3. Maxwell's equations are the universe of electromagnetic (electrodynamic) phenomena

Richard Feynman (from Feynman's Lectures, Vol. 2, Chap. 21): "So here is the center of the universe of electromagnetism, the complete theory of electricity and magnetism, and of light; a complete description of the fields produced by any moving charges; and more. It is all here. Here is the structure built by Maxwell, complete in all its power and beauty. It is probably one of the greatest accomplishments of physics. To remind you of its importance, we will put it all together in a nice frame:"

$$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon\_0} (\rho\_\mathbf{f} - \nabla \cdot \mathbf{P}) \tag{1}$$

$$\nabla \cdot \mathbf{B} = \mathbf{0} \tag{2}$$

$$\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial \mathbf{t}} = \mathbf{0} \tag{3}$$

$$\mathbf{V} \times \mathbf{B} - \frac{1}{\mathbf{c}^2} \frac{\partial \mathbf{E}}{\partial \mathbf{t}} = \mu\_0 \left( \mathbf{J}\_\mathbf{f} + \frac{\partial \mathbf{P}}{\partial \mathbf{t}} + \nabla \times \mathbf{M} \right) \tag{4}$$

He states such definite statements once he obtains the general solutions for the electric potential, V, and the vector potential, A, for charges in motion. We have changed Feynman's notation φ for V to denote electric potential, and we use MKS units. In these equations ρ's are charge densities, J's are free currents, P is the polarization vector, M the magnetization vector, and E and B the electric and magnetic fields. We already defined ϵ, μ, and σ above.

We are not concerned here in the detail deduction of Maxwell's equations, nor the deduction of electromagnetic waves from them. Reflection and refraction and absorption of microwaves in conducting magnetic and dielectric materials are

presented in graphical form along with the principal equations that govern them. The deduction of such results is too long, and they are developed in a number of excellent, well-established textbooks [27–33]. The deduction of electromagnetic fields and waves from Maxwell's equations does not impose any restriction on the wavelength-frequency, as shown in Figure 1. The result is quite general, but we are interested and focused on microwaves interacting with ϵ, μ, and σ matter. But real materials, the air, ocean, hurricanes, semiconductors, ferrites, conductors, submarines, cars, planes, ferrous materials, steel, and magnets, which are of technological relevance, are dispersive, anisotropic, and absorb microwave energy [19, 28, 31]; then the adequate ϵ and μ become complex (dispersion and dissipation effects taken into account) and tensorial (anisotropy effects taken into account) [28, 31, 32]:

$$\mathbf{e}\_{\mathbf{i}\mathbf{j}} = \begin{bmatrix} \mathbf{e}\_{11} & \mathbf{e}\_{12} & \mathbf{e}\_{13} \\ \mathbf{e}\_{21} & \mathbf{e}\_{22} & \mathbf{e}\_{23} \\ \mathbf{e}\_{31} & \mathbf{e}\_{32} & \mathbf{e}\_{33} \end{bmatrix} = \begin{bmatrix} \mathbf{e}\_{\mathbf{x}} & \mathbf{0} & \mathbf{j}\mathbf{e}\_{13} \\ \mathbf{0} & \mathbf{e}\_{\mathbf{y}} & \mathbf{0} \\ -\mathbf{j}\mathbf{e}\_{31} & \mathbf{0} & \mathbf{e}\_{\mathbf{z}} \end{bmatrix} \tag{5}$$

$$
\boldsymbol{\mu}\_{\mathbf{ij}} = \begin{bmatrix}
\boldsymbol{\mu}\_{11} & \boldsymbol{\mu}\_{12} & \boldsymbol{\mu}\_{13} \\
\boldsymbol{\mu}\_{21} & \boldsymbol{\mu}\_{22} & \boldsymbol{\mu}\_{23} \\
\boldsymbol{\mu}\_{31} & \boldsymbol{\mu}\_{32} & \boldsymbol{\mu}\_{33}
\end{bmatrix} = \begin{bmatrix}
\boldsymbol{\mu} & \mathbf{0} & \mathbf{jk} \\
\mathbf{0} & \mathbf{1} & \mathbf{0} \\
\end{bmatrix} \tag{6}
$$

