For the optical fiber shown in the figure, find the minimum angle of incidence (0;) that will result in total internal reflection if the refractive index for the cladding (n2) is 1.44 and the refractive index of the core is (A.) n¡ = 1.59 and (B.) n¡ = 1.79. 0. Cone of acceptance Core Cladding (n,) A. O; > В. 0, 2

For the optical fiber shown in the figure, find the minimum angle of incidence (0;) that will result in total internal reflection if the refractive index for the cladding (n2) is 1.44 and the refractive index of the core is (A.) n¡ = 1.59 and (B.) n¡ = 1.79. 0. Cone of acceptance Core Cladding (n,) A. O; > В. 0, 2

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter38: Refraction And Images Formed By Refraction

Section: Chapter Questions

Problem 124PQ

See similar textbooks

Similar questions

If a microscope can accept light from objects at angles as large as =70 , what is the smallest structure that can be resolved when illuminated with light of wavelength 500 nm and (a) the specimen is in air? (b) When the specimen is immersed in oil, with index of refraction of 1.52?
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore = 1.482 ) and the cladding (ncladding = 1.44). Numerically, what is the largest angle (in degrees) a ray will make with respect to the interface of the fiber θmax, and still experience total internal reflection?
The index of refraction of the core of a typical fiber optic is ncore = 1.46; the cladding has nclad = 1.4. Calculate the critical angles for the total internal reflection icrit and αcrit
What is the smallest possible feature size that can be resolved by using a 193-nm optical source and the highest possible NA that can be obtained by a lens when exposing the image in air? Note that NA can be determined by the equation NA = n∙sin(θ) where n is the index of refraction of the medium in which the lens is working (air in this instance) and θ is the maximal half-angle of the cone of light that can enter or exit the lens and has a maximum possible value of 90˚ for a very large lens.
In an optical fiber, what can be said about the index of refraction of the core versus the cladding and the angle of reflection, θ, inside the fiber relative the critical angle, θc? Question 9 options: ncore > ncladding ; θ > θc ncore > ncladding ; θ < θc ncore < ncladding ; θ > θc ncore < ncladding ; θ < θc
What is the critical angle ic at the interface glass-cladding of an optical fiber whose core has a refractive index equal to 1.5 and cladding with a refractive index 1.45?
A light ray (red) enters a concentrically cladded optical fiber at its center and an angle of 23 o to the axis of the fiber. The diameter of the fiber is 1mm and its refractive index is nf=1.35. What must be the value of ncladding to guarantee total internal reflection? (nair=1.00)
. A plastic-coated window consists of Perspex and glass. At the Perspex to glassinterface the angle of refraction is 30.00. and the angle of incidence is 20.00. Ifthe refractive index of the glass is 1.75, what is the refractive index of thePerspex?
Calculate the angle of refraction at the air/core interface, r . critical angle, c and incident angle at the core/cladding interface, i . Will this light ray propagate down the fiber? You have the following data: nair = 1, ncore = 1.46, ncladding =1.43, incident =12o r = 8.2o, c = 78.4o, i = 81.8olight will propagate Formula summery Index of Refraction n= n sin 01= n, sin 0, o = sin1(2) Snell's Law Critical Angle Acceptance angle a= sin1 Numerical Aperture NA = sin a = ni- nị
When the D line of sodium light impinges an air-diamond interface at an angle of incidence of 25.0°, theangle of refraction is 10.1°. What is nD for diamond?
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore = 1.482 ) and the cladding (ncladding = 1.44). Suppose you wanted the largest angle at which total internal reflection occurred to be θmax = 5 degrees. What index of refraction does the cladding need if the core is unchanged?
What is the critical angle for the interface between water and diamond? The index of refraction of diamond is 2.42. To be internally reflected, the light must start in which material?