Boundary Behavior of Waves and Polarization
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AP Physics 2 › Boundary Behavior of Waves and Polarization
A wave pulse on a string reaches a boundary where the string is attached to a frictionless ring that can move vertically. The boundary condition changes to a free end (nonzero displacement allowed). Which statement best describes the reflected pulse?
It is completely blocked because a free end cannot reflect waves.
It reflects upright, with the same speed in the original string.
It reflects upright, and its speed increases because the end is free.
It reflects inverted, and its frequency decreases after reflection.
Explanation
This question tests understanding of boundary behavior of waves and polarization. At a free boundary (soft boundary), the string end can move freely, creating a boundary condition where the restoring force must be zero. To satisfy this condition, the reflected wave must have the same polarity (upright) as the incident wave, doubling the displacement at the boundary. The wave speed depends only on string properties and remains constant in the original string after reflection. Choice A incorrectly suggests inversion, which occurs only at fixed boundaries where displacement must be zero. At free boundaries, waves reflect without inversion to maintain zero force at the boundary.
Linearly polarized light with intensity $I_0$ passes through two ideal polarizers whose transmission axes differ by $90^\circ$. The filters change polarization direction but not frequency. Which statement best describes the transmitted intensity after the second polarizer?
It is $I_0$.
It is $I_0\cos^2(90^\circ)/2$.
It is $\tfrac{1}{2}I_0$.
It is $0$.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When linearly polarized light passes through two polarizers with perpendicular transmission axes (90° apart), no light emerges. After the first polarizer, light is polarized along its axis, but this polarization is perpendicular to the second polarizer's axis. Applying Malus's Law: I = I₀cos²(90°) = I₀(0)² = 0. This configuration is called "crossed polarizers" and blocks all light. Choice B incorrectly suggests some light passes through, not recognizing that cos²(90°) = 0. When polarizers are crossed at 90°, complete extinction occurs—no light is transmitted.
A light wave in air strikes a glass surface. At the boundary, the wave speed decreases because the refractive index increases. Which statement best describes the transmitted light in the glass?
It has a lower frequency and a shorter wavelength.
It has the same frequency and a shorter wavelength.
It has the same frequency and a longer wavelength.
It has a higher frequency and the same wavelength.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When light enters a denser medium like glass from air, the wave speed decreases due to the higher refractive index (v = c/n). Since frequency remains constant at boundaries (determined by the source), and v = fλ, the wavelength must decrease proportionally with the speed. If n_glass ≈ 1.5, then both speed and wavelength are reduced to about 2/3 their values in air. Choice C incorrectly suggests wavelength increases when speed decreases, violating the fundamental wave equation. Remember: when light slows down in a denser medium, its wavelength decreases proportionally.
A pulse travels from a light string into a heavier string tied directly to it. At the boundary, the wave speed decreases due to greater linear mass density. Which statement best describes the reflected pulse on the light string?
It is inverted relative to the incident pulse.
It travels back faster than the incident pulse traveled forward.
It has a lower frequency than the incident pulse.
It is not inverted relative to the incident pulse.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a pulse travels from a lighter string to a heavier string, part of the wave is transmitted and part is reflected. The reflected pulse experiences phase reversal (inversion) because the heavier string acts somewhat like a fixed boundary—it's harder to displace due to its greater inertia. This is analogous to reflection from a fixed end, though not as extreme. Choice B incorrectly suggests no inversion occurs, which would only be true for reflection from a lighter string. When waves encounter denser media, expect inverted reflections due to the impedance mismatch.
A sound wave in air enters a region of helium. At the boundary, the wave speed increases while the source maintains a constant frequency. Which statement best describes the transmitted sound wave in helium?
It has the same frequency and a longer wavelength.
It has the same frequency and a shorter wavelength.
It has a lower frequency and the same wavelength.
It has a higher frequency and a longer wavelength.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When sound waves cross from air into helium, the wave speed increases because helium has lower density than air. Since the source maintains constant frequency, and v = fλ, the wavelength must increase proportionally with the speed increase. The frequency cannot change at a boundary—it's determined solely by the source. Choice B incorrectly suggests frequency increases at the boundary, which would require the source to change its vibration rate. For boundary problems, apply the rule: frequency stays constant, speed and wavelength change together.
