Question

A uniform plane wave in region 1 is normally incident on the planar boundary separating regions 1 and 2. If ${ϵ}_1^{\prime \prime }={ϵ}_2^{\prime \prime }=0$, while ${ϵ}_{r1}^{\prime }={\mu }_{r1}^3$ and ${ϵ}_{r2}^{\prime }={\mu }_{r2}^3$, find the ratio ${ϵ}_{r2}^{\prime }/{ϵ}_{r1}^{\prime }$ if $20\mathrm{\%}$ of the energy in the incident wave is reflected at the boundary. There are two possible answers.

Short answer

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 $\frac{{ϵ}_{r2}^{\prime }}{{ϵ}_{r1}^{\prime }}$ 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

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.

Write down the transmission coefficient formula

The transmission coefficient $T$ is given by: $$T=1-R$$

Solve for the refractive indices ratio

Since we are given that $20\mathrm{\%}$ 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 $\frac{n_2}{n_1}$.

Use given conditions relating permittivity and permeability

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

Solve for the desired ratio

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

Key concepts

Reflection Coefficient

In the context of electromagnetic waves, the reflection coefficient is a crucial parameter. It essentially measures how much of an incident wave is reflected back at the boundary of two different media. For a wave normally incident at a boundary, the reflection coefficient $R$ can be described by the formula:

• $R={\left(\frac{n_2-n_1}{n_2+n_1}\right)}^2$

Here, $n_1$ and $n_2$ represent the refractive indices of the two regions. A reflection coefficient of 0.2, for instance, implies that 20% of the incident energy is reflected.
This concept is pivotal in understanding how waves behave when encountering interfaces, which is particularly relevant in optical applications like coatings and fiber optics.

Transmission Coefficient

While the reflection coefficient focuses on the reflected part of a wave, the transmission coefficient $T$ tells us about the transmitted portion of the wave as it passes through the boundary. It is inherently linked to the reflection coefficient by the relation:

• $T=1-R$

This means that the transmission and reflection coefficients together account for the entirety of the wave's energy. If a wave has a reflection coefficient of 0.2, the transmission coefficient would naturally be 0.8, indicating that 80% of the incident energy is transmitted.
The transmission coefficient is vital for designing systems where maximum energy transfer across boundaries is needed, such as in lens design.

Permittivity

Permittivity is a material property that measures how an electric field affects, and is affected by, a dielectric medium. It's denoted by $ϵ$ and indicates how much electric charge is stored in a material when a voltage is applied.
In the exercise, permittivity is represented as ${ϵ}_{r1}^{\prime }={\mu }_{r1}^3$ and ${ϵ}_{r2}^{\prime }={\mu }_{r2}^3$. This shows that permittivity is related to the material's permeability and refractive index. Knowing the permittivity helps in calculating the refractive index as $n=\sqrt{ϵ\mu }$, a key parameter in wave propagation.
Understanding permittivity is essential for applications ranging from antenna design to understanding materials in capacitors.

Permeability

Permeability is another material property, symbolized by $\mu$, which describes how a material responds to a magnetic field. Essentially, it reflects the ability of a material to support the formation of magnetic fields within itself.
In electromagnetic theory, permeability affects the propagation of electromagnetic waves through a medium. In the given exercise, permeability is related to permittivity, with ${ϵ}_{r1}^{\prime }={\mu }_{r1}^3$, illustrating an intrinsic relationship to refractive characteristics.

• This links the material's permeability directly to its electromagnetic behavior, as $n={\mu }_r^2$.

Engineers often leverage these properties to tailor materials for specific electromagnetic applications, like inductors in circuits.