Constructive vs Destructive Interference

Q: What is the difference between constructive and destructive interference?

The relative direction of the two displacements where the waves overlap. AP Physics 2 essential knowledge 14.6.A.4.i says that when the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference, and 14.6.A.4.ii says that when they are in opposite directions it is called destructive interference. Both are the same operation underneath, the superposition of 14.6.A.3, which says the resulting displacement is found by adding the individual displacements. Constructive gives a larger result than either wave alone; destructive gives a smaller one.

Q: What path length difference gives constructive interference?

For two sources emitting the same wavelength in step, a whole number of wavelengths. If one wave has travelled an extra one, two or three full wavelengths, it arrives at the same point in its cycle as the other, so the two displacements point the same way and the interference is constructive. A path length difference of a half-integer number of wavelengths, such as half, one and a half, or two and a half, puts the waves at opposite points in their cycles and gives destructive interference. AP Physics 2 essential knowledge 14.7.A.4.ii and 14.8.A.1.iii both state that the amount of interference between two wavefronts depends on the path length difference. Two conditions are hidden in this: the sources must emit the same wavelength, and they must start in step, because a reflection off a higher-index medium introduces a 180-degree phase change under 14.9.A.2.i that swaps the two lists.

Q: Does destructive interference destroy the waves?

No. It reduces the displacement at a particular place at a particular time, and nothing more. AP Physics 2 essential knowledge 14.6.A.2 states that when two or more wave pulses or waves interact with each other, they travel through each other and overlap rather than bouncing off each other, so two pulses that momentarily flatten a rope emerge afterwards with their original heights, still moving in their original directions. Complete cancellation also requires the two amplitudes to be equal: a large pulse meeting a small inverted one leaves a visible remainder.

Q: Why does the same equation give dark bands for a single slit and bright bands for a double slit?

Because the two relations describe different features despite having the same right-hand side. AP Physics 2 essential knowledge 14.7.A.4.iv writes the single-slit relation with y-min, the distance from the middle of the central bright fringe to the m-th order of minimum brightness. Essential knowledge 14.8.A.1.v writes the double-slit relation with y-max, the distance to the m-th order of maximum brightness. Both come out as slit dimension times y over L is approximately m times lambda, and both use the small-angle approximation for angles under 10 degrees, but one locates a dark band and the other a bright one. The other trap in the same pair of equations is that the single-slit letter a is the width of one opening while the double-slit letter d is the separation between two.

Q: Is interference a property of a wave or of a place?

Of a place, and of a moment. The definitions in AP Physics 2 essential knowledge 14.6.A.4.i and 14.6.A.4.ii both describe the displacements where the waves are superposed, so the answer changes as you move across a screen and, for waves of slightly different frequency, as you wait. Essential knowledge 14.6.A.6.i makes that explicit for beats: waves with different frequencies are sometimes in phase and sometimes out of phase at locations along the waves, causing periodic amplitude changes in the resultant wave. A question asking whether interference is constructive or destructive has to specify a location, and often a time.

Q: How do antireflection coatings use destructive interference?

They arrange for the two reflected waves to cancel. AP Physics 2 essential knowledge 14.9.A.5.ii says antireflection coatings eliminate reflected light by applying the relationships between indices of refraction, phase shift, and wave interference to create destructive interference of the light reflected from the two surfaces of the coating. 14.9.A.5.iii gives the simplest design: a thickness equal to one quarter of the wavelength of the light in the coating, with the coating's index greater than that of air and less than that of the surface it covers, and it states that this assumes incident light is normal to the surface. Both reflections then undergo the same 180-degree phase change from 14.9.A.2.i, so the phase shifts cancel, and the round trip through the coating adds exactly half a wavelength of path, which is the destructive condition. Topic 14.9's boundary statement limits quantitative analysis of thin-film interference to waves normal to the incident surface.

Q: Are constructive and destructive interference the only two possibilities?

They are the two ends of a range rather than a pair of switches. AP Physics 2 essential knowledge 14.7.A.4.ii and 14.8.A.1.iii both say the amount of interference depends on the path length difference, and between a maximum and the next minimum on a screen the path difference passes through every value in between, so the brightness slides continuously from full to zero. Essential knowledge 14.6.A.4.iii covers a related case: two or more traveling wave pulses or waves can interact in such a way as to produce amplitude variations in the resultant wave pulse or wave, which is what beats are.

