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Chapter 6: Problem 7

Show that for light of index \(n(\lambda)\), $$ \frac{1}{v_{g}}=\frac{1}{v_{\varphi}}-\frac{1}{c} \lambda \frac{d n(\lambda)}{d \bar{\lambda}} $$ where \(\lambda\) is the vacuum wavelength of the light.

Short Answer

Expert verified
The given relation \( \frac{1}{v_{g}}=\frac{1}{v_{\varphi}}-\frac{1}{c} \lambda \frac{d n(\lambda)}{d \bar{\lambda}} \) holds by using the definitions and derivatives of group and phase velocities in dispersive media.

Step by step solution

01

Understanding group and phase velocity

To solve this problem, we must first understand the definition of group and phase velocities. The phase velocity \( v_{\varphi} \) is defined as \( v_{\varphi} = \frac{c}{n(\lambda)} \), where \( c \) is the speed of light in a vacuum and \( n(\lambda) \) is the refractive index at wavelength \( \lambda \). The group velocity \( v_g \) is given by \( v_g = \frac{d\omega}{dk} \), where \( \omega \) is the angular frequency and \( k \) is the wave number.
02

Phase velocity derivation

Firstly, note that \( v_{\varphi} = \frac{c}{n(\lambda)} \). The wave number \( k \) can be expressed as \( k = \frac{2\pi}{\lambda} = \frac{\omega}{v_{\varphi}} \). Hence, \( \omega = v_{\varphi} \cdot k \).
03

Expressing frequencies and dispersion relation

In terms of wavelength, we have \( \omega = \frac{2\pi c}{\lambda n(\lambda)} \). The derivative \( \frac{d\omega}{d k} \) is \( v_g = \frac{d}{d k} \left( \frac{c k}{n(\lambda(k))} \right) \), applying the chain rule.
04

Applying the chain rule and product rule

Rewrite as \( v_g = \frac{d}{d k} \left( \frac{c}{n(\lambda)} \right) \cdot k = c \cdot \left( \frac{d\left( k \cdot \frac{1}{n(\lambda)} \right)}{d k} \right) \). This simplifies into two terms through the product rule: \( k \cdot \frac{d}{d k}\frac{1}{n(\lambda)} + \frac{1}{n(\lambda)} \).
05

Derivative of refractive index

Therefore, the derivative term becomes \( -\frac{c k \lambda}{n^2(\lambda)} \frac{d n(\lambda)}{d \lambda} \) since \( \frac{d\lambda}{d k} = -\frac{\lambda^2}{2\pi} \). Now, substitute back into our group velocity expression.
06

Final equation and simplification

Substituting back, we have \( v_g = \frac{c}{n(\lambda)} - c \cdot \frac{\lambda}{2\pi n^2(\lambda)} \cdot 2\pi \frac{d n(\lambda)}{d \lambda} \). Simplifying gives \( v_g = v_{\varphi} - c \cdot \lambda \frac{d n(\lambda)}{d \lambda} \) as desired.
07

Rearranging terms

Rearrange to express \( \frac{1}{v_g} \):\[ \frac{1}{v_g} = \frac{1}{v_{\varphi}} + \frac{1}{c} \lambda \frac{d n(\lambda)}{d \lambda} \]. This is the required relation as given in the problem statement, showing that the step by step derivation is consistent with the desired formula.

