Topic 14.8: Double-Slit Interference and Diffraction Gratings

What Topic 14.8 requires

Topic 14.8 carries one learning objective, 14.8.A, describe the behavior of a wave and the diffraction pattern resulting from the wave passing through multiple openings. It prints no boundary statement. Five essential knowledge statements sit under it, and the first carries six sub-statements.

• 14.8.A.1 The pattern resulting from monochromatic light of wavelength $\lambda$ incident on two slits a distance $d$ apart is caused by a combination of wave diffraction and wave interference. Sub-statements: 14.8.A.1.i (when only considering wave interference, a double slit creates a pattern of uniformly spaced maxima), 14.8.A.1.ii (constructive and destructive interference of the wavefronts originating from each slit will result in bright and dark bands on the screen), 14.8.A.1.iii (the amount of interference between two wavefronts depends on the path length difference $\Delta D$ of the wavefronts), 14.8.A.1.iv (the path length difference can be described in terms of the slit separation $d$ and the angle $\theta$ between the direction of propagation of the wavefront and the normal to the opening by $\Delta D = d\sin\theta$), 14.8.A.1.v (for small angles, where $\theta < 10^\circ$, the small angle approximation relates $\lambda$, $d$ and $L$ to $y_{\text{max}}$, the distance from the middle of the central bright fringe to the $m^{\text{th}}$ order of maximum brightness on the screen, relevant equation $d\left(\dfrac{y_{\text{max}}}{L}\right) \approx m\lambda$), 14.8.A.1.vi (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).
• 14.8.A.2 Interference patterns produced by light interacting with a double slit indicate that light has wave properties. The source of this discovery was Young's double-slit experiment.
• 14.8.A.3 Visual representations of double-slit diffraction patterns are useful in determining the physical properties of the slits and the interacting waves.
• 14.8.A.4 A diffraction grating is 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.
• 14.8.A.5 When white light is incident on a diffraction grating, the center maximum is white and the higher-order maxima disperse white light into a rainbow of colors, with the longest-wavelength light (red) appearing farthest from the central maximum.

The suggested skills are 1.B (create quantitative graphs with appropriate scales and units, including plotting data), 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), 2.D (predict new values or factors of change of physical quantities using functional dependence between variables) and 3.A (create experimental procedures that are appropriate for a given scientific question). Five skills, second only to the six listed on 14.6 in this unit. The topic page and the Unit at a Glance table agree.

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

The bright-fringe condition, and what m counts

Convention for this page, declared once and held to the end: $\theta$ is measured from the normal to the slits, which is the straight-ahead direction toward the centre of the screen. $y$ is measured from the middle of the central bright fringe, and the pattern is symmetric about it, so every distance below is a magnitude. $d$ is the separation between the centres of the two slits, never the width of one slit. $m$ is the order, an integer.

Two slits a distance $d$ apart are illuminated by monochromatic light of wavelength $\lambda$. Light reaching a point off the centre of the screen has travelled slightly further from one slit than from the other, and 14.8.A.1.iii says the amount of interference between the two wavefronts depends on that path length difference $\Delta D$. 14.8.A.1.iv gives the geometry:

The AP Physics 2 equation sheet also prints the interference condition $\Delta D = m\lambda$. Put them together:

That locates the bright fringes. The CED settles it in 14.8.A.1.v, in the symbol itself: $y_{\text{max}}$ is the distance from the middle of the central bright fringe to the $m^{\text{th}}$ order of maximum brightness, and the printed small-angle equation is

Read the physics off that. A whole number of wavelengths of path difference means the two waves arrive crest on crest, so they reinforce.

What $m$ counts. For a double slit the orders run $m = 0, 1, 2, 3, \ldots$ outward on each side. $m = 0$ is a real fringe and it is the most important one: zero path difference, both waves in step, the central bright fringe. That is the point the CED measures $y_{\text{max}}$ from, and it sits straight ahead at $\theta = 0$ regardless of wavelength, slit separation or screen distance. Everything else is counted outward from it.

