Interface Rules Worksheet Boundary Logic
Electromagnetic Interface Analysis
Physics: EM Waves
Lesson 1: Boundary Conditions
Researcher:
Date:
1. The Integral Foundation
Boundary conditions are derived by applying the integral forms of Maxwell's equations to infinitesimal Gaussian surfaces and Ampèrian loops at the interface between two media.
Gauss's Law for E
\[ \oint_S \mathbf{D} \cdot d\mathbf{a} = Q_{f,enc} \]
Used for the normal component of the electric displacement field.
Gauss's Law for B
\[ \oint_S \mathbf{B} \cdot d\mathbf{a} = 0 \]
Used for the normal component of the magnetic flux density.
Faraday's Law
\[ \oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \]
Used for the tangential component of the electric field.
Ampère-Maxwell Law
\[ \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{f,enc} + \frac{d\Phi_D}{dt} \]
Used for the tangential component of the magnetic field H.
2. Derivation Analysis: Normal Components
Consider an interface between Medium 1 (\(\epsilon_1, \mu_1\)) and Medium 2 (\(\epsilon_2, \mu_2\)). A thin Gaussian pillbox of height \(h \rightarrow 0\) straddles the boundary.
Q1: Starting from Gauss's Law for B, prove that the normal component \(B^\perp\) is continuous across the boundary (\(B_1^\perp = B_2^\perp\)). Why does the flux through the sides of the pillbox vanish as \(h \rightarrow 0\)?
Q2: For the D-field, the condition is \(D_2^\perp - D_1^\perp = \sigma_f\). Under what conditions can we assume \(D^\perp\) is continuous? How does this relate to the concept of "linear, isotropic dielectrics"?
3. Tangential Components & Surface Currents
Now consider a rectangular Ampèrian loop of length \(L\) and height \(h \rightarrow 0\) oriented perpendicular to the interface.
Q3: Faraday's Law yields \(E_1^{||} = E_2^{||}\). Explain physically why the magnetic flux term \(\frac{d\Phi_B}{dt}\) does not contribute to the line integral in the limit where the loop area goes to zero.
Q4: The magnetic field H obeys \(\mathbf{n} \times (\mathbf{H}_2 - \mathbf{H}_1) = \mathbf{K}_f\). In most optical materials (non-conductors), we assume the free surface current density \(\mathbf{K}_f = 0\). What does this imply for the tangential H components? Is this also true for the tangential B components if the materials have different permeabilities?
Mastery Check
An EM wave hits a perfect conductor. Inside the conductor, \(\mathbf{E} = 0\) and \(\mathbf{B} = 0\). Use the boundary conditions to describe the fields just outside the surface. Are there surface charges or currents present?
Boundary Logic Slides Boundary Logic
Deriving Electromagnetic Boundary Conditions from Maxwell's Equations
Physics 301
Unit: Wave Optics
The Core Question
"What happens to the electric and magnetic fields exactly at the surface of a material interface?"
Is light reflected? Transmitted? Absorbed?
The answer lies in the continuity of the fields.
Maxwell's Integral Forms
Gauss's Law
\[ \oint_S \mathbf{D} \cdot d\mathbf{a} = Q_{f,enc} \]
Relates normal D to surface charge.
Faraday's Law
\[ \oint_C \mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi_B}{dt} \]
Relates tangential E across interface.
Gauss's Law (B)
\[ \oint_S \mathbf{B} \cdot d\mathbf{a} = 0 \]
Relates normal B across interface.
Ampère-Maxwell
\[ \oint_C \mathbf{H} \cdot d\mathbf{l} = I_{f,enc} + \frac{d\Phi_D}{dt} \]
Relates tangential H to surface current.
Mastering the Interface
Component Boundary Condition Simplification (Non-conductors) Normal D \( D_2^\perp - D_1^\perp = \sigma_f \) \( \epsilon_2 E_2^\perp = \epsilon_1 E_1^\perp \) Normal B \( B_2^\perp = B_1^\perp \) Always Continuous Tangential E \( E_2^{ Tangential H \( H_2^{
Maxwell's Logic Teacher Guide Maxwell's Logic
Teacher Guide | Lesson 1: Boundary Conditions
Faculty Resource
Instructional Goal
Students should be able to derive the four fundamental electromagnetic boundary conditions from Maxwell’s integral equations and apply them to simple dielectric and metallic interfaces.
