Interference vs Diffraction: What Is the Difference?

Diffraction is what a wave does when it meets an obstacle or an opening: it spreads. Interference is what happens where two or more waves overlap: their displacements add. They are not rival explanations. Every diffraction pattern you can see is interference between parts of one wavefront.

AP Physics: Unit 14 (topics 14.6 Wave Interference and Standing Waves, 14.7 Diffraction, 14.8 Double-Slit Interference and Diffraction Gratings). The two definitions are AP Physics 2 essential knowledge 14.7.A.1 (diffraction is the spreading of a wave around the edges of an obstacle or through an opening) and 14.6.A.1 (wave interference is the interaction of two or more wave pulses or waves), with 14.6.A.3 supplying superposition as the arithmetic behind the second. The link between them is 14.7.A.3, that diffraction of multiple wavefronts through a single opening leads to observable interference patterns, and 14.7.A.4.i, that constructive and destructive interference of those wavefronts produces the bright and dark bands. 14.7.A.2 makes the amount of diffraction depend on the ratio of opening size to wavelength, while 14.7.A.4.ii and 14.8.A.1.iii both make the amount of interference depend on the path length difference. The double slit carries both effects by 14.8.A.1, and 14.8.A.1.vi places the interference maxima and minima inside an envelope created by single-slit diffraction. 14.7.A.5 adds that the pattern depends on the shape of the opening. Topics 14.6, 14.7 and 14.8 print no boundary statement. The equation sheet's exam conventions box calls the two-slit case diffraction, and 14.8.A.4 calls a grating's output an interference pattern made of superimposed diffraction patterns, so the naming in the source material is not consistent. Unit 14 carries 12 to 15 percent of the multiple-choice section across about 14 to 23 class periods. This material is unique to AP Physics 2 among the four courses: the words interference, diffraction, standing wave and antinode appear nowhere in the AP Physics 1, C: Mechanics or C: E&M frameworks.

The distinction, stated once

The two words answer different questions, and once you see which question each one answers the pair stops competing.

Diffraction answers "what did the wave do when it met something?" AP Physics 2 essential knowledge 14.7.A.1: diffraction is the spreading of a wave around the edges of an obstacle or through an opening. That is a statement about propagation. One wave, one barrier, and a shape of wavefront on the far side that is wider than the gap it came through.

Interference answers "what happens where waves land on top of each other?" Essential knowledge 14.6.A.1: wave interference is the interaction of two or more wave pulses or waves. That is a statement about superposition, and 14.6.A.3 supplies the arithmetic: when two or more wave pulses or waves overlap, the resulting displacement can be determined by adding the individual displacements.

So one is a cause and the other is an effect, not two competing causes. Diffraction puts wave energy where geometry says it should not be. Interference decides what happens once several routes deliver waves to the same place.

The CED then makes the link between them explicit, in a sentence students usually skim. Essential knowledge 14.7.A.3: diffraction of multiple wavefronts through a single opening leads to observable interference patterns. And 14.7.A.4.i spells out the consequence for the screen: constructive and destructive interference of multiple wavefronts originating from the opening will result in bright and dark bands on the screen.

Read those two statements carefully. A single slit produces a banded pattern, and the CED attributes those bands to interference, not to diffraction on its own. Diffraction is what turns one opening into a source of many wavefronts; interference between those wavefronts is what puts the bands where they are. The word "diffraction pattern" names the pattern by its cause and describes it by its mechanism, and the mechanism is interference.

That is why the honest one-line answer to the title question is not "they are opposites". It is: diffraction is why there is anything to interfere, in the geometries where a single opening is the only source you have.