Linearly polarized light with intensity $I_0$ passes through an ideal polarizer. The light’s polarization is $60^\circ$ to the polarizer’s transmission axis; the polarizer changes polarization direction but not speed. Which statement best describes the transmitted intensity?
It is $I_0$ because polarization does not affect intensity.
It is $I_0\cos^2(60^\circ)$.
It is $I_0\cos(60^\circ)$.
It is zero because the polarizer blocks all light.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When linearly polarized light passes through a polarizer, Malus's Law determines the transmitted intensity: I = I₀cos²θ, where θ is the angle between the incident polarization and the polarizer's transmission axis. With θ = 60°, the transmitted intensity is I₀cos²(60°) = I₀(1/2)² = I₀/4. The polarizer only transmits the component of the electric field aligned with its axis, and intensity is proportional to the square of the electric field amplitude. Choice B incorrectly uses cos(60°) instead of cos²(60°), forgetting that intensity depends on the square of the field amplitude. For polarizer problems, always apply Malus's Law: I = I₀cos²θ.
Unpolarized light of intensity $I_0$ passes through an ideal polarizing filter. The filter changes the light’s polarization by transmitting only one polarization component. Which statement best describes the transmitted light?
It is polarized with intensity $\tfrac{1}{2}I_0$.
It is blocked completely with intensity $0$.
It is polarized with intensity $I_0$.
It is unpolarized with intensity $\tfrac{1}{2}I_0$.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When unpolarized light passes through an ideal polarizer, the polarizer transmits only the component of light aligned with its transmission axis. Since unpolarized light has equal intensity in all polarization directions, on average half the intensity is transmitted, resulting in I = I₀/2. The transmitted light becomes linearly polarized along the polarizer's axis. Choice B incorrectly claims the light remains unpolarized after passing through a polarizer, which contradicts the fundamental function of a polarizing filter. Remember: unpolarized light through one polarizer yields polarized light with half the original intensity.
A wave pulse on a string reaches a rigidly fixed end. At the boundary, the end cannot move, forcing the displacement there to remain zero. Which statement best describes the reflected pulse?
It returns with a greater speed than the incident pulse.
It is not inverted and returns upright.
It returns with a lower frequency than the incident pulse.
It is inverted relative to the incident pulse.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a wave pulse reaches a fixed boundary, the boundary condition requires zero displacement at that point. To satisfy this condition, the reflected pulse must be inverted (flipped upside down) so that the incident and reflected pulses cancel at the boundary. This phase reversal occurs because the fixed end exerts an equal and opposite force on the string. Choice B incorrectly suggests no inversion occurs, which would violate the fixed boundary condition of zero displacement. At fixed boundaries, always expect phase reversal (inversion) of the reflected wave.
A wave pulse on a string reaches a free end attached to a low-friction ring. At the boundary, the end is free to move, so the transverse force there is approximately zero. Which statement best describes the reflected pulse?
It returns with a greater speed than the incident pulse.
It is not inverted relative to the incident pulse.
It is inverted relative to the incident pulse.
It returns with a lower frequency than the incident pulse.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a wave pulse reaches a free boundary (like a ring that can slide freely), the boundary condition requires zero transverse force at that point. This means the reflected pulse returns without inversion, maintaining the same orientation as the incident pulse. The free end can move up and down, so there's no restoring force to cause phase reversal. Choice B incorrectly predicts inversion, which only occurs at fixed boundaries where displacement must be zero. At free boundaries, reflected waves maintain their original orientation without phase reversal.
A plane sound wave of frequency $500\ \text{Hz}$ travels from air into water. At the boundary, the wave speed increases from $343\ \text{m/s}$ to $1500\ \text{m/s}$. Which statement best describes the transmitted wave in water?
It has the same frequency and a longer wavelength.
It has the same frequency and the same wavelength.
It has a higher frequency and a shorter wavelength.
It has a lower frequency and a longer wavelength.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a wave crosses a boundary between two media, the frequency remains constant because it's determined by the source, not the medium. Since wave speed v = fλ, and the speed increases from 343 m/s to 1500 m/s while frequency stays at 500 Hz, the wavelength must increase proportionally: λ = v/f = 1500/500 = 3 m in water versus 343/500 = 0.686 m in air. Choice B incorrectly assumes frequency changes at boundaries, which violates the principle that the source controls frequency. When waves cross boundaries, remember: frequency stays constant, but speed and wavelength change together.