AP Physics 2 Quiz: Reflection

What this quiz covers

This quiz focuses on Reflection, 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. The reflected pulse is inverted relative to the incident pulse. (correct answer)
2. The reflected pulse is not inverted because the string's speed is unchanged.
3. The reflected pulse travels along the wall instead of back on the string.
4. The reflected pulse amplitude must increase because the wall is rigid.

Explanation: This question tests understanding of reflection of mechanical waves at boundaries. When a wave pulse on a string encounters a rigid boundary (fixed end), the reflected pulse is inverted relative to the incident pulse because the boundary cannot move. The string must have zero displacement at the wall, so when the incident pulse tries to lift the string upward at the boundary, the reflected pulse must pull it downward to maintain this condition. Choice B incorrectly claims no inversion occurs, confusing wave speed (which doesn't change) with the boundary condition that determines pulse orientation. For mechanical wave reflections, remember that rigid boundaries invert pulses while free boundaries do not.

Question 2

1. The normal bisects the angle between the incident and reflected rays. (correct answer)
2. The mirror surface bisects the angle between the incident and reflected rays.
3. The reflected ray must be parallel to the surface for any smooth mirror.
4. The reflected angle increases when light slows down near the mirror.

Explanation: This question tests understanding of reflection. The law of reflection requires that the angle of incidence equals the angle of reflection, both measured from the normal. This means the normal line bisects the angle between the incident and reflected rays—it acts as the axis of symmetry for the reflection. The incident ray, normal, and reflected ray all lie in the same plane. Choice B incorrectly suggests the mirror surface bisects this angle, which would violate the law of reflection except when rays are perpendicular to the surface. Visualize reflection as a symmetric process about the normal, not about the surface.

Question 3

1. They are equal when measured from the normal to the surface. (correct answer)
2. They are equal when measured from the surface itself.
3. They differ because reflection depends on the light's frequency.
4. They differ because the mirror changes the light's speed.

Explanation: This question tests understanding of reflection angle measurement. The law of reflection states that the incident angle equals the reflected angle when both are measured from the normal (the line perpendicular to the surface at the point of incidence). The incident ray traveling down and to the right and the reflected ray traveling up and to the right make equal angles with the normal, which points vertically from the horizontal mirror. Choice B incorrectly suggests measuring angles from the surface itself, a common error that violates the law of reflection. Always identify the normal first—it's perpendicular to the surface—then measure both angles from this reference line.

Question 4

1. The reflected angle equals $\theta_i$ measured from the normal. (correct answer)
2. The reflected angle equals $90^\circ-\theta_i$ measured from the normal.
3. The reflected ray must travel along the mirror when $\theta_i\neq 0$.
4. The reflected angle depends on the mirror's material and thickness.

Explanation: This question tests understanding of reflection angles. The law of reflection states that the angle of incidence equals the angle of reflection, with both angles measured from the normal to the reflecting surface. Since the incident ray makes angle θᵢ with the normal, the reflected ray must also make angle θᵢ with the normal, traveling on the opposite side. Choice B incorrectly suggests the reflected angle is 90°-θᵢ, which would be the complementary angle—this error often arises from measuring angles relative to the surface instead of the normal. Always establish the normal at the point of incidence and measure both incident and reflected angles from this reference line.

Question 5

1. The reflected wavefront makes the same angle to the barrier as the incident wavefront.
2. The reflected wavefront makes the same angle to the normal as the incident wavefront. (correct answer)
3. The reflected angle increases because the barrier is rigid.
4. The reflected wave travels along the barrier for any nonzero incident angle.

Explanation: This question tests understanding of reflection of water waves. The law of reflection applies to all wave types: the angle of incidence equals the angle of reflection, with both angles measured from the normal to the reflecting surface. When a water wavefront strikes a barrier at an angle, the reflected wavefront makes the same angle with the normal as the incident wavefront, traveling on the opposite side of the normal. Choice A incorrectly refers to angles measured from the barrier rather than the normal, a common misconception that leads to incorrect angle relationships. For any wave reflection, establish the normal (perpendicular to the barrier) and measure all angles from it.