Key Skills
Vector integral manipulation
Limiting case analysis (\(h \rightarrow 0\))
Physical interpretation of fields
Pacing
50-75 Minute Lecture/Workshop
Intro Hook: 10m
Derivation Demo: 25m
Collaborative Solve: 25m
Common Pitfalls
Flux vs. Line Integrals: Students often confuse which field components (normal vs tangential) come from which Maxwell equation.
The Zero Limit: Failure to conceptualize why the flux through the "sides" of the pillbox vanishes as height goes to zero.
Surface Charges: Forgetting that \(\sigma_f\) is zero in most dielectric optics problems but vital for metals.
Facilitation Notes
1
The Hook: Perfect Conductors
Ask students: "If E=0 inside a conductor, what is E just outside?" Use this to lead into the derivation. It forces them to realize that the normal and tangential components behave differently.
2
Mathematical Rigor vs. Physical Intuition
Don't just write the result. Draw the Pillbox and the Ampèrian Loop on the board. Color-code the Normal (red) and Tangential (blue) vectors. Students need to visualize the geometry to understand the dot products.
3
Worksheet Implementation
The "Interface Rules Worksheet" is designed for small groups (2-3 students). Walk around and check specifically for their reasoning in Q3 regarding the time-varying magnetic flux term. It's a classic nuance that separates surface-level understanding from mastery.
Solution Briefs
Normal B:
Continuous. \(\nabla \cdot \mathbf{B} = 0\) implies no magnetic monopoles exist on the surface.
Tangential E:
Continuous. Faraday's Law loop shows potential difference across surface is zero as area \(\rightarrow 0\).
Normal D:
Discontinuity = Surface Charge (\(\sigma_f\)). For optical glass, assume \(\sigma_f=0\).
Tangential H:
Discontinuity = Surface Current (\(K_f\)). Only relevant for conductors or superconducting surfaces.
Fresnel Formula Slides Fresnel Formulae
Predicting Reflection and Transmission at a Boundary
Interface Physics
TE vs. TM Polarizations
The Setup
To solve for reflection/transmission, we decompose light into two fundamental polarizations relative to the **plane of incidence**:
INTERFACE Incident (n1) Reflected Transmitted (n2)
TE Mode (s-polarization)
Reflection Coefficient (r_s):
\[ r_s = \frac{n_1 \cos\theta_i - n_2 \cos\theta_t}{n_1 \cos\theta_i + n_2 \cos\theta_t} \]
Note: Uses continuity of \(E_{\text{tan}}\) and \(B_{\text{tan}}\).
Physics Insight
For \(n_1 < n_2\), \(r_s\) is always negative at all angles. This implies a 180° phase shift upon reflection from a denser medium.
TM Mode (p-polarization)
Reflection Coefficient (r_p):
\[ r_p = \frac{n_2 \cos\theta_i - n_1 \cos\theta_t}{n_2 \cos\theta_i + n_1 \cos\theta_t} \]
Note the reversal of the refractive indices compared to \(r_s\).
The Zero Point
Look closely at the numerator!
Is there an angle where \(n_2 \cos\theta_i = n_1 \cos\theta_t\)? If so, reflection vanishes entirely for this mode.
Reflection Amplitudes Problem Set Reflection Amplitudes
Problem Set: Fresnel Applications
Physics 301
Lesson 2: Fresnel Equations
Student Name:
Date:
Useful Equations
\( r_s = \frac{n_1 \cos\theta_i - n_2 \cos\theta_t}{n_1 \cos\theta_i + n_2 \cos\theta_t} \)
\( r_p = \frac{n_2 \cos\theta_i - n_1 \cos\theta_t}{n_2 \cos\theta_i + n_1 \cos\theta_t} \)
\( t_s = 1 + r_s \)
\( t_p = (1 + r_p)\frac{\cos\theta_i}{\cos\theta_t} \)
1
Normal Incidence Limits
At normal incidence (\(\theta_i = 0\)), show that both \(r_s\) and \(r_p\) reduce to the same expression (up to a sign convention). Calculate the reflection coefficient \(r\) for light passing from air (\(n \approx 1\)) into glass (\(n = 1.5\)). What percentage of the power is reflected?
