Single Slit vs Double Slit: What Is the Difference?

The two printed conditions look identical and mean opposite things. For one slit of width a, a times y-min over L is about m lambda, and it locates a dark band. For two slits a distance d apart, the same form with y-max locates a bright one. Check the subscript before you substitute.

AP Physics: Unit 14 (topics 14.7 Diffraction, 14.8 Double-Slit Interference and Diffraction Gratings). The two printed relations are AP Physics 2 essential knowledge 14.7.A.4.iv, which relates lambda, a and L to y-min, the distance from the middle of the central bright fringe to the m-th order of minimum brightness, and 14.8.A.1.v, which relates lambda, d and L to y-max, the distance to the m-th order of maximum brightness. Both are stated for angles below 10 degrees, and the exam conventions box on the AP Physics 2 sheet confirms that the small angle approximation is valid for single- and double-slit diffraction. The path length forms are 14.7.A.4.iii for one opening of width a and 14.8.A.1.iv for two openings a distance d apart, and the sheet's symbol list defines a as width, d as separation, L as distance and m as order or mass. Uniform spacing of double-slit maxima is 14.8.A.1.i; the single-slit envelope that shapes them is 14.8.A.1.vi. 14.7.A.5 adds that the pattern depends on the shape of the opening, and 14.7.A.2 that diffraction is most pronounced when the opening is comparable to the wavelength. Gratings are 14.8.A.4 and 14.8.A.5, and the historical claim about wave properties is 14.8.A.2. Neither Topic 14.7 nor Topic 14.8 prints a boundary statement. Unit 14 carries 12 to 15 percent of the multiple-choice section across about 14 to 23 class periods.

The distinction, stated once

The AP Physics 2 equation sheet prints these two lines four lines apart, in the Waves, Sound, and Optics group:

a\left(\frac{y_{\text{min}}}{L}\right) \approx m\lambda \qquad d\left(\frac{y_{\text{max}}}{L}\right) \approx m\lambda

They have the same right-hand side, the same shape, the same approximation sign, and the same list of symbols underneath. They locate opposite features.

Single slit. Essential knowledge 14.7.A.4.iv defines the left-hand equation's terms: for small angles, where $\theta < 10$ degrees, 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 $m$ th order of minimum brightness on the screen. Minimum brightness. A dark band.

Double slit. Essential knowledge 14.8.A.1.v uses the identical sentence with three words changed: for small angles, where $\theta < 10$ degrees, the small angle approximation can be used to relate $\lambda$ , $d$ and $L$ to $y_{\text{max}}$ , the distance from the middle of the central bright fringe to the $m$ th order of maximum brightness on the screen. Maximum brightness. A bright band.

The subscript on $y$ is the entire distinction, and the sheet prints it. Nothing else on either line tells you which feature you have found.

The same split runs through the two path-length equations one line above each of them. 14.7.A.4.iii gives $\Delta D = a\sin\theta$ with $a$ the width of the one opening; 14.8.A.1.iv gives $\Delta D = d\sin\theta$ with $d$ the separation between the two openings. Those are different lengths measured on different parts of an apparatus, and the sheet's symbol list keeps them apart in four words: $a$ is width, $d$ is separation.

So the whole comparison reduces to two questions you can answer by looking at the apparatus. How many openings, and which length is quoted? One opening and a width means $a$ , $y_{\text{min}}$ , and dark bands. Two openings and a separation means $d$ , $y_{\text{max}}$ , and bright bands.

Interference against diffraction handles the conceptual question of why one opening produces a pattern at all. This page is about the two experiments and the arithmetic that tells them apart.

Side by side

Property	Single slit	Double slit
Apparatus	One opening of width $a$	Two openings a distance $d$ apart
CED topic	14.7 Diffraction	14.8 Double-Slit Interference and Diffraction Gratings
Length symbol on the sheet	$a$ , defined as width	$d$ , defined as separation
Path length difference	$\Delta D = a\sin\theta$ (14.7.A.4.iii)	$\Delta D = d\sin\theta$ (14.8.A.1.iv)
Small-angle relation	$a(y_{\text{min}}/L) \approx m\lambda$ (14.7.A.4.iv)	$d(y_{\text{max}}/L) \approx m\lambda$ (14.8.A.1.v)
What that relation locates	Minimum brightness, a dark band	Maximum brightness, a bright band
Lowest valid order	$m = 1$ ; there is no $m = 0$ dark band	$m = 0$ , the central maximum
Centre of the pattern	Bright, and the widest feature	Bright, and the same width as the rest
Spacing of the located features	Uniform at $\lambda L/a$ , except the central band	Uniform at $\lambda L/d$ (14.8.A.1.i)
Brightness across the pattern	Falls off sharply away from the centre	Uniform if interference alone; in reality shaped by the single-slit envelope (14.8.A.1.vi)
Effect of halving the length symbol	Pattern spreads to twice the size	Fringes spread to twice the separation
Small-angle limit	$\theta < 10$ degrees	$\theta < 10$ degrees
Historical significance named by the CED	None stated	Young's double-slit experiment showed light has wave properties (14.8.A.2)