All the elements of both material electrodynamic tensors are in general dependent on frequency, ϵijð Þ ω and μijð Þ ω . In what follows we will continue to use just the symbols ϵ and μ. However, we understand their complexity. Using all four field vectors E, D, B and H,

$$\nabla \cdot \mathbf{D} = \rho\_{\mathbf{f}} \tag{7}$$

$$\nabla \cdot \mathbf{B} = \mathbf{0} \tag{8}$$

$$
\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial \mathbf{t}} = \mathbf{0} \tag{9}
$$

$$\nabla \times \mathbf{H} - \frac{\partial \mathbf{D}}{\partial \mathbf{t}} = \mathbf{J}\_{\mathbf{f}} \tag{10}$$

 Remember that D is ϵ0EþP and P ¼ ϵ0χ E, while B is μ0ðHþMÞ, and M ¼ χ H, <sup>e</sup> <sup>m</sup> and μ ¼ μ0μ<sup>r</sup> ¼ μ0ð1þχmÞ. Notice that the material's electrodynamics response is within D and B. In fact, the fields D and B are valued inside the material bodies. We are interested in microwave fields that in free space and in ε, μ, and σ matter, the field components E and H are sinusoidal functions of time and orthogonal to each other. In such case we can replace the time derivative by factors of jω. Setting also D ¼ ϵE and B ¼ μH,

$$\nabla \cdot \epsilon \mathbf{E} = \rho\_{\mathbf{f}} \tag{11}$$

$$\nabla \cdot \mu \mathbf{H} = \mathbf{0} \tag{12}$$

$$\nabla \times \mathbf{E} + \text{joo}\mu \mathbf{H} = \mathbf{0} \tag{13}$$

The Interaction of Microwaves with Materials of Different Properties DOI: http://dx.doi.org/10.5772/intechopen.83675

$$\mathbf{V} \times \mathbf{H} - \text{joo} \mathbf{E} = \mathbf{J}\_{\mathbf{f}} \tag{14}$$

From Maxwell' equations inhomogeneous wave equations are obtained for the fields E and H. The procedure is taking the rot (∇�) of the two rot Maxwell equations, rearranging terms and using the divergence equations; we find for a medium ϵ, μ, and σ nearby (ρf, Jf) microwave source:

$$
\nabla^2 \mathbf{E} - \epsilon \mu \frac{\partial^2 \mathbf{E}}{\partial \mathbf{t}^2} = \nabla \left(\frac{\rho\_\mathbf{f}}{\epsilon}\right) + \mu \frac{\partial \mathbf{J}\_\mathbf{f}}{\partial \mathbf{t}} \tag{15}
$$

$$
\nabla^2 \mathbf{B} - \epsilon \mu \frac{\partial^2 \mathbf{B}}{\partial \mathbf{t}^2} = \mu \nabla \times \mathbf{J}\_\mathbf{f} \tag{16}
$$

Far from the microwave sources (centimeters, meters, kilometers, and, even, light years), we can safely drop the "source" terms, and homogeneous partial differential equations are left:

$$
\nabla^2 \mathbf{E} - \epsilon \mu \frac{\partial^2 \mathbf{E}}{\partial \mathbf{t}^2} - \mu \sigma \frac{\partial \mathbf{E}}{\partial \mathbf{t}} = \mathbf{0} \tag{17}
$$

$$
\nabla^2 \mathbf{H} - \epsilon \mu \frac{\partial^2 \mathbf{H}}{\partial \mathbf{t}^2} - \mu \sigma \frac{\partial \mathbf{H}}{\partial \mathbf{t}} = \mathbf{0} \tag{18}
$$

It should be stressed that these identical wave equations describe attenuated waves. This comes naturally from the third term that includes magnetic permeability, electric conductivity, and the first time derivative of each field. The solutions of these equations are readily obtained by separation of variables [27]. The detailed deduction of these mathematical expressions is lengthy and elaborated; after such work, we find the fields to be transverse and plane-polarized for which the xdirection for E is chosen; hence H ( j ð ωμÞH ¼ ∇ � E) goes along the y-direction and propagates along the z-direction with a flux of energy density S ¼ E � H [27] and an impedance Z = E/H. The solutions are summarized in Table 1.

Here we show in the form of a table (deductions are given in [27, 28, 31]) the solutions for E and H for propagation in free space and in a dielectric material and in a conducting material. In this concise form, we highlight the most relevant features and parameters of the solutions. When we put back the solutions found for a σ, μ, and ϵ material into the wave equations, we obtain the dispersion relation:

˛ ˝ �k<sup>2</sup> <sup>þ</sup> <sup>ω</sup><sup>2</sup>ϵμ � <sup>j</sup>ωσμ <sup>E</sup> <sup>¼</sup> 0, which is shown in the third column for each


#### Table 1. Wave solutions in different media (no bounds considered).

electrodynamic medium. A summary of E and H solutions, along with the dispersion relations, impedance, attenuation, and skin depth, is presented in Table 1 for different materials characterized by particular sets of electrodynamic properties (σ, μ, ϵ).