Both are superposition, adding the displacements of overlapping waves. It is constructive where the two displacements point the same way, so the result is larger, and destructive where they point opposite ways, so the result is smaller. The path length difference decides which.

AP Physics: Unit 14 (topics 14.6 Wave Interference and Standing Waves, 14.7 Diffraction, 14.8 Double-Slit Interference and Diffraction Gratings, 14.9 Thin-Film Interference). The definitions are AP Physics 2 essential knowledge 14.6.A.4.i (when the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference) and 14.6.A.4.ii (when they are in opposite directions, the interaction is called destructive interference), sitting under 14.6.A.4 and resting on 14.6.A.3, superposition. 14.6.A.2 is what makes the pair non-destructive: interacting pulses travel through each other and overlap rather than bouncing off each other. The quantitative side is filed in two later topics that phrase the control identically: 14.7.A.4.ii and 14.8.A.1.iii both make the amount of interference depend on the path length difference. The two small-angle relations differ in what they locate, 14.7.A.4.iv giving the distance to the m-th order of minimum brightness for a single slit of width a and 14.8.A.1.v the distance to the m-th order of maximum brightness for two slits a distance d apart, both restricted to angles under 10 degrees. Thin films add the phase rules 14.9.A.2.i and 14.9.A.2.ii and the quarter-wavelength antireflection coating of 14.9.A.5.iii. Topics 14.6, 14.7 and 14.8 print no boundary statement; Topic 14.9's reads that quantitative analysis of thin-film interference is limited to waves that are normal to the incident surface. Unit 14 carries 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

The distinction, stated once

There is only one rule underneath both words, and AP Physics 2 essential knowledge 14.6.A.3 states it: when two or more wave pulses or waves overlap, the resulting displacement can be determined by adding the individual displacements. That is superposition, and it is arithmetic, not a choice between two mechanisms.

The two names are what you call the answer depending on the signs of the terms. 14.6.A.4 says wave interference may be constructive or destructive, and the two statements under it are deliberately parallel:

14.6.A.4.i: when the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference.
14.6.A.4.ii: when the displacements of the superposed wave pulses or waves are in opposite directions, the interaction is called destructive interference.

Notice what the CED does not make the definition. It does not say constructive means the waves add and destructive means they cancel, and it does not define either one by a phase difference or a path difference. Those are consequences. The definition is the relative direction of the two displacements at the point and instant you are looking at.

That matters because it makes the labels local and momentary. The same pair of waves can interfere constructively at one place on a screen and destructively a millimetre away, and two pulses on a rope can be in one condition as they meet and the other as they pass. Nothing is a property of a wave; everything is a property of a location.

Constructive against destructive, row by row

Property	Constructive	Destructive
CED condition	displacements in the same direction (14.6.A.4.i)	displacements in opposite directions (14.6.A.4.ii)
Resulting displacement	sum of the two, larger than either alone	difference of the two, smaller than the larger one alone
Complete version	amplitudes add exactly when the waves are in phase	full cancellation only if the two amplitudes are also equal
Path length difference for two identical in-phase sources	a whole number of wavelengths	a half-integer number of wavelengths
On a screen with light	a bright band	a dark band
With sound	a loud spot	a quiet spot
In a standing wave	at an antinode, always at maximum amplitude	at a node, always zero (14.6.B.1.i)
In beats	the loud moments	the quiet moments (14.6.A.6)
In a thin film	one film thickness	a different film thickness, which is how antireflection coatings work (14.9.A.5.ii)
Governed by	the path length difference $\Delta D$ (14.7.A.4.ii and 14.8.A.1.iii)	the same $\Delta D$

The last row is the one to hold on to. Both topics that quantify interference say the same thing about what controls it: 14.7.A.4.ii for a single opening and 14.8.A.1.iii for two, in identical words, that the amount of interference between two wavefronts depends on the path length difference $\Delta D$ of the wavefronts. So constructive and destructive are not two rules to learn. They are two values of one variable.