Side by side

Property	Diffraction	Interference
CED definition	Spreading of a wave around the edges of an obstacle or through an opening (14.7.A.1)	The interaction of two or more wave pulses or waves (14.6.A.1)
What kind of statement it is	About propagation, how a wave travels	About superposition, what overlapping waves do
Needs an obstacle or opening	Yes, by definition	No
Needs two or more waves	No, one wave is enough	Yes, at least two contributions
Underlying rule	None printed; it is a described behaviour	Adding displacements, 14.6.A.3
What controls how much you get	The ratio of wavelength to opening size (14.7.A.2)	The path length difference $\Delta D$ (14.7.A.4.ii and 14.8.A.1.iii)
Visible result on its own	Wave energy where a shadow was expected	Bright and dark bands, loud and quiet spots
Can it occur without the other	Yes, when the pattern is too fine or too broad to resolve	Yes, whenever two separate sources overlap in open space
AP Physics 2 topic	14.7 Diffraction	14.6 Wave Interference and Standing Waves
Printed equations that name it	$\Delta D = a\sin\theta$ and $a(y_{\text{min}}/L) \approx m\lambda$	$\Delta D = m\lambda$ , and the double-slit pair

Three rows repay a second look.

"Needs two or more waves" is the cleanest test. Ask how many wave sources the situation contains. Two loudspeakers in an open field is interference with nothing in the way, so no diffraction is involved. One water wave meeting a harbour wall is diffraction, and whether you see interference afterwards depends on whether the resulting pattern is fine enough to resolve.

"What controls how much you get" is where the two are genuinely independent. Diffraction is governed by a size comparison, 14.7.A.2: it is most pronounced when the size of the opening is comparable to the wavelength of the wave. Interference is governed by a distance comparison, the path length difference, which both slit topics phrase identically. Those are different variables, and a question can change one without touching the other.

"Can it occur without the other" is the row that stops the pair collapsing into one word. The middle two examples of this page are built on it: an interference pattern with no obstacle anywhere, and a diffraction so slight that no pattern is visible at all.

Why every diffraction pattern is an interference pattern

Here is the piece that resolves the confusion rather than restating it.

A single slit has one wave going into it. That looks like it should rule interference out, since 14.6.A.1 requires two or more waves. The resolution is in 14.7.A.3's plural: wavefronts. Once the wave has spread through the opening, different parts of the same opening act as separate starting points, and light arriving at a point on the screen from the top of the slit has travelled a different distance than light from the bottom. Two different path lengths to one point is exactly the condition interference needs, so the wave interferes with itself.

14.7.A.4.ii states what then decides the outcome: the amount of interference between two wavefronts depends on the path length difference $\Delta D$ of the wavefronts. 14.8.A.1.iii says the same sentence about two slits. The CED uses identical wording for the one-opening case and the two-opening case, which is the strongest hint available that it regards them as the same physics with different geometry.

So the sequence for a single slit runs:

A wave meets an opening comparable to its wavelength, and diffracts (14.7.A.1, 14.7.A.2).
The opening now behaves as many wavefront sources rather than one (14.7.A.3).
Those wavefronts reach each screen point by different path lengths (14.7.A.4.iii).
They interfere, constructively at some points and destructively at others (14.7.A.4.i).
The bands you see are step 4. The reason there is anything at step 4 is step 1.

Every one of those five steps is a separate CED statement, which is unusual, and it tells you the College Board considers the chain worth building explicitly.

The double slit makes the layering visible. Essential knowledge 14.8.A.1 says the pattern resulting from monochromatic light incident on two slits a distance $d$ apart is caused by a combination of wave diffraction and wave interference. And 14.8.A.1.vi finishes the picture: when considering wave interference and wave diffraction, a double slit creates an interference pattern of maxima and minima superimposed within the envelope created by single-slit diffraction.

Two effects, one screen. The closely spaced fringes come from interference between the two slits; the slow fade in brightness across the screen comes from diffraction at each individual slit. Worked example three finds the place where the two collide and a fringe disappears.

The names get looser still one topic later. 14.8.A.4 describes a diffraction grating as a collection of evenly spaced parallel slits or openings that produce an interference pattern that is the combination of numerous diffraction patterns superimposed on each other. And the exam conventions box on the AP Physics 2 equation sheet states that the small angle approximation is valid for single- and double-slit diffraction, using the word diffraction for the two-slit experiment whose own topic title calls it interference. The vocabulary in the source material is not consistent, so do not try to make your answer hinge on which word a question used. Hinge it on the geometry.