The lowest valid order row is the one that catches people out in the middle of an otherwise correct answer. Substituting $m = 0$ into the double-slit relation gives $y_{\text{max}} = 0$ , the central bright fringe, which is a real feature and the one every other distance is measured from. Substituting $m = 0$ into the single-slit relation gives $y_{\text{min}} = 0$ , which would put a dark band at the centre of a pattern whose centre 14.7.A.4.iv itself calls the central bright fringe. There is no zeroth-order minimum. Single-slit orders start at $m = 1$ .

The centre of the pattern row is the visual tell. Both patterns have a bright middle, but a single-slit central band is twice as wide as every other bright band, while double-slit fringes are all the same width. If you are handed a photograph rather than an apparatus, that is how you tell which experiment produced it.

What a and d actually measure

These two letters are measured on different parts of the equipment, and swapping them changes an answer by whatever factor the apparatus happens to supply.

$a$ is a width: the gap across one opening, edge to edge. It is a property of a single slit and it exists whether or not there is a second slit anywhere.

$d$ is a separation: centre to centre between two openings. It is a property of the pair and it says nothing at all about how wide either slit is.

A real double slit therefore has both numbers, and they are independent. You can widen the slits without moving them, or move them without widening them. The AP Physics 2 sheet's symbol list keeps this straight in the shortest possible way, printing $a$ = width and $d$ = separation on consecutive lines.

Three consequences follow, and all three appear in exam questions.

A double slit always has an $a$ as well as a $d$ . Essential knowledge 14.8.A.1 says the pattern is caused by a combination of wave diffraction and wave interference, and 14.8.A.1.vi says the maxima and minima sit inside an envelope created by single-slit diffraction. The $d$ sets where the fringes are; the $a$ of each individual slit sets how bright they are as you move outward. A question that gives you both numbers is asking about both effects.

$d$ is always larger than $a$ on real apparatus. Two openings separated by less than their own width would overlap into one opening. So if a problem quotes two lengths for a double slit, the larger is the separation.

Neither letter is the slit-to-screen distance. That is $L$ , printed in the symbol list as distance. Three lengths, three letters, and the two small ones are easy to swap under time pressure. Write them down with units before substituting.

One naming trap sits on top of this. The sheet's symbol list defines $m$ as "order or mass", so in both slit equations $m$ is an integer counting features outward, not a mass in kilograms. And it defines $f$ as "frequency or focal length", which matters because the same equation group carries the thin-lens equation.

The case that separates them: identical arithmetic, opposite brightness

Run the two experiments with the same light, the same screen distance, and numerically equal values of $a$ and $d$ . The arithmetic is then character for character the same, and the answers are a dark band and a bright band at the same place.

Take $\lambda = 650 \ \mathrm{nm}$ , $L = 3.00 \ \mathrm{m}$ , and set both the single slit's width and the double slit's separation to $0.200 \ \mathrm{mm}$ . Worked example one does it in full; the result is $9.75 \ \mathrm{mm}$ from the centre in both cases, dark for the single slit and bright for the double.

That is the single most expensive confusion in AP Physics 2 optics, and it is expensive precisely because nothing in the calculation warns you. The numbers are right. The units are right. The small-angle check passes. Only the classification at the end is wrong, and it is wrong by the maximum possible amount, since dark and bright are the two extremes of the quantity being asked about.

The defence is a habit rather than a piece of understanding: read the subscript on $y$ before you substitute, and write the word "dark" or "bright" next to the answer as you get it. The sheet has already done the work of distinguishing the two cases. All that is required is not to discard the distinction while copying the equation down.

One caveat about the numbers above, because it matters if you try to build that apparatus. Setting $d$ equal to $a$ describes two separate experiments that happen to share a number, not one double slit whose separation equals its slit width. On a real double slit, two openings of width $a$ separated by $d = a$ would touch and merge into a single opening of width $2a$ . Whenever a problem gives one apparatus both numbers, expect $d$ to be several times $a$ , as in worked example three of the interference against diffraction page.

The shape of each pattern

The two relations locate features. What they do not tell you directly is what the whole pattern looks like, and questions ask about that.

The double-slit pattern is regular. 14.8.A.1.i: when only considering wave interference, a double slit creates a pattern of uniformly spaced maxima. The spacing between neighbouring maxima is the difference between consecutive $m$ values, $\lambda L/d$ , and it is the same everywhere across the screen. Dark bands sit midway between them. The central maximum at $m = 0$ is no wider and no different in kind from the rest.

The single-slit pattern is not regular. Its minima are uniformly spaced at $\lambda L/a$ , so the bright bands between them are all $\lambda L/a$ wide, except for the one in the middle. The central bright fringe runs from the first minimum on one side to the first minimum on the other, so it is $2\lambda L/a$ across, twice the width of every other band, and it is far brighter than them.

That factor of two is not printed anywhere on the AP Physics 2 sheet and is not stated in the CED. It follows in one line from 14.7.A.4.iv, since the first minima sit at $\pm\lambda L/a$ and the central band is everything between them, so you can derive it in the margin rather than remember it.

A real double slit shows both patterns at once. 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. Fine, evenly spaced fringes from the $d$ ; a broad brightness envelope from the $a$ of each slit. Where the envelope reaches zero, a fringe vanishes.

Neither pattern depends on the shape of the openings being rectangular. 14.7.A.5 says the diffraction pattern produced by a wave passing through an opening depends on the shape of the opening, so a circular aperture gives rings rather than bands. Everything on this page assumes slits, because that is what the printed relations describe.

Gratings are the double slit taken further. 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. The same $d$ and the same $\Delta D = d\sin\theta$ apply, with $d$ now the spacing between neighbouring lines. 14.8.A.5 adds what white light does to it: the centre 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.

What happens when you change one variable

Both relations rearrange to the same form, $y \approx m\lambda L / (\text{the length symbol})$ , so their responses are identical in shape and differ only in which feature moves.

Change	Single-slit minima	Double-slit maxima
Double $\lambda$	Every $y_{\text{min}}$ doubles	Every $y_{\text{max}}$ doubles
Double the frequency	$\lambda$ halves, so every $y_{\text{min}}$ halves	Every $y_{\text{max}}$ halves
Double $L$	Everything doubles	Everything doubles
Double the length symbol	$a$ doubles, pattern shrinks by half	$d$ doubles, fringes close up by half
Halve the length symbol	Pattern spreads to twice the size	Fringes spread to twice the separation
Switch red light for blue	Pattern contracts	Fringes close up

Two of those rows are worth extra care.

The frequency row is a CED question in disguise. The Unit 14 overview says 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, which is 2.D, predicting new values using functional dependence between variables. Frequency does not appear in either printed relation, so the route is $\lambda = v/f$ first, then the slit equation. Doubling $f$ halves $\lambda$ , which halves every distance in the pattern. Worked example three runs it.

The "double the length symbol" row runs backwards from intuition. A wider slit gives a narrower pattern and a wider slit separation gives more tightly packed fringes. Both are the same algebra: the length symbol is in the denominator once the relation is rearranged. 14.7.A.2 states the physical version for the single slit, that diffraction is most pronounced when the size of the opening is comparable to the wavelength.

There is one asymmetry hidden in the table. Changing $a$ on a double slit does not move the fringes at all, because the fringe positions depend only on $d$ . It changes the envelope, so it changes which fringes are bright and which have vanished. Changing $d$ moves the fringes without touching the envelope.

When it costs a mark

Using $y_{\text{max}}$ reasoning on a single slit, or $y_{\text{min}}$ reasoning on a double slit. This is the headline error. The printed relations differ only in that subscript, and the classification of the answer flips entirely.
Putting a dark band at the centre of a single-slit pattern. There is no $m = 0$ minimum. 14.7.A.4.iv measures $y_{\text{min}}$ from the middle of the central bright fringe, so the middle is bright by the definition of the variable, and the minima start at $m = 1$ .
Using $a$ where $d$ belongs. The sheet defines $a$ as width and $d$ as separation. They are different lengths on the same apparatus and they are not interchangeable even when a problem quotes only one of them.
Applying the small-angle relation outside its range. Both 14.7.A.4.iv and 14.8.A.1.v state $\theta < 10$ degrees, and the equation sheet's exam conventions box confirms that the small angle approximation is valid for single- and double-slit diffraction. Beyond that, go back to $\Delta D = a\sin\theta$ or $\Delta D = d\sin\theta$ with $\Delta D = m\lambda$ .
Forgetting that $\sin\theta$ cannot exceed one. If $m\lambda$ comes out larger than the slit width or separation, that order does not exist. This is not an error message, it is the answer: it is why a doorway-sized opening produces no dark direction for audible sound.
Assuming the double-slit fringes are equally bright. 14.8.A.1.i gives uniform spacing, not uniform brightness. 14.8.