The group velocity of electromagnetic waves traveling through a gas is given by:w dK1-VE2K doWhere K is the dielectric constant of the gas and is related to the refractive index(n) as K = n. Thefrequency dependence of the dielectric constant of the gas can be expressed asA1K =1+-A= 1+(@ - a*)27x3x101.06x10-62ncWhere @.=17.78x104radians (i.e. 2 = 1.06 um)A0.8982 → A=0.8982o, rada) Determine the wavelength (in nanometers) at which the group velocity equals the phasevelocityb) Determine the group velocity (in meters/second) for the gas at the wavelength of part a.c) Determine the refractive index of the gas at the wavelength of part a

Given: The group velocity of electromagnetic waves traveling through a gas is given by: vg=cK1-ω2KdKdω where K is the dielectric constant related to refractive index as K=n2 and its frequency dependence is given by: K=1+Aωo2-ω2=1+Aωo211-ω2ωo2 where ωo=2πcλ=2π×3×1081.06×10-6=17.78×1014 rad i.e. λ=1.06 μm Aωo2=0.8982→A=0.8982ωo2 rad2 To find: (a) Determine the wavelength (in nanometers) at which the group velocity equals phase velocity. (b) Determine group velocity (in meters/sec) for the gas at the wavelength of part a. (c) Determine the refractive index of the gas at the wavelength of part a.(a) When the group velocity is equal to the phase velocity vp=vg The wavelength is doesn't get changed as it doesn't depend upon the phase velocity. So, the wavelength will be: λ=1.06 μm=1.06×10-6 m=1060×10-9 m=1060 nm So, the wavelength (in nanometers) at which the group velocity equals phase velocity is 1060 nm.(b) The phase velocity is given by: vp=λc=λωo2π=1060×10-9×17.78×10142π=0.029 m/s At this wavelength, the group velocity is equal to the phase velocity. So, group velocity at this wavelength will be: vg=vp=0.029 m/s The group velocity (in meters/sec) for the gas at this wavelength is 0.029 m/s.(c) The refractive index of the gas has the following relation: K=n2∴n=K and K=2πλ=2π1060×10-9=5.9×106 So, the refractive index will be: n=5.9×106=2.43×103 The refractive index of the gas at this wavelength will be 2.43×103.

Engineering

Electrical Engineering

The group velocity of electromagnetic waves traveling through a gas is given by: o dK 1 VK 2K do. Where K is the dielectric constant of the gas and is related to the refractive index(n) as K =n² . The frequency dependence of the dielectric constant of the gas can be expressed as A 1 K =1+ =1+ (@; - o*) 2лс 2лх3x10° Where w. = 17.78x10"radians (i.e. 2 1.06um) 1.06x10- A =0.8982 → A =0.8982o, rad a) Determine the wavelength (in nanometers) at which the group velocity equals the phase velocity b) Determine the group velocity (in meters/second) for the gas at the wavelength of part a. c) Determine the refractive index of the gas at the wavelength of part a

The group velocity of electromagnetic waves traveling through a gas is given by: o dK 1 VK 2K do. Where K is the dielectric constant of the gas and is related to the refractive index(n) as K =n² . The frequency dependence of the dielectric constant of the gas can be expressed as A 1 K =1+ =1+ (@; - o*) 2лс 2лх3x10° Where w. = 17.78x10"radians (i.e. 2 1.06um) 1.06x10- A =0.8982 → A =0.8982o, rad a) Determine the wavelength (in nanometers) at which the group velocity equals the phase velocity b) Determine the group velocity (in meters/second) for the gas at the wavelength of part a. c) Determine the refractive index of the gas at the wavelength of part a

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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