14.3 Boundary Behavior of Waves and Polarization

1. AThe distance between its adjacent compressions
   A student who thinks the wavelength is kept when a wave enters a new medium picks this. The distance between compressions is the wavelength, λ = v/f; the frequency is unchanged, but sound travels about four times as fast in water as in air, so the wavelength in water is about four times as long.
2. BThe distance it travels in each second
   A student who thinks a wave keeps the speed its source gave it picks this. The speed of sound is set by the medium: sound travels about four times as fast in water as in air.
3. CThe amount of energy it carries each second
   A student who thinks all of the sound passes into the water picks this. Part of the sound is reflected at the surface (for air and water, most of it), so the energy carried into the water each second is less than the energy arriving at the surface each second.
4. DThe number of oscillations it makes each second Correct
   A wave keeps the same frequency when it passes from one medium into another. Each compression that reaches the surface produces one compression in the water, so the water oscillates as many times each second as the air does.

CED 14.3.A.1.iv · Read this in Fix

1. AThe wave travels in a single direction, away from its source.
   A student who thinks polarization describes the direction of travel picks this. Every wave on a rope travels along the rope; polarization describes the direction of the oscillations, which for a polarized wave is one direction perpendicular to the rope.
2. BEach point of the rope oscillates back and forth along the rope's length.
   A student who thinks a polarized wave oscillates along its direction of travel picks this. Oscillation along the rope would make a longitudinal wave, which cannot be polarized; a polarized transverse wave oscillates in one direction perpendicular to the rope.
3. CThe rope oscillates in a single plane, for example up and down only. Correct
   A polarized transverse wave oscillates in a single plane: every point of the rope moves back and forth along the same direction (for example, only up and down), perpendicular to the direction in which the wave travels.
4. DThe wave has a single frequency, so its crests are equally spaced.
   A student who confuses polarization with having a single frequency picks this. A wave of one frequency can oscillate in many directions, and a wave oscillating in one plane can contain several frequencies; polarization refers only to the direction of oscillation.

CED 14.3.A.2.i · Read this in Fix

1. AI₁ > I₂ > I₃ = 0 Correct
   P₁ transmits only the part of the unpolarized light that oscillates along its vertical axis, so I₂ is less than I₁ and the light at point 2 is polarized vertically. P₂'s axis is horizontal, perpendicular to that oscillation, so P₂ transmits none of it: I₃ = 0.
2. BI₁ > I₂ > I₃ > 0
   A student who thinks each polarizing filter dims light by a fixed fraction, like tinted glass, picks this. After P₁ the light oscillates only vertically; P₂ passes only horizontal oscillation, so it blocks all of that light rather than dimming it.
3. CI₁ > I₂ = I₃ > 0
   A student who thinks that light, once polarized, is unaffected by further filters picks this. How much P₂ transmits depends on the angle between the light's direction of oscillation and P₂'s axis; at 90°, it transmits nothing.
4. DI₁ > I₂ = I₃ = 0
   A student who thinks a polarizing filter blocks unpolarized light picks this. P₁ transmits the vertically oscillating part of the unpolarized light, so I₂ is not zero; it is P₂, crossed with P₁, that blocks the light.

Working Point 1: unpolarized light, intensity I₁. P₁ (axis vertical) passes only the vertical part of the oscillation, so 0 < I₂ < I₁ and the light at 2 is polarized vertically. P₂ (axis horizontal) is perpendicular to that oscillation, so I₃ = 0. Ranking: I₁ > I₂ > I₃ = 0.

CED 14.3.A.3 · Read this in Fix

1. AE_T is less than E₀, since part of the energy goes into a reflected pulse. Correct
   At the knot the pulse is partly reflected and partly transmitted. Energy is conserved, so E₀ is shared: the reflected pulse carries some of it back along the light string, and the transmitted pulse carries the rest, so E_T < E₀.
2. BE_T is less than E₀, since the pulse slows down on the heavy string.
   A student who treats the pulse like an object whose energy falls when it slows down picks this. The transmitted pulse does travel more slowly, but that is not why E_T is less than E₀: the missing energy is carried back along the light string by the reflected pulse. A pulse passing from a heavy string into a light one speeds up, yet it too transmits less than all of its energy, for the same reason.
3. CE_T is equal to E₀, since the whole pulse passes into the heavy string.
   A student who thinks joined strings pass the whole wave on picks this. Wherever the wave speed changes, part of the wave is reflected, so the heavy string receives only part of the energy.
4. DE_T is zero, since the heavy string reflects the entire pulse straight back.
   A student who pictures the heavy string as a wall picks this. The heavy string can move, so a pulse is transmitted along it as well as reflected; only an end that cannot move reflects the whole pulse.