In the ideal case of propagation of microwaves in air, the phenomenon of attenuation (absorption of microwave energy, E ¼ hν, in the form of photons) does not take place. In reality, quite a substantial attenuation of microwaves takes place when they propagate in air (molecules that constitute the air readily absorb microwaves. Such energy excites efficiently rotation and vibration degrees of freedom). The Wi-Fi technology [5, 6, 16] knows very well this fact, that is why there exist in the market potentiators of Wi-Fi signal to cover adequately a room of several meters square. The problem of attenuated signal becomes worse if panels, wall, wood, and metals come in the way of the propagation. The microwave engineer, designer, and scientist know that the main factor which causes attenuation in microwave propagation is water, H2O. War technology for decades has faced the problem of the no radar signal penetration in the oceans to a depth enough to detect submarines. So, in order to propagate microwaves in the atmosphere or in the ocean, for at least 1 m, or longer distances, it is mandatory to guide such electromagnetic GHz waves or increase considerably their power content, Poynting vector, S ¼ E � H being proportional to its power content per unit area.

### 3.1 Microwaves in spaces with obstacles: propagation, reflection, refraction, and absorption

In the modern microwave Wi-Fi, military, radar communication systems, the microwaves have to travel through "free space" to reach detectors and receivers, but it is pretty common in urban areas that the propagation encounters obstacles, such as buildings, windows, walls, bridges, metallic structures, and fog. At the surface of those obstacles, reflection, absorption, refraction, and dispersion of the microwaves occur. When microwaves hit obstacles, energy is irremediably lost at these obstacles. This fact could be used to detect that precisely an obstacle is present and fine measurements of the reflected or refracted microwave energy could give valuable information on the electromagnetic character (ϵ, μ, σ) of such obstacle. This is how radar in air and other atmospheric microwave instruments detect objects (the obstacles themselves). But many other times, obstacles, like walls, wood floors, and concrete structures, attenuate, disperse, and deviate the microwaves, and reception at the desired site is deficient or completely lost. In such cases midway potentiators are used to increase the signal power to compensate the losses at the obstacles. These examples are meant to show how important are the reflection, absorption, refraction, and dispersion of microwaves in real-world applications. The laws that govern these phenomena are Snell law and Fresnel laws of reflection and refraction. They are obtained as a consequence of the boundary conditions applied to the interface, any interface between two media as shown in Figure 4. The reflected (transmitted) waves carry an energy R(T) per unit of incident energy. Appropriate detectors (common radiometers, photocells, diodes) measure with ease R and T, also depicted in Figure 4. These radar and Wi-Fi microwaves hit all the time buildings, hot air, street floors, and so forth, and, at all times, they are reflected, absorbed, and refracted [16–20] as illustrated in Figure 4.

## 3.2 Boundary conditions that microwaves fulfill at the interface between any two media

The general boundary conditions the electromagnetic fields, E, D, H and B, must satisfy at the interface of any two (σ, μ, ε) media which are easily obtained The Interaction of Microwaves with Materials of Different Properties DOI: http://dx.doi.org/10.5772/intechopen.83675

#### Figure 4.

Microwaves bounce back from virtually any material. (a) The Wi-Fi signal being reflected and refracted from anything can be in an urban city area. Wi-Fi and radar signals are commonly present in the air in any city. These microwaves hit buildings, windows, fog, hot air, street floors, and so forth; at all times they are reflected (R), absorbed (A), and refracted (T) [16–20], (b) satellite weather radar detecting and following 2005 Katrina hurricane trajectory just about to land [21], and (c) abstraction of any process in which microwaves hit a body of matter.

from applying Gaus law to a small cylinder placed through the two media and from applying Ampere's circuital law to a closed (rectangular) trajectory, again, crossing through both media. The results are [27–29] E1t ¼ E2t; H1t ¼ H2t, i.e., the tangential components of E and H must be continuous through the interface. And D1n ¼ D2n; B1n ¼ B2n, i.e., the normal components of D must be continuous, provided the interface does not carry any charge and the normal components of B must be continuous. All this is represented in Figure 5.

