# 8.4 Wave Reflection and Transmission at Oblique Incidence

In this section, we develop a rigorous theory of reflection and refraction of plane waves obliquely incident upon planar boundaries between different media. Our treatment parallels that in Section 8-1 for the normal-incidence case and goes beyond that in Section 8-2 on Snell’s laws, which yielded information on only the angles of reflection and refraction.

For normal incidence, the reflection and transmission coefficients Γ and τ at a boundary between two media are independent of the polarization of the incident wave, because both the electric and magnetic fields of a normally incident plane wave are tangential to the boundary regardless of the wave polarization. This is not the case for obliquely incident waves traveling at an angle θ i≠0 with respect to the normal to the interface.

A wave of arbitrary polarization may be described as the superposition of two orthogonally polarized waves: one with its electric field parallel to the plane of incidence (parallel polarization) and the other with its electric field perpendicular to the plane of incidence (perpendicular polarization). These two polarization configurations are shown in Fig. 8-14, where the plane of incidence is coincident with the x–z plane. Polarization with E perpendicular to the plane of incidence is also called transverse electric (TE) polarization because E is perpendicular to the plane of incidence and with E parallel to the plane of incidence is called transverse magnetic (TM) polarization because in that case it is the magnetic field that is perpendicular to the plane of incidence.

# Figure 8-14 The plane of incidence is the plane containing the direction of wave travel, k ˆ i, and the surface normal to the boundary. In the present case, the plane of incidence containing k ˆ i and z ˆ coincides with the plane of the paper. A wave is (a) perpendicularly polarized when its electric field vector is perpendicular to the plane of incidence and (b) parallel polarized when its electric field vector lies in the plane of incidence.

(a) Perpendicular polarization

Figure 8 14-a Full Alternative Text

(b) Parallel polarization

Figure 8 14-b Full Alternative Text

For the general case of a wave with an arbitrary polarization, it is common practice to decompose the incident wave (E i,H i) into a perpendicularly polarized component (E⊥i,H⊥i) and a parallel polarized component (E∥i,H∥i). Then, after determining the reflected waves (E⊥r,H⊥r) and (E∥r,H∥r) due to the two incident components, the reflected waves are added together to give the total reflected wave (E r,H r) corresponding to the original incident wave. A similar process can be used to determine the total transmitted wave (E t,H t).

# 8.4.1 Perpendicular Polarization

Figure 8-15 shows a perpendicularly polarized incident plane wave propagating along the x i direction in dielectric medium 1. The electric field phasor E˜⊥ i points along the y direction, and the associated magnetic field phasor H˜⊥ i is along the y i axis. The directions of E˜⊥ i and H˜⊥ i are such that E˜⊥ i×H˜⊥ i points along the propagation direction x ˆ i. The electric and magnetic fields of such a plane wave are given by

E˜⊥ i=y ˆ E⊥0 i e−j k 1 x i, (8.46a)
H˜⊥ i=y ˆ i E⊥0 i η 1 e−j k 1 x i, (8.46b)

where E⊥0 i is the amplitude of the electric field phasor at x i=0 and k 1=ω μ 1 ε 1 and η 1=μ 1/ε 1 are the wavenumber and intrinsic impedance of medium 1. From Fig. 8-15, the distance x i and the unit vector y ˆ i may be expressed in terms of the (x, y, z) global coordinate system as

x i=x sin θ i+z cos θ i, (8.47a)
y ˆ i=−x ˆ cos θ i+z ˆ sin θ i. (8.47b)

# Figure 8-15 Perpendicularly polarized plane wave incident at an angle θ i upon a planar boundary.

Figure 8 15 Full Alternative Text

Substituting Eqs. (8.47a) and (8.47b) into Eqs. (8.46a) and (8.46b) gives

Incident Wave

E˜⊥ i=y ˆ E⊥0 i e−j k 1(x sin θ i+z cos θ i), (8.48a)
H˜⊥ i=(−x ˆ cos θ i+z ˆ sin θ i)E⊥0 i η 1 e−j k 1(x sin θ i+z cos θ i). (8.48b)

With the aid of the directional relationships given in Fig. 8-15 for the reflected and transmitted waves, these fields are given by the following:

Reflected Wave

E˜⊥ r=y ˆ E⊥0 r e−j k 1 x r=y ˆ E⊥0 r e−j k 1(x sin θ r−z cos θ r), (8.49a)
H˜⊥ r=y ˆ r E⊥0 r η 1 e−j k 1 x r=(x ˆ cos θ r+z ˆ sin θ r)E⊥0 r η 1 e−j k 1(x sin θ r−z cos θ r), (8.49b)

