14.3 Boundary Behavior of Waves and Polarization

Wave-Boundary Interaction

Transmission and Reflection

When waves encounter a boundary between two different media, they can be transmitted, reflected, or both, depending on the properties of the boundary and the media on either side.

• When a wave travels from a low-mass string to a high-mass string, part of the wave energy is reflected back while some is transmitted into the new medium
• The reflected wave may undergo a phase change (inversion) depending on the relative wave speeds in the two media
• If the wave speed decreases in the new medium, the reflected wave will be inverted (180° phase shift)
• If the wave speed increases in the new medium, the reflected wave maintains its original orientation (no phase shift)
• The frequency of the wave remains constant across boundaries, though wavelength changes with wave speed

This behavior is analogous to what happens when a pulse travels along a rope that's connected to another rope of different density. The boundary conditions determine how the wave energy is distributed between reflection and transmission.

Wave Polarization

Polarization is a phenomenon that applies specifically to transverse waves, where the oscillations occur perpendicular to the direction of wave propagation.

• Transverse waves can oscillate in any direction perpendicular to their travel direction
• Polarization restricts these oscillations to a single plane
• Polarization can occur through:
  1. Reflection off certain surfaces
  2. Passing through polarizing filters
  3. Transmission through certain crystalline materials
  4. Scattering

Light waves are a common example of transverse waves that can be polarized. When unpolarized light passes through a polarizing filter, only the component of the wave oscillating in the allowed direction passes through.

Polarization and Intensity

Wave intensity represents the energy carried by a wave across a unit area per unit time, and polarization can significantly affect this value.

• Wave intensity is defined as the average power transferred per unit area over one complete wave period
• Mathematically: $I = \frac{P_{avg}}{A}$ where $I$ is intensity, $P_{avg}$ is average power, and $A$ is area
• When unpolarized light passes through a polarizing filter, approximately 50% of the original intensity is transmitted
• If a second polarizing filter is placed with its axis perpendicular to the first (crossed polarizers), the intensity drops to zero

The relationship between intensity and polarizer orientation follows Malus's Law: $I = I_0\cos^2\theta$, where $I_0$ is the initial intensity and $θ$ is the angle between the polarization direction and the polarizer axis.

Practice Problem 1: Wave Reflection at Boundaries

Solution

First, we need to determine whether the reflected pulse will be inverted or maintain its orientation:

• The wave is moving from a light string (higher wave speed) to a heavy string (lower wave speed)
• When a wave moves to a medium where its speed decreases (8 m/s → 4 m/s), the reflected pulse will be inverted (180° phase shift)

To calculate the percentage of energy reflected, we can use the reflection coefficient formula: $R = \left(\frac{v_1 - v_2}{v_1 + v_2}\right)^2$

Where $v_1$ is the wave speed in the first medium and $v_2$ is the wave speed in the second medium.

$R = \left(\frac{8 - 4}{8 + 4}\right)^2 = \left(\frac{4}{12}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \approx 0.111$

Therefore, approximately 11.1% of the incident wave energy is reflected back, while the remaining 88.9% is transmitted into the heavier string.

Practice Problem 2: Polarization and Intensity

Solution

When unpolarized light passes through a polarizing filter, the intensity is reduced by half: $I_1 = \frac{I_0}{2} = \frac{100 \text{ W/m²}}{2} = 50 \text{ W/m²}$

When this polarized light passes through a second polarizer at an angle θ to the first, Malus's Law applies: $I_2 = I_1\cos^2\theta$

With θ = 30°: $I_2 = 50 \text{ W/m²} \times \cos^2(30°) = 50 \text{ W/m²} \times (0.866)^2 = 50 \text{ W/m²} \times 0.75 = 37.5 \text{ W/m²}$

Therefore, the final intensity after passing through both polarizers is 37.5 W/m².