Question 6

1. The reflected pulse returns with the same angle to the normal as the incident pulse. (correct answer)
2. The reflected pulse returns at a larger angle because concrete is denser than air.
3. The reflected pulse must travel along the wall due to friction with the surface.
4. The reflected direction depends on the speed of sound at the wall surface.

Explanation: This question tests understanding of reflection of sound waves. The law of reflection applies to all types of waves, including sound waves: the angle of incidence equals the angle of reflection, both measured from the normal to the reflecting surface. When a sound pulse reflects from a flat wall, it returns at the same angle to the normal as it arrived, regardless of the wall material's density. Choice B incorrectly claims the angle increases due to the concrete's density, confusing reflection (which follows a simple geometric law) with refraction (where material properties matter). For any wave reflection from a smooth surface, the incident and reflected angles are always equal when measured from the normal.

Question 7

1. The reflected angle is $15^\circ$ when measured from the normal. (correct answer)
2. The reflected angle is $75^\circ$ because angles are measured from the surface.
3. The reflected ray travels along the surface since the incident angle is small.
4. The reflected angle depends on the surface material's color.

Explanation: This question tests understanding of reflection. According to the law of reflection, when the incident ray approaches at 15° from the normal, the reflected ray must leave at 15° from the normal on the opposite side. This angle relationship is independent of surface color, material properties, or the size of the angle. Choice A incorrectly measures from the surface instead of the normal - if the ray makes 15° with the normal, it makes 75° with the surface, but we always use normal angles for the reflection law. The fundamental principle to remember is that reflection angles are always measured from the normal, and the angle in equals the angle out.

Question 8

1. The reflected angle depends on the wall material and is not predictable
2. The reflected wave travels parallel to the wall regardless of incidence angle
3. The reflected angle equals the incident angle, both measured from the normal (correct answer)
4. The reflected angle is larger because sound travels slower near the wall

Explanation: This question tests understanding of reflection. The law of reflection applies to all types of waves, including sound waves, stating that the angle of incidence equals the angle of reflection when measured from the normal to the reflecting surface. This law holds regardless of the wave type or the material of the reflecting surface, as long as the surface is smooth compared to the wavelength. Choice B incorrectly suggests the angle depends on wave speed near the wall, confusing reflection with refraction. When analyzing any wave reflection, apply the same geometric principle: incident angle equals reflected angle, both measured from the normal.

Question 9

1. The reflected direction depends on wave amplitude, not on the incident direction
2. The angle between the incident direction and the normal equals the angle between the reflected direction and the normal (correct answer)
3. The reflected direction makes a larger angle because water waves slow near the wall
4. The reflected direction is always along the wall regardless of incidence

Explanation: This question tests understanding of reflection. When water waves reflect from a straight seawall, the law of reflection applies: the angle between the incident direction and the normal equals the angle between the reflected direction and the normal. The incident and reflected directions are perpendicular to their respective wavefronts, maintaining the same angular relationship with the normal. Choice B incorrectly suggests that wave speed affects the reflection angle, confusing reflection with refraction phenomena. For any wave reflection from a straight barrier, visualize the wavefronts as parallel lines and remember that the perpendicular directions follow the equal-angle rule.

Question 10

1. The reflected wavefront spacing increases because the wall adds energy.
2. The reflected wavefront spacing equals the incident spacing. (correct answer)
3. The reflected wavefronts become parallel to the wall regardless of incidence.
4. The reflected wavefront spacing decreases because the speed changes at the wall.

Explanation: This question tests understanding of reflection. When water waves reflect from a vertical seawall, the law of reflection applies: the angle of incidence equals the angle of reflection, measured from the normal to the wall. The wavelength (and thus wavefront spacing) remains constant during reflection because the wave stays in the same medium (water) with the same frequency and speed. The reflected wavefronts maintain their original spacing but travel in a new direction determined by the reflection angle. The wave energy is redirected, not added to or removed. Choice B incorrectly suggests the wall adds energy to increase spacing, but walls only redirect waves without changing their properties in the same medium. When analyzing wave reflection, remember that frequency, wavelength, and speed remain constant in the same medium.

