Physics of Reflections and Refractions

Snell's Laws

Illustrates the behavior of a wave (light) as it transitions between two media with different properties.
Key parameters:
- \theta_i: Angle of incidence.
- \theta_r: Angle of reflection.
- \theta_t: Angle of refraction.
- k_i: Wave vector of the incident wave.
- k_r: Wave vector of the reflected wave.
- k_t: Wave vector of the transmitted wave.
- \epsilon1, \mu1: Permittivity and permeability of medium 1.
- \epsilon2, \mu2: Permittivity and permeability of medium 2.

Angles of Incidence, Reflection & Refraction

Snell's Law of Reflection: The angle of reflection equals the angle of incidence.
- \thetai = \thetar (8.28a)
Snell's Law of Refraction: Relates the sine of the angles of incidence and refraction to the ratio of phase velocities in the two media.
- \frac{\sin \thetat}{\sin \thetai} = \sqrt{\frac{\mu1 \epsilon1}{\mu2 \epsilon2}} = \frac{v{p1}}{v{p2}} (8.28b)

Nonmagnetic Media

Index of Refraction (n): Defined as the ratio of the speed of light in vacuum (c) to the phase velocity (v_p) in the medium.
- n = \frac{c}{vp} = \sqrt{\mur \epsilonr} = \sqrt{\frac{\mu \epsilon}{\mu0 \epsilon_0}} (8.29)
Rewriting Snell's Law of Refraction: Using the index of refraction.
- \frac{\sin \thetat}{\sin \thetai} = \frac{n1}{n2} = \sqrt{\frac{\mu{r1} \epsilon{r1}}{\mu{r2} \epsilon{r2}}} (8.30)
For Nonmagnetic Materials (\mu{r1} = \mu{r2} = 1):
- \frac{\sin \thetat}{\sin \thetai} = \frac{n1}{n2} = \sqrt{\frac{\epsilon{r1}}{\epsilon{r2}}} (8.31)

Refraction Behavior

Inward Refraction:
- Occurs when a wave enters a more dense medium (n1 < n2).
- The angle of refraction is smaller than the angle of incidence (\thetat < \thetai).
Outward Refraction:
- Occurs when a wave enters a less dense medium (n1 > n2).
- The angle of refraction is larger than the angle of incidence (\thetat > \thetai).

Total Internal Reflection

Critical Angle: The angle of incidence at which the refraction angle is 90 degrees.
Total Internal Reflection: When the angle of incidence exceeds the critical angle, the wave is totally reflected back into the original medium.
- Surface waves exist on the transmission side.
- The reflection coefficient (\Gamma) equals 1 (\Gamma = 1).

Optical Fiber

Components:
- Fiber core (index of refraction n_f).
- Cladding (index of refraction n_c).
- Surrounding medium (index of refraction n_0).
Acceptance Cone: Defines the range of angles within which light can enter the fiber and be guided through it via total internal reflection.
Waveguiding: Waves can be guided along optical fibers due to successive internal reflections.

Example: Light Beam Passing Through a Slab

Setup: A dielectric slab (index of refraction n2) surrounded by a medium with index of refraction n1.
Condition: If the incidence angle (\thetai) is less than the critical angle (\thetac), the emerging beam is parallel to the incident beam.
Solution:
- At the upper surface: \frac{\sin \theta2}{\sin \theta1} = \frac{n1}{n2} (8.33)
- At the lower surface: \frac{\sin \theta3}{\sin \theta2} = \frac{n2}{n1} (8.34)
- Combining the equations: \sin \theta3 = \frac{n2}{n1} \sin \theta2 = \frac{n2}{n1} \cdot \frac{n1}{n2} \sin \theta1 = \sin \theta1
- Therefore: \theta3 = \theta1
Conclusion: The slab displaces the beam's position, but the beam's direction remains unchanged.

Oblique Incidence: Snell's Law

Inward Refraction:
- Occurs when n1 < n2.
- \thetat < \thetai
Outward Refraction:
- Occurs when n1 > n2.
- \thetat > \thetai

EM Polarized Waves

Traveling and standing waves in different mediums.
Medium 1: (\mu1, \epsilon1, \sigma1, \eta1, \beta_1)
Medium 2: (\mu2, \epsilon2, \sigma2, \eta2, \beta_2)

Oblique Incidence

Plane of Incidence: Defined by the normal to the boundary and the direction of propagation of the incident wave (x-y plane).