2
Phase Shifts at a Denser Interface
A light wave in air (\(n_1=1.0\)) hits a water interface (\(n_2=1.33\)) at an angle of \(\theta_i = 45^\circ\).
a) Use Snell's Law to find the angle of transmission \(\theta_t\).
b) Calculate the numerical values for \(r_s\) and \(r_p\). Which coefficient is larger in magnitude? Explain the physical meaning of the signs you obtained.
3
Conservation of Energy
Reflectivity \(R\) and Transmittivity \(T\) are defined as the ratios of the power (irradiance) reflected and transmitted. For s-polarization: \(R_s = |r_s|^2\) and \(T_s = \frac{n_2 \cos\theta_t}{n_1 \cos\theta_i} |t_s|^2\).
Prove that \(R_s + T_s = 1\) using the Fresnel coefficients for any angle \(\theta_i\). (Hint: Algebraic substitution is your friend here).
The Grazing Angle Challenge
What happens to \(r_s\) and \(r_p\) as \(\theta_i \rightarrow 90^\circ\) (grazing incidence)? Explain why this makes almost any surface look like a mirror when viewed from a very shallow angle.
Reflection Amplitudes Answer Key Answer Key
Reflection Amplitudes: Problem Set
Confidential
1. Normal Incidence Limits
As \(\theta_i \rightarrow 0\), \(\cos\theta_i \approx 1\) and \(\cos\theta_t \approx 1\). Both formulas yield:
\( r = \frac{n_1 - n_2}{n_1 + n_2} \) (Note: \(r_p\) usually includes a negative sign in literature due to coordinate choice).
For air to glass: \( r = \frac{1 - 1.5}{1 + 1.5} = \frac{-0.5}{2.5} = -0.2 \).
Power Reflected: \( R = |r|^2 = (-0.2)^2 = 0.04 \) or 4% .
2. Phase Shifts at a Denser Interface
a) Snell's Law:
\( \sin(45^\circ) = 1.33 \sin\theta_t \Rightarrow 0.707 = 1.33 \sin\theta_t \Rightarrow \sin\theta_t = 0.531 \Rightarrow \mathbf{\theta_t \approx 32.1^\circ} \).
b) Coefficients:
\( r_s = \frac{1(0.707) - 1.33(0.847)}{0.707 + 1.126} \approx \frac{-0.420}{1.833} \approx \mathbf{-0.229} \).
\( r_p = \frac{1.33(0.707) - 1(0.847)}{0.940 + 0.847} \approx \frac{0.093}{1.787} \approx \mathbf{0.052} \).
Magnitude: \( |r_s| > |r_p| \). The s-polarized light reflects more strongly. The negative sign for \(r_s\) indicates a \(\pi\) phase shift.
3. Conservation of Energy
Substitute \(t_s = 1 + r_s\):
\( T_s = \frac{n_2 \cos\theta_t}{n_1 \cos\theta_i} (1 + r_s)^2 \). Let \(\alpha = \frac{n_2 \cos\theta_t}{n_1 \cos\theta_i}\). Note that \(r_s = \frac{1 - \alpha}{1 + \alpha}\).
\( R_s + T_s = (\frac{1 - \alpha}{1 + \alpha})^2 + \alpha (1 + \frac{1 - \alpha}{1 + \alpha})^2 \)
\( = \frac{(1 - \alpha)^2}{(1 + \alpha)^2} + \alpha (\frac{1 + \alpha + 1 - \alpha}{1 + \alpha})^2 = \frac{1 - 2\alpha + \alpha^2}{(1 + \alpha)^2} + \alpha \frac{4}{(1 + \alpha)^2} = \frac{1 + 2\alpha + \alpha^2}{(1 + \alpha)^2} = \mathbf{1} \). Q.E.D.