CED 14.3.A.1.i · Read this in Fix

1. AIt is upright, as the transmitted pulse travels more slowly.
   A student who has the rule the wrong way round picks this. A reflected pulse is upright when the transmitted wave speeds up; here the transmitted pulse slows down, so the reflected pulse is inverted.
2. BIt is inverted, as the transmitted pulse travels more slowly. Correct
   A reflected wave is inverted if the transmitted wave travels into a medium in which the wave speed decreases. The pulse passes from the light string into the heavy string, where it travels more slowly, so the reflected pulse is displaced downward.
3. CIt is inverted, as a reflection flips a pulse at any boundary.
   A student who thinks every reflection inverts a pulse picks this. The pulse here is inverted, but the reason is wrong: a pulse reflected where the wave speeds up, as from a heavy string into a light one, is upright. The inversion here happens because the wave slows down.
4. DThere is none, as the whole pulse passes into the heavy string.
   A student who thinks joined strings pass the whole wave on picks this. Where the wave speed changes, part of the pulse is reflected, and the reflected pulse travels back along the light string.

CED 14.3.A.1.ii · Read this in Fix

1. AWaves travel more slowly on Y, as its reflected pulse is not inverted.
   A student who has the inversion rule the wrong way round picks this. An upright reflected pulse shows that the wave sped up on entering Y. The distances agree: in the same time, the transmitted pulse has moved twice as far from J as the reflected pulse.
2. BWaves travel at the same speed on both strings, as their source set it.
   A student who thinks a wave keeps the speed its source gave it picks this. The speed is set by the medium, and the diagram shows it changing: in the same time the transmitted pulse moved twice as far from J as the reflected pulse.
3. CWaves travel faster on Y, as Y is the heavier of the two strings.
   A student who thinks waves travel faster on heavier strings picks this. Waves are faster on Y, but at the same tension waves travel faster on the LIGHTER string, v_string = √(F_T/(m/ℓ)), so Y is the lighter string, not the heavier one.
4. DWaves travel faster on Y, since the reflected pulse is upright. Correct
   The reflected pulse is on the same side of the string as the incident pulse, so it is not inverted. A reflected wave is not inverted when the transmitted wave travels into a medium in which the speed increases, so waves travel faster on Y. The transmitted pulse agrees: in the same time it has moved twice as far from J as the reflected pulse, and it is twice as wide.

CED 14.3.A.1.iii · Read this in Fix

1. AIt is the same, as the frequency changes to suit the new speed.
   A student who thinks the frequency changes at a boundary while the wavelength is kept picks this. Each crest that reaches the join goes on into Y, so crests leave the join as often as they arrive: the frequency stays f, and the wavelength changes instead.
2. BIt is longer, as a wave that slows down gets a longer wavelength.
   A student who thinks wavelength is inversely related to speed picks this. At a fixed frequency λ = v/f, so the wavelength is proportional to the speed: the slower wave on Y has the shorter wavelength.
3. CIt is shorter, as the frequency stays f while the speed is less. Correct
   The frequency of a wave does not change when it passes into another medium: the join oscillates at frequency f and drives string Y at that frequency. With λ = v/f, a lower speed at the same frequency gives a shorter wavelength.
4. DIt is shorter, as some of the wave's energy is reflected at the join.
   A student who thinks a wave with less energy has a shorter wavelength picks this. The energy carried off by the reflected wave affects the amplitude of the transmitted wave, not its wavelength; the wavelength is shorter because the wave travels more slowly on Y.