The case that separates them: two situations, one word each

Interference with no diffraction at all. Put two radio antennas eight metres apart on an open field and drive them from one transmitter. There is no obstacle, no opening, and nothing for a wave to spread around, so 14.7.A.1 has no purchase on the situation. Yet a receiver walked in a circle around the pair finds strong signal in some directions and almost none in others, because the two paths differ by a whole number of wavelengths in some directions and by a half-integer number in others. That is pure interference, and worked example two counts the directions.

The same is true of two loudspeakers in a room, of two water-wave dippers in a tank, and of two pulses sent from opposite ends of a rope. If you can name two sources, you have interference whether or not anything is in the way. Constructive against destructive interference works that side of the pair in detail.

Diffraction with no visible interference. Now send sound and light through the same open doorway, $0.90$ metres wide. Both are waves, both diffract, and 14.7.A.2 says how much: the effect is most pronounced when the opening is comparable to the wavelength.

For a $343 \ \mathrm{Hz}$ tone the wavelength is a metre, which is larger than the doorway, and worked example one shows that the equation for the first dark direction has no solution at all. The sound simply fills the space beyond, which is why the unit's own essential question asks why you can hear a person around a corner but cannot see them.

For green light the wavelength is about $5.5 \times 10^{-7}$ metres, six million times smaller than the doorway. The light still diffracts, and there is still an interference pattern on the far wall, but worked example one puts its first dark band $3.5 \times 10^{-5}$ degrees off axis. No eye resolves that, so what you observe is a sharp-edged patch of light, and geometry with rays describes it perfectly well. Essential knowledge 13.1.A.1.i licenses exactly that shortcut, allowing light to be modeled as a ray when the wave nature of light can be neglected, and 13.1.A.1.ii names the cases where it cannot: rays are not sufficient to understand the spreading of light, and in interference and diffraction the wave nature of the light is important.

Same physics in both halves of the doorway experiment. The only thing that changed was the ratio of wavelength to opening, and it changed the answer from "spreads into the whole room" to "does not measurably spread at all".

When it costs a mark

Answering "diffraction" to a question about a double slit, or "interference" to one about a single slit. Both are present in both. 14.8.A.1 says the two-slit pattern is caused by a combination of wave diffraction and wave interference, and 14.7.A.4.i attributes the single-slit bands to constructive and destructive interference. Name the geometry and the mechanism, not one word.
Saying a single slit cannot produce interference because there is only one wave. 14.7.A.3 is the answer: diffraction of multiple wavefronts through a single opening leads to observable interference patterns. The opening turns one wave into many wavefronts, and those interfere with each other.
Treating diffraction as something that only light does. 14.7.A.1 says "a wave", with no restriction. Sound diffracts around corners more visibly than light does, and water waves in a ripple tank are the standard demonstration.
Predicting that a narrower slit gives a narrower pattern. It gives a wider one. With $y_{\text{min}} \approx m\lambda L/a$ from 14.7.A.4.iv, halving $a$ doubles every distance in the pattern. 14.7.A.2's "most pronounced when the opening is comparable to the wavelength" points the same way.
Assuming the double-slit fringes are all equally bright. 14.8.A.1.i says that considering interference alone the maxima are uniformly spaced, and 14.8.A.1.vi then places them inside an envelope created by single-slit diffraction. Uniform spacing, non-uniform brightness, and occasionally a fringe missing entirely.
Forgetting that a real slit has a shape. 14.7.A.5: the diffraction pattern produced by a wave passing through an opening depends on the shape of the opening. A rectangular slit and a circular aperture do not give the same figure.
Using the word "diffraction" as if it named an equation. It does not. The printed relations $\Delta D = a\sin\theta$ and $a(y_{\text{min}}/L) \approx m\lambda$ are interference conditions applied to the wavefronts a single opening produces. What diffraction contributes is the existence of those wavefronts off the axis in the first place.
Deciding on the basis of the word a question used. The equation sheet's own conventions box says the small angle approximation is valid for single- and double-slit diffraction, while Topic 14.8 is titled double-slit interference. The source material uses both words for the same apparatus.

