AP Physics 2 · Topic 14.3

Topic 14.3: Boundary Behavior of Waves and Polarization

Unit 14: Waves, Sound, and Physical Optics12-15% of the multiple-choice section

At a boundary a wave splits: part reflects and part transmits. The reflected part is inverted when the transmitted wave moves into a medium where the wave speed decreases, and upright when the speed increases. Frequency never changes across a boundary. Only transverse waves can be polarized.

AP Physics: Unit 14 (topics 14.3 Boundary Behavior of Waves and Polarization). AP Physics 2 Unit 14, Topic 14.3. One learning objective, 14.3.A, describe the interaction between a wave and a boundary. Three essential knowledge statements. 14.3.A.1 (a wave that travels from one medium to another can be transmitted or reflected, depending on the properties of the boundary separating the two media) carries four sub-statements: 14.3.A.1.i (a wave traveling from one medium to another, for example a wave traveling between low-mass and high-mass strings, will result in reflected and transmitted waves), 14.3.A.1.ii (a reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases), 14.3.A.1.iii (a reflected wave is not inverted if the transmitted wave travels into a medium in which the speed of the wave increases) and 14.3.A.1.iv (the frequency of a wave does not change when it travels from one medium to another). 14.3.A.2 (transverse waves that are reflected from a surface, refracted through a medium, or pass through specific openings may be polarized) carries 14.3.A.2.i (transverse waves can be polarized and oscillate in a single plane) and 14.3.A.2.ii (longitudinal waves cannot be polarized). 14.3.A.3 (polarization of a wave may result in a reduction of the wave's intensity) carries 14.3.A.3.i (intensity is a measure of the amount of power transferred per unit area) and 14.3.A.3.ii (the intensity of a wave is the average power per unit area over one period of the wave). The inversion rule is stated in terms of the wave speed in the second medium, not in terms of density; the widespread denser-medium paraphrase is a consequence of the string speed equation at fixed tension rather than the CED's rule. The topic prints no boundary statement; Unit 14's only three sit under Topics 14.4, 14.5 and 14.9. No relevant equation is attached to any statement in this topic, and no intensity equation appears among the 129 on the AP Physics 2 equation sheet; nothing corresponding to a polarizer intensity law appears in the framework. Suggested skills are 1.A, 2.C, 3.B and 3.C, listed identically on the topic page and in the Unit at a Glance table; 2.B is not among them. Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

What Topic 14.3 requires

One learning objective, 14.3.A: describe the interaction between a wave and a boundary. Three essential knowledge statements, and the topic bolts two subjects together, boundary behavior and polarization, because reflection and refraction are among the things that can polarize a wave.

14.3.A.1. A wave that travels from one medium to another can be transmitted or reflected, depending on the properties of the boundary separating the two media.

14.3.A.1.i. A wave traveling from one medium to another (for example, a wave traveling between low-mass and high-mass strings) will result in reflected and transmitted waves.
14.3.A.1.ii. A reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases.
14.3.A.1.iii. A reflected wave is not inverted if the transmitted wave travels into a medium in which the speed of the wave increases.
14.3.A.1.iv. The frequency of a wave does not change when it travels from one medium to another.

14.3.A.2. Transverse waves that are reflected from a surface, refracted through a medium, or pass through specific openings may be polarized.

14.3.A.2.i. Transverse waves can be polarized and oscillate in a single plane.
14.3.A.2.ii. Longitudinal waves cannot be polarized.

14.3.A.3. Polarization of a wave may result in a reduction of the wave's intensity.

14.3.A.3.i. Intensity is a measure of the amount of power transferred per unit area.
14.3.A.3.ii. The intensity of a wave is the average power per unit area over one period of the wave.

Suggested skills: 1.A create diagrams, tables, charts, or schematics to represent physical situations, 2.C compare physical quantities between two or more scenarios or at different times and locations in a single scenario, 3.B apply an appropriate law, definition, theoretical relationship, or model to make a claim, and 3.C justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws. Identical on the topic page and in the Unit at a Glance table.

Topic 14.3 prints no boundary statement. The three in Unit 14 sit under Topics 14.4, 14.5 and 14.9. Note also what is absent from the list above: no equation of any kind. Topic 14.3 is the only place in Unit 14 where intensity is defined, and it defines it in words.

At a boundary, a wave does both things

14.3.A.1 says a wave arriving at a boundary can be transmitted or reflected, depending on the properties of the boundary separating the two media. 14.3.A.1.i then makes the practical point: a wave traveling from one medium to another will result in reflected and transmitted waves.