The third row is the one that gets skipped. Destructive interference reduces the displacement; it only removes it if the two waves are the same size. A $3\ \text{cm}$ pulse meeting an inverted $1\ \text{cm}$ pulse leaves $2\ \text{cm}$ , which is destructive interference with a visible result.

The case that separates them: two pulses on one rope

Send a pulse from each end of a long rope at the same speed, with the same width and the same height, one upward and one downward. Watch what happens as they meet.

At the instant of exact overlap the rope is straight. Every point has one wave pushing it up by some amount and the other pushing it down by the same amount, so 14.6.A.3's addition gives zero everywhere. If the two pulses had both been upward, the same instant would instead show a single pulse twice as tall.

Same rope, same two pulses, same arithmetic, opposite result. The only thing that changed was the sign of one of the terms, which is exactly what 14.6.A.4.i and 14.6.A.4.ii distinguish.

Now watch a moment longer, because this is where the page earns its place. The pulses come out the other side unchanged. 14.6.A.2 is explicit: when two or more wave pulses or waves interact with each other, they travel through each other and overlap rather than bouncing off each other. So the flat rope was not the end of either pulse. It was one instant in the middle of both of them.

That rules out the reading most students carry into the topic, which is that destructive interference destroys a wave. It does not. At the instant the rope looks flat, no point is displaced but every point is moving, and a moment later the two pulses reappear with their original heights, still traveling in their original directions.

So the pairing to remember is not "constructive builds and destructive destroys". It is: constructive and destructive are two readings of the same sum, taken at one place at one time, and neither one does anything permanent to the waves involved.

How the path length difference decides which you get

For two sources emitting the same wavelength in step, the question at any point is how much further one wave travelled than the other. That is the path length difference, written $\Delta D$ , and the AP Physics 2 equation sheet prints exactly three lines that name it: $\Delta D = m\lambda$ , $\Delta D = a\sin\theta$ for a single opening of width $a$ , and $\Delta D = d\sin\theta$ for two openings a distance $d$ apart. Two further printed lines, the small-angle relations in the next section, work with the same idea without naming $\Delta D$ .

The reasoning is short. If one wave has travelled a whole extra number of wavelengths, it arrives at the same point in its cycle as the other, both displace the medium the same way, and 14.6.A.4.i applies. If it has travelled a whole number of wavelengths plus a half, it arrives at the opposite point in its cycle, the displacements oppose, and 14.6.A.4.ii applies.

$\Delta D = 0, \lambda, 2\lambda, 3\lambda, \ldots$ gives constructive interference.
$\Delta D = 0.5\lambda, 1.5\lambda, 2.5\lambda, \ldots$ gives destructive interference.

Two conditions are baked into that and both get tested. The sources have to emit the same wavelength, or there is no fixed relationship to maintain; 14.6.A.6 covers the case where they differ slightly, and the result is beats rather than a fixed pattern. And the sources have to start in step, because a source that emits half a cycle behind the other swaps the two lists above completely.

That second condition is not hypothetical. Reflection can supply exactly such a half-cycle shift. 14.9.A.2.i says 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, and 14.9.A.2.ii says no phase change occurs when it reflects from a medium with a lower index. 14.9.A.3 adds that the phase of a wave does not change when it is refracted as it passes from one medium into another. Count the reflections, count the phase flips, and only then apply the path-difference lists.