Perpendicular/TE Polarization

Electric field is perpendicular to the plane of incidence.
Magnetic Field components: Hi, Hr, H_t
Electric Field components: Ei, Er, E_t

Parallel/TM Polarization

Electric field is parallel to the plane of incidence.
Magnetic Field components: Hi, Hr, H_t
Electric Field components: Ei, Er, E_t

General Polarization

Electric and Magnetic Field components.
\vec{k} = k(\cos\phi \hat{x} - \sin\phi \hat{y})
\vec{r} = x\hat{x} + y\hat{y} + z\hat{z}
\vec{E} = Ex + Ey
\vec{H}

Perpendicular Polarization

Detailed illustration of incident, reflected, and transmitted waves with perpendicular polarization.
Relationships between electric and magnetic fields, angles, and media properties.
Incident Wave: \vec{Ei} = \hat{y} E{i0} e^{-jk1(x\sin\thetai + z\cos\thetai)} \vec{Hi} = (-\hat{x} \cos\thetai + \hat{z} \sin\thetai) \frac{E{i0}}{\eta1} e^{-jk1(x\sin\thetai + z\cos\theta_i)}
Reflected Wave: \vec{Er} = \hat{y} E{r0} e^{-jk1(x\sin\thetar - z\cos\thetar)} \vec{Hr} = (\hat{x} \cos\thetar + \hat{z} \sin\thetar) \frac{E{r0}}{\eta1} e^{-jk1(x\sin\thetar - z\cos\theta_r)}
Transmitted Wave: \vec{Et} = \hat{y} E{t0} e^{-jk2(x\sin\thetat + z\cos\thetat)} \vec{Ht} = (-\hat{x} \cos\thetat + \hat{z} \sin\thetat) \frac{E{t0}}{\eta2} e^{-jk2(x\sin\thetat + z\cos\theta_t)}

Applying Boundary Conditions

Tangential E Continuous:
Tangential H Continuous:

Solution of Boundary Equations

Exponents must be equal for all values of x.
Remaining terms become expressions for reflection and transmission coefficients.

Parallel Polarization

Reflection coefficient: r\parallel = \frac{\eta2 \cos \thetat - \eta1 \cos \thetai}{\eta2 \cos \thetat + \eta1 \cos \theta_i}
Transmission coefficient: t\parallel = \frac{2\eta2 \cos \thetai}{\eta2 \cos \thetat + \eta1 \cos \theta_i}
Relationship: t\parallel= (1 + r\parallel) \frac{\cos \thetai}{\cos \thetat} .

General Polarization

General case with both perpendicular and parallel components.
\vec{Ei} = \vec{E{\parallel i}} + \vec{E{\perp i}} \vec{Er} = \vec{E{\parallel r}} + \vec{E{\perp r}}
\vec{Et} = \vec{E{\parallel t}} + \vec{E_{\perp t}}

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

A plane wave from a distant antenna is incident on a soil surface at z = 0.
- \vec{E_i} = \hat{y} 100 \cos( \omega t - \pi x - 1.73\pi z) \quad (V/m)
- Soil: lossless dielectric with \epsilon_r = 4
Objectives:
- Determine k1, k2, and the incidence angle \theta_i.
- Obtain expressions for the total electric fields in air and soil.
- Determine the average power density in the soil.

Example 8-6 (cont.)

[ k r E x y     j x y ˆE ˆE e ~

Example 8-6 (cont.)

Wavelength in air (medium 1):
- \lambda_1 = \frac{2\pi}{\pi} = 2 \quad m
Wavelength in soil (medium 2):
- \lambda2 = \frac{\lambda1}{\sqrt{\epsilon_{r2}}} = \frac{2}{\sqrt{4}} = 1 \quad m
Wave number in medium 2:
- k2 = \frac{2\pi}{\lambda2} = \frac{2\pi}{0.5} = 4\pi \quad (rad/m)
Incidence angle:
- \sin \thetai = \frac{\pi}{k1} = \frac{\pi}{2\pi} = 0.5
- \theta_i = 30^\circ
Transmission angle:
- \sin \thetat = \frac{k1}{k2} \sin \thetai = \frac{2\pi}{4\pi} \sin 30^\circ = 0.25
- \theta_t = 14.5^\circ
Reflection and transmission coefficients (perpendicular polarization):
- r\perp = \frac{\cos \thetai - \sqrt{(\epsilon2/\epsilon1) - \sin^2 \thetai}}{\cos \thetai + \sqrt{(\epsilon2/\epsilon1) - \sin^2 \theta_i}} = -0.38
- t\perp = 1 + r\perp = 0.62

Example 8-6 (cont.)