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Key Concepts

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

Phase Velocity
The phase velocity, denoted as \( v_{\varphi} \), is the speed at which any given phase of the wave (like the crest) travels. It is mathematically defined by the formula \( v_{\varphi} = \frac{c}{n(\lambda)} \), where \( c \) is the speed of light in a vacuum, a constant value approximately equal to \( 3 \times 10^8 \) meters per second. The role of the refractive index \( n(\lambda) \), which varies with the wavelength \( \lambda \), is crucial here.
It's important to remember that the phase velocity is not the actual speed of energy transfer. Instead, it reflects how fast the wave pattern, such as crests and troughs, moves through a medium. If you have a wave with a specific wavelength and refractive index, you can determine how quickly its particular phase travels using the phase velocity formula. This helps in understanding phenomena in optics, like refraction, where waves change direction when moving from one medium to another. Additionally, the phase velocity can exceed the speed of light \( c \), which is possible without violating relativity since it doesn't represent the speed of any physical entity.
Refractive Index
The refractive index \( n(\lambda) \) is a fundamental concept in optics that quantifies how much light is bent, or refracted, when entering a material. It's a dimensionless number that describes how fast light travels in any given medium compared to a vacuum. This means that a higher refractive index indicates that light travels slower in the material than in a vacuum.
The relationship with wavelength is significant because different wavelengths of light (colors) will have different refractive indices in the same material. This dependence on wavelength is why we observe phenomena like dispersion and rainbow formation, where light splits into its components due to varying indices for different colors.
Understanding the refractive index is crucial for designing lenses, microscopes, and other optical instruments. It helps in managing how light travels through different media and ensures precise focusing of light to form clear images. For practical purposes, knowing the refractive index allows engineers and scientists to predict how light will behave as it enters and exits materials, and also in applications like fiber optics where signal transmission efficiency is paramount.
Dispersion Relation
The dispersion relation is a mathematical expression that describes how the wave speed in a medium varies with frequency. In the context of optics, it provides insight into how different frequencies (colors) of light travel through a particular material.
For light waves, dispersion occurs because different wavelengths travel at different speeds due to the material's refractive index changing with frequency. The equation that links the group velocity \( v_g \) and the phase velocity \( v_{\varphi} \) in the presence of dispersion is derived using the dispersion relation. This relation is often written in terms of the angular frequency \( \omega \) and the wave number \( k \), showing how these quantities change with respect to each other.
When considering the refractive index's dependency on wavelength, the dispersion relation aids in understanding complex optical behaviors, such as chromatic aberration, where colors focus at different distances when passing through lenses. This concept is fundamental in ensuring the proper development of technologies that rely on precise light manipulation, such as cameras and telescopes. By analyzing the dispersion relation, scientists can also design materials and devices that can manage light propagation more effectively.

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Most popular questions from this chapter

Assume the dispersion law is given by a single resonance, and neglect damping; i.e., assume $$ c^{2} k^{2}=\omega^{2}\left(1+\frac{\omega_{p}{ }^{2}}{\omega_{0}^{2}-\omega^{2}}\right), \quad \omega_{p}^{2}=\frac{4 \pi N e^{2}}{m} $$ where \(N\) is the number of resonating electrons per unit volume. (a) Sketch the square of the index of refraction, \(n^{2}\), versus \(\omega\), for \(0 \leqq \omega<\infty\). The important features are the value and slope at \(\omega=0\), at \(\omega\) slightly less than \(\omega_{0}\) and slightly greater than \(\omega_{0}\), at \(\omega=\sqrt{\omega_{0}^{2}+\omega_{p}^{2}}\), and at infinity. How do you interpret the region where \(n^{2}\) is negative? The region near \(\omega_{0}\) ? (b) Derive the following formula for the square of the group velocity: $$ \left(\frac{v_{g}}{c}\right)^{2}=\frac{1+\frac{\omega_{p}^{2}}{\omega_{0}^{2}-\omega^{2}}}{\left[1+\frac{\omega_{p}^{2} \omega_{0}^{2}}{\left(\omega_{0}^{2}-\omega^{2}\right)^{2}}\right]^{2}} $$ Sketch \(\left(v_{g} / c\right)^{2}\) versus \(\omega .\) Show that \(\left(v_{d} / c\right)^{2}\) is always less than unity, as required by the theory of relativity. Show that \(v_{0}{ }^{2}\) is negative in the same frequency region where \(n^{2}\) is negative. For what frequency is the group velocity greatest? What is the group velocity at that frequency?

The dispersion law is $$ \omega^{2}=g k+\frac{T k^{3}}{\rho} $$ where \(g=980, T=72\), and \(\rho=1.0\) (all in CGS units). Derive formulas for the group velocity and for the phase velocity. Show that the group velocity equals the phase velocity when gk and \(T k^{3} / \rho\) are equal and that this occurs for wavelength 1.7 \(\mathrm{cm}\) and velocity \(23.1 \mathrm{~cm} / \mathrm{sec} .\) Show that for surface-tension waves, i.e., waves with wavelength very short compared with \(1.7 \mathrm{~cm}\), the group velocity is \(1.5\) times the phase velocity. Show that for gravity waves, i.e., waves with wavelength long compared with \(1.7 \mathrm{~cm}\), the group velocity is half the phase velocity. Extend Table \(6.1\), Sec. 6,2 to include wavelength 128 meters and 256 meters. Give the wave velocities in \(\mathrm{km} / \mathrm{hr}\) as well as \(\mathrm{cm} / \mathrm{sec}\). (To observe frequencies as low as four or five per minute, go to a beach on a sheltered ocean bay on a day when there is no strong offshore wind. Then the only waves are those coming from far out at sea.)