When it costs a mark

Using $\Delta D = m\lambda$ for maxima in every geometry. This is the single most expensive error on this topic, because the same right-hand side means opposite things in the two places the CED prints it. 14.7.A.4.iv relates $\lambda$ , $a$ and $L$ to $y_{\text{min}}$ , the distance from the middle of the central bright fringe to the $m$ th order of minimum brightness, through $a(y_{\text{min}}/L) \approx m\lambda$ . 14.8.A.1.v relates $\lambda$ , $d$ and $L$ to $y_{\text{max}}$ , the distance to the $m$ th order of maximum brightness, through $d(y_{\text{max}}/L) \approx m\lambda$ . Single slit, that is a dark band. Double slit, that is a bright band. Check the subscript on the sheet before substituting.
Reading $a$ and $d$ as the same quantity. In 14.7 the letter $a$ is the width of one opening. In 14.8 the letter $d$ is the separation between two openings. They appear in identically shaped equations and they are different lengths of the same apparatus.
Saying destructive interference cancels the waves. It reduces the displacement at that point at that time. Full cancellation needs equal amplitudes, and 14.6.A.2 says the waves travel through each other regardless.
Treating interference as a property of a wave rather than of a place. A double-slit screen shows both conditions at once, alternating across it. Asking whether the interference is constructive or destructive without naming a location has no answer.
Forgetting the reflection phase flip in a thin film. 14.9.A.2.i's 180-degree change applies when the reflection is off a higher-index medium, so a film with air on one side and glass on the other can flip one reflection and not the other, which swaps the thicknesses that produce a bright film.
Ignoring the boundary statement on thin films. Topic 14.9's boundary statement reads: quantitative analysis of thin-film interference is limited to waves that are normal to the incident surface. 14.9.A.5.iii repeats the condition for the quarter-wavelength coating in its own last sentence, which assumes incident light is normal to the surface. An angled ray is a qualitative question only.
Assuming the pattern is uniform. 14.8.A.1.i says that when only considering wave interference, a double slit creates a pattern of uniformly spaced maxima, but 14.8.A.1.vi then says that when wave diffraction is considered too, the double slit creates maxima and minima superimposed within the envelope created by single-slit diffraction. The spacing is uniform; the brightness is not.

Where each one shows up in Unit 14

Both conditions appear in every interference topic the unit covers, which is why the pair is worth separating once rather than relearning per topic.

Topic	Constructive shows up as	Destructive shows up as
14.6 interference and standing waves	antinodes, and the loud moments of a beat	nodes, and the quiet moments of a beat
14.7 diffraction	the central bright fringe and the bright bands between minima	the minima located by $a(y_{\text{min}}/L) \approx m\lambda$
14.8 double slit and gratings	the maxima located by $d(y_{\text{max}}/L) \approx m\lambda$	the dark bands between them
14.9 thin films	the colours seen in soap and oil films (14.9.A.5.i)	the suppressed reflection from an antireflection coating (14.9.A.5.ii)

Two entries in that table are worth reading closely.

14.7.A.4.i says constructive and destructive interference of multiple wavefronts originating from the opening will result in bright and dark bands on the screen, so a single slit produces both, not just the minima the printed equation locates. And 14.9.A.5.ii describes antireflection coatings as eliminating reflected light by applying the relationships between indices of refraction, phase shift, and wave interference to create destructive interference of the light reflected from the two surfaces of the coating. That is destructive interference used as a design tool, which is the clearest sign that it is not a failure mode.

When they coincide, and why that lulls you

They never coincide, exactly, but they blur, and the blurring is what makes the pair feel harder than it is.

Most points are neither cleanly one nor the other. Between a maximum and the next minimum on a double-slit screen the path difference passes through every value in between, so the brightness slides continuously from full to zero. 14.7.A.4.ii and 14.8.A.1.iii both phrase it as the amount of interference depending on $\Delta D$ , not as a switch between two states. Constructive and destructive name the two ends of a continuum.

The second blur is in the wording of a question. "Are the waves in phase?" and "is the interference constructive?" are the same question only when the two sources started in step and no reflection has flipped one of them. Add a phase change from 14.9.A.2.i and they come apart, and a thin-film question is built on exactly that gap.

The third is the one worth being most careful about. Both conditions are momentary. 14.6.A.6.i describes beats as waves with different frequencies that are sometimes in phase and sometimes out of phase at locations along the waves, causing periodic amplitude changes in the resultant wave. So at a fixed listening position the interference is constructive some of the time and destructive the rest of it, several times a second, with nothing about either wave changing. A question that asks which condition holds has to tell you where and when, and if it does not, that is the question.

Two speakers, one listener, and the path difference that decides

Two speakers driven in step emit a pure $686\ \text{Hz}$ tone into air where the speed of sound is $343\ \text{m/s}$ . (a) Find the wavelength. (b) A listener stands $4.00\ \text{m}$ from one speaker and $4.75\ \text{m}$ from the other. Is the interference at that point constructive or destructive? (c) The listener moves to a point $4.00\ \text{m}$ from the first speaker and $5.00\ \text{m}$ from the second. Answer again. (d) What is the smallest change in path length difference that would take the listener from a loud spot to a quiet one?

