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A right-circularly polarized plane wave is normally incident from air onto a semi-infinite slab of plexiglas \(\left(\epsilon_{r}^{\prime}=3.45, \epsilon_{r}^{\prime \prime}=0\right) .\) Calculate the fractions of the incident power that are reflected and transmitted. Also, describe the polarizations of the reflected and transmitted waves.

Short Answer

Expert verified
Answer: The reflected wave is left-circularly polarized, and the transmitted wave remains right-circularly polarized.

Step by step solution

01

Convert right-circularly polarized wave to linearly polarized waves

Right-circularly polarized wave can be represented as a combination of two linearly polarized waves: one is s-polarized (parallel to the plane of incidence) and the other is p-polarized (perpendicular to the plane of incidence). This will allow us to evaluate the reflection and transmission coefficients for each individual polarization using Fresnel equations.
02

Apply the Fresnel equations

The Fresnel equations relate the amplitudes of reflected and transmitted electric fields to the incident electric field amplitude. They are given by: For s-polarized waves (TE waves): \(r_s = \frac{n_1 \cos{\theta_i} - n_2 \cos{\theta_t}}{n_1 \cos{\theta_i} + n_2 \cos{\theta_t}}\) For p-polarized waves (TM waves): \(r_p = \frac{n_2 \cos{\theta_i} - n_1 \cos{\theta_t}}{n_2 \cos{\theta_i} + n_1 \cos{\theta_t}}\) In this problem, the wave is normally incident, so \(\theta_i = 0\) and \(\theta_t = 0\). Also, since the incident medium is air, we have \(n_1=1\). Since the slab is plexiglass and \(\epsilon_{r}^{\prime}=3.45\), we can calculate the refractive index of the plexiglass as \(n_2=\sqrt{\epsilon_{r}^{\prime}}\).
03

Calculate the Fresnel reflection and transmission coefficients

By applying the Fresnel equations, we can find the reflection and transmission coefficients for both s- and p-polarized waves respectively: \(r_s = \frac{1- \sqrt{3.45}}{1 + \sqrt{3.45}}\) \(r_p = \frac{\sqrt{3.45} - 1}{\sqrt{3.45} + 1}\)
04

Calculate the fractions of the incident power that are reflected and transmitted

We can obtain the fractions of the incident power that are reflected and transmitted using the square of the reflection and transmission coefficients: \(R_s = |r_s|^2\) and \(R_p = |r_p|^2\). Since both s- and p-polarized components have equal power, we can average the two values to obtain the overall reflected and transmitted fractional power : \(R = \frac{R_s + R_p}{2}\), where R is the reflected power fraction \(T = 1 - R\), where T is the transmitted power fraction
05

Describe the polarizations of the reflected and transmitted waves

The Fresnel coefficients and polarizations will now give us the polarization state of the reflected and transmitted waves. As both components have reverse signs of the reflection coefficient, the reflected wave will be left-circularly polarized. Since both components have the same signs for the transmission coefficient, the transmitted wave will remain right-circularly polarized.