Where the CED files each of them

Topic	What it is called	Which effect the CED names in it
14.6 Wave interference and standing waves	Interference	Interference only. No obstacle appears anywhere in the topic.
14.7 Diffraction	Diffraction	Both. Diffraction supplies the wavefronts (14.7.A.3), interference supplies the bands (14.7.A.4.i).
14.8 Double-slit interference and diffraction gratings	Interference in the title, both in the content	Both, stated outright in 14.8.A.1 and located in 14.8.A.1.vi.
14.9 Thin-film interference	Interference	Interference only. Two reflections, no opening.

Two entries in that table settle arguments.

Topic 14.6 contains no obstacle at all. Its interference comes from two pulses meeting on a rope, two sources of slightly different frequency producing beats (14.6.A.6), or a wave meeting its own reflection to make a standing wave (14.6.B.1). None of those involves anything diffracting, which is the cleanest proof that interference is not a special case of diffraction.

Topic 14.9 is the same point in a different setting. A thin film splits one beam into two reflections and recombines them. There is no opening and no edge to spread around, and the result is a bright or dark film. Interference, and only interference.

Going the other way, no topic in the CED contains diffraction without interference, and that is not an oversight. Once a wave has spread through an opening, its own wavefronts overlap downstream, so an interference pattern exists whether or not the apparatus can resolve it. The doorway case in worked example one is a pattern too fine to see, not a pattern that is absent.

When they coincide, and why that lulls you

They coincide in the case almost every exam question is set in, which is the whole difficulty. Slit experiments are the standard apparatus, and slit experiments contain both effects at once, so a student can work them correctly for months without ever needing the distinction and then meet a question that asks for it directly.

The first lull is that the equations look interchangeable. $a(y_{\text{min}}/L) \approx m\lambda$ and $d(y_{\text{max}}/L) \approx m\lambda$ have the same shape and both are conventionally called diffraction equations. They are both statements about path length difference, so they are both interference conditions. Nothing in either equation refers to spreading. Single slit against double slit works the two conditions and the subscripts that separate them.

The second lull is scale. Diffraction is either overwhelming or invisible, rarely in between, because the controlling ratio $\lambda/a$ is usually enormous or tiny. Sound in a room is almost all diffraction; light in a room is almost none. So the effect tends to be either the entire answer or absent from it, and a student who has only met one regime has no reason to suspect the other exists.

The third lull is the vocabulary of the sources themselves. The equation sheet says double-slit diffraction. Topic 14.8 says double-slit interference. 14.8.A.4 calls a grating's output an interference pattern made of superimposed diffraction patterns. If the CED will not commit to one word for one apparatus, do not build an answer on which word appeared in the stem.

What to do instead. Answer with the geometry and the mechanism in the same sentence: how many openings or sources there are, and what the path length difference is doing. "The bands come from interference between wavefronts that diffraction has spread across the whole opening" is a complete answer to a single-slit question and does not depend on guessing which word the examiner had in mind.

One doorway, two waves: why you hear round the corner and cannot see round it

A doorway is $a = 0.90 \ \mathrm{m}$ wide. (a) Find the wavelength of a $343 \ \mathrm{Hz}$ tone in air where the speed of sound is $343 \ \mathrm{m/s}$ , and use $\Delta D = a\sin\theta = m\lambda$ to find the angle of the first minimum. (b) Repeat for green light of wavelength $550 \ \mathrm{nm}$ . (c) Explain the everyday consequence in terms of 14.7.A.2.