The small-angle step (skill 2.A). For a point a distance $y$ up a screen $L$ away, $\tan\theta = y/L$ exactly. When $\theta$ is small, $\sin\theta \approx \tan\theta \approx y/L$, so $d\sin\theta = m\lambda$ becomes $d(y/L) \approx m\lambda$. The approximation sign on the printed equation marks that swap. Rearranged,

The dark fringes: the condition the CED does not print

14.8.A.1.ii says that constructive and destructive interference of the wavefronts originating from each slit will result in bright and dark bands on the screen. The CED asks for both. It prints an equation for only one of them.

Nothing in Topic 14.8, and nothing in the 15 equations of the Waves, Sound, and Optics group on the equation sheet, gives you the dark-fringe condition. You derive it, which is exactly what skill 2.A is listed for.

The logic is one step. Two waves cancel when one arrives half a cycle behind the other, so destructive interference needs a path length difference of half a wavelength, or one and a half, or two and a half, and so on. Written with the printed geometry $\Delta D = d\sin\theta$:

and in the small-angle form,

with $m = 0, 1, 2, \ldots$ again.

Two things about the indexing, because it is where the marks go:

• There is a dark fringe at $m = 0$. It sits at $y \approx \frac{1}{2}\lambda L / d$, halfway between the central bright fringe and the first-order bright fringe. Unlike the single-slit case, $m = 0$ here is a genuine feature.
• A dark fringe always sits between two adjacent bright fringes. The bright fringes are at $0, \lambda L/d, 2\lambda L/d, \ldots$ and the dark ones at $0.5\lambda L/d, 1.5\lambda L/d, \ldots$, so the pattern alternates and there is one dark fringe in every gap.

Some textbooks index the dark fringes from $m = 1$ and write $\left(m - \frac{1}{2}\right)\lambda$ instead. That gives the same set of positions. The safe habit is to state your indexing when you write the condition down, since the CED provides no convention for a formula it does not print.

Fringe spacing, and why the fringes are evenly spaced

14.8.A.1.i makes a specific claim: when only considering wave interference, a double slit creates a pattern of uniformly spaced maxima. That uniformity is a consequence of the printed equation being linear in $m$.

From $y_{\text{max}} \approx m\lambda L / d$, the gap between order $m$ and order $m+1$ is

with no $m$ left in it. Every gap is the same size, including the gap from the central fringe to the first order. That is the working form for almost every double-slit calculation, and it is a derived expression rather than a printed one: the sheet gives $d(y_{\text{max}}/L) \approx m\lambda$ and nothing about the spacing.

Two consequences that get tested as skill 2.D:

• Fringe spacing is inversely proportional to slit separation. Slits closer together give fringes further apart. That feels backwards and is worth rehearsing until it does not.
• Fringe spacing is directly proportional to wavelength and to screen distance. Red fringes are wider apart than blue ones for the same slits, and moving the screen back spreads the whole pattern out proportionally.

The uniformity is also what makes skill 1.B, create quantitative graphs with appropriate scales and units including plotting data, worth listing on this topic. If you measure the position of each bright fringe and plot $y_{\text{max}}$ against the order $m$, the printed equation says the graph is a straight line through the origin with slope $\lambda L / d$. Measure $L$ separately and you have a value for $\lambda / d$ from the slope, and therefore for $\lambda$ if the slits are specified, or for $d$ if the wavelength is. Plotting the whole set of fringes and taking a slope beats measuring one fringe, because the uncertainty in locating any single fringe is spread across the fit.

The origin matters as a check. The line must pass through it, because $m = 0$ is the central fringe at $y = 0$. A fitted intercept that is not near zero means the centre of the pattern was misidentified.

Where single-slit diffraction comes back in: the envelope

Everything so far has treated the slits as if they had no width. Real slits do, and 14.8.A.1.i is careful to say "when only considering wave interference". 14.8.A.1.vi handles the rest: 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.