CED 14.3.A.1.iv · Read this in Fix

1. AReflection partly polarizes the light, and the lenses are oriented to block its main oscillation direction. Correct
   Transverse waves reflected from a surface may be polarized. Light reflected from a road is partly polarized, oscillating mostly parallel to the road surface, and the transmission axes of the lenses are set perpendicular to that direction. Much of the glare is therefore blocked, while unpolarized light from elsewhere is only partly dimmed.
2. BReflection changes the frequency of the light, and the lenses absorb light of that changed frequency.
   A student who thinks a wave's frequency changes at a boundary picks this. Reflection does not change the frequency (the color) of light; the glare is reduced because of its direction of oscillation.
3. CReflection leaves the light unpolarized, and polarizing lenses are made to block unpolarized light.
   A student who thinks polarizing filters pass polarized light and block unpolarized light picks this. A polarizing filter transmits the part of any light that oscillates along its axis, and reflection tends to polarize light, not to unpolarize it.
4. DReflected light oscillates along its direction of travel, and the lenses block light oscillating that way.
   A student who thinks polarized light oscillates along its direction of travel picks this. The polarized part of the reflected light oscillates in a single direction perpendicular to its travel, mostly parallel to the road, and that is the direction the lenses block.

CED 14.3.A.2 · Read this in Fix

1. AThe wave passes through unchanged, since it oscillates across the slot.
   A student who thinks a slot passes oscillations perpendicular to it picks this. The rope can move only along the length of the slot; moving across the slot, it is stopped by the edges.
2. BAlmost no wave passes, as the slot stops the rope moving sideways. Correct
   The rope oscillates horizontally, across the narrow slot, and the edges of the slot stop it moving that way. The oscillation is blocked, so almost no wave reaches the far side. Only oscillations along the slot's length (vertical) could pass.
3. CA vertical wave of the same amplitude passes, since the slot turns it.
   A student who thinks a slot turns a wave's oscillation into line with itself picks this. The slot removes the part of the oscillation that it does not allow; it cannot make the rope oscillate vertically, and a horizontal oscillation has no vertical part to pass.
4. DA horizontal wave of smaller amplitude passes, since the slot weakens it.
   A student who thinks a slot weakens every wave by the same fraction, like tinted glass, picks this. Whether a wave passes depends on its direction of oscillation: across the slot, it is stopped, not just weakened.

CED 14.3.A.2.i · Read this in Fix

1. ANo, as polarization is a property of electromagnetic waves, not mechanical ones.
   A student who thinks only light and other electromagnetic waves can be polarized picks this. Sound is indeed a mechanical wave, but a transverse mechanical wave, such as a wave on a rope, can be polarized; sound cannot be polarized because it is longitudinal.
2. BYes, as the air oscillates perpendicular to the direction of travel.
   A student who pictures sound as a transverse wave, like a wave on a string, picks this. Sound in air is longitudinal: the air oscillates along the direction of travel, so there is no transverse oscillation for a filter or slot to select.
3. CNo, as the air oscillates back and forth along the direction of travel. Correct
   Sound in air is a longitudinal wave: the air oscillates along the direction of travel, forming compressions and rarefactions. There is no oscillation across the direction of travel whose direction a filter or slot could select, so longitudinal waves cannot be polarized.
4. DNo, as sound is a stream of air carried to the ear, rather than an oscillation.
   A student who thinks sound is a stream of air moving from the source to the listener picks this. The air in a sound wave oscillates about fixed positions; it is energy, not air, that travels to the listener.

CED 14.3.A.2.ii · Read this in Fix

1. A6.7 × 10⁴ W
   A student who takes intensity as power times area, and so divides the intensity by the area, picks this: (1.0 × 10³)/(1.5 × 10⁻²) = 6.7 × 10⁴. Intensity is power per unit area, so the power is the intensity TIMES the area.
2. B1.0 × 10³ W
   A student who treats intensity and power as the same quantity picks this. 1.0 × 10³ W/m² is the power reaching each square meter; the cell's area, 1.5 × 10⁻² m², is much smaller than 1 m², so it receives only 15 W.
3. C1.5 × 10³ W
   A student who converts 150 cm² to m² by dividing by 100, the factor for centimeters to meters, gets 1.5 m² and picks this. An area conversion factor must be squared: 1 m² = (100 cm)² = 10⁴ cm², so 150 cm² = 1.5 × 10⁻² m².
4. D1.5 × 10¹ W Correct
   Intensity is power per unit area, so P = IA. Converting the area, 150 cm² = 1.5 × 10⁻² m² because 1 m² = 10⁴ cm², so P = (1.0 × 10³ W/m²)(1.5 × 10⁻² m²) = 15 W.