(a) The printed relationship is $\lambda = \dfrac{v}{f} = \dfrac{343\ \text{m/s}}{686\ \text{Hz}} = 0.500\ \text{m}$ .
(b) The path length difference is $\Delta D = 4.75 - 4.00 = 0.75\ \text{m}$ . In wavelengths, $\dfrac{0.75}{0.500} = 1.5$ .
That is a half-integer number of wavelengths, so one wave arrives at the opposite point of its cycle from the other. The two displacements oppose, and by 14.6.A.4.ii the interference is destructive. Because the two speakers are identical, the amplitudes arriving are close to equal and the listener hears a marked quiet spot.
(c) Now $\Delta D = 5.00 - 4.00 = 1.00\ \text{m}$ , which is $\dfrac{1.00}{0.500} = 2.0$ wavelengths, a whole number. The waves arrive at the same point in their cycles, the displacements are in the same direction, and by 14.6.A.4.i the interference is constructive. The listener hears a loud spot.
(d) Moving from a whole number of wavelengths to a half-integer number takes half a wavelength, so $\Delta D$ must change by $\lambda/2 = 0.250\ \text{m}$ .
Check the direction of the reasoning against 14.8.A.1.iii, that the amount of interference between two wavefronts depends on the path length difference. Nothing here needed the speaker separation, the distance to a screen, or an angle: only the two path lengths and the wavelength.
One condition to state explicitly, because it is doing work: the speakers are driven in step. If one were wired half a cycle behind the other, the two answers in (b) and (c) would swap.

(a) $\lambda = 0.500\ \text{m}$ . (b) $\Delta D = 0.75\ \text{m} = 1.5\lambda$ , so destructive interference and a quiet spot. (c) $\Delta D = 1.00\ \text{m} = 2.0\lambda$ , so constructive interference and a loud spot. (d) A change of $0.250\ \text{m}$ , one half wavelength.

The same arithmetic, opposite brightness: single slit against double slit

Monochromatic light of wavelength $550\ \text{nm}$ falls on a screen $L = 2.00\ \text{m}$ away. In experiment A it first passes through a single slit of width $a = 0.100\ \text{mm}$ . In experiment B it passes through two slits separated by $d = 0.100\ \text{mm}$ . For each, find the distance from the middle of the central bright fringe to the first-order feature the sheet's small-angle relation locates, and say whether that feature is bright or dark.

Convert first. $\lambda = 550\ \text{nm} = 5.50 \times 10^{-7}\ \text{m}$ , and $a = d = 0.100\ \text{mm} = 1.00 \times 10^{-4}\ \text{m}$ .
Experiment A, single slit. 14.7.A.4.iv gives $a\left(\dfrac{y_{\text{min}}}{L}\right) \approx m\lambda$ , where $y_{\text{min}}$ is the distance from the middle of the central bright fringe to the $m$ th order of minimum brightness. Rearranging, $y_{\text{min}} \approx \dfrac{m\lambda L}{a}$ .
For $m = 1$ : $y_{\text{min}} \approx \dfrac{(1)(5.50 \times 10^{-7})(2.00)}{1.00 \times 10^{-4}} = \dfrac{1.100 \times 10^{-6}}{1.00 \times 10^{-4}} = 1.10 \times 10^{-2}\ \text{m}$ , which is $11.0\ \text{mm}$ . That is a dark band.
Experiment B, double slit. 14.8.A.1.v gives $d\left(\dfrac{y_{\text{max}}}{L}\right) \approx m\lambda$ , where $y_{\text{max}}$ is the distance to the $m$ th order of maximum brightness. Rearranging, $y_{\text{max}} \approx \dfrac{m\lambda L}{d}$ .
For $m = 1$ , with $d$ numerically equal to $a$ : $y_{\text{max}} \approx 1.10 \times 10^{-2}\ \text{m}$ , the same $11.0\ \text{mm}$ . That is a bright band.
Check that the small-angle condition holds, since 14.7.A.4.iv and 14.8.A.1.v both require it. $\tan\theta = 11.0/2000 = 5.50 \times 10^{-3}$ , so $\theta = 0.315$ degrees, comfortably inside the $\theta < 10$ degrees the CED states.
Note where $m = 0$ sits. In experiment B, $m = 0$ gives $y_{\text{max}} = 0$ , the central maximum. In experiment A, the middle of the pattern is the central bright fringe by 14.7.A.4.iv's own wording, so the minima start at $m = 1$ ; substituting $m = 0$ into the single-slit relation does not locate a dark band.
The physical difference behind the identical arithmetic: in A the $11.0\ \text{mm}$ point is where wavefronts from across one opening arrive with path differences that pair up destructively, while in B it is where the two wavefronts differ by exactly one wavelength.