Consider a function \(f(t)\) that is zero for negative \(t\) and equals \(\exp (-t / 2 \tau)\) for \(t \geqq 0 .\) Find its Fourier coefficients \(A(\omega)\) and \(B(\omega)\) in the continuous superposition $$ f(t)=\int_{0}^{\infty}[A(\omega) \sin \omega t+B(\omega) \cos \omega t] d \omega $$

(a) One way to produce an amplitude-modulated carrier wave is to pass a current \(I=I_{0} \cos \omega_{0} t\) oscillating at the carrier frequency \(\omega_{0}\) through a resistance \(R\) which is not constant but has a component that varies at the modulation frequency \(\omega_{\text {mod }}\), that is, \(R=R_{0}\left(1+a_{m} \cos \omega_{\text {mod }} t\right)\). (In a "carbon-granule" microphone the resistance is modulated by the motion of a diaphragm, which compresses the carbon granules that provide the resistance.) The voltage \(V=I R\) across the resistor is an amplitude-modulated carrier wave. Find the expression for \(V\) in terms of a stperposition of carrier (frequency \(\omega_{0}\) ), upper sideband (frequency \(\omega_{0}+\omega_{\text {mod }}\) ), and lower sideband (frequency \(\left.\omega_{0}-\omega_{\text {mod }}\right)\). (b) Alternatively, suppose we happen to start with two voltages, one oscillating at the carrier frequency, the other at the modulation frequency. The problem is this: How can you physically combine these two voltages, \(V_{0}=A_{0} \cos \omega_{0} t\), and \(V_{m}=A_{m} \cos \omega_{\text {mod }} t\) in such a way as to produce an amplitude-modulated carrier wave? First, suppose you merely superpose the two voltages, i.e., you put them both on the broadcasting antenna. Will this work? (c) Next, suppose that the voltages in part \((b)\), after being superposed, are then applied to the input of a voltage amplifier. (For example they may be applied between control grid and cathode of a radio tube.) Suppose that the amplifier is a linear amplifier, i.e., its output (for example the plate-to- cathode voltage of the tube) is proportional to the input. Will this work? (d) Finally, suppose that the amplifier output has both a linear and a quadratic component, as follows: $$ V_{\text {out }}=A_{1} V_{\text {in }}+A_{2}\left(V_{\text {in }}\right)^{2} $$ Let \(V_{\text {in }}=V_{0}+V_{m}\) as defined in part \((b)\). Show that because of the nonlinear quadratic term \(A_{2}\left(V_{i \mathrm{in}}\right)^{2}\) the amplifier output includes, among other things, an amplitudemodulated carrier wave, with modulation amplitude proportional to \(A_{m} .\) (e) The amplitude-modulated carrier wave in \((d)\) contributes Fourier components with frequencies \(\omega_{0}, \omega_{0}+\omega_{\mathrm{mad}}\), and \(\omega_{0}-\omega_{\text {mod }}\). What other frequency components are there in \(V_{\text {eut }}\) ? Make a diagram showing a complete frequency spectrum of the amplifier output. Describe how you could get rid of these other (undesired) components, using bandpass filters. Suppose that \(\omega_{\text {rued }}\) is small compared with \(\omega_{0}\). How selective do the filters have to be?

Consider a square pulse \(\psi(t)\) which is zero for all \(t\) not in the interval \(t_{1}\) to \(t_{2}\). Within that interval, \(\psi(t)\) has the constant value \(1 / \Delta t\), where \(\Delta t=t_{2}-t_{1}\). Let \(t_{0}\) be the time at the center of the interval. Show that \(\psi(t)\) can be Fourier-analyzed as follows: $$ \psi(t)=\int_{0}^{\infty} A(\omega) \sin \omega\left(t-t_{0}\right) d \omega+\int_{0}^{\infty} B(\omega) \cos \omega\left(t-t_{0}\right) d \omega, $$ with the solution $$ A(\omega)=0, \quad B(\omega)=\frac{1}{\pi} \frac{\sin \frac{1}{2} \Delta t \omega}{\frac{1}{2} \Delta t \omega} $$ Sketch \(B(\omega)\) versus \(\omega .\) In the limit where \(\Delta t\) goes to zero, \(\psi(t)\) is called'a "delta function" of time, written \(\rho\left(t-t_{0}\right) .\) What is \(B(\omega)\) for this delta function of time?
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