Since medium 1 sustains the incident and reflected fields, E<sup>1</sup> ¼ E<sup>i</sup> þ Er; D<sup>1</sup> ¼ D<sup>i</sup> þ Dr; H<sup>1</sup> ¼ H<sup>i</sup> þ Hr; and B<sup>1</sup> ¼ B<sup>i</sup> þ Br. In medium 1 a sum of vector fields must be formed. In medium 2 there are only the transmitted fields Et, Dt, Ht, and Bt.

When boundaries are present on purpose or as an inevitable obstacle, the incident microwaves reflect and refract, and the formal treatment will conduct us to Snell law and the Fresnel equations for incidence with H parallel to the plane of incidence as shown in Figures 6 and 7 to the Fresnel equations with H perpendicular to the plane of incidence as shown in Figure 8.

In Figure 6a, medium 1 is the space where we want to propagate some microwave signal and medium 2 being a window, a wall, water, wood, concrete, metallic sheets, a building in front of us, etc. A very common situation is when sunlight shines and goes through a glass window; in this case most of the light goes through and illuminates the room. And yet, part of it is reflected, and part is absorbed (the energy of which heats up the glass window itself). A second very common situation generated by the modern wireless communication technology is the Wi-Fi signal (it plays exactly the same role as the sunshine in the above illustration). The typical Wi-Fi signal is around 2.3 GHz, and the newest cable boxes are delivering to our home rooms ≈5.2GHz with about 1 W power.

When the obstacle (medium 2) is made of materials that are also ferromagnetic or ferrimagnetic (ferrites), the absorption will increase considerably because the magnetic moments of the material absorb energy from the Wi-Fi waves.

#### Figure 5.

Boundary conditions for the magnetic field H and for the magnetic induction. The condition B1<sup>n</sup> ¼ B2<sup>n</sup> is obtained from ∇ � B ¼ 0, and H1<sup>t</sup> ¼ H2<sup>t</sup> is obtained from Ampere's circuital law. Similar vector diagrams for the D and E vectors are considered, and the conditions, D1<sup>n</sup> ¼ D2<sup>n</sup> and H1<sup>t</sup> ¼ H2t, are obtained.

#### Figure 6.

Incident, reflected, and transmitted microwaves at any kind of interface. (a) H parallel to the plane of incidence. ni, nr, and n<sup>t</sup> are unitary vectors normal to the respective front-plane waves and pointing in the propagation direction, k ¼ kn. θi, θr, and θ<sup>t</sup> are, respectively, the incident, the reflected, and the transmitted direction of the waves with respect to the dotted line axis. (b) Feynman's Lintern. A white light beam incident on a glass that has been coated with red paint refracts red light and reflects the green component; hence ω<sup>t</sup> 6¼ ωr.

Conductors (materials with free electrons) will absorb Wi-Fi wave energy by virtue of the term JE which becomes Joule heating. Abstracting the cases mentioned above, we consider, as the first case, the incidence of a plane electromagnetic wave on the interface of two media as shown in Figure 7; the incident electric field is of the form

$$\mathbf{E}\_{\mathbf{i}} = \mathbf{E}\_{0\mathbf{i}} \exp\left[\mathbf{u}\_{\mathbf{i}} \left(\mathbf{t} - \frac{\mathbf{n}\_{\mathbf{i}} \cdot \mathbf{r}}{\mathbf{u}\_{1}}\right) \mathbf{n}\_{\mathbf{i}}\right] \tag{19}$$

˛ ˝ <sup>t</sup> � nr �<sup>r</sup> The reflected and the refracted waves are Er <sup>¼</sup> E0r exp <sup>j</sup>ω<sup>r</sup> and u1 ˛ ˝ <sup>t</sup> � nt�<sup>r</sup> Et <sup>¼</sup> E0t exp <sup>j</sup>ω<sup>t</sup> , where u1and u2 are the velocities of the waves in mediums u2 1 and 2, respectively. The E0i, E0r, and E0r are the amplitudes of the respective

The Interaction of Microwaves with Materials of Different Properties DOI: http://dx.doi.org/10.5772/intechopen.83675

#### Figure 7.

 The incident wave with magnetic vector in the incidence plane. Snell laws demand that the reflected and transmitted magnetic vectors are also in the same plane. The arrows indicate the directions of the vectors involved. The Poynting vectors Si,r,t = Ei,r,t � Hi,r,t point always in the direction of propagation of that particular wave, and its magnitude indicates how much electromagnetic energy is flowing in that direction.

waves. For linear, isotropic, homogeneous media (LIH), and nondispersive, it is readily shown that the three vector fields Ei, Er, and E<sup>t</sup> are identical functions of time [27–29, 32–34], ω<sup>i</sup> ¼ ω<sup>r</sup> ¼ ωt. For dispersive media, at least [27–29, 32–34] two frequencies are different ω<sup>i</sup> 6¼ ω<sup>t</sup> 6¼ ωr.