(a) The sheet prints $\lambda = v/f$ , so $\lambda = \dfrac{343 \ \mathrm{m/s}}{343 \ \mathrm{Hz}} = 1.00 \ \mathrm{m}$ .
The first minimum of a single-opening pattern is at $m = 1$ , so 14.7.A.4.iii's $\Delta D = a\sin\theta$ combined with $\Delta D = m\lambda$ requires $\sin\theta = \dfrac{m\lambda}{a} = \dfrac{1.00}{0.90} = 1.11$ .
No angle has a sine of $1.11$ . There is no first minimum, so the pattern has no dark direction anywhere in the half-space beyond the doorway: the sound spreads into all of it. Note also that the small-angle form is useless here, since there is no small angle to speak of.
(b) For green light, $\lambda = 550 \ \mathrm{nm} = 5.50 \times 10^{-7} \ \mathrm{m}$ , and $\sin\theta = \dfrac{5.50 \times 10^{-7}}{0.90} = 6.11 \times 10^{-7}$ .
That angle is $\theta = 3.5 \times 10^{-5}$ degrees, which is $0.13$ arcseconds. Nothing in ordinary vision resolves that, so the bright central patch beyond the doorway is the size the shadow geometry predicts, to any precision the eye can check.
(c) 14.7.A.2 states the controlling comparison: diffraction is most pronounced when the size of the opening is comparable to the wavelength of the wave. For the sound, $\lambda/a = 1.11$ , so the wavelength and the opening are the same size and the spreading is total. For the light, $\lambda/a = 6.1 \times 10^{-7}$ , so the opening is millions of wavelengths wide and the spreading is unmeasurable.
Both cases are diffraction, and both produce an interference pattern of bright and dark bands by 14.7.A.4.i. The difference is entirely whether the bands are wide enough to notice. That is Unit 14's own essential question, why you can hear a person around a corner but cannot see them, answered by one ratio.
One caution on the sound result. Real doorways sit in walls of finite thickness, and real rooms reflect, so the full acoustic picture is more than one slit. The calculation shows that diffraction alone is enough to explain the effect, not that nothing else contributes.

(a) $\lambda = 1.00 \ \mathrm{m}$ , and $\sin\theta = 1.11$ has no solution, so there is no minimum and the sound spreads through the whole space beyond. (b) $\theta = 3.5 \times 10^{-5}$ degrees, far too small to see, so the light appears to travel in straight lines. (c) The ratio of wavelength to opening is $1.11$ for the sound and $6.1 \times 10^{-7}$ for the light, which is 14.7.A.2 deciding the outcome.

Interference with nothing in the way: two radio antennas

Two antennas $d = 8.00 \ \mathrm{m}$ apart on open ground are driven in step at $100 \ \mathrm{MHz}$ . There is no obstacle, no opening and no screen. (a) Find the wavelength. (b) Find every direction, measured from the perpendicular bisector of the pair, in which a distant receiver picks up a maximum. (c) Say which of the two effects is present.

(a) Radio waves are electromagnetic, so $v = c = 3.00 \times 10^8 \ \mathrm{m/s}$ from the constants box, and $\lambda = \dfrac{c}{f} = \dfrac{3.00 \times 10^8}{1.00 \times 10^8} = 3.00 \ \mathrm{m}$ .
(b) Two sources a distance $d$ apart give a path length difference $\Delta D = d\sin\theta$ to a distant point, which is 14.8.A.1.iv's relation used for antennas instead of slits. Maxima need $\Delta D = m\lambda$ , so $\sin\theta = \dfrac{m\lambda}{d} = \dfrac{3.00m}{8.00} = 0.375m$ .
$m = 0$ : $\sin\theta = 0$ , so $\theta = 0$ degrees, straight out along the bisector.
$m = 1$ : $\sin\theta = 0.375$ , so $\theta = 22.0$ degrees.
$m = 2$ : $\sin\theta = 0.750$ , so $\theta = 48.6$ degrees.
$m = 3$ : $\sin\theta = 1.125$ , which is impossible, so the list stops. Counting both sides of the bisector, there are five directions of maximum signal in total: one at $0$ degrees and two each at $22.0$ and $48.6$ degrees.
Check the small-angle form is not being misused. $22.0$ and $48.6$ degrees are both far above the $10$ degrees that 14.8.A.1.v requires, so $d(y_{\text{max}}/L) \approx m\lambda$ does not apply here and $\Delta D = d\sin\theta$ is the correct relation to use.
(c) Interference, and nothing else. There is no obstacle and no opening anywhere in the problem, so 14.7.A.1's definition of diffraction has nothing to attach to. Two sources plus superposition (14.6.A.3) is the complete account. This is the cleanest available demonstration that interference is not a special case of diffraction.