Both, not one or the other. Send a pulse down a light string tied to a heavy one and you get two pulses out of the junction: one continuing into the heavy string and one heading back the way it came. The energy of the original divides between them.

The CED's chosen illustration, in the parenthesis of 14.3.A.1.i, is a wave traveling between low-mass and high-mass strings. That example is chosen carefully, because it is the case where you can compute the speeds. From Topic 14.1, 14.1.A.3.ii gives:

v_{\text{string}} = \sqrt{\frac{F_T}{m/\ell}}

Two strings tied together are under the same tension, so $F_T$ is common to both and only the mass per length differs. A higher mass per length means a lower speed. That single fact is what turns the CED's low-mass and high-mass strings into a statement about speed, which is what the inversion rule is actually written in terms of.

The extreme cases are worth naming. A string tied to a wall is a boundary into something so heavy the wave speed there is effectively zero: nothing transmits, everything reflects. A string ending in a light ring free to slide on a pole is the opposite extreme. The CED does not name either case here; Topic 14.6 refers to strings with fixed or loose ends when it sets up standing waves.

Inverted or not: the CED's rule is about speed, not density

This is the part of Topic 14.3 most worth getting word-perfect, because the version in circulation is not the version the CED prints.

14.3.A.1.ii. A reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases.
14.3.A.1.iii. A reflected wave is not inverted if the transmitted wave travels into a medium in which the speed of the wave increases.

The criterion is the wave speed in the second medium. Not its density, not its mass, not whether it feels heavier. Textbooks and revision notes routinely paraphrase this as "reflecting off a denser medium inverts the pulse", and for two strings at the same tension that paraphrase happens to give the right answer, because a larger mass per length does mean a slower wave. It is a consequence, not the rule.

Answer in the CED's terms and the reasoning transfers. A justification that says "the second string is heavier, so the wave speed in it is lower, so by 14.3.A.1.ii the reflected pulse is inverted" is a complete chain. A justification that stops at "the second string is heavier" is missing the step that makes it work, and it will fail outright in any situation where the denser medium is the faster one.

Three things follow.

The transmitted pulse is never inverted. Both statements are about the reflected wave. Whatever the boundary does, the part that carries on into the new medium stays on the same side of equilibrium as the incoming pulse.

Only the reflected pulse can flip. So "does it invert?" is always a question about one of the two outgoing pulses, and you should say which.

Inversion means a 180 degree phase change. Topic 14.9 states the light version of exactly this idea, that a phase change of 180 degrees occurs when a light ray is reflected from a medium with a greater index of refraction than the medium through which the ray is traveling. A greater index of refraction means a lower speed of light in that medium, since $n = c/v$ from Topic 13.3. So 14.9's rule and 14.3's rule agree: reflect off the slower medium, and the wave flips.

Frequency is what survives

14.3.A.1.iv: the frequency of a wave does not change when it travels from one medium to another.

This is the single most useful sentence in the topic, and it is the one to reach for whenever a question moves a wave between media.

The reason is a counting argument. The boundary is one point. Wavefronts arrive there at a certain rate and, since nothing accumulates or disappears at a point, they leave at the same rate. Frequency belongs to whatever is driving the wave, and driving does not stop at the junction.

Now combine it with 14.2.A.3 from Topic 14.2, the printed relation $\lambda = v/f$ . Cross a boundary and $v$ changes, by 14.1.A.3. And $f$ does not, by 14.3.A.1.iv. So $\lambda$ has to move, and it moves by exactly the same factor as the speed.

Crossing into a slower medium	Crossing into a faster medium
Speed decreases	Speed increases
Frequency unchanged	Frequency unchanged
Wavelength decreases by the same factor	Wavelength increases by the same factor
Reflected wave inverted (14.3.A.1.ii)	Reflected wave not inverted (14.3.A.1.iii)

Three consequences students trip on:

A note does not change pitch as it passes through a wall. Pitch is related to frequency (14.2.A.1.v), and frequency is what survives.
Light does not change color entering glass. Same reason. Its speed drops and its wavelength shortens in the glass, and it comes out the far side at the wavelength it went in with.
Wavelength and speed change together, in the same direction and by the same factor. They are locked, because their ratio is the unchanged frequency.

That last point is the fastest sanity check in the unit. If a worked answer has a wave speeding up and its wavelength shortening, something has gone wrong.

Polarization: what it is, and why sound cannot do it

14.3.A.2.i: transverse waves can be polarized and oscillate in a single plane. 14.3.A.2.ii: longitudinal waves cannot be polarized.

