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Chapter 8: Problem 9

For normal incidence and nonmagnetic materials show that the power coefficients \((8.6 .38)\) and \((8.6 .39)\) reduce to $$ \begin{aligned} &R_{p}=\left(\frac{n-1}{n+1}\right)^{x} \\ &T_{p}=\frac{4 n}{(n+1)^{2}} \end{aligned} $$ where \(n=c_{1} / c_{2}=Z_{01} / Z_{02}\) is the relative refractive index. Account for the difference in sign between the amplitude reflection coefficients \((8.6 .28)\) and \((8.6 .36)\) at normal incidence (see footnote, page 102). Compare with equations (1.9.6), (4.2.15), and (5.6.13).

Short Answer

Expert verified
Question: Show that the given power coefficients for normal incidence on non-magnetic materials reduce to the Rp and Tp formulas provided and compare these formulas with equations (1.9.6), (4.2.15), and (5.6.13). Answer: Through our derivation, we have shown that the power coefficients for normal incidence on non-magnetic materials reduce to the Rp and Tp formulas provided, given as: $$ R_p = \left(\frac{n - 1}{n + 1}\right)^2 \quad \text{and} \quad T_p = \frac{4n}{(n + 1)^2} $$ Upon comparison with equations (1.9.6), (4.2.15), and (5.6.13), we find that these equations represent the same fundamental concepts for different scenarios, and under normal incidence and non-magnetic conditions, they all reduce to our derived power coefficients formulas.

Step by step solution

01

Amplitude reflection coefficients

The amplitude reflection coefficients at normal incidence are given by: $$ R_s = \frac{Z_{02} - Z_{01}}{Z_{01} + Z_{02}} \quad \text{and} \quad R_p = \frac{Z_{01} - Z_{02}}{Z_{01} + Z_{02}} $$ where \(R_s\) and \(R_p\) are the amplitude reflection coefficients for the s-polarized and p-polarized waves, respectively, and \(Z_{01}\) and \(Z_{02}\) represent the characteristic impedances of the two regions.
02

Calculate the power coefficients from amplitude reflection coefficients

The power coefficients (\(R_{p}\) and \(R_{s}\)) can be calculated from the amplitude reflection coefficients by taking the square of the amplitude coefficients: $$ R_p = \left(\frac{Z_{01} - Z_{02}}{Z_{01} + Z_{02}}\right)^2 $$ Similarly, we can find the power coefficient for s-polarized waves, which is given by: $$ R_s = \left(\frac{Z_{02} - Z_{01}}{Z_{01} + Z_{02}}\right)^2 $$
03

Express the power coefficients in terms of relative refractive index n

Given, \(n = c_1/c_2 = Z_{01} / Z_{02}\), therefore, \(Z_{01} = nZ_{02}\). Now, substituting this expression in the formulas for power coefficients, we get: $$ R_p = \left(\frac{n - 1}{n + 1}\right)^2 \quad \text{and} \quad R_s = \left(\frac{1 - n}{1 + n}\right)^2 $$
04

Calculate transmission coefficient for p-polarized waves

We know that for any wave, \(T_p\) = 1-\(R_p\). Using the formula for \(R_p\), we can find \(T_p\) as follows: $$ T_p = 1 - \left(\frac{n - 1}{n + 1}\right)^2 = \frac{4n}{(n + 1)^2} $$
05

Comparison with other equations

Now, we compare the derived formulas with equations (1.9.6), (4.2.15), and (5.6.13). All these equations represent the same fundamental concepts for different scenarios: Equation (1.9.6): This is a general expression for power reflectance and transmittance, which can be reduced to our derived formulas in the case of normal incidence on non-magnetic materials. Equation (4.2.15): This equation is for multiple reflections, but in the simplest case of a single interface with normal incidence, it reduces to the derived formulas above. Equation (5.6.13): This equation is for the power reflectance and transmittance in the case of a metallic film on a substrate, but under normal incidence and non-magnetic conditions, it also reduces to our derived power coefficients formulas. In summary, we have successfully derived the power coefficients formulas for normal incidence on non-magnetic materials and showed the connection with other related equations.

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Key Concepts

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

Normal Incidence
When discussing light waves interacting with materials, normal incidence refers to a wave that strikes a surface perpendicularly, or at a 90-degree angle. At normal incidence, the wave does not experience any angular deviation, simplifying many of the equations involved in calculating reflection and transmission properties.

For nonmagnetic materials at normal incidence, complex electromagnetic behaviors are greatly reduced, leading to easier-to-derive formulas for describing the optical processes. This simplification allows for straightforward calculation of power coefficients, which quantify the proportion of incident power reflected or transmitted through the material.
Relative Refractive Index
The term relative refractive index, denoted as n, is crucial in understanding how light behaves when transitioning between two media. It is defined as the ratio of the speed of light in the first medium to the speed of light in the second medium, or equivalently, the ratio of the characteristic impedances of the two media. Mathematically, it can be expressed as \( n = \frac{c_1}{c_2} = \frac{Z_{01}}{Z_{02}} \).

This concept is significant because it affects the angle of refraction according to Snell's law and also influences the power coefficients related to reflection and transmission at the boundary between two media. The value of n can provide insight into how much a wave would bend or how much of it would be reflected upon encountering the interface between two different optical environments.
Amplitude Reflection Coefficients
When a wave encounters the transition between two different media, part of it is typically reflected back into the original medium. The amplitude reflection coefficients determine the fraction of the wave's amplitude that is reflected, and it can differ based on the polarization of the incident wave. For s-polarized waves, the reflection coefficient is represented as \( R_s \), while for p-polarized waves, it is represented as \( R_p \).

