Topic 14.7: Diffraction

What Topic 14.7 requires

Topic 14.7 carries one learning objective, 14.7.A, describe the behavior of a wave and the diffraction pattern resulting from a wave passing through a single opening. It prints no boundary statement. Six essential knowledge statements sit under it, and one of them carries four sub-statements.

• 14.7.A.1 Diffraction is the spreading of a wave around the edges of an obstacle or through an opening.
• 14.7.A.2 Diffraction is most pronounced when the size of the opening is comparable to the wavelength of the wave.
• 14.7.A.3 Diffraction of multiple wavefronts through a single opening leads to observable interference patterns.
• 14.7.A.4 Diffraction is commonly demonstrated by monochromatic light of wavelength $\lambda$ incident on a narrow opening of width $a$ that is a distance $L$ from a screen. Sub-statements: 14.7.A.4.i (constructive and destructive interference of multiple wavefronts originating from the opening will result in bright and dark bands on the screen), 14.7.A.4.ii (the amount of interference between two wavefronts depends on the path length difference $\Delta D$ of the wavefronts), 14.7.A.4.iii (the path length difference $\Delta D$ can be described in terms of the opening width $a$ and the angle $\theta$ between the direction of propagation of the wavefront and the normal to the opening by the equation $\Delta D = a\sin\theta$), 14.7.A.4.iv (for small angles, where $\theta < 10^\circ$, the small angle approximation can be used to relate $\lambda$, $a$ and $L$ to $y_{\text{min}}$, the distance from the middle of the central bright fringe to the $m^{\text{th}}$ order of minimum brightness on the screen, with relevant equation $a\left(\dfrac{y_{\text{min}}}{L}\right) \approx m\lambda$).
• 14.7.A.5 The diffraction pattern produced by a wave passing through an opening depends on the shape of the opening.
• 14.7.A.6 Visual representations of single-slit diffraction patterns are useful in determining the physical properties of the slit and the interacting waves.

The suggested skills are 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.B (calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway) and 2.D (predict new values or factors of change of physical quantities using functional dependence between variables). Three skills, and all three come from Science Practice 2. Reading down the Unit at a Glance table, 14.7 is the only topic in the unit whose skill list has no Science Practice 1 skill and no Science Practice 3 skill in it. The topic page and the Unit at a Glance table list the same three.

That skill list tells you what the questions look like before you read another word: derive, calculate, predict a factor of change. There is no "sketch the apparatus" skill and no "design the experiment" skill on this topic.

Unit 14 is weighted at 12 to 15 percent of the multiple-choice section across a suggested 14 to 23 class periods.

Spreading, and the scale that decides whether you notice

14.7.A.1 is a one-line definition covering two situations: diffraction is the spreading of a wave around the edges of an obstacle or through an opening. A wave meeting an obstacle bends around its edges. A wave passing through a gap spreads out on the far side instead of casting a sharp shadow of the gap.

14.7.A.2 supplies the condition that decides whether you can see it happen: diffraction is most pronounced when the size of the opening is comparable to the wavelength of the wave.

"Comparable" is doing real work in that sentence. It is a ratio, not an absolute size. A one metre doorway does essentially nothing visible to light, because visible wavelengths are a few hundred nanometres and the opening is millions of wavelengths across. The same doorway spreads a one metre sound wave straight out into the corridor.

That ratio answers one of the unit's own essential questions, printed on the Unit 14 opener: why can you hear a person around a corner, but you can't see them? Audible sound covers wavelengths from centimetres to metres, so a doorway or a corner is comparable in size and the sound spreads around it. Visible light sits near $5 \times 10^{-7}$ m, so the same corner is enormous by comparison and the light keeps going in what look like straight lines. Nothing about sound is special here. The wavelength is.

The same reasoning sizes the apparatus. Worked example 2 below recovers a slit width from a measured pattern and gets about 84 micrometres, roughly 130 wavelengths of red light: wide enough that the spreading stays a small angle, narrow enough that the pattern is centimetres across a metre away.

