Boundary Behavior of Waves and Polarization
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AP Physics 2 › Boundary Behavior of Waves and Polarization
A transverse wave on a string enters a second string where the wave speed decreases from $4.0\ \text{m/s}$ to $2.0\ \text{m/s}$ at a knot. The source frequency stays constant. Which statement best describes the transmitted wave on the second string?
It has the same frequency and a longer wavelength.
It has the same frequency and a shorter wavelength.
It has a lower frequency and a shorter wavelength.
It has a higher frequency and the same wavelength.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a wave on a string encounters a boundary where the wave speed changes, the frequency remains constant because it's determined by the vibrating source. Since v = fλ and the speed decreases from 4.0 m/s to 2.0 m/s while frequency stays constant, the wavelength must decrease proportionally to maintain the relationship. Choice C incorrectly suggests the wavelength increases when speed decreases, which would violate the wave equation v = fλ. To solve boundary problems, use the principle that frequency never changes at boundaries—only speed and wavelength adjust together.
A transverse wave on a string passes from a light string into a heavier string. At the boundary, the wave speed decreases. Which statement best describes the transmitted wave in the heavier string?
It has the same frequency and a shorter wavelength than before.
It has the same frequency and a longer wavelength than before.
It has a lower frequency and the same wavelength as before.
It has a higher frequency and a shorter wavelength than before.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When a wave passes from a lighter to heavier string, the wave speed decreases because v = √(T/μ) and the linear mass density μ increases. The frequency remains constant at the boundary because it's determined by the source driving the wave. Since v = fλ and speed decreases while frequency stays constant, the wavelength must decrease proportionally. Choice B incorrectly suggests frequency changes at boundaries, which would require the source to somehow change its oscillation rate. At any boundary, frequency is conserved while speed and wavelength adjust to the new medium.
Linearly polarized light of intensity $I_0$ passes through an ideal polarizer whose axis is perpendicular ($90^\circ$) to the light’s polarization direction. The filter changes allowed polarization direction. Which statement best describes the transmitted light?
It is transmitted with intensity $I_0/2$ and unchanged polarization.
It has zero transmitted intensity (blocked) for ideal polarizers.
It is transmitted with intensity $I_0$ but slowed by the polarizer.
It is transmitted with reduced frequency and the same intensity.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When linearly polarized light encounters a polarizer perpendicular to its polarization direction, there is no component of the electric field along the polarizer's transmission axis. According to Malus's law, I = I₀cos²(90°) = I₀ × 0 = 0, meaning no light is transmitted through an ideal polarizer at 90°. The light is completely blocked, not because polarizers slow light or change frequency, but because there's no electric field component to transmit. Choice A incorrectly suggests 50% transmission, which only occurs for unpolarized light or 45° alignment. For crossed polarizers (90° angle), transmission is always zero for ideal polarizers.
A wave pulse on a string reaches a boundary where the string is fixed to a wall. The boundary condition changes the end’s displacement to zero. Which statement best describes the reflected pulse?
It reflects inverted, with a higher speed because it hits a wall.
It is not reflected because fixed ends absorb all wave energy.
It reflects upright, with a lower frequency than the incident pulse.
It reflects inverted, with the same speed in the original string.
Explanation
This question tests understanding of boundary behavior of waves and polarization. At a fixed boundary (hard boundary), the string cannot move, creating a boundary condition where displacement must be zero. To satisfy this condition, the reflected wave must have opposite polarity (inverted) to cancel the incident wave's displacement at the boundary. The wave speed in the string depends only on string properties (v = √(T/μ)) and doesn't change upon reflection. Choice B incorrectly suggests the pulse reflects upright, which would violate the zero-displacement boundary condition at a fixed end. At fixed boundaries, waves always reflect with inversion to maintain zero displacement at the boundary.
A light wave in water ($n=1.33$) enters glass ($n=1.50$) at normal incidence. The refractive index changes at the boundary. Which statement best describes the transmitted light in glass?
It has a lower frequency and the same wavelength as in water.
It has the same frequency and a shorter wavelength than in water.
It has a higher frequency and a longer wavelength than in water.
It is completely reflected because the refractive index increases.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When light enters a medium with higher refractive index, its speed decreases according to v = c/n, where n is the refractive index. The frequency of light remains constant at boundaries because it's an intrinsic property of the wave determined by the source. Since v = fλ and speed decreases (from c/1.33 to c/1.50) while frequency stays constant, the wavelength must decrease proportionally. Choice C incorrectly claims both frequency and wavelength increase, violating conservation of frequency at boundaries. Remember that at optical boundaries: frequency stays constant, speed decreases in denser media, and wavelength follows speed.
A $800,\text{Hz}$ sound wave in air enters water at normal incidence. The wave speed increases from $343,\text{m/s}$ to $1480,\text{m/s}$ at the boundary. Which statement best describes the transmitted wave in water?
It has the same frequency and a shorter wavelength than in air.
It has the same frequency and a longer wavelength than in air.
It has a lower frequency and a longer wavelength than in air.
It has a higher frequency and the same wavelength as in air.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When sound crosses from air to water, the frequency remains constant at 800 Hz because it's determined by the source, not the medium. The wave speed increases from 343 m/s to 1480 m/s due to water's different mechanical properties. Using v = fλ, the wavelength in air is λ₁ = 343/800 = 0.429 m, and in water λ₂ = 1480/800 = 1.85 m, showing wavelength increases. Choice C incorrectly claims wavelength decreases when speed increases, violating the wave equation v = fλ. At any boundary, frequency is conserved while speed and wavelength change 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 has a lower frequency than the incident pulse.
It travels back faster than the incident pulse traveled forward.
It is inverted relative to 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.
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 zero because the polarizer blocks all light.
It is $I_0\cos^2(60^\circ)$.
It is $I_0$ because polarization does not affect intensity.
It is $I_0\cos(60^\circ)$.
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.
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 the same wavelength.
It has the same frequency and a longer 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.
A light wave travels from air ($n=1.00$) into water ($n=1.33$) at normal incidence. The refractive index increases at the boundary, changing wave speed. Which statement best describes the transmitted light in water?
It is completely reflected because the refractive index increases.
It has a lower frequency and a shorter wavelength than in air.
It has the same frequency and a longer wavelength than in air.
It has the same frequency and a shorter wavelength than in air.
Explanation
This question tests understanding of boundary behavior of waves and polarization. When light enters a denser medium (higher n), its speed decreases according to v = c/n, going from c to c/1.33. The frequency of electromagnetic waves remains constant at boundaries because it's an intrinsic property of the wave. Since v = fλ and speed decreases while frequency stays constant, the wavelength must decrease proportionally (by a factor of 1.33). Choice C incorrectly suggests wavelength increases when entering a denser medium, contradicting the relationship between speed and wavelength. In optics, remember: higher refractive index means lower speed and shorter wavelength, but frequency never changes.