Working 150 cm² = 150 × 10⁻⁴ m² = 1.5 × 10⁻² m². I = P/A, so P = IA = (1.0 × 10³ W/m²)(1.5 × 10⁻² m²) = 15 W = 1.5 × 10¹ W.

CED 14.3.A.3.i · Read this in Fix

1. A3.0 × 10⁻³ W/m²
   A student who thinks intensity is inversely proportional to distance picks this, halving the intensity. The area over which the power spreads grows as r², so doubling the distance divides the intensity by four.
2. B1.5 × 10⁻³ W/m² Correct
   Intensity is power per unit area. The same power passes through every sphere around the source, and a sphere's area, 4πr², is four times as large at 4.0 m as at 2.0 m. The same power over four times the area gives one fourth of the intensity: (6.0 × 10⁻³ W/m²)/4 = 1.5 × 10⁻³ W/m².
3. C6.0 × 10⁻³ W/m²
   A student who treats intensity as the power of the sound picks this; the power passing through each sphere is indeed the same. But intensity is that power per unit area, and at 4.0 m the power is spread over four times the area.
4. D2.4 × 10⁻² W/m²
   A student who takes intensity as power multiplied by area picks this, making it four times as great. Intensity is power DIVIDED by area; spreading the same power over four times the area makes it one fourth as great.

Working The power P passes through spheres of area 4πr², so I = P/(4πr²) ∝ 1/r². I₂ = I₁(r₁/r₂)² = (6.0 × 10⁻³ W/m²)(2.0 m/4.0 m)² = 1.5 × 10⁻³ W/m².

CED 14.3.A.3.i · Read this in Fix

1. A2.0 × 10⁻³ W/m² Correct
   Intensity is the average power per unit area over one period. The power rises and falls between 0 and 2.0 × 10⁻⁶ W along a cosine-shaped curve centered on 1.0 × 10⁻⁶ W, whose parts above that level exactly fill the gaps below it, so the average power is 1.0 × 10⁻⁶ W. I = (1.0 × 10⁻⁶ W)/(5.0 × 10⁻⁴ m²) = 2.0 × 10⁻³ W/m².
2. B4.0 × 10⁻³ W/m²
   A student who uses the greatest power, 2.0 × 10⁻⁶ W, picks this: (2.0 × 10⁻⁶ W)/(5.0 × 10⁻⁴ m²) = 4.0 × 10⁻³ W/m². The intensity is the AVERAGE power per unit area over a period, which here is half the peak value.
3. C1.0 × 10⁻⁶ W/m²
   A student who takes the average power itself as the intensity picks this. 1.0 × 10⁻⁶ W is the average power delivered to the whole diaphragm; dividing by its area, 5.0 × 10⁻⁴ m², gives the power per unit area.
4. D4.0 × 10⁻⁶ W/m²
   A student who takes the area under the graph, the energy delivered in one period (1.0 × 10⁻⁶ W × 2.0 × 10⁻³ s = 2.0 × 10⁻⁹ J), and divides it by the diaphragm's area picks this. That is energy per unit area, in J/m²; intensity is a rate, the average power per unit area, in W/m².

Working The graph oscillates between 0 and 2.0 × 10⁻⁶ W with the shape of a cosine curve about 1.0 × 10⁻⁶ W, so P_avg over one period (2.0 ms) = 1.0 × 10⁻⁶ W. I = P_avg/A = (1.0 × 10⁻⁶ W)/(5.0 × 10⁻⁴ m²) = 2.0 × 10⁻³ W/m².