Question 11

1. The reflected ray makes a $35^\circ$ angle with the normal. (correct answer)
2. The reflected ray makes a $55^\circ$ angle with the normal.
3. The reflected ray travels along the mirror surface.
4. The reflected angle depends on the light's speed in the mirror.

Explanation: This question tests understanding of reflection. The law of reflection states that when a wave encounters a smooth surface, the angle of incidence equals the angle of reflection, both measured from the normal (a line perpendicular to the surface at the point of incidence). Since the incident ray makes a 35° angle with the normal, the reflected ray must also make a 35° angle with the normal, traveling on the opposite side of the normal. Choice B incorrectly suggests a 55° angle, which would be the angle to the surface rather than the normal—a common misconception. When solving reflection problems, always identify the normal first and measure all angles from it, not from the surface itself.

Question 12

1. $20^\circ$
2. $90^\circ$
3. $40^\circ$
4. $70^\circ$ (correct answer)

Explanation: This question tests understanding of reflection. The law of reflection requires that incident and reflected angles be measured from the normal, not from the surface. If the incident ray makes 20° with the mirror surface, it makes 90° - 20° = 70° with the normal. By the law of reflection, the reflected ray also makes 70° with the normal on the opposite side. Choice A (20°) represents the misconception of directly using the surface angle instead of converting to the normal angle. When working with reflection problems, always convert surface angles to normal angles using the complementary relationship: angle from normal = 90° - angle from surface.

Question 13

1. $30^\circ$ (correct answer)
2. $15^\circ$
3. $60^\circ$
4. $90^\circ$

Explanation: This question tests understanding of reflection. The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal to the surface. The problem states the incident direction is 60° above the surface, which means it makes a 30° angle with the normal (since 90° - 60° = 30°). By the law of reflection, the reflected ray must also make a 30° angle with the normal on the opposite side. The reflection is symmetric about the normal, not about the surface. Choice A (60°) represents the common error of measuring from the surface instead of the normal. Always convert surface angles to normal angles first: if given an angle from the surface, subtract from 90° to find the angle from the normal.

Question 14

1. The reflected pulse is inverted relative to the incident pulse. (correct answer)
2. The reflected pulse travels faster because it reflects from a heavier rope.
3. The reflected pulse stays upright because the boundary is not a wall.
4. The reflected pulse must propagate along the boundary instead of back on the rope.

Explanation: This question tests understanding of reflection. When a wave pulse travels from a lighter rope to a boundary with a much heavier rope, the reflected pulse is inverted. This occurs because the heavier rope acts similarly to a rigid boundary - it has much greater inertia and resists the motion imposed by the incident pulse. The boundary exerts an opposing force that creates an inverted reflected pulse traveling back on the lighter rope. Some wave energy also transmits into the heavier rope, but the reflection is inverted. Choice C incorrectly suggests the reflected pulse travels faster, but wave speed depends only on the medium properties, not the reflection process. To predict reflection behavior, compare the impedances: reflection from higher impedance (heavier rope or rigid wall) inverts the pulse.

Question 15

1. The reflected pulse in air is not inverted in pressure. (correct answer)
2. The reflected pulse is inverted because rigid walls always invert waves.
3. The reflected pulse must travel faster because it reverses direction.
4. The reflected pulse travels along the wall due to friction with the surface.

Explanation: This question tests understanding of reflection. When a sound pulse in air reflects from a rigid wall, the pressure pulse is not inverted during reflection. Unlike mechanical waves on strings where displacement inverts at rigid boundaries, sound waves are pressure waves where the wall creates a pressure antinode (maximum). The incident compression reflects as a compression, and the incident rarefaction reflects as a rarefaction, maintaining the same phase. The reflected sound travels at the same speed in air but in the opposite direction. Choice B incorrectly generalizes that all waves invert at rigid boundaries, but this only applies to displacement waves, not pressure waves. To distinguish between wave types, remember that pressure waves (sound) don't invert at rigid boundaries while displacement waves (strings) do.