Ray optics, which is the whole of Unit 13, is the limit where this spreading is small enough to ignore. Essential knowledge 13.1.A.1.iii says a laser is a common source of a single coherent, monochromatic beam of light that can be modeled as a ray, then adds that the wave nature of lasers will be considered in Unit 14. Topic 14.7 is where that gets cashed in.

Why one opening produces a pattern at all

14.7.A.3 is the sentence students read past: diffraction of multiple wavefronts through a single opening leads to observable interference patterns.

Interference normally needs two sources. Here there is one slit. The CED's answer is in the words "multiple wavefronts". The opening is not a point. Every part of it, across the whole width $a$, sends a wave toward the screen, and those parts sit at different distances from any off-centre point on the screen. They arrive out of step with each other and interfere. One opening, many wavefronts.

Worth knowing what the CED does not do here: the name Huygens appears nowhere in the AP Physics 2 CED. You are not expected to cite a principle by name. You are expected to accept that a wide opening behaves as many sources across its width, which is exactly what 14.7.A.3 states.

14.7.A.4 then fixes the standard arrangement for the rest of the topic: monochromatic light of wavelength $\lambda$ incident on a narrow opening of width $a$ that is a distance $L$ from a screen. The three sub-statements that follow build the equation one piece at a time.

• 14.7.A.4.i: constructive and destructive interference of multiple wavefronts originating from the opening will result in bright and dark bands on the screen.
• 14.7.A.4.ii: the amount of interference between two wavefronts depends on the path length difference $\Delta D$ of the wavefronts.
• 14.7.A.4.iii: the path length difference can be written

with $\theta$ the angle between the direction of propagation of the wavefront and the normal to the opening.

Convention for this page, declared once and held to the end: $\theta$ is measured from the normal to the opening, which is the straight-ahead direction toward the centre of the pattern. $y$ is measured from the middle of the central bright fringe. The pattern is symmetric about that centre, so every $y$ below is a magnitude and there are no negative distances anywhere on this page. And $a$ is the width of one opening. It is never a separation between openings.

Geometrically, $a\sin\theta$ is the extra distance travelled by light from one edge of the opening compared with light from the other edge, for a direction $\theta$ off the normal. The CED gives the equation without saying which two wavefronts it compares, but the width $a$ in it is the full span of the opening, so the two edges are what it measures.

The variable list on the AP Physics 2 equation sheet backs that reading up. In the Waves, Sound, and Optics group it defines $a$ as width, $d$ as separation, $D$ as path length, $L$ as distance, and $m$ as "order or mass", and the CED states that variables used within the course framework follow the definitions given on the equation sheet.

The condition that locates the dark fringes

Two of the printed equations combine into the one relationship this topic is built on. The sheet prints $\Delta D = m\lambda$ and it prints $\Delta D = a\sin\theta$. Setting them equal:

Now the question that decides whether you get the whole topic right: does that locate the bright fringes or the dark ones?

Dark. The CED answers it in 14.7.A.4.iv, in the definition of the symbol. $y_{\text{min}}$ is the distance from the middle of the central bright fringe to the $m^{\text{th}}$ order of minimum brightness on the screen, and the relevant equation printed alongside it is

So for a single slit, an integer number of wavelengths of path difference gives you darkness.

The small-angle step (skill 2.A). Going from $a\sin\theta = m\lambda$ to the printed form takes one line. For a point a distance $y$ from the centre of a screen $L$ away, $\tan\theta = y/L$ exactly. When $\theta$ is small, $\sin\theta \approx \tan\theta$, so $\sin\theta \approx y/L$, and substituting gives $a(y/L) \approx m\lambda$. The approximation sign that the sheet prints is not decoration: it marks the step where $\sin\theta$ was swapped for $\tan\theta$. Rearranged for the position of a dark fringe,

What $m$ counts, and why it does not start at zero. Put $m = 0$ into the printed equation and you get $y_{\text{min}} = 0$, which is the middle of the central bright fringe, the point the CED measures $y_{\text{min}}$ from. Zero path difference means every wavefront from across the opening arrives in step, which is the brightest point on the screen and not a dark one. So the orders of minimum brightness run $m = 1, 2, 3, \ldots$, counted outward on each side from the centre. The CED does not spell that restriction out in a sentence. It follows from what $y_{\text{min}}$ is measured from.