Take those together with the definitions from Topic 14.1 and the reason falls out. In a transverse wave the disturbance is perpendicular to the direction of propagation (14.1.A.4), and perpendicular is not one direction. A wave travelling along a line can be disturbed vertically, horizontally, or at any angle in the plane at right angles to its travel, so there is a genuine choice to constrain. Polarizing it means restricting the oscillation to a single plane.

In a longitudinal wave the disturbance is parallel to the direction of propagation (14.1.A.5), and parallel is one direction. There is nothing left to choose, so there is nothing to polarize. Sound is a mechanical longitudinal wave (14.1.A.5.i), which is why sound cannot be polarized.

So "can this be polarized?" is the same question as "is this transverse?", and it makes the transverse and longitudinal split from 14.1 into something with consequences rather than vocabulary. Light can be polarized because electromagnetic waves are transverse, which Topic 14.4 states in 14.4.A.1.i.

14.3.A.2 also names what can do the polarizing, and the list is short and specific. Transverse waves that are reflected from a surface, refracted through a medium, or pass through specific openings may be polarized. Two of those three are the mechanisms of Unit 13, which is why polarization sits in the same topic as boundary behavior rather than in a topic of its own.

Note the word may. The CED says these processes may polarize a transverse wave, not that they always do. Reflected light is not automatically fully polarized.

Intensity: defined in the CED, absent from the equation sheet

14.3.A.3: polarization of a wave may result in a reduction of the wave's intensity. Then two definitions.

14.3.A.3.i. Intensity is a measure of the amount of power transferred per unit area.
14.3.A.3.ii. The intensity of a wave is the average power per unit area over one period of the wave.

The second refines the first. Power in a wave is not steady from moment to moment, since the wave oscillates, so the CED specifies the average taken over one full period. Intensity therefore comes out in watts per square metre.

Now the part worth counting rather than assuming. The AP Physics 2 equation sheet prints 129 equations across seven groups, of which the Waves, Sound, and Optics group holds 15. None of them is an intensity equation, and neither of the two statements above is labeled a relevant equation in the framework. Nothing resembling a polarizer intensity law, relating transmitted intensity to the angle between a filter and the incoming polarization, appears anywhere in Topic 14.3 or on the sheet.

What that means in practice:

You can use the definition. Average power divided by area is what the CED says intensity is, so a question that gives you a power and an area is answerable.
You cannot compute how much a polarizing filter cuts the intensity from an angle. AP Physics 2 does not give you the relationship, so a question that wants a number has to give you the transmitted fraction directly.
The supportable claim about polarization and intensity is 14.3.A.3's own wording: polarization may result in a reduction of the wave's intensity. Not always, and not by a stated amount.

If you have met a cosine-squared law elsewhere, park it. It is real physics and it is outside this course's framework, and an AP Physics 2 justification is graded against the framework.

Drawing the boundary (skill 1.A)

Skill 1.A, create diagrams, tables, charts, or schematics to represent physical situations, heads this topic's list. A boundary diagram that earns credit has a fixed set of parts.

Draw the two media side by side with the junction marked, and label which side is which: for strings, the mass per length or at least "light" and "heavy"; for light, the index of refraction. Then draw three things and label all three:

The incident pulse, with an arrow for its direction of travel, on the incoming side.
The transmitted pulse, on the far side, on the same side of equilibrium as the incident pulse, with an arrow away from the boundary.
The reflected pulse, back on the incoming side, with an arrow away from the boundary, drawn either upright or inverted according to 14.3.A.1.ii and 14.3.A.1.iii.

Two features of a good diagram are easy to leave out and are exactly what a marker looks for.

The wavelengths should differ. The transmitted pulse is in a medium with a different wave speed, so if the wave is periodic its wavelength there is different, by 14.2.A.3 with frequency held fixed. The reflected wave is back in the original medium, so its wavelength matches the incident one.

The amplitudes should be smaller than the incident amplitude. The incoming energy divides between two outgoing waves, so neither can be as large as what arrived.

For polarization, the diagram is different in kind: draw the propagation direction as an arrow, then show the oscillation direction as a double-headed arrow perpendicular to it, and mark the single plane the polarized wave is confined to. Showing the oscillation parallel to the propagation arrow is the standard error, and it draws a longitudinal wave, which by 14.3.A.2.ii cannot be polarized at all.