Transmitted Wave

E˜⊥ t=y ˆ E⊥0 t e−j k 2 x t=y ˆ E⊥0 t e−j k 2(x sin θ t+z cos θ t), (8.49c)
H˜⊥ t=y ˆ t E⊥0 t η 2 e−j k 2 x t=(−x ˆ cos θ t+z ˆ sin θ t)E⊥0 t η 2 e−j k 2(x sin θ t+z cos θ t), (8.49d)

where θ r and θ t are the reflection and transmission angles shown in Fig. 8-15 and k 2 and η 2 are the wavenumber and intrinsic impedance of medium 2. Our goal is to describe the reflected and transmitted fields in terms of the parameters that characterize the incident wave, namely the incidence angle θ i and the amplitude E⊥0 i. The four expressions given by Eqs. (8.49a) through (8.49d) contain four unknowns: E⊥0 r, E⊥0 t, θ r, and θ t. Even though angles θ r and θ t are related to θ i by Snell’s laws (Eqs. (8.28a) and (8.28b)), here we choose to treat them as unknown for the time being because we intend to show that Snell’s laws also can be derived by applying field boundary conditions at z=0. The total electric field in medium 1 is the sum of the incident and reflected electric fields: E˜⊥1=E˜⊥ i+E˜⊥ r; a similar statement holds true for the total magnetic field in medium 1: H˜⊥1=H˜⊥ i+H˜⊥ r. Boundary conditions state that the tangential components of E˜ and H˜ each must be continuous across the boundary between the two media. Field components tangential to the boundary extend along x ˆ and y ˆ. Since the electric fields in media 1 and 2 have y ˆ components only, the boundary condition for E˜ is

(E˜⊥y i+E˜⊥y r)z=0=E˜⊥y t z=0. (8.50)

Upon using Eqs. (8.48a), (8.49a), and (8.49c) in Eq. (8.50) and then setting z=0, we have

E⊥0 i e−j k 1 x sin θ i+E⊥0 r e−j k 1 x sin θ r=E⊥0 t e−j k 2 x sin θ t. (8.51)

Since the magnetic fields in media 1 and 2 have no y ˆ components, the boundary condition for H˜ is

(H˜⊥x i+H˜⊥x r)z=0=H˜⊥x t z=0, (8.52)

or

−E⊥0 i η 1 cos θ i e−j k 1 x sin θ i+E⊥0 r η 1 cos θ r e−j k 1 x sin θ r=−E⊥0 t η 2 cos θ t e−j k 2 x sin θ t. (8.53)

To satisfy Eqs. (8.51) and (8.53) for all possible values of x (i.e., all along the boundary), it follows that the arguments of all three exponentials must be equal. That is,

k 1 sin θ i=k 1 sin θ r=k 2 sin θ t, (8.54)

which is known as the phase-matching condition. The first equality in Eq. (8.54) leads to

while the second equality leads to

The results expressed by Eqs. (8.55) and (8.56) are identical with those derived previously in Section 8-2 through consideration of the ray path traversed by the incident, reflected, and transmitted wavefronts.

In view of Eq. (8.54), the boundary conditions given by Eqs. (8.51) and (8.53) reduce to

E⊥0 i+E⊥0 r=E⊥0 t, (8.57a)
cos θ i η 1(−E⊥0 i+E⊥0 r)=−cos θ t η 2 E⊥0 t. (8.57b)

These two equations can be solved simultaneously to yield the following expressions for the reflection and transmission coefficients in the perpendicular polarization case:

These two coefficients are known formally as the Fresnel reflection and transmission coefficients for perpendicular polarization and are related by

If medium 2 is a perfect conductor (η 2=0), Eqs. (8.58a and b) reduce to Γ⊥=−1 and τ⊥=0, respectively, which means that the incident wave is totally reflected by the conducting medium.

For nonmagnetic dielectrics with μ 1=μ 2=μ 0 and with the help of Eq. (8.56), the expression for Γ⊥ can be written as

Since (ε 2/ε 1)=(n 2/n 1)2, this expression also can be written in terms of the indices of refraction n 1 and n 2.

# Example 8-6: Wave Incident Obliquely on a Soil Surface

Using the coordinate system of Fig. 8-15, a plane wave radiated by a distant antenna is incident in air upon a plane soil surface located at z=0. The electric field of the incident wave is given by

E i=y ˆ 100 cos(ω t−π x−1.73 π z)(V/m), (8.61)

and the soil medium may be assumed to be a lossless dielectric with a relative permittivity of 4.