(a) $\lambda = 3.00 \ \mathrm{m}$ . (b) Maxima at $0$ , $22.0$ and $48.6$ degrees on each side, five directions in all; $m = 3$ would need a sine greater than one. (c) Pure interference, because nothing in the arrangement diffracts.

Both at once: the double-slit fringe that vanishes

Monochromatic light of $\lambda = 650 \ \mathrm{nm}$ falls on two slits each of width $a = 0.200 \ \mathrm{mm}$ , separated by $d = 0.600 \ \mathrm{mm}$ , with a screen $L = 3.00 \ \mathrm{m}$ away. (a) Find the spacing of the interference maxima. (b) Find the position of the first single-slit minimum. (c) Identify which interference maxima are missing and explain why in terms of 14.8.A.1.vi.

Convert first: $\lambda = 650 \ \mathrm{nm} = 6.50 \times 10^{-7} \ \mathrm{m}$ , $a = 2.00 \times 10^{-4} \ \mathrm{m}$ , $d = 6.00 \times 10^{-4} \ \mathrm{m}$ .
(a) 14.8.A.1.v gives $d\left(\dfrac{y_{\text{max}}}{L}\right) \approx m\lambda$ , so $y_{\text{max}} \approx \dfrac{m\lambda L}{d} = m\dfrac{(6.50 \times 10^{-7})(3.00)}{6.00 \times 10^{-4}} = m(3.25 \times 10^{-3}) \ \mathrm{m}$ .
The maxima therefore sit at $3.25 \ \mathrm{mm}$ intervals: $0$ , $3.25$ , $6.50$ , $9.75$ , $13.0 \ \mathrm{mm}$ and so on. This is 14.8.A.1.i's uniformly spaced maxima.
(b) 14.7.A.4.iv gives $a\left(\dfrac{y_{\text{min}}}{L}\right) \approx m\lambda$ for one slit of width $a$ , so $y_{\text{min}} \approx \dfrac{m\lambda L}{a} = m\dfrac{(6.50 \times 10^{-7})(3.00)}{2.00 \times 10^{-4}} = m(9.75 \times 10^{-3}) \ \mathrm{m}$ .
The first single-slit minimum is at $9.75 \ \mathrm{mm}$ , the second at $19.5 \ \mathrm{mm}$ , the third at $29.25 \ \mathrm{mm}$ .
(c) Compare the two lists. The $m = 3$ interference maximum is predicted at $3(3.25) = 9.75 \ \mathrm{mm}$ , which is exactly where the first diffraction minimum sits. At that point each individual slit is sending no light in that direction at all, so there is nothing for the two slits to interfere constructively with. The third-order fringe is missing.
The pattern repeats with the ratio $d/a = 0.600/0.200 = 3$ , so orders $m = 3, 6, 9$ all coincide with diffraction minima at $9.75$ , $19.5$ and $29.25 \ \mathrm{mm}$ and are all absent.
This is 14.8.A.1.vi made numerical: the double slit creates an interference pattern of maxima and minima superimposed within the envelope created by single-slit diffraction. The interference sets the $3.25 \ \mathrm{mm}$ spacing; the diffraction sets the envelope that dims the outer fringes and zeroes three of them.
Check the angles. $9.75 \ \mathrm{mm}$ over $3.00 \ \mathrm{m}$ gives $\tan\theta = 3.25 \times 10^{-3}$ , so $\theta = 0.186$ degrees, comfortably inside the $10$ degrees both 14.7.A.4.iv and 14.8.A.1.v require.
A note on scope. The AP Physics 2 CED asks for the envelope qualitatively rather than for missing-order arithmetic, so treat this as a check on what 14.8.A.1.vi means rather than as a procedure to reproduce under time pressure.

(a) Maxima every $3.25 \ \mathrm{mm}$ . (b) First diffraction minimum at $9.75 \ \mathrm{mm}$ . (c) Orders $m = 3$ , $6$ and $9$ are missing, because $d/a = 3$ puts every third interference maximum on top of a single-slit minimum, where the envelope of 14.8.A.1.vi is zero.