Why an integer number of wavelengths cancels. The CED gives the result and not the derivation, so here is the standard argument, which is also a clean run at skill 2.A. Take the direction where light from one edge of the opening is exactly one wavelength behind light from the other edge, so $a\sin\theta = \lambda$. Split the opening down the middle into a top half and a bottom half. Pair the very top of the opening with the point exactly $a/2$ below it, then the next point down with the point $a/2$ below that, and so on. Every one of those pairs is separated by $a/2$, so every pair has a path difference of exactly $\lambda/2$ and cancels. The whole opening cancels in pairs and the screen is dark. Push to $a\sin\theta = 2\lambda$ and the same trick works with quarters instead of halves. That is why integers give minima here, and it is the reason a single slit behaves back to front compared with two slits.

The collision: single-slit minima look exactly like double-slit maxima

This is the single most confusable pair in Unit 14, and it is worth meeting head on rather than discovering it in an exam. Both topics print an equation of the form "spacing times $\sin\theta$ equals $m\lambda$". They do not mean the same thing.

The College Board built the tell into the symbols. The sheet prints $y_{\text{min}}$ on the single-slit line and $y_{\text{max}}$ on the double-slit line, and the variable list distinguishes $a$ as width from $d$ as separation. If you can read those two subscripts off the sheet under exam conditions, you cannot get this backwards, because the sheet is in front of you the whole time.

The physical reason for the flip is worth one sentence each way.

• Two narrow slits. Straight ahead, both slits are the same distance from the centre of the screen, the two waves arrive in step, and you get the brightest fringe of the pattern. Zero path difference means bright. Add one whole wavelength and they are back in step, so integers stay bright.
• One wide slit. Straight ahead, everything across the opening arrives in step, so the centre is bright there too. But the equation is not comparing two sources at the centre. It is comparing the two edges of one opening, and when those edges are one wavelength apart the whole opening cancels in pairs, as the previous section showed. Zero path difference is not what $m = 1$ describes.

The pattern shapes differ as well, and that difference is a fast visual check. A double slit considered on interference alone gives uniformly spaced maxima of similar brightness, which is what essential knowledge 14.8.A.1.i states. A single slit gives one broad bright band in the middle with much fainter bands either side that fade away quickly. If a diagram shows one obviously dominant central band, you are looking at single-slit diffraction.

One last thing that stops the two from being independent: a real double slit has slits of finite width, so both effects are present at once. Essential knowledge 14.8.A.1.vi says so directly, and Topic 14.8 covers what that superposition does to the pattern.

Reading the pattern: the central fringe is twice as wide

Skill 2.A wants a symbolic result out of the printed relationship, and the cleanest one in this topic is the width of the central bright band.

The central bright fringe runs from the first minimum on one side to the first minimum on the other. Each of those sits at $y_{\text{min}} \approx \lambda L / a$, taking $m = 1$. So its total width is

Every other bright band is bounded by two consecutive minima, at $m$ and $m+1$, so its width is $\lambda L / a$. The central fringe is twice as wide as the rest of them. That is a derived result, not something printed anywhere: the sheet gives you $a(y_{\text{min}}/L) \approx m\lambda$ and stops.

Three more features of the pattern:

• The centre is the brightest point. All the wavefronts arrive in step there.
• The side bands get fainter as you move out. No intensity formula appears among the 15 equations in the Waves, Sound, and Optics group, so treat brightness as something to describe rather than to calculate.
• The minima are evenly spaced in the small-angle region, $\lambda L/a$ apart, because $y_{\text{min}} \approx m\lambda L/a$ is linear in $m$. That is what makes the pattern measurable: count several fringes and divide, rather than trying to pin down one fringe precisely.

14.7.A.5 adds a caution that is easy to skip: the diffraction pattern produced by a wave passing through an opening depends on the shape of the opening. The equation you have been given is written for a slit, with $a$ defined on the sheet as a width. It does not carry over unchanged to a round hole, and the CED does not give you a version that does. If a question shows a circular aperture, it is asking you to describe, not to substitute.