Total electric field in medium 1 (air):
- \vec{E1} = \hat{y} E{i0} e^{-j k1 (x \sin \thetai + z \cos \thetai)} + \hat{y} r\perp E{i0} e^{-j k1 (x \sin \thetai - z \cos \thetai)}
- \vec{E_1} = \hat{y} 100 e^{-j(\pi x + 1.73 \pi z)} - \hat{y} 38 e^{-j(\pi x - 1.73 \pi z)}
Instantaneous electric field in medium 1:
- E1(x, z, t) = Re[\vec{E1} e^{j \omega t}] = \hat{y} [100 \cos(\omega t - \pi x - 1.73 \pi z) - 38 \cos(\omega t - \pi x + 1.73 \pi z)] \quad (V/m)

Example 8-6 (cont.)

Electric field in medium 2 (soil):
- \vec{Et} = \hat{y} t\perp E{i0} e^{-j k2 (x \sin \thetat + z \cos \thetat)} = \hat{y} 62 e^{-j(\pi x + 3.87 \pi z)}
Instantaneous electric field in medium 2:
- Et(x, z, t) = Re[\vec{Et} e^{j \omega t}] = \hat{y} 62 \cos(\omega t - \pi x - 3.87 \pi z) \quad (V/m)
Average power density in medium 2:
- \eta2 = \frac{\eta0}{\sqrt{\epsilon_{r2}}} = \frac{120 \pi}{\sqrt{4}} = 60 \pi \quad (\Omega)
- S{av} = \frac{|E{t0}|^2}{2 \eta_2} = \frac{(62)^2}{2 \cdot 60 \pi} \approx 10.2 \quad (W/m^2)

Reflection Coefficient vs. Angle

Plots of the magnitude of the reflection coefficient (\Gamma) for dry soil, wet soil, and water surfaces vs. incidence angle.
Brewster angle is shown for each surface, where \Gamma_{\parallel} = 0.

Brewster Angle

Perpendicular Polarization: Does not occur
Parallel Polarization: Perfect Transmission occurs at Brewster angle (\theta_B).

Summary For Reflection and Transmission

Table summarizing reflection coefficient (\Gamma), transmission coefficient (t), reflectivity (R), and transmissivity (T) for both normal and oblique incidence, for both perpendicular and parallel polarization.
Normal Incidence:
- \thetai = \thetat = 0
- \Gamma = \frac{\eta2 - \eta1}{\eta2 + \eta1}
- t = \frac{2 \eta2}{\eta2 + \eta_1}
- t = 1 + \Gamma
- R = |\Gamma|^2
- T = |t|^2 \frac{\eta1}{\eta2}
- T = 1 - R
Perpendicular Polarization:
- \Gamma\perp = \frac{\eta2 \cos \thetai - \eta1 \cos \thetat }{\eta2 \cos \thetai + \eta1 \cos \theta_t}
- t\perp = \frac{2 \eta2 \cos \thetai}{\eta2 \cos \thetai + \eta1 \cos \theta_t}
- t\perp = 1 + \Gamma\perp
- R\perp = |\Gamma\perp|^2
- T\perp = |t\perp|^2 \frac{\eta1 \cos \thetat }{\eta2 \cos \thetai}
- T\perp = 1 - R\perp
Parallel Polarization:
- \Gamma\parallel = \frac{\eta2 \cos \thetat - \eta1 \cos \thetai}{\eta2 \cos \thetat + \eta1 \cos \theta_i}
- t\parallel = \frac{2 \eta2 \cos \thetai}{\eta2 \cos \thetat + \eta1 \cos \theta_i}
- t\parallel = (1 + \Gamma\parallel) \frac{\cos \thetai}{\cos \thetat}
- R\parallel = |\Gamma\parallel|^2
- T\parallel = |t\parallel|^2 \frac{\eta1 \cos \thetat}{\eta2 \cos \thetai}
- T\parallel = 1 - R\parallel