Frequently asked questions

What is the difference between interference and diffraction?

Diffraction describes how a wave travels past an obstacle; interference describes what happens where waves overlap. AP Physics 2 essential knowledge 14.7.A.1 defines diffraction as the spreading of a wave around the edges of an obstacle or through an opening, and 14.6.A.1 defines wave interference as the interaction of two or more wave pulses or waves. They are not alternatives. Essential knowledge 14.7.A.3 connects them: diffraction of multiple wavefronts through a single opening leads to observable interference patterns. So diffraction is what makes several paths to the same point available, and interference is what decides whether that point ends up bright or dark.

Can interference happen without diffraction?

Yes, and it happens whenever two separate sources overlap in open space. Two loudspeakers in a room, two radio antennas driven in step, or two pulses sent from opposite ends of a rope all produce interference with no obstacle and no opening anywhere, so nothing diffracts. AP Physics 2 Topic 14.6 is built entirely on cases like these, and Topic 14.9 on thin films is another: two reflections from the two surfaces of a film recombine, with no edge to spread around. Interference needs two or more contributions arriving at one place, which is all that 14.6.A.1 requires.

Can diffraction happen without interference?

Not really, though it can happen without a visible pattern. Once a wave has spread through an opening, different parts of that opening send wavefronts to the same point by different path lengths, so interference is unavoidable, and AP Physics 2 essential knowledge 14.7.A.3 says as much: diffraction of multiple wavefronts through a single opening leads to observable interference patterns. What varies is whether the resulting bands are wide enough to detect. Green light through a metre-wide doorway diffracts, and its first dark band sits about 3 times 10 to the minus 5 degrees off axis, which no eye resolves. The pattern is there; the apparatus is wrong for seeing it.

Is a single slit diffraction or interference?

Both, and the CED says so in two consecutive statements. AP Physics 2 essential knowledge 14.7.A.3 says diffraction of multiple wavefronts through a single opening leads to observable interference patterns, and 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. Diffraction is why the single opening acts as many wavefront sources rather than one; interference between those wavefronts is what puts the dark bands at the positions the printed relation locates. Calling the result a diffraction pattern names it by its cause and describes it by its mechanism.

Why do the double-slit fringes get dimmer towards the edges?

Because both effects are acting at once. AP Physics 2 essential knowledge 14.8.A.1 says the two-slit pattern is caused by a combination of wave diffraction and wave interference, and 14.8.A.1.vi says a double slit creates an interference pattern of maxima and minima superimposed within the envelope created by single-slit diffraction. Interference between the two slits sets the positions of the fringes and 14.8.A.1.i makes their spacing uniform. Diffraction at each individual slit sets how much light is going in each direction at all, and that falls off away from the centre, which dims the outer fringes. Where a diffraction minimum lands exactly on an interference maximum, that fringe disappears entirely.

Does a narrower slit make the diffraction pattern narrower?

No, it makes it wider, which is the reverse of what intuition suggests. AP Physics 2 essential knowledge 14.7.A.4.iv gives the small-angle relation as slit width times y-min over L is approximately m times lambda, so rearranged, the distance to the m-th dark band is m lambda L divided by the slit width. Halving the width doubles every distance in the pattern. Essential knowledge 14.7.A.2 gives the qualitative version: diffraction is most pronounced when the size of the opening is comparable to the wavelength of the wave, so squeezing the opening towards the wavelength maximises the spreading.

Why does the AP equation sheet call the double slit diffraction when the topic is called interference?

Because the vocabulary in the source material is genuinely inconsistent, and it is safer to know that than to guess. The AP Physics 2 exam conventions box states that the small angle approximation is valid for single- and double-slit diffraction, while Topic 14.8 is titled Double-Slit Interference and Diffraction Gratings, and essential knowledge 14.8.A.4 describes a grating as producing an interference pattern that is the combination of numerous diffraction patterns superimposed on each other. Every one of those phrasings is defensible, because slit experiments contain both effects. When a question asks what is happening, answer with the geometry and the mechanism rather than with a single word: how many openings there are, and what the path length difference is doing.