Challenge: Grazing Angle
As \(\theta_i \rightarrow 90^\circ\), \(\cos\theta_i \rightarrow 0\). The Fresnel equations both approach -1 (or +1 depending on convention), meaning 100% of light is reflected regardless of polarization or material. This explains why asphalt or calm water becomes mirror-like at a distance.
Glint Gone Slides Glint Gone
Brewster's Angle and the Physics of Polarization by Reflection
Wave Optics
The Case of the Missing Reflection
The Photographer's Trick
"How can you see to the bottom of a lake through bright surface glare without using software?"
The answer isn't just a filter; it's a specific geometry of light waves.
Polarizing Filters
Brewster's Angle (\(\theta_B\))
The Condition:
\( n_1 \sin\theta_B = n_2 \sin\theta_t \)
Where \(\theta_B + \theta_t = 90^\circ\)
At this specific angle, the reflected and transmitted rays are perpendicular.
Final Equation
\( \tan\theta_B = \frac{n_2}{n_1} \)
Why does it vanish?
1. Incident light makes atoms in the material oscillate (dipole radiation).
2. These dipoles radiate light in all directions except along their axis of oscillation .
3. For p-polarized light at Brewster's angle, the reflected direction aligns exactly with the dipole axis.
Result: No energy can be radiated into the reflected ray!
Dipole Radiation Pattern
Polarization Hunt Activity Polarization Hunt
Inquiry Activity: Interface Glare Analysis
Physics Lab
Lesson 3: Brewster's Angle
Investigator:
Station No.:
Objective
Determine the unknown refractive index of a material by finding the angle where reflected light is completely polarized.
Procedure
Direct an unpolarized light source at the material surface.
Place a linear polarizer in front of your detector (or eye).
Rotate the polarizer to the **horizontal** position (blocking s-polarized light, allowing p-polarized light).
Vary the angle of incidence \(\theta_i\) until the reflected light intensity reaches a **minimum**.
Record this angle as \(\theta_B\).
Data Collection
Material Sample Measured \(\theta_B\) (degrees) Calculated \(n = \tan\theta_B\) Predicted Material A (Transparent Slab) B (Black Plastic) C (Liquid Interface)
Analysis
Q1: Why must the polarizer be set to the horizontal position to find Brewster's angle? What would happen if it were set to the vertical position?
Q2: If you use a laser instead of unpolarized white light, how would you ensure you can actually see the "null" point? Does the wavelength of the light matter?
Q3: Look at Sample B. Does Brewster's angle work for opaque (absorbing) materials? Explain your reasoning based on your measurements.
Engineering Extension
In laser cavity design, "Brewster Windows" are used to ensure that the light passes through the ends of the laser medium with zero reflection loss for one polarization. Sketch a laser tube with windows oriented at the Brewster angle. Which way should the window be tilted?
Beyond Critical Slides Beyond Critical
Total Internal Reflection and the Mysterious Evanescent Wave
Physics 301
Boundary Tunnelling
The 90° Limit
When light travels from a **denser** to a **less dense** medium (\(n_1 > n_2\)), the refracted ray bends away from the normal.
The Critical Condition:
\( \sin\theta_c = \frac{n_2}{n_1} \)
At \(\theta_i = \theta_c\), the refracted ray skims the surface (\(\theta_t = 90^\circ\)).
n2 (Air) n1 (Glass) Θt = 90°
What happens when \(\theta_i > \theta_c\)?
Mathematically, Snell's Law requires:
\( \sin\theta_t = \frac{n_1}{n_2} \sin\theta_i > 1 \)
This implies \(\theta_t\) is a **complex number**. But complex angles have physical meaning!
Specifically, \(\cos\theta_t = \sqrt{1 - \sin^2\theta_t}\) becomes **purely imaginary**.
Energy Result:
Reflectivity \(R = 1\)
Every single photon is reflected. No energy is lost to the second medium... or is it?
The Field That "Leaks"
Spatial Profile
\[ E(z) = E_0 e^{-\alpha z} e^{i(kx - \omega t)} \]
It decays exponentially into the second medium. This is the Evanescent Wave .