Question 16

1. The reflected angle equals $\theta_i$, measured from the normal. (correct answer)
2. The reflected angle is smaller because light travels slower in water.
3. The reflected angle depends on mirror composition more than $\theta_i$.
4. The reflected ray travels along the surface because water reduces reflection.

Explanation: This question tests understanding of reflection. The law of reflection applies regardless of the medium in which reflection occurs. When light reflects from a mirror underwater, the angle of incidence equals the angle of reflection, both measured from the normal to the mirror surface. The fact that the light is traveling through water instead of air doesn't change the reflection law - the reflected angle still equals θᵢ. The speed of light in water affects refraction at interfaces between different media, but not reflection within the same medium. Choice B incorrectly conflates the slower speed of light in water with reflection angles, which are purely geometric. When dealing with reflection, the surrounding medium doesn't matter - only the angles relative to the normal determine the reflected path.

Question 17

1. $10^\circ$
2. $5^\circ$
3. $20^\circ$ (correct answer)
4. $0^\circ$

Explanation: This question tests understanding of reflection. When a plane mirror rotates by an angle θ, the reflected ray rotates by 2θ. This occurs because rotating the mirror changes the normal direction by θ, which affects both the incident and reflected angles equally. Since the incident beam direction is fixed, the incident angle relative to the new normal changes by θ, and by the law of reflection, the reflected angle also changes by θ in the same rotational sense. The total angular change of the reflected ray is θ + θ = 2θ. For a 10° mirror rotation, the reflected ray rotates by 20°. Choice A (10°) represents the common misconception that the ray rotation equals the mirror rotation. Always remember: mirror rotation causes double the angular deviation in the reflected beam.

Question 18

1. The reflected wave must travel along the boundary instead of back on the string.
2. The reflected wave speed increases because the end is free.
3. The reflected wave is not inverted relative to the incident wave. (correct answer)
4. The reflected wave is inverted because all reflections invert waves.

Explanation: This question tests understanding of reflection. When a transverse wave on a string reflects from a free boundary (like a light ring that can move freely), the reflected wave is not inverted relative to the incident wave. At a free boundary, the string end can move without constraint, so an upward incident pulse creates an upward force that moves the ring up, generating an upward reflected pulse traveling back. This contrasts with rigid boundaries where inversion occurs. The wave maintains the same speed in the same medium but reverses direction. Choice B incorrectly claims all reflections invert waves, but inversion depends on boundary type: rigid boundaries invert, free boundaries don't. Remember the boundary condition determines reflection behavior: fixed ends invert pulses, free ends preserve orientation.

Question 19

1. The reflected ray makes a $35^\circ$ angle with the normal. (correct answer)
2. The reflected ray makes a larger angle because light slows near mirrors.
3. The reflected ray travels along the surface because the mirror is smooth.
4. The reflected ray makes an angle set by the mirror material, not the incident angle.

Explanation: This question tests understanding of reflection. The law of reflection states that when a wave encounters a smooth surface, the angle of incidence equals the angle of reflection, both measured from the normal (perpendicular) to the surface. Since the incident ray makes a 35° angle with the normal, the reflected ray must also make a 35° angle with the normal, but on the opposite side. The reflected ray leaves at the same angle because reflection conserves the angle relative to the normal, not because of any speed changes or material properties. Choice B incorrectly suggests that light slows near mirrors, which would affect refraction, not reflection angles. When solving reflection problems, always identify the normal first and measure angles from it, not from the surface itself.

Question 20

1. $50^\circ$ (correct answer)
2. $40^\circ$
3. $25^\circ$
4. $90^\circ$

Explanation: This question tests understanding of reflection. The problem states the incident beam makes a 50° angle with the surface, which means it makes a 40° angle with the normal (since 90° - 50° = 40°). By the law of reflection, the reflected ray also makes a 40° angle with the normal on the opposite side. To find the reflected angle with the surface, we calculate 90° - 40° = 50°. The reflected ray makes the same angle with the surface as the incident ray, creating a symmetric V-shape about the normal. Choice A (40°) represents confusing the angle from the normal with the angle from the surface. When given surface angles, always convert to normal angles for applying the reflection law, then convert back if needed.