Both come out at $11.0\ \text{mm}$ from the centre. For the single slit that point is a dark band, the first-order minimum. For the double slit it is a bright band, the first-order maximum. Identical numbers, opposite brightness, and the only way to tell them apart is which subscript the sheet's relation carries.

Designing destructive interference: an antireflection coating

A glass lens is to be coated so that reflected light of vacuum wavelength $550\ \text{nm}$ is suppressed. The coating has an index of refraction of $1.38$ , which is greater than that of air and less than that of the glass. Light arrives normal to the surface. (a) Find the wavelength of the light inside the coating. (b) Find the thickness the simplest coating should have. (c) Show that the two reflected waves interfere destructively.

(a) The printed relationship $n = c/v$ and the printed $\lambda = v/f$ combine to give the wavelength in a medium as the vacuum wavelength divided by the index, because 14.3.A.1.iv keeps the frequency unchanged when the light enters the coating while the speed drops by a factor of $n$ .
So $\lambda_{\text{coating}} = \dfrac{550\ \text{nm}}{1.38} = 398.55\ \text{nm}$ , which is $399\ \text{nm}$ to three significant figures.
(b) 14.9.A.5.iii states that the simplest antireflection coating has a thickness equal to one quarter of the wavelength of the light in the coating. So $t = \dfrac{398.55}{4} = 99.64\ \text{nm}$ , which is $99.6\ \text{nm}$ .
(c) Count the phase changes with 14.9.A.2. At the air-to-coating surface the light reflects off a medium of greater index, $1.38$ against about $1.00$ , so 14.9.A.2.i gives a 180-degree phase change. At the coating-to-glass surface the light again reflects off a medium of greater index, since the problem states the glass index exceeds $1.38$ , so 14.9.A.2.i gives a second 180-degree change.
Two identical phase changes shift both reflected waves by the same amount, so they contribute nothing to the difference between them. 14.9.A.3 confirms that the refraction on the way in and out changes no phase.
That leaves the extra distance travelled by the second wave, which goes down through the coating and back: $2t = 2(99.64) = 199.3\ \text{nm}$ . Compare it with the wavelength inside the coating: $\dfrac{199.3}{398.55} = 0.500$ , exactly half a wavelength.
A path difference of half a wavelength puts the two reflected waves' displacements in opposite directions, so by 14.6.A.4.ii the interference is destructive and the reflection is suppressed. This is 14.9.A.5.ii's mechanism stated in numbers.
State the limits. Topic 14.9's boundary statement restricts quantitative analysis of thin-film interference to waves that are normal to the incident surface, and 14.9.A.5.iii carries the same assumption. Tilt the incoming light and the extra path is no longer $2t$ , and the calculation above stops being available.

(a) $\lambda_{\text{coating}} = 399\ \text{nm}$ . (b) $t = 99.6\ \text{nm}$ , one quarter of that wavelength. (c) Both reflections undergo a 180-degree phase change, so they cancel out of the comparison, and the round trip through the coating adds exactly half a wavelength of path. Half a wavelength puts the displacements in opposite directions, which is destructive interference by 14.6.A.4.ii.

Frequently asked questions

What is the difference between constructive and destructive interference?