How Topic 14.3 shows up on the exam, and where it goes wrong

Topic 14.3's skills are 1.A, 2.C, 3.B and 3.C, and 2.B, calculate an unknown quantity, is not among them. Nor is any equation attached to the topic. So questions here are diagram-reading, comparison and justification: two strings and a pulse, asked what comes back; a wave crossing into a new medium, asked what changed; a claim about polarizing sound, asked to be evaluated. Skill 1 is not assessed on the multiple-choice section, so the diagram work lands mostly on the free-response side, where the Translation Between Representations question lives.

The errors that cost marks:

Saying the wave either reflects or transmits. 14.3.A.1.i says a wave crossing between media results in reflected and transmitted waves. Both.
Using density instead of speed for the inversion rule. 14.3.A.1.ii and 14.3.A.1.iii are stated in terms of whether the speed of the wave decreases or increases. Get to speed before you conclude.
Inverting the transmitted pulse. Only the reflected wave can invert.
Changing the frequency at a boundary. 14.3.A.1.iv says it does not change. Wavelength absorbs the difference.
Changing the wavelength but not the speed, or the other way round. They move together, by the same factor, because their ratio is the fixed frequency.
Polarizing sound. 14.3.A.2.ii is explicit that longitudinal waves cannot be polarized, and 14.1.A.5.i models sound as longitudinal.
Computing a polarized intensity. No intensity equation is printed on the AP Physics 2 sheet, and none is given in the framework.
Reading "may" as "will". Both 14.3.A.2 and 14.3.A.3 use it.

Where to go next. Topic 14.1 supplies the wave speed equation and the transverse and longitudinal definitions this topic runs on, and Topic 14.2 supplies $\lambda = v/f$ . Topic 14.4 explains why light is transverse and therefore polarizable, and Topic 14.9 reuses the inversion rule as a 180 degree phase change on reflection. The unit hub shows the whole sequence.

A pulse crossing between two strings, both directions

A light string of mass per length $5.00 \times 10^{-3} \text{ kg/m}$ is tied to a heavy string of mass per length $2.00 \times 10^{-2} \text{ kg/m}$ , and the whole arrangement is held at a tension of $45.0 \text{ N}$ . (a) Find the wave speed on each string. (b) An upward pulse travels along the light string toward the junction. Is the reflected pulse upright or inverted? What about the transmitted pulse? (c) The source is now a periodic wave of frequency $25.0 \text{ Hz}$ on the light string. Find the wavelength on each string. (d) Repeat part (b) for a pulse sent the other way, from the heavy string toward the light one.

(a) Both strings carry the same tension, so use 14.1.A.3.ii on each. Light string: $v_1 = \sqrt{\dfrac{45.0 \text{ N}}{5.00 \times 10^{-3} \text{ kg/m}}} = \sqrt{9.00 \times 10^{3}} = 94.9 \text{ m/s}$ .
Heavy string: $v_2 = \sqrt{\dfrac{45.0 \text{ N}}{2.00 \times 10^{-2} \text{ kg/m}}} = \sqrt{2.25 \times 10^{3}} = 47.4 \text{ m/s}$ .
Check the ratio rather than trusting two square roots: $\dfrac{v_2}{v_1} = \sqrt{\dfrac{5.00 \times 10^{-3}}{2.00 \times 10^{-2}}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}$ exactly. The heavy string's speed is exactly half.
(b) Apply 14.3.A.1.ii in the CED's own terms. The transmitted wave travels into the heavy string, where the speed is $47.4 \text{ m/s}$ against $94.9 \text{ m/s}$ , so the speed decreases. The reflected pulse is therefore inverted. The transmitted pulse is upright: the inversion rule applies only to the reflected wave.
(c) Frequency does not change at the boundary (14.3.A.1.iv), so $f = 25.0 \text{ Hz}$ on both strings. Use $\lambda = v/f$ from 14.2.A.3 twice. Light string: $\lambda_1 = \dfrac{94.9 \text{ m/s}}{25.0 \text{ Hz}} = 3.79 \text{ m}$ . Heavy string: $\lambda_2 = \dfrac{47.4 \text{ m/s}}{25.0 \text{ Hz}} = 1.90 \text{ m}$ .
The wavelength ratio is $1/2$ , matching the speed ratio exactly, which is the check to run every time: with $f$ fixed, $\lambda$ and $v$ scale together.
(d) Now the pulse arrives from the heavy side. The transmitted wave travels into the light string, where the speed rises from $47.4$ to $94.9 \text{ m/s}$ , so by 14.3.A.1.iii the reflected pulse is not inverted. The transmitted pulse is upright, as always. Nothing about the strings changed between (b) and (d); only the direction of travel did, and that is enough to reverse the answer.