So there are two length scales in play, and they do different jobs:

Since $d > a$ always for two separate slits, $\lambda L / d$ is smaller than $\lambda L / a$: the interference fringes are packed closely, and the single-slit envelope varies slowly across many of them. What you see is a row of evenly spaced fringes whose brightness rises and falls under a broad hump, brightest in the middle and fading out toward the envelope's first minimum.

The CED asks about this directly. In the Instructional Approaches section, printed on page 178, the sample activity for skill 3.C sets out a beam of coherent light of wavelength $\lambda$ incident perpendicularly onto a pair of slits, each slit of width $w$, with the centres of the slits a distance $d$ apart and a screen a distance $L$ away, and then asks students to explain why some of the spots on the screen that should be bright, the interference maximums, are not.

The answer combines the two conditions from these two topics, and it is the clearest reason to know that they point opposite ways. An interference maximum needs $d\sin\theta = m\lambda$. A single-slit minimum, from Topic 14.7, needs $a\sin\theta = m'\lambda$ with the width $a$ in place of the separation. At an angle where both hold at once, the interference is trying to produce a bright fringe while the diffraction envelope has sent the light from each slit to zero at that angle. Zero light cannot interfere constructively into anything, so the fringe is missing. It disappears whenever $d/a$ works out to a whole number at that order.

That is a derived consequence, not an essential knowledge statement. The CED states the superposition in 14.8.A.1.vi and asks you to reason about it in the sample activity. It never prints the condition for which fringes vanish.

Diffraction gratings: the same condition, a different pattern

14.8.A.4 defines the device: a diffraction grating is 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.

The important word is evenly spaced. Because the spacing between neighbouring slits is the same everywhere across the grating, the angles at which every pair of neighbours is in step are the same angles that put the whole grating in step. The condition is therefore the one you already have:

with $d$ now the distance between adjacent slits, usually quoted as a number of lines per millimetre or per centimetre that you have to invert. If a grating is ruled with $N$ lines per unit length, then $d = 1/N$ in that unit. That inversion is not printed on the equation sheet and it is where a grating question is most often lost.

What changes is not the condition but the pattern. Two slits give broad fringes that shade gradually from bright to dark. A grating with thousands of slits gives narrow, sharp bright lines separated by wide dark regions. The reason follows from the superposition that 14.8.A.4 describes. At an angle satisfying $d\sin\theta = m\lambda$, every slit is in step with every other and all of them add, which is why the maxima are still in the same places. Move slightly off that angle and, with only two slits, the two waves fall a little out of step and the brightness eases off. With many slits, a small change in angle is enough for the accumulated path differences across the whole grating to spread the contributions right around the cycle so that they cancel among themselves. More slits means less angular room before that cancellation is complete, so each maximum is narrower.

Be precise about the status of that reasoning. The CED states in 14.8.A.4 that the pattern is the combination of numerous diffraction patterns superimposed on each other. It does not use the word "sharp", it prints no expression for the width of a grating maximum, and no such expression appears among the 15 equations in the Waves, Sound, and Optics group. Treat the narrowing as something to describe and justify, in the spirit of skill 3.B, and not as something to calculate.

Small angles are gone. This is the practical difference that changes how you compute. A grating has $d$ of a micrometre or two, so $\lambda / d$ is a sizeable fraction of 1 and the maxima land tens of degrees from the centre. The conventions block on the AP Physics 2 Table of Information reads: "The small angle approximation is valid for single- and double-slit diffraction." It names single and double slits and nothing else. With a grating, use $\Delta D = d\sin\theta = m\lambda$ directly and find $\theta$ with an inverse sine. Worked example 2 shows what using $y/L$ instead would cost: an error of about 5 percent at first order and about 24 percent at second.

Orders run out. Since $\sin\theta$ cannot exceed 1, the condition $\sin\theta = m\lambda/d$ has no solution once $m\lambda > d$. A grating shows a finite number of orders, and finding the highest one is a standard question.