The relative direction of the two displacements where the waves overlap. AP Physics 2 essential knowledge 14.6.A.4.i says that when the displacements of the superposed wave pulses or waves are in the same direction, the interaction is called constructive interference, and 14.6.A.4.ii says that when they are in opposite directions it is called destructive interference. Both are the same operation underneath, the superposition of 14.6.A.3, which says the resulting displacement is found by adding the individual displacements. Constructive gives a larger result than either wave alone; destructive gives a smaller one.

What path length difference gives constructive interference?

For two sources emitting the same wavelength in step, a whole number of wavelengths. If one wave has travelled an extra one, two or three full wavelengths, it arrives at the same point in its cycle as the other, so the two displacements point the same way and the interference is constructive. A path length difference of a half-integer number of wavelengths, such as half, one and a half, or two and a half, puts the waves at opposite points in their cycles and gives destructive interference. AP Physics 2 essential knowledge 14.7.A.4.ii and 14.8.A.1.iii both state that the amount of interference between two wavefronts depends on the path length difference. Two conditions are hidden in this: the sources must emit the same wavelength, and they must start in step, because a reflection off a higher-index medium introduces a 180-degree phase change under 14.9.A.2.i that swaps the two lists.

Does destructive interference destroy the waves?

No. It reduces the displacement at a particular place at a particular time, and nothing more. AP Physics 2 essential knowledge 14.6.A.2 states that when two or more wave pulses or waves interact with each other, they travel through each other and overlap rather than bouncing off each other, so two pulses that momentarily flatten a rope emerge afterwards with their original heights, still moving in their original directions. Complete cancellation also requires the two amplitudes to be equal: a large pulse meeting a small inverted one leaves a visible remainder.

Why does the same equation give dark bands for a single slit and bright bands for a double slit?

Because the two relations describe different features despite having the same right-hand side. AP Physics 2 essential knowledge 14.7.A.4.iv writes the single-slit relation with y-min, the distance from the middle of the central bright fringe to the m-th order of minimum brightness. Essential knowledge 14.8.A.1.v writes the double-slit relation with y-max, the distance to the m-th order of maximum brightness. Both come out as slit dimension times y over L is approximately m times lambda, and both use the small-angle approximation for angles under 10 degrees, but one locates a dark band and the other a bright one. The other trap in the same pair of equations is that the single-slit letter a is the width of one opening while the double-slit letter d is the separation between two.

Is interference a property of a wave or of a place?

Of a place, and of a moment. The definitions in AP Physics 2 essential knowledge 14.6.A.4.i and 14.6.A.4.ii both describe the displacements where the waves are superposed, so the answer changes as you move across a screen and, for waves of slightly different frequency, as you wait. Essential knowledge 14.6.A.6.i makes that explicit for beats: waves with different frequencies are sometimes in phase and sometimes out of phase at locations along the waves, causing periodic amplitude changes in the resultant wave. A question asking whether interference is constructive or destructive has to specify a location, and often a time.

How do antireflection coatings use destructive interference?

They arrange for the two reflected waves to cancel. AP Physics 2 essential knowledge 14.9.A.5.ii says antireflection coatings eliminate reflected light by applying the relationships between indices of refraction, phase shift, and wave interference to create destructive interference of the light reflected from the two surfaces of the coating. 14.9.A.5.iii gives the simplest design: a thickness equal to one quarter of the wavelength of the light in the coating, with the coating's index greater than that of air and less than that of the surface it covers, and it states that this assumes incident light is normal to the surface. Both reflections then undergo the same 180-degree phase change from 14.9.A.2.i, so the phase shifts cancel, and the round trip through the coating adds exactly half a wavelength of path, which is the destructive condition. Topic 14.9's boundary statement limits quantitative analysis of thin-film interference to waves normal to the incident surface.

Are constructive and destructive interference the only two possibilities?

They are the two ends of a range rather than a pair of switches. AP Physics 2 essential knowledge 14.7.A.4.ii and 14.8.A.1.iii both say the amount of interference depends on the path length difference, and between a maximum and the next minimum on a screen the path difference passes through every value in between, so the brightness slides continuously from full to zero. Essential knowledge 14.6.A.4.iii covers a related case: two or more traveling wave pulses or waves can interact in such a way as to produce amplitude variations in the resultant wave pulse or wave, which is what beats are.