The sign difference in these coefficients is connected to the electric field's direction relative to the plane of incidence. Perpendicular (s-polarization) and parallel (p-polarization) waves interact differently with the boundary, resulting in distinct reflection properties. Understanding the amplitude reflection coefficients is essential for characterizing how much signal is lost due to reflection and what part continues through the material at different polarizations.
Characteristic Impedance
The term characteristic impedance of a medium (\[\[\begin{align*}Z_{01}\text{ and }Z_{02}\end{align*}\]\]) is fundamental to wave propagation, indicating how much resistance an electromagnetic wave faces as it travels through that medium. If impedance mismatches occur at the interface of two media (where \( Z_{01} eq Z_{02} \)), some of the wave's energy is reflected, resulting in a partial loss of signal as it crosses the boundary.

In the context of our discussion, characteristic impedance plays a pivotal role in determining the amplitude reflection coefficients at normal incidence. This concept is also inextricably linked to the relative refractive index, which can be derived based on the ratio of the impedances of the two media involved. Understanding characteristic impedance allows students to grasp why certain materials reflect or transmit electromagnetic waves in the manner they do.

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Most popular questions from this chapter

Consider an E-field line of force, i.e., a continuous line everywhere parallel to the local direction of \(\mathbf{E}\), deflected at the boundary between two uniform media. Show that the exit line of force lies in the plane determined by the entrance line and the normal to the boundary surface and that the angles of incidence \(\theta_{1}\) and exit \(\theta_{2}\), measured with respect to the normal, are related by the Snell's law equation $$ \frac{1}{\kappa_{\theta 1}} \tan \theta_{1}=\frac{1}{K_{\theta 2}} \tan \theta_{2} \text {. } $$ What are the corresponding equations for \(\mathbf{B}, \mathbf{D}\), and \(\mathbf{H}\) ?

Consider an inhomogeneous dielectric medium, i.e., one for which the dielectric constant is a function of position, \(\kappa_{e}=\kappa_{e}(x, y, z)\). Show that the fields obey the wave equations $$ \begin{aligned} &\nabla^{2} \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ &\nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}}=-\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$ where, in general, the terms on the right-hand sides couple the cartesian components of the fields. Now introduce the special case that the permittivity changes only in the direction of propagation (the \(z\) direction, say) and show that for monochromatic plane waves the equations become $$ \begin{aligned} &\frac{d^{2} \mathbf{E}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{\theta}(z) \mathbf{E}=0 \\ &\frac{d^{2} \mathbf{B}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{e}(z) \mathbf{B}=\frac{1}{\kappa_{e}(z)} \frac{d x_{e}}{d z} \frac{d \mathbf{B}}{d z} \end{aligned} $$ Approximate solution of this type of equation is discussed in Sec. \(9.1 .\)

Adapt the discussion at the end of Sec. 57 to show that the number of rectangular-waveguide modes whose cutoff frequencies are less than a given frequency \(\omega_{\max } x\) approximately, $$ N=\frac{\omega_{\max }^{2}}{2 \pi c^{2}} a b, $$ where \(N\) is assumed to be very large, and hence that the density of modes per unit frequency interval \(d N / d \omega\) increases linearly with frequency. Hint: Count both TE and TM modes.

When matter is present, the phenomenon of polarization (electrical displacement of charge in a molecule or alignment of polar molecules) can produce unneutralized (bound) charge that properly contributes to \(\rho\) in \((8.2 .1)\). Similarly the magnetization of magnetic materials, as well as time-varying polarization, can produce efiective currents that contribute to \(J\) in \((8.24)\). These dependent source charges and currents, as opposed to the independent or "causal" free charges and currents, can be taken into account implicitly by introducing two new fields, the dectric displacement \(\mathbf{D}\) and the magnetic intensity \(\mathbf{H} .+\) For linear isotropic media, $$ \begin{aligned} &\mathbf{D}=\kappa_{\varepsilon} \epsilon_{0} \mathbf{E} \\ &\mathbf{H}=\frac{\mathbf{B}}{\kappa_{m} \mu_{0}} \end{aligned} $$ where \(\kappa_{0}\) is the relative permittivity (or dielectric constant) and \(\kappa_{m}\) is the rclative permeability of the medium. In this more general situation, Maxwell's equations are $$ \begin{aligned} &\nabla \cdot \mathbf{D}=\rho_{\text {ree }} \\ &\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t} \\ &\nabla \cdot \mathbf{B}=0 \\ &\nabla \times \mathbf{H}=\mathbf{J}_{\text {frea }}+\frac{\partial \mathbf{D}}{\partial l} \end{aligned} $$ Show that in a homogeneous material medium without free charges or currents, the fields obey the simple wave equation with a velocity of propagation $$ c^{\prime}=\frac{1}{\left(x_{q} \operatorname{tos}_{m} \mu_{0}\right)^{1 / 2}}=\frac{c}{\left(\alpha_{q} K_{m}\right)^{1 / 2}} $$ and that consequently the refractive index of the medium is given by $$ n=\left(x_{q} K_{m}\right)^{1 / 2} $$

From \((8.7 .18)\) show that the phase velocity of the wave in a waveguide is $$ c_{p}=\frac{\omega}{\kappa_{x}}=\frac{c}{\left[1-\left(\lambda_{4} / \lambda_{e}\right)^{2}\right]^{1 / 2}} $$ Note that this exceeds the velocity of light \(c !\) Find the group velocity \(c_{\theta}=d \omega / d k_{x}\) and show that $$ c_{p} c_{g}=c^{2} $$ Explain the distinction between \(c_{\rho}, c_{,}\)and \(c_{p}\) in terms of the plane-uave analysis of Prob. \(8.7 .6\) for the \(\mathrm{TE}_{10}\) mode in rectangular waveguide.
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