1. Determine k 1, k 2, and the incidence angle θ i.

2. Obtain expressions for the total electric fields in air and in the soil.

3. Determine the average power density carried by the wave traveling in soil.

Solution: (a) We begin by converting Eq. (8.61) into phasor form, akin to the expression given by Eq. (8.46a):

E˜i=y ˆ 100 e−j π x−j 1.73 π z=y ˆ 100 e−j k 1 x i(V/m), (8.62)

where x i is the axis along which the wave is traveling and

k 1 x i=π x+1.73 π z. (8.63)

Using Eq. (8.47a), we have

k 1 x i=k 1 x sin θ i+k 1 z cos θ i. (8.64)

Hence,

k 1 sin θ i=π,k 1 cos θ i=1.73 π,

which together give

k 1=π 2+(1.73 π)2=2 π(rad/m),θ i=tan−1 π 1.73 π=30°.

The wavelength in medium 1 (air) is

λ 1=2 π k 1=1 m,

and the wavelength in medium 2 (soil) is

λ 2=λ 1 ε r 2=1 4=0.5 m.

The corresponding wavenumber in medium 2 is

k 2=2 π λ 2=4 π(rad/m).

Since E˜i is along y ˆ, it is perpendicularly polarized (y ˆ is perpendicular to the plane of incidence containing the surface normal z ˆ and the propagation direction x ˆ i).

(b) Given that θ i=30°, the transmission angle θ t is obtained with the help of Eq. (8.56):

sin θ t=k 1 k 2 sin θ i=2 π 4 π sin 30°=0.25

or

θ t=14.5°.

With ε 1=ε 0 and ε 2=ε r 2 ε 0=4 ε 0, the reflection and transmission coefficients for perpendicular polarization are determined with the help of Eqs. (8.59) and (8.60),

Γ⊥=cos θ i−(ε 2/ε 1)−sin 2 θ i cos θ i+(ε 2/ε 1)−sin 2 θ i=−0.38,τ⊥=1+Γ⊥=0.62.

Using Eqs. (8.48a) and (8.49a) with E⊥0 i=100 V/m and θ i=θ r, the total electric field in medium 1 is

E˜⊥1=E˜⊥ i+E˜⊥ r=y ˆ E⊥0 i e−j k 1(x sin θ i+z cos θ i)+y ˆ Γ E⊥0 i e−j k 1(x sin θ i−z cos θ i)=y ˆ 100 e−j(π x+1.73 π z)−y ˆ 38 e−j(π x−1.73 π z),

and the corresponding instantaneous electric field in medium 1 is

E⊥1(x,z,t)=R e E˜⊥1 e j ω t=y ˆ[100 cos(ω t−π x−1.73 π z)−38 cos(ω t−π x+1.73 π z)](V/m).

In medium 2, using Eq. (8.49c) with E⊥0 t=τ⊥E⊥0 i gives

E˜⊥ t=y ˆ τ E⊥0 i e−j k 2(x sin θ t+z cos θ t)=y ˆ 62 e−j(π x+3.87 π z)

and, correspondingly,

E⊥t(x,z,t)=R e E˜⊥ t e j ω t=y ˆ 62 cos(ω t−π x−3.87 π z)(V/m).

(c) In medium 2, η 2=η 0/ε r 2≈120 π/4=60 π (Ω), and the average power density carried by the wave is

S av t=|E⊥0 t|2 2 η 2=(62)2 2×60 π=10.2(W/m 2).

# 8.4.2 Parallel Polarization

If we interchange the roles played by E and H in the perpendicular polarization scenario covered in the preceding subsection, while keeping in mind the requirement that E×H must point in the direction of propagation for each of the incident, reflected, and transmitted waves, we end up with the parallel polarization scenario shown in Fig. 8-16. Now the electric fields lie in the plane of incidence, while the associated magnetic fields are perpendicular to the plane of incidence. With reference to the directions indicated in Fig. 8-16, the fields of the incident, reflected, and transmitted waves are given by

Incident Wave

E˜∥i=y ˆ i E∥0 i e−j k 1 x i=(x ˆ cos θ i−z ˆ sin θ i)E∥0 i e−j k 1(x sin θ i+z cos θ i), (8.65a)
H˜∥i=y ˆ E∥0 i η 1 e−j k 1 x i=y ˆ E∥0 i η 1 e−j k 1(x sin θ i+z cos θ i), (8.65b)

# Figure 8-16 Parallel-polarized plane wave incident at an angle θ i upon a planar boundary.

Figure 8 16 Full Alternative Text

Reflected Wave

E˜∥r=y ˆ r E∥0 r e−j k 1 x r=(x ˆ cos θ r+z ˆ sin θ r)E∥0 r e−j k 1(x sin θ r−z cos θ r), (8.65c)
H˜∥r=−y ˆ E∥0 r η 1 e−j k 1 x r=−y ˆ E∥0 r η 1 e−j k 1(x sin θ r−z cos θ r), (8.65d)