As the simple example illustrated in Figure 6b, given by Feynman [29], shows "A large imaginary part of the index of refraction (or equivalently a large imaginary component of k, k") means a strong absorption. So there is a general rule that if any material gets to be a very good absorber at any frequency (let's say red), the waves are strongly reflected at the surface, and very little gets inside to be absorbed. You can see this effect with strong dyes. Pure crystals of the strongest dyes have a metallic shine. Red ink absorbs out the greens of transmitted light, so if the ink is very concentrated, it will exhibit a strong surface reflection for the frequencies of green light (see Figure 6b).

But in our case the three fields must be identical functions of space and time at any point on the interface [27–29, 32, 33]. From such requirement, it follows that θ<sup>i</sup> ¼ θ<sup>r</sup> the angle of incidence is equal to the angle of reflection. This is the law of reflection. It is also obtained that ½sin θt= sin θi� ¼ ½n1=n2�, with n = k/λ. This equation is the well-known Snell law. With these results at hand, we now write: E<sup>i</sup> ¼ E0i exp j½ωt � k1ðxsinθ<sup>i</sup> � zcosθiÞ�, E<sup>r</sup> ¼ E0r exp j½ωt � k1ðxsinθ<sup>i</sup> � zcosθiÞ�, Et ¼ E0t exp j½ωt � k2ðxsinθ<sup>t</sup> � zcosθtÞ�; the respective magnetic fields are Hp ¼ ðk=ωμÞEp, with p ¼ i,r,t.

Now we proceed to determine the quantities E0i, E0r, E0t that assure the continuity of the tangential components of E and H at the interface:

˜ ° ˜ ° ˜ ° <sup>E</sup> <sup>¼</sup> E0 exp j <sup>ω</sup><sup>t</sup> � <sup>k</sup><sup>η</sup> � r n<sup>η</sup> and <sup>H</sup> <sup>¼</sup> <sup>k</sup>η=ωμ1, <sup>2</sup> E0 exp <sup>j</sup> <sup>ω</sup><sup>t</sup> � <sup>k</sup><sup>η</sup> � r nη, with kη¼ki, kr, k<sup>t</sup> with η = i, r, t the propagation vectors and μ1, <sup>2</sup> the magnetic permeability of medium 1 or medium 2.

We have four unknowns: (E0r, H0r) and (E0t, H0t). And we have four equations, which are the four boundary conditions. E0i and H0i are taken as known since they are the primary waves that we send-propagate from a "controllable" source. We would find the unknowns normalized by E0i and H0i. Of course the problem can be inverted, and we could start knowing the transmitted waves and would like to determine the initial fields, Ei, Hi, that come from an unknown (potentially fundamental) source. Present cosmological problems are exactly of this type [3, 35, 36]. Continuity of the tangential components of E requires

$$\mathbf{E\_{0i}} + \mathbf{E\_{0r}} = \mathbf{E\_{0t}} \tag{20}$$

At any point and at any time at the interface. Likewise, continuity of the tangential component of the magnetic field requires H0i cos θ<sup>i</sup> � H0r cos θ<sup>i</sup> ¼ H0t cos θ<sup>t</sup> which becomes k1 <sup>ð</sup>E0i � E0r<sup>Þ</sup> cos <sup>θ</sup><sup>i</sup> <sup>¼</sup> k2 E0t cos <sup>θ</sup>t. But <sup>k</sup> <sup>¼</sup> <sup>n</sup>=λ, so ωμ<sup>1</sup> ωμ<sup>2</sup>

$$\frac{\mathbf{n}\_1}{\mu\_1}(\mathbf{E}\_{0\mathbf{i}} - \mathbf{E}\_{0\mathbf{r}})\cos\theta\_{\mathbf{i}} = \frac{\mathbf{n}\_2}{\mu\_2}\mathbf{E}\_{0\mathbf{t}}\cos\theta\_{\mathbf{t}}\tag{21}$$