(a) $v_1 = 94.9 \text{ m/s}$ on the light string and $v_2 = 47.4 \text{ m/s}$ on the heavy one, exactly half. (b) Light to heavy: the speed decreases, so the reflected pulse is inverted (14.3.A.1.ii) and the transmitted pulse is upright. (c) $\lambda_1 = 3.79 \text{ m}$ and $\lambda_2 = 1.90 \text{ m}$ , at $25.0 \text{ Hz}$ on both. (d) Heavy to light: the speed increases, so the reflected pulse is not inverted (14.3.A.1.iii).

Three strings, no calculator (skill 2.C)

Three strings are joined end to end and held at a single common tension. Their mass per length values are string A at $8.0 \times 10^{-3} \text{ kg/m}$ , string B at $2.0 \times 10^{-3} \text{ kg/m}$ , and string C at $3.2 \times 10^{-2} \text{ kg/m}$ , in that order along the line. A pulse starts on A and travels toward C. At each junction, state whether the reflected pulse is inverted, and give the speed on each string as a multiple of the speed on A. Do it without computing any speed in metres per second.

The tension is common to all three, so 14.1.A.3.ii reduces to $v \propto \dfrac{1}{\sqrt{m/\ell}}$ . Every comparison is then a ratio of mass per length under a square root, and no tension value is needed.
A to B: $\dfrac{v_B}{v_A} = \sqrt{\dfrac{8.0 \times 10^{-3}}{2.0 \times 10^{-3}}} = \sqrt{4} = 2$ . String B carries waves at twice the speed of A.
So at the A-to-B junction the transmitted wave travels into a medium in which the speed increases, and by 14.3.A.1.iii the reflected pulse is not inverted.
B to C: $\dfrac{v_C}{v_B} = \sqrt{\dfrac{2.0 \times 10^{-3}}{3.2 \times 10^{-2}}} = \sqrt{\dfrac{1}{16}} = \dfrac{1}{4}$ . String C carries waves at a quarter of B's speed.
At the B-to-C junction the speed decreases, so by 14.3.A.1.ii the reflected pulse is inverted.
Speeds relative to A: $v_A = v_A$ , $v_B = 2 v_A$ , and $v_C = \dfrac{1}{4} v_B = \dfrac{1}{4}(2 v_A) = \dfrac{1}{2} v_A$ . Cross-check directly: $\sqrt{\dfrac{8.0 \times 10^{-3}}{3.2 \times 10^{-2}}} = \sqrt{\dfrac{1}{4}} = \dfrac{1}{2}$ . The two routes agree.
Note what the ratios exposed. String C is the heaviest of the three and the pulse entering it does invert on reflection, but string A is heavier than B and the pulse entering B does not. Sorting the strings by mass tells you nothing on its own; the ratio at each junction is what the CED's rule is written in terms of.

At the A-to-B junction the speed increases (B is twice as fast as A), so the reflected pulse is not inverted. At the B-to-C junction the speed decreases (C is a quarter of B), so the reflected pulse is inverted. Relative to A: $v_B = 2 v_A$ and $v_C = 0.5 v_A$ .

Intensity, and where AP Physics 2 stops

A laser delivers an average power of $4.00 \text{ mW}$ into a circular spot of radius $1.00 \text{ mm}$ . (a) Find the intensity of the beam at that spot. (b) The beam is passed through a polarizing filter that a data sheet says transmits $40.0$ percent of the incident power. Find the transmitted intensity. (c) A second, identical filter is then placed after the first, rotated to an unstated angle. What can you predict about the intensity after it?