CED 14.3.A.3.ii · Read this in Fix

1. A4E/(πD²Δt) Correct
   The average power is E/Δt, and the beam's cross-sectional area is π(D/2)² = πD²/4. Intensity is power per unit area, so I = (E/Δt)/(πD²/4) = 4E/(πD²Δt), in W/m².
2. B(E/Δt)πD²/4
   A student who takes intensity to be power multiplied by area picks this. Intensity is power divided by area: spreading the same power over a larger area lowers the intensity. This expression has units W·m², not W/m².
3. CE/Δt
   A student who treats intensity as the same thing as power divides the energy by the time and stops, picking this. E/Δt is the power, in watts; intensity is power per unit area, so it must still be divided by the beam's area.
4. DE/(πD²Δt)
   A student who puts the diameter in place of the radius in A = πr² uses an area of πD² and picks this. The radius is D/2, so the area is πD²/4 and the intensity is four times this expression.

Working Average power delivered: P = E/Δt. Cross-sectional area of the beam: A = π(D/2)² = πD²/4. Intensity is power per unit area: I = P/A = (E/Δt)/(πD²/4) = 4E/(πD²Δt). Units: J/(m²·s) = W/m².

CED 14.3.A.3.i · Read this in Fix

1. A2.00vt
   A student who thinks a wave keeps the speed its source gave it lets the transmitted pulse go on at v and picks this. The speed of a pulse is set by the string it is on: on the heavy string, at the same tension, it is √(F_T/(4(m/ℓ))) = v/2, so the transmitted pulse covers only vt/2.
2. B1.25vt
   A student who takes the wave speed as inversely proportional to the mass per length gives the transmitted pulse a speed of v/4, covering vt/4, and picks this. The speed depends on the square root of the mass per length: √(1/4) = 1/2, so the transmitted pulse moves at v/2 and covers vt/2.
3. C1.50vt Correct
   The reflected pulse stays on the light string and moves back from the knot at v, covering vt. On the heavy string the speed is √(F_T/(4(m/ℓ))) = v/2, so the transmitted pulse covers vt/2 in the other direction. The pulses are on opposite sides of the knot, so they are vt + vt/2 = 1.50vt apart.
4. D3.00vt
   A student who thinks waves travel faster in a heavier medium turns the ratio over, gives the transmitted pulse a speed of 2v, and picks this. At the same tension a larger mass per length gives a smaller speed: √(F_T/(4(m/ℓ))) = v/2.

Working Reflected pulse: it stays on the light string, so it moves at v back away from the knot and covers vt. Transmitted pulse: on the heavy string, at the same tension, v₂ = √(F_T/(4(m/ℓ))) = (1/2)√(F_T/(m/ℓ)) = v/2, so it moves forward away from the knot and covers (v/2)t. The two pulses move in opposite directions from the knot, so their separation is vt + (1/2)vt = 1.50vt.

CED 14.3.A.1.i · Read this in Fix

1. A2.2 m
   A student who thinks the wavelength stays the same at a boundary, with the frequency changing instead, keeps the heavy string's wavelength, (20 m/s)/(9.0 Hz) = 2.2 m, and picks this. Each crest that reaches the knot goes on into the light string, so the frequency stays 9.0 Hz; the speed doubles, so the wavelength doubles.
2. B4.4 m Correct
   The frequency is set by the oscillator and does not change at the knot, so it is 9.0 Hz on the light string too. At the same tension, one-fourth the mass per length gives √4 = 2 times the speed, 40 m/s, so λ = v/f = (40 m/s)/(9.0 Hz) = 4.4 m.
3. C1.1 m
   A student who thinks a faster wave has a shorter wavelength halves the heavy string's wavelength of 2.2 m and picks this. At a fixed frequency λ = v/f, so the wavelength is proportional to the speed: doubling the speed doubles the wavelength.
4. D8.9 m
   A student who takes the wave speed as inversely proportional to the mass per length makes the speed on the light string four times 20 m/s, 80 m/s, and gets (80 m/s)/(9.0 Hz) = 8.9 m. The speed depends on the square root of the mass per length: √4 = 2, so the speed is 40 m/s.

Working Speed on the light string (same tension): v = √(F_T/(m/ℓ)), so v_light = v_heavy √((m/ℓ)_heavy/(m/ℓ)_light) = (20 m/s)√(0.040/0.010) = (20 m/s)(2) = 40 m/s. The frequency does not change at the knot: f = 9.0 Hz on both strings. λ = v/f = (40 m/s)/(9.0 Hz) = 4.4 m.

CED 14.3.A.1.iv · Read this in Fix