It propagates along the interface (\(x\)), not into the medium (\(z\)).
Average energy flow (Poynting vector) in the \(z\)-direction is zero .
It allows for "Optical Tunnelling" if another dense medium is brought close.
Tunneling Light Quiz Tunneling Light
Concept Check: TIR and Evanescent Fields
Physics 301
Lesson 4: TIR
Name:
Date:
Part 1: Conceptual Core
1. Total Internal Reflection (TIR) can only occur when light travels from:
A) A lower index to a higher index medium.
B) A higher index to a lower index medium.
C) Vacuum to any dielectric medium.
D) Any two media with different indices.
2. The evanescent wave exists beyond the interface in the lower-index medium. Which statement about this wave is FALSE?
A) It decays exponentially as a function of distance from the boundary.
B) It carries net power away from the interface into the second medium.
C) Its phase changes along the direction of the interface.
D) It is a direct result of satisfying boundary conditions for Maxwell's equations.
Part 2: Mathematical Analysis
3. If the angle of incidence \(\theta_i\) increases further beyond the critical angle \(\theta_c\), what happens to the penetration depth (the distance at which the field drops to \(1/e\)) of the evanescent wave? Explain using physical reasoning.
4. Frustrated Total Internal Reflection (FTIR): Suppose you have two glass prisms separated by a small air gap. If the gap is 1 mm, no light crosses. If the gap is 50 nm, light is transmitted. Explain this phenomenon using the concept of the evanescent wave.
The Phase Shift
In TIR, the reflection coefficient \(r\) is a complex number with magnitude 1. This means the reflected wave has a phase shift \(\delta\). Why is this phase shift different for s-polarized and p-polarized light? (Consider the different boundary equations for E and H).
Tunneling Light Answer Key Answer Key
Tunneling Light: Concept Check
Instructor Only
Part 1: Multiple Choice
1. Correct Answer: B (Higher index to lower index). Refraction must bend "away" from the normal to reach 90°.
2. Correct Answer: B (False statement). The evanescent wave carries no net power in the z-direction (it is "purely reactive" like the field in a capacitor).
Part 2: Mathematical Analysis
3. Penetration Depth Trend:
The penetration depth \(\delta = \frac{1}{\alpha}\) where \(\alpha = \frac{\omega}{c}\sqrt{n_1^2 \sin^2\theta_i - n_2^2}\). As \(\theta_i\) increases further beyond \(\theta_c\), the term inside the square root gets larger, making \(\alpha\) larger and thus the penetration depth smaller (shorter) . The field becomes more tightly confined to the interface.
4. Frustrated TIR (FTIR):
The evanescent wave from the first prism extends into the air gap. Since it decays exponentially (\(e^{-\alpha z}\)), if the second prism is placed within a few penetration depths (\(z \approx \text{nanometers}\)), the field "couples" into the second prism. This allows energy to propagate through the gap as a real transmitted wave, effectively "tunneling" through the forbidden air region.
Visualization: Phase Shift
The phase shift \(\delta\) occurs because the reflection coefficients \(r_s\) and \(r_p\) become complex numbers with magnitude 1. They have different phase factors because the boundary conditions for the E-field (s-pol) and the H-field (p-pol) involve different refractive index ratios (\(n_1/n_2\) vs \(n_2/n_1\)). This phase difference (\(\delta_s - \delta_p\)) is the principle behind the Fresnel Rhomb , which converts linear polarization to circular polarization via two internal reflections.
Glass Threads Slides Glass Threads
Engineering Wave Propagation in Optical Fibers
Applied Optics
TIR in Action
The Step-Index Fiber
Cladding (n2) Core (n1)
Condition: n1 > n2
A fiber consists of a high-index **Core** surrounded by a lower-index **Cladding**.
Light is trapped in the core via successive Total Internal Reflections .
Critical Angle Requirement:
\(\theta_{\text{bounce}} > \sin^{-1}(n_2 / n_1)\)
The Acceptance Angle
Not all light entering the end of the fiber will be trapped. Only light within the **Acceptance Cone** (\(\theta_a\)) undergoes TIR.