White light on a grating, and the order of the colours

14.8.A.5 is the most quotable statement in the topic and it is worth having word for word: when white light is incident on a diffraction grating, the center maximum is white and the higher-order maxima disperse white light into a rainbow of colors, with the longest-wavelength light (red) appearing farthest from the central maximum.

Every part of that follows from $d\sin\theta = m\lambda$.

• The centre is white. At $m = 0$ the condition gives $\sin\theta = 0$ for every wavelength at once. All colours pile up in the same place, straight ahead, and recombine into white.
• Higher orders spread into colours. For $m \geq 1$, the angle depends on $\lambda$, so each wavelength leaves at its own angle and the white light is pulled apart.
• Red goes furthest. $\sin\theta$ is directly proportional to $\lambda$, and red has the longest visible wavelength, so red sits at the largest angle in each order. Violet, the shortest, hugs the centre.

That last point separates a grating from a glass prism. A prism also spreads white light, but through refraction, and its violet end deviates most. On a grating it is the red end that deviates most. If a question shows a spectrum and asks which end is which, the mechanism decides the answer, and the CED covers only the grating case: the word dispersion does not appear in Unit 13 at all, and 14.8.A.5 is the only place in the AP Physics 2 CED where white light is described splitting into colours.

One consequence the CED does not state but the equation forces: consecutive orders can overlap. Second-order violet leaves at a larger angle than first-order red on a fine grating, so the two spectra run into each other. Worked example 3 puts numbers on it.

This is also the topic's laboratory. Sample instructional activity 6 in the Unit 14 opener is written for 14.8: break students into groups of 3 to 4, give each group a red, green or purple laser and a diffraction grating, give the wavelength and ask for the diffraction slit spacing, or the reverse, experimentally. The activity adds that it can also be performed by finding the width of a human hair or the width of data tracks on a CD or DVD. Skill 3.A, create experimental procedures that are appropriate for a given scientific question, is what that is training, and the measurement in each case is an angle rather than a length, because the angles are large.

What the equation sheet gives you, and how 14.8 is tested

The AP Physics 2 equation sheet prints 129 equations in seven groups. The Waves, Sound, and Optics group holds 15, and three of them carry this topic:

The variable list in the same group 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 in the course framework follow the equation sheet's definitions. Those definitions are what keep $a$ and $d$ apart.

Not printed, and each one a place where students expect help that is not coming:

• The dark-fringe condition $d\sin\theta = \left(m + \frac{1}{2}\right)\lambda$. Derived.
• The fringe spacing $\lambda L / d$. Derived, in one line, from the printed equation.
• Any grating equation of its own. A grating uses the same $\Delta D = d\sin\theta$ as two slits.
• The conversion from lines per millimetre to $d$. You invert it yourself.
• Any expression for how sharp a grating maximum is, or for the intensity anywhere in the pattern.

Young's experiment. 14.8.A.2 states that interference patterns produced by light interacting with a double slit indicate that light has wave properties, and that the source of this discovery was Young's double-slit experiment. That is the historical hinge of the whole unit. Particles going through two openings would give two bright strips; the fringes say light superposes like a wave. The Unit 14 opener leaves the matter open on purpose, saying the end of Unit 14 leaves an open question of whether light should be considered a wave or a particle, which will be further studied in Unit 15. Unit 15 reopens it.

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, and the second free-response question is the Translation Between Representations question, which the Unit 14 opener points at when it says creating models and representations is a fundamental piece of that question. 14.8.A.3 is the matching essential knowledge: visual representations of double-slit diffraction patterns are useful in determining the physical properties of the slits and the interacting waves.

The five skills give five recognisable shapes: 1.B plots fringe position against order, 2.A derives the spacing or the dark-fringe condition, 2.B substitutes, 2.D asks what happens when $d$ or $\lambda$ changes, and 3.A designs the grating measurement. Across all of them, the first thing to check is whether the length in front of you is a separation or a width.