Algebraic Eqs. (20) and (21) are readily solved for unknowns E0r and E0t:

$$\frac{\mathbf{n}\_1}{\mu\_1} \mathbf{E}\_{0\mathbf{i}} \cos \theta\_{\mathbf{i}} - \frac{\mathbf{n}\_1}{\mu\_1} \mathbf{E}\_{0\mathbf{r}} \cos \theta\_{\mathbf{i}} = \frac{\mathbf{n}\_2}{\mu\_2} \mathbf{E}\_{0\mathbf{r}} \cos \theta\_{\mathbf{i}} \tag{22}$$

Using (20) E0i þ E0r ¼ E0t and after some algebra, we obtain [27]: the amplitude of the fields E0r/E0i and E0t/E0i:

$$\left(\frac{\mathbf{E\_{0r}}}{\mathbf{E\_{0i}}}\right)\_{\mathbf{N}} = \frac{\frac{\mathbf{n\_i}}{\mu\_{r1}}\cos\Theta\_{\mathbf{i}} - \frac{\mathbf{n\_2}}{\mu\_{r2}}\cos\Theta\_{\mathbf{t}}}{\frac{\mathbf{n\_i}}{\mu\_{r1}}\cos\Theta\_{\mathbf{i}} + \frac{\mathbf{n\_2}}{\mu\_{r2}}\cos\Theta\_{\mathbf{t}}}\tag{23}$$

$$\left(\frac{\text{E}\_{\text{0t}}}{\text{E}\_{\text{0i}}}\right)\_{\text{N}} = \frac{2\frac{\text{n}\_{\text{1}}}{\mu\_{\text{1}}}\cos\Theta\_{\text{i}}}{\frac{\text{n}\_{\text{1}}}{\mu\_{\text{1}}}\cos\Theta\_{\text{i}} + \frac{\text{n}\_{\text{2}}}{\mu\_{\text{2}}}\cos\Theta\_{\text{t}}}\tag{24}$$

where N indicates that E is normal to the incidence plane and/or H is in the plane of incidence.

In the case the three H vectors are perpendicular to the plane of incidence, Figure 8, we have: H0i � H0r ¼ H0t. In terms of E fields

$$\frac{\mathbf{n}\_1}{\mu\_1}(\mathbf{E}\_{0\mathbf{i}} - \mathbf{E}\_{0\mathbf{r}}) = \frac{\mathbf{n}\_2}{\mu\_2}\mathbf{E}\_{0\mathbf{t}} \tag{25}$$

$$(\mathbf{E\_{0i}} + \mathbf{E\_{0r}})\cos\theta\_{\mathbf{i}} = \mathbf{E\_{0t}}\cos\theta\_{\mathbf{t}}\tag{26}$$

After some algebra and rearranging terms, we obtain E0t/E0i. Hence:

$$\left(\frac{\mathbf{E\_{0r}}}{\mathbf{E\_{0i}}}\right)\_{\mathbf{p}} = \frac{-\frac{\mathbf{n\_2}}{\mu\_{\rm r2}}\cos\Theta\_{\mathbf{i}} + \frac{\mathbf{n\_1}}{\mu\_{\rm r1}}\cos\Theta\_{\mathbf{i}}}{\frac{\mathbf{n\_1}}{\mu\_{\rm r1}}\cos\Theta\_{\mathbf{i}} + \frac{\mathbf{n\_2}}{\mu\_{\rm r2}}\cos\Theta\_{\mathbf{i}}}\tag{27}$$

$$\left(\frac{\mathbf{E\_{0t}}}{\mathbf{E\_{0i}}}\right)\_{\mathbf{p}} = \frac{2\frac{\mathbf{n\_1}}{\mu\_{\rm r1}}\cos\Theta\_{\rm i}}{\frac{\mathbf{n\_1}}{\mu\_{\rm r1}}\cos\Theta\_{\rm t} + \frac{\mathbf{n\_2}}{\mu\_{\rm r2}}\cos\Theta\_{\rm i}}\tag{28}$$

The Interaction of Microwaves with Materials of Different Properties DOI: http://dx.doi.org/10.5772/intechopen.83675

Here P denotes that the Hi, Hr, H<sup>t</sup> are all parallel to the interface. In the case of normal incidence with all the H vectors parallel to the interface, θ<sup>i</sup> ¼ θ<sup>t</sup> ¼ 0 (Figure 8), and cos θ<sup>i</sup> = cos θ<sup>t</sup> = 1 in the above equations.