14.7.A.6 says visual representations of single-slit diffraction patterns are useful in determining the physical properties of the slit and the interacting waves. That is the reverse direction, and it is the one experiments actually use: measure the pattern, then solve for $a$ or for $\lambda$. Worked example 2 does exactly that.

Predicting factors of change (skill 2.D)

Skill 2.D, predict new values or factors of change of physical quantities using functional dependence between variables, is listed on this topic, and the Unit 14 opener names the exact question type. Under Building the Science Practices it says that students might be asked to determine the new distance between bright fringes of a diffraction pattern if the frequency of the light through the single slit is doubled.

Work from the small-angle result $y_{\text{min}} \approx m\lambda L / a$. Every horizontal distance in the pattern, the fringe spacing, the width of the central band, the position of the fourth minimum, scales the same way, so the answer does not depend on which feature you were asked to measure.

The frequency row is the one the CED singles out, and it needs one extra step that costs marks when it is skipped. Frequency is not in the diffraction equation at all. You get there through $\lambda = v/f$, printed at the top of the same group on the equation sheet, so doubling $f$ halves $\lambda$, and halving $\lambda$ halves every distance in the pattern. The pattern contracts toward the centre. Anyone who reasons directly from "frequency up, so spreading up" gets it backwards.

The inverse dependence on $a$ is worth saying out loud because it feels wrong: a narrower slit gives a wider pattern. It is the same statement as 14.7.A.2. Squeezing the opening toward the wavelength is what makes the spreading pronounced.

The CED's Instructional Approaches section has a Quickwrite activity built on this, printed on page 171: after students have learned how to analyse single- and double-slit experiments, have them list as many ways as possible to change the interference pattern on the screen and explain how each change affects the pattern. The table above is that list for a single slit.

What the equation sheet gives you, and how 14.7 is tested

The AP Physics 2 equation sheet prints 129 equations in seven groups. The Waves, Sound, and Optics group holds 15 of them, and exactly three are the ones Topic 14.7 runs on:

Only the first of those three is shared with Topic 14.8: $\Delta D = m\lambda$ is common to both, while $\Delta D = d\sin\theta$ and $d(y_{\text{max}}/L) \approx m\lambda$ are the double-slit pair. Also printed in the same group and useful here: $\lambda = v/f$, which is how a frequency change enters, and $T = 1/f$. The wave speed guide covers that relationship separately.

Not printed anywhere on the sheet, and worth listing because students look for them:

• The width of the central maximum, $2\lambda L/a$. You derive it.
• Any intensity formula for the pattern.
• Any statement of the range of $m$, or that $m$ starts at 1 for a single slit.
• Any version of the equation for an opening that is not a slit.

The small-angle convention comes from the Table of Information, not from a habit. The conventions block on the AP Physics 2 Table of Information opens with "The following conventions are used in this exam unless otherwise stated" and its final bullet reads: "The small angle approximation is valid for single- and double-slit diffraction." That is printed with the constants, on the same appendix page you get in the exam booklet. Read it precisely. It licenses the approximation for single and double slits, and it names no other geometry, which matters once you reach diffraction gratings in Topic 14.8, where the angles run well past $10^\circ$. Essential knowledge 14.7.A.4.iv puts a number on "small": $\theta < 10^\circ$.

How it shows up. The AP Physics 2 exam is 3 hours long: 42 multiple-choice questions in 85 minutes for half the score, and 4 free-response questions in 95 minutes for the other half, with a calculator allowed on both sections. Unit 14's 12 to 15 percent weighting applies to the multiple-choice section. The three suggested skills give three shapes of question. 2.B is a substitution, usually with a unit conversion buried in it, since wavelengths arrive in nanometres and slit widths in millimetres or micrometres. 2.A asks for a symbolic expression, most often the central width or $a$ in terms of measured quantities. 2.D asks for a factor of change with no numbers at all.

None of those is the costliest error, though. That one is answering a single-slit question with the double-slit rule. Check the subscript on $y$ before you substitute.