.14^\circ\).
06

Determine Largest Angle with Longitudinal Axis

The largest angle \(\theta\) between the light ray and the axis is the complement of the critical angle, given by \(\theta = 90^\circ - \theta_c = 90^\circ - 80.14^\circ = 9.86^\circ\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Index of Refraction
Light behaves differently depending on the medium through which it travels. This behavior is quantified by the index of refraction, a fundamental concept in optics. The index of refraction, denoted as \( n \), is a dimensionless number that tells us how much a substance can bend or "refract" a beam of light. It is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium:
  • \( n = \frac{c}{v} \)
where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium.
Each material has its unique index of refraction, which determines how much it bends light. For example, in our exercise, the core of the optical fiber has an index of refraction of 1.46, while the cladding has an index of 1.44.
This tiny difference is enough to trap light within the core using total internal reflection, allowing data to travel long distances through the fiber with minimal loss.
Critical Angle
The critical angle is a key concept to understand how light is guided within optical fibers. The critical angle is the minimum angle of incidence for which total internal reflection occurs. When light travels from a medium with a higher index of refraction to one with a lower index, it bends away from the normal.
If the angle of incidence is greater than this critical angle, the light is not refracted out of the first medium; instead, it is completely reflected back. This critical angle \( \theta_c \) can be found using the formula:
  • \( \sin(\theta_c) = \frac{n_b}{n_a} \)
In our case, \( n_a = 1.46 \) (core) and \( n_b = 1.44 \) (cladding), which gives \( \sin(\theta_c) = 0.9863 \). The critical angle was calculated as approximately \( 80.14^\circ \).
If a light ray hits the core/cladding boundary at an angle larger than this critical angle, it will reflect entirely back into the core, ensuring the light stays within the fiber.
Optical Fiber
Optical fibers are thin, flexible strands made of glass or plastic that are used for transmitting data in the form of light. They rely on a principle called total internal reflection to trap light within the core of the fiber. The core, made of a material with a higher index of refraction than the surrounding cladding, directs light down the fiber with minimal loss.
Here's a simple breakdown of how optical fibers work:
  • Core and Cladding: The core is the inner glass center, which has a higher index of refraction, and the cladding is the outer optical material surrounding the core, with a lower refractive index.
  • Total Internal Reflection: Light enters the fiber at a small angle and, because of the higher index of refraction in the core, is reflected repeatedly within the core instead of escaping through the sides.
  • Data Transmission: By sending coded light pulses, optical fibers can transmit data over long distances with high bandwidth.
This efficient method of data transmission is crucial for modern telecommunications, providing fast internet, cable TV, and long-distance phone communications, among other uses. By leveraging total internal reflection, optical fibers avoid energy loss and interference, making them highly effective for data transport.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The critical angle for total internal reflection at a liquid-air interface is \(42.5^{\circ} .\) (a) If a ray of light traveling in the liquid has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of \(35.0^{\circ}\) at the interface, what angle does the refracted ray in the liquid make with the normal?

A geostationary communications satellite orbits the earth directly above the equator at an altitude of \(35,800 \mathrm{~km}\). Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back. Would this delay be noticeable in a conversation?

A sinusoidal electromagnetic wave emitted by a cellular phone has a wavelength of \(35.4 \mathrm{~cm}\) and an electric-field amplitude of \(5.40 \times 10^{-2} \mathrm{~V} / \mathrm{m}\) at a distance of \(250 \mathrm{~m}\) from the antenna. Calculate (a) the frequency of the wave; (b) the magnetic-field amplitude; (c) the intensity of the wave.

Plane-polarized light passes through two polarizers whose axes are oriented at \(35.0^{\circ}\) to each other. If the intensity of the original beam is reduced to \(15.0 \%,\) what was the polarization direction of the original beam, relative to the first polarizer?

The vitreous humor, a transparent, gelatinous fluid that fills most of the eyeball, has an index of refraction of \(1.34 .\) Visible light ranges in wavelength from \(400 \mathrm{nm}\) (violet) to \(700 \mathrm{nm}\) (red), as measured in air. This light travels through the vitreous humor and strikes the rods and cones at the surface of the retina. What are the ranges of (a) the wavelength, (b) the frequency, and (c) the speed of the light just as it approaches the retina within the vitreous humor?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.