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Key Concepts

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

Polarization
Polarization refers to the orientation of the oscillations of electromagnetic waves. When a wave is polarized, its oscillations are confined to a particular direction or pattern. There are several types of polarization, such as linear, circular, and elliptical.
For linear polarization, oscillations follow a straight line, while circular polarization involves rotation. Circular polarization can be further divided into right-circular and left-circular polarization. In right-circular polarization, the electric field rotates in the direction of a right-handed screw with respect to the direction of propagation, which is critical for understanding reflection and transmission behaviors in electromagnetic theory.
Fresnel Equations
The Fresnel equations are used to describe how light reflects and transmits at an interface between two different media. These equations provide the reflection and transmission coefficients for electromagnetic waves impinging on a surface.
There are separate equations for s-polarized and p-polarized light. For s-polarized light (TE waves), the Fresnel reflection coefficient is calculated using the formula: \(r_s = \frac{n_1 \cos{\theta_i} - n_2 \cos{\theta_t}}{n_1 \cos{\theta_i} + n_2 \cos{\theta_t}}\).
Similarly, for p-polarized light (TM waves), the coefficient is: \(r_p = \frac{n_2 \cos{\theta_i} - n_1 \cos{\theta_t}}{n_2 \cos{\theta_i} + n_1 \cos{\theta_t}}\).
These coefficients are crucial for determining how much of the wave's energy is reflected versus transmitted. In normal incidence, where the angle of incidence is zero, these equations simplify significantly.
Reflection and Transmission Coefficients
Reflection and transmission coefficients quantify how much of an incoming electromagnetic wave is reflected back and how much is transmitted through an interface.
They are derived from the Fresnel equations, and involve the refractive indices of the two media. These coefficients can then be squared to find the fractions of the incident power reflected and transmitted.
A key aspect of these coefficients is their dependence on the polarization of the wave. They describe varying behaviors for s-polarized and p-polarized light, which, in turn, affect the change in polarization state of the wave after interaction with a medium.
Refractive Index
The refractive index, often denoted as \(n\), is a measure of how much light slows down in a particular medium. It is a fundamental property of optical materials, affecting the speed and direction of light propagation.
The refractive index is calculated with the formula: \( n = \sqrt{\epsilon_r} \).
In this context, \(\epsilon_r\) represents the dielectric constant of the medium (assuming no magnetic permeability).
Refractive index differences between media are responsible for phenomena such as refraction and reflection. Understanding \(n\) is crucial for calculating the Fresnel coefficients and predicting how electromagnetic waves behave when crossing boundaries, as demonstrated in the exercise.
Right-Circular Polarization
Right-circular polarization describes a type of circular polarization where the electric field vector rotates clockwise when looking in the direction of propagation. This involves both the x and y components of the electric field being equal in magnitude but having a phase difference of 90 degrees.
When such a wave encounters an interface, it can be decomposed into linear polarizations, allowing the use of Fresnel equations to determine reflections and transmissions. At the interface, the polarization can change, turning into left-circular polarization in reflections due to changes in phase relationships.
Understanding and recognizing right-circular polarization is important in optics and telecommunications, helping us understand how signals can be maintained, altered, or potentially lost when traveling between different media.

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Most popular questions from this chapter

A \(10 \mathrm{MHz}\) uniform plane wave having an initial average power density of \(5 \mathrm{~W} / \mathrm{m}^{2}\) is normally incident from free space onto the surface of a lossy material in which \(\epsilon_{2}^{\prime \prime} / \epsilon_{2}^{\prime}=0.05, \epsilon_{r 2}^{\prime}=5\), and \(\mu_{2}=\mu_{0} .\) Calculate the distance into the lossy medium at which the transmitted wave power density is down by \(10 \mathrm{~dB}\) from the initial \(5 \mathrm{~W} / \mathrm{m}^{2}\).

A uniform plane wave in free space is normally incident onto a dense dielectric plate of thickness \(\lambda / 4\), having refractive index \(n\). Find the required value of \(n\) such that exactly half the incident power is reflected (and half transmitted). Remember that \(n>1\).

Region 1, \(z<0\), and region \(2, z>0\), are both perfect dielectrics ( \(\mu=\mu_{0}\), \(\epsilon^{\prime \prime}=0\) ). A uniform plane wave traveling in the \(\mathbf{a}_{z}\) direction has a radian frequency of \(3 \times 10^{10} \mathrm{rad} / \mathrm{s}\). Its wavelengths in the two regions are \(\lambda_{1}=\) \(5 \mathrm{~cm}\) and \(\lambda_{2}=3 \mathrm{~cm}\). What percentage of the energy incident on the boundary is \((a)\) reflected; \((b)\) transmitted? \((c)\) What is the standing wave ratio in region \(1 ?\)

You are given four slabs of lossless dielectric, all with the same intrinsic impedance, \(\eta\), known to be different from that of free space. The thickness of each slab is \(\lambda / 4\), where \(\lambda\) is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally incident. The slabs are to be arranged such that the airspaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and airspaces such that \((a)\) the wave is totally transmitted through the stack, and \((b)\) the stack presents the highest reflectivity to the incident wave. Several answers may exist.

A uniform plane wave in region 1 is normally incident on the planar boundary separating regions 1 and 2. If \(\epsilon_{1}^{\prime \prime}=\epsilon_{2}^{\prime \prime}=0\), while \(\epsilon_{r 1}^{\prime}=\mu_{r 1}^{3}\) and \(\epsilon_{r 2}^{\prime}=\mu_{r 2}^{3}\), find the ratio \(\epsilon_{r 2}^{\prime} / \epsilon_{r 1}^{\prime}\) if \(20 \%\) of the energy in the incident wave is reflected at the boundary. There are two possible answers.

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