Numerical Aperture (NA)
\[ NA = n_0 \sin\theta_a = \sqrt{n_1^2 - n_2^2} \]
Higher NA = More "Light-Gathering" Power
But higher NA also leads to more modal dispersion (blurring of signals).
The Transatlantic Pipeline
Optical fibers carry 99% of international data. Submarine cables span thousands of miles.
Loss: Signal is boosted every ~100km.Speed: Speed of light in glass is ~2/3 \(c\).Bandwidth: Terabits per second per fiber.
Embedded media
Inside a Submarine Cable Landing Station
Deep Sea Data Case Study Deep Sea Data
Case Study: Submarine Fiber Engineering
Applied Physics
Lesson 5: Waveguides
Engineer:
Project Date:
The Context
The "MAREA" cable connects Virginia Beach, USA, to Bilbao, Spain. It contains eight pairs of optical fibers. Each fiber must maintain total internal reflection over distances of thousands of kilometers while minimizing signal loss and dispersion. You are tasked with analyzing the optical properties of a standard single-mode fiber used in such a cable.
Fiber Specifications
Core Index (\(n_1\)) 1.468
Cladding Index (\(n_2\)) 1.463
Task 1: The Critical Angle
Calculate the critical angle \(\theta_c\) for the core-cladding interface. What happens to light that hits the cladding at an angle of 85° relative to the normal?
Task 2: Acceptance and Numerical Aperture
Assuming the fiber is in air (\(n_0 = 1.0\)), calculate the Numerical Aperture (NA) and the maximum acceptance angle \(\theta_a\) for light entering the fiber core from the end-face.
Task 3: Dispersion and Path Length
Consider a "ray" that bounces just at the critical angle vs. a ray that travels straight down the center of the core. In a 100 km segment of fiber, what is the difference in path length traveled by these two rays? (This is the origin of intermodal dispersion).
Final Inquiry
Modern high-speed fibers use "graded-index" profiles instead of the "step-index" profile analyzed above. Why might a continuous variation in refractive index be better for signal integrity than a sharp boundary? (Think about the speed of light in different parts of the fiber).
Deep Sea Data Answer Key Answer Key
Case Study: Deep Sea Data
Faculty Reference
Task 1: The Critical Angle
\( \sin\theta_c = \frac{n_2}{n_1} = \frac{1.463}{1.468} \approx 0.99659 \)
\( \theta_c = \sin^{-1}(0.99659) \approx \mathbf{85.24^\circ} \)
Analysis: If light hits at 85°, it is **below** the critical angle (closer to the normal than \(\theta_c\)). Therefore, it will **not** undergo TIR; it will refract into the cladding and eventually be lost. This demonstrates how narrow the "containment" angle is in real fibers.
Task 2: Acceptance and NA
\( NA = \sqrt{n_1^2 - n_2^2} = \sqrt{1.468^2 - 1.463^2} = \sqrt{2.155024 - 2.140369} = \sqrt{0.014655} \approx \mathbf{0.121} \)
For entrance from air (\(n_0 = 1\)): \( NA = \sin\theta_a \Rightarrow \theta_a = \sin^{-1}(0.121) \approx \mathbf{6.95^\circ} \).
Meaning: Light must enter the fiber end-face within a cone of half-angle ~7° to be guided.
Task 3: Dispersion
Straight ray length: \( L = 100 \text{ km} \).
Bouncing ray (at \(\theta_c\)) length: \( L' = \frac{L}{\sin\theta_c} = \frac{100 \text{ km}}{0.99659} \approx \mathbf{100.342 \text{ km}} \).
Path Difference: \(\Delta L \approx \mathbf{342 \text{ meters}} \).
Over 100 km, the bouncing ray travels 342 meters further than the center ray, causing a short pulse of light to "smear" out in time (modal dispersion).
Reflection Brief
In graded-index fibers, the index is highest at the center and drops off smoothly. This means rays further from the center travel through lower-index material, where they move faster (\(v = c/n\)). By carefully shaping the index profile, we can make the outer "longer" paths take the same amount of time as the inner "shorter" paths, effectively eliminating modal dispersion.
