AP Physics 2 Quiz: Boundary Behavior Of Waves And Polarization

What this quiz covers

This quiz focuses on Boundary Behavior Of Waves And Polarization, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It has the same frequency and a shorter wavelength. (correct answer)
2. It has a lower frequency and a shorter wavelength.
3. It has the same frequency and a longer wavelength.
4. 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.

Question 2

1. It is transmitted with reduced frequency and the same intensity.
2. It is transmitted with intensity $I_0/2$ and unchanged polarization.
3. It has zero transmitted intensity (blocked) for ideal polarizers. (correct answer)
4. It is transmitted with intensity $I_0$ but slowed by the polarizer.

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.

Question 3

1. It has the same frequency and a longer wavelength than in air. (correct answer)
2. It has the same frequency and a shorter wavelength than in air.
3. It has a lower frequency and a longer wavelength than in air.
4. 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.

Question 4

1. It is inverted relative to the incident pulse. (correct answer)
2. It is not inverted relative to the incident pulse.
3. It has a lower frequency than the incident pulse.
4. It travels back faster than the incident pulse traveled forward.

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.

Question 5

1. It is $I_0\cos^2(60^\circ)$. (correct answer)
2. It is $I_0\cos(60^\circ)$.
3. It is zero because the polarizer blocks all light.
4. It is $I_0$ because polarization does not affect intensity.

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.

Question 6

1. It has the same frequency and a longer wavelength. (correct answer)
2. It has a lower frequency and a longer wavelength.
3. It has the same frequency and the same wavelength.
4. It has a higher frequency and a shorter 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.

Question 7

1. It has a lower frequency and a shorter wavelength than in air.
2. It has the same frequency and a longer wavelength than in air.
3. It has the same frequency and a shorter wavelength than in air. (correct answer)
4. 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 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.

Question 8

1. It reflects upright, and its speed increases because the end is free.
2. It reflects upright, with the same speed in the original string. (correct answer)
3. It reflects inverted, and its frequency decreases after reflection.
4. It is completely blocked because a free end cannot reflect waves.

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.

Question 9

1. It is $0$. (correct answer)
2. It is $\tfrac{1}{2}I_0$.
3. It is $I_0$.
4. It is $I_0\cos^2(90^\circ)/2$.

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.

Question 10

1. It has the same frequency and a longer wavelength. (correct answer)
2. It has a higher frequency and a longer wavelength.
3. It has the same frequency and a shorter wavelength.
4. It has a lower frequency and the same 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.

Question 11

1. It is polarized with intensity $\tfrac{1}{2}I_0$. (correct answer)
2. It is unpolarized with intensity $\tfrac{1}{2}I_0$.
3. It is polarized with intensity $I_0$.
4. It is blocked completely with intensity $0$.

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.

Question 12

1. It is not inverted relative to the incident pulse. (correct answer)
2. It is inverted relative to the incident pulse.
3. It returns with a lower frequency than the incident pulse.
4. It returns with a greater speed 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.

Question 13

1. $12\,\text{W/m}^2$
2. $6.0\,\text{W/m}^2$
3. $9.0\,\text{W/m}^2$
4. $3.0\,\text{W/m}^2$ (correct answer)

Explanation: This question tests understanding of boundary behavior of waves and polarization. When linearly polarized light passes through a polarizer at angle θ to the light's polarization, Malus's law applies: I = I₀cos²θ. With θ = 60°, we have I = 12 × cos²(60°) = 12 × (0.5)² = 12 × 0.25 = 3.0 W/m². The polarizer only transmits the component of the electric field aligned with its axis, reducing the intensity. Choice B incorrectly suggests I = I₀/2, which only applies to unpolarized light, not already-polarized light at an angle. For polarized light through a polarizer, always use Malus's law: I = I₀cos²θ.

Question 14

1. It is linearly polarized along the filter axis, with reduced intensity. (correct answer)
2. It remains unpolarized, but travels faster due to the filter.
3. It becomes circularly polarized and keeps the same intensity.
4. It is completely blocked because unpolarized light cannot pass.

Explanation: This question tests understanding of boundary behavior of waves and polarization. Unpolarized light contains electric field components oscillating in all directions perpendicular to propagation. A linear polarizer only transmits the component of the electric field aligned with its polarization axis, blocking all perpendicular components. The transmitted light becomes linearly polarized along the filter axis with intensity reduced to half the original (I = I₀/2 for unpolarized incident light). Choice D incorrectly assumes unpolarized light is completely blocked, but every ray has some component along the polarizer axis. For unpolarized light through a polarizer, remember: output is linearly polarized with half the input intensity.

Question 15

1. It has a higher frequency and a longer wavelength than in air.
2. It has the same frequency and a longer wavelength than in air. (correct answer)
3. It has a lower frequency and the same wavelength as in air.
4. It is completely blocked because helium has a different wave speed.

Explanation: This question tests understanding of boundary behavior of waves and polarization. When a wave crosses a boundary between two media, its 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 970 m/s while frequency stays at 500 Hz, the wavelength must increase proportionally (λ = v/f gives 0.686 m in air and 1.94 m in helium). Choice C 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.

Question 16

1. It has a lower frequency and the same wavelength as before.
2. It has a higher frequency and a shorter wavelength than before.
3. It has the same frequency and a shorter wavelength than before. (correct answer)
4. It has the same frequency and a longer 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.

Question 17

1. It reflects inverted, with the same speed in the original string. (correct answer)
2. It reflects inverted, with a higher speed because it hits a wall.
3. It is not reflected because fixed ends absorb all wave energy.
4. It reflects upright, with a lower frequency than the incident pulse.

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.

Question 18

1. It has a lower frequency and the same wavelength as in water.
2. It has a higher frequency and a longer wavelength than in water.
3. It has the same frequency and a shorter wavelength than in water. (correct answer)
4. 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.

Question 19

1. It has the same frequency and a shorter wavelength. (correct answer)
2. It has a lower frequency and a shorter wavelength.
3. It has the same frequency and a longer wavelength.
4. 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.

Question 20

1. It is inverted relative to the incident pulse. (correct answer)
2. It is not inverted and returns upright.
3. It returns with a lower frequency than the incident pulse.
4. It returns with a greater speed than 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.