(a) Use the CED's definition. 14.3.A.3.i says intensity is a measure of the amount of power transferred per unit area, and 14.3.A.3.ii specifies the average power per unit area over one period. So divide the average power by the illuminated area.
Area of the spot: $A = \pi r^2 = \pi (1.00 \times 10^{-3} \text{ m})^2 = 3.14 \times 10^{-6} \text{ m}^2$ .
Intensity: $I = \dfrac{P}{A} = \dfrac{4.00 \times 10^{-3} \text{ W}}{3.14 \times 10^{-6} \text{ m}^2} = 1.27 \times 10^{3} \text{ W/m}^2$ .
(b) The transmitted fraction was given, which is the only reason this part is answerable. $I' = 0.400 \times 1.27 \times 10^{3} \text{ W/m}^2 = 5.09 \times 10^{2} \text{ W/m}^2$ . The spot size did not change, so the area cancels out of the comparison and the intensity falls in the same ratio as the power.
(c) Only the direction, not a number. 14.3.A.3 says polarization of a wave may result in a reduction of the wave's intensity, so the intensity after the second filter will be no greater than $5.09 \times 10^{2} \text{ W/m}^2$ and in general lower.
Say why you cannot go further, because that is what a 3.C justification wants. AP Physics 2 gives no relationship between the angle of a polarizing filter and the fraction of intensity it transmits: no such statement appears in Topic 14.3, and no intensity equation appears among the 129 printed on the AP Physics 2 equation sheet. Without a stated transmitted fraction, there is no number to produce.
The same reasoning is why part (b) had to hand you the $40.0$ percent. On this exam, any polarizer question with a numerical answer will supply the fraction.

(a) $I = 1.27 \times 10^{3} \text{ W/m}^2$ . (b) $I' = 5.09 \times 10^{2} \text{ W/m}^2$ . (c) The intensity can only fall or stay the same (14.3.A.3 says polarization may result in a reduction), and no number can be predicted: AP Physics 2 supplies no relationship between filter angle and transmitted intensity, and no intensity equation is printed on the sheet.

Frequently asked questions

What happens when a wave reaches a boundary between two media?

It does both things at once. Essential knowledge 14.3.A.1 says a wave that travels from one medium to another can be transmitted or reflected depending on the properties of the boundary, and 14.3.A.1.i states that a wave traveling from one medium to another, for example between low-mass and high-mass strings, will result in reflected and transmitted waves. The energy of the incoming wave divides between the two, so both outgoing waves have smaller amplitudes than the wave that arrived.

Is a reflected wave always inverted?

No, and the AP Physics 2 CED states the condition in terms of speed rather than density. Essential knowledge 14.3.A.1.ii says a reflected wave is inverted if the transmitted wave travels into a medium in which the speed of the wave decreases, and 14.3.A.1.iii says it is not inverted if the speed increases. For two strings under the same tension a heavier string is the slower one, so the common paraphrase about reflecting off a denser medium gives the right answer there, but it is a consequence of the speed rule and not the rule itself.

Why does wavelength change at a boundary but frequency does not?

Because frequency belongs to the source and speed belongs to the medium. Essential knowledge 14.3.A.1.iv states that the frequency of a wave does not change when it travels from one medium to another, while 14.1.A.3 makes wave speed a property of the medium. Since the printed relation lambda = v/f ties the three together, with f fixed and v changed the wavelength has to move, by exactly the same factor as the speed. Wavelength is the quantity that absorbs the change.

Why can't sound be polarized?

Because it is longitudinal. Essential knowledge 14.3.A.2.ii states plainly that longitudinal waves cannot be polarized, and 14.1.A.5.i models sound waves as mechanical longitudinal waves. In a longitudinal wave the disturbance is parallel to the direction of propagation, which is a single direction, so there is nothing to restrict. In a transverse wave the disturbance is perpendicular to the direction of propagation and could point anywhere in that perpendicular plane, so 14.3.A.2.i says it can be polarized and confined to a single plane.

What can polarize a wave according to the AP Physics 2 CED?

Essential knowledge 14.3.A.2 names three situations: transverse waves that are reflected from a surface, refracted through a medium, or pass through specific openings may be polarized. Two of those, reflection and refraction, are the mechanisms of Unit 13, which is why polarization shares a topic with boundary behavior. Note the word may. The CED says these processes can polarize a transverse wave, not that they always fully do.

Is Malus's law on the AP Physics 2 exam?

It does not appear in the course framework. No essential knowledge statement in Topic 14.3 relates the intensity transmitted by a polarizing filter to the angle between the filter and the incoming polarization, and no intensity equation of any kind is printed among the 129 equations on the AP Physics 2 equation sheet. What the CED gives is 14.3.A.3, that polarization of a wave may result in a reduction of the wave's intensity, plus the definition of intensity itself. Any polarizer question wanting a number has to supply the transmitted fraction.

What is intensity in AP Physics 2?

Essential knowledge 14.3.A.3.i defines intensity as a measure of the amount of power transferred per unit area, and 14.3.A.3.ii sharpens that to the average power per unit area over one period of the wave. The averaging matters because a wave's power varies through its cycle. Intensity is measured in watts per square metre. The AP Physics 2 CED gives the definition in words only and labels neither statement a relevant equation, and no intensity formula appears on the equation sheet.