Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 3

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.

Short Answer

Expert verified
Short Answer: To find the two possible values for the ratio, follow these steps: 1) Use the given reflection coefficient and transmission coefficient to find the possible values for the refractive indices ratio, \(\frac{n_2}{n_1}\). 2) Use the given relationships between permittivity and permeability to express the refractive indices in terms of permeabilities. 3) Calculate the desired ratio \(\frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}}\) using the two possible values of \(\frac{n_2}{n_1}\). The two possible values for the ratio will be the cubes of the values of \(\frac{n_2}{n_1}\) found in step 1.

Step by step solution

01

Write down the reflection coefficient formula

For normal incidence, the reflection coefficient \(R\) is given by: $$ R = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2 $$ where \(n_1\) and \(n_2\) are the refractive indices of regions 1 and 2, respectively.
02

Write down the transmission coefficient formula

The transmission coefficient \(T\) is given by: $$ T = 1 - R $$
03

Solve for the refractive indices ratio

Since we are given that \(20\%\) of the energy is reflected, \(R = 0.2\) and \(T = 0.8\). Plugging these values into the reflection coefficient formula and solving for the ratio \(\frac{n_2}{n_1}\), we get: $$ 0.2 = \left(\frac{n_2 - n_1}{n_2 + n_1}\right)^2 $$ Solving this equation, we find two possible values for \(\frac{n_2}{n_1}\).
04

Use given conditions relating permittivity and permeability

Now we need to use the given relationships for regions 1 and 2: $$ \epsilon_{r1}^{\prime} = \mu_{r1}^3 $$ $$ \epsilon_{r2}^{\prime} = \mu_{r2}^3 $$ The refractive index is given by \(n = \sqrt{\epsilon_r \mu_r}\), therefore: $$ n_1 = \sqrt{\epsilon_{r1}^{\prime} \mu_{r1}} = \sqrt{(\mu_{r1}^3)\mu_{r1}} = \sqrt{\mu_{r1}^4} = \mu_{r1}^2 $$ Similarly, we obtain \(n_2 = \mu_{r2}^2\). Now, we have: $$ \frac{n_2}{n_1} = \frac{\mu_{r2}^2}{\mu_{r1}^2} $$
05

Solve for the desired ratio

The desired ratio in the problem is \(\frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}}\), which can be written as: $$ \frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}} = \frac{\mu_{r2}^3}{\mu_{r1}^3} $$ Using the two possible values of \(\frac{n_2}{n_1}\) obtained in step 3, we can find the two possible values of \(\frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}}\) by cubing the values of \(\frac{n_2}{n_1}\) as follows: $$ \frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}} = \left(\frac{n_2}{n_1}\right)^3 $$ Hence, the two possible values for the ratio \(\frac{\epsilon_{r2}^{\prime}}{\epsilon_{r1}^{\prime}}\) are obtained.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that \(\phi\) in Figure \(12.17\) is Brewster's angle, and that \(\theta_{1}\) is the critical angle. Find \(n_{0}\) in terms of \(n_{1}\) and \(n_{2}\).

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 \(T=20\) ps transform-limited pulse propagates through \(10 \mathrm{~km}\) of a dispersive medium for which \(\beta_{2}=12 \mathrm{ps}^{2} / \mathrm{km}\). The pulse then propagates through a second \(10 \mathrm{~km}\) medium for which \(\beta_{2}=-12 \mathrm{ps}^{2} / \mathrm{km}\). Describe the pulse at the output of the second medium and give a physical explanation for what happened.

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 \(n>1\).

Show how a single block of glass can be used to turn a p-polarized beam of light through \(180^{\circ}\), with the light suffering (in principle) zero reflective loss. The light is incident from air, and the returning beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles and use \(n=1.45\) for glass. More than one design is possible here.
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Wave Optics

Read Explanation

Space Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Medical Physics

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free