Transmitted Wave

E˜∥t=y ˆ t E∥0 t e−j k 2 x t=(x ˆ cos θ t−z ˆ sin θ t)E∥0 t e−j k 2(x sin θ t+z cos θ t), (8.65e)
H˜∥t=y ˆ E∥0 t η 2 e−j k 2 x t=y ˆ E∥0 t η 2 e−j k 2(x sin θ t+z cos θ t). (8.65f)

By matching the tangential components of E˜ and H˜ in both media at z=0, we again obtain the relations defining Snell’s laws, as well as the following expressions for the Fresnel reflection and transmission coefficients for parallel polarization:

The preceding expressions can be shown to yield the relation

We noted earlier in connection with the perpendicular-polarization case that, when the second medium is a perfect conductor with η 2=0, the incident wave gets totally reflected at the boundary. The same is true for the parallel polarization case; setting η 2=0 in Eqs. (8.66a and b) gives Γ∥=−1 and τ∥=0.

For nonmagnetic materials, Eq. (8.66a) becomes

To illustrate the angular variations of the magnitudes of Γ⊥ and Γ∥, Fig. 8-17 shows plots for waves incident in air onto three different types of dielectric surfaces: dry soil (ε r=3), wet soil (ε r=25), and water (ε r=81). For each of the surfaces, (1) Γ⊥=Γ∥ at normal incidence (θ i=0), as expected; (2) |Γ⊥|=|Γ∥|=1 at grazing incidence (θ i=90°); and (3) Γ∥ goes to zero at an angle called the Brewster angle in Fig. 8-17. Had the materials been magnetic too (μ 1≠μ 2), it would have been possible for Γ⊥ to vanish at some angle as well. However, for nonmagnetic materials, the Brewster angle exists only for parallel polarization, and its value depends on the ratio (ε 2/ε 1), as we see shortly.

# Figure 8-17 Plots for |Γ⊥| and |Γ∥| as a function of θ i for a dry soil surface, a wet soil surface, and a water surface. For each surface, |Γ∥|=0 at the Brewster angle.

Figure 8 17 Full Alternative Text

# 8.4.3 Brewster Angle

The Brewster angle θ B is defined as the incidence angle θ i at which the Fresnel reflection coefficient Γ=0.

# Perpendicular Polarization

For perpendicular polarization, the Brewster angle θ B⊥ can be obtained by setting the numerator of the expression for Γ⊥, given by Eq. (8.58a), equal to zero. This happens when

η 2 cos θ i=η 1 cos θ t. (8.69)

By (1) squaring both sides of Eq. (8.69), (2) using Eq. (8.56), (3) solving for θ i, and then denoting θ i as θ B⊥, we obtain

sin θ B⊥=1−(μ 1 ε 2/μ 2 ε 1)1−(μ 1/μ 2)2. (8.70)

Because the denominator of Eq. (8.70) goes to zero when μ 1=μ 2, θ B⊥ does not exist for nonmagnetic materials.

# Parallel Polarization

For parallel polarization, the Brewster angle θ B∥ at which Γ∥=0 can be found by setting the numerator of Γ∥ in Eq. (8.66a) equal to zero. The result is identical to Eq. (8.70), but with μ and ε interchanged. That is,

sin θ B∥=1−(ε 1 μ 2/ε 2 μ 1)1−(ε 1/ε 2)2. (8.71)

For nonmagnetic materials,

The Brewster angle is also called the polarizing angle. This is because, if a wave composed of both perpendicular and parallel polarization components is incident upon a nonmagnetic surface at the Brewster angle θ B∥, the parallel polarized component is totally transmitted into the second medium, and only the perpendicularly polarized component is reflected by the surface. Natural light, including sunlight and light generated by most manufactured sources, is unpolarized because the direction of the electric field of the light waves varies randomly in angle over the plane perpendicular to the direction of propagation. Thus, on average, half of the intensity of natural light is perpendicularly polarized and the other half is parallel polarized. When unpolarized light is incident upon a surface at the Brewster angle, the reflected wave is strictly perpendicularly polarized. Hence, the surface acts as a polarizer.

Concept Question 8-4: Can total internal reflection take place for a wave incident from medium 1 (with n 1) onto medium 2 (with n 2) when n 2>n 1?

Concept Question 8-5: What is the difference between the boundary conditions applied in Section 8-1.1 for normal incidence and those applied in Section 8-4.1 for oblique incidence with perpendicular polarization?

Concept Question 8-6: Why is the Brewster angle also called the polarizing angle?

Concept Question 8-7: At the boundary, the vector sum of the tangential components of the incident and reflected electric fields has to equal the tangential component of the transmitted electric field. For ε r 1=1 and ε r 2=16, determine the Brewster angle and then verify the validity of the preceding statement by sketching to scale the tangential components of the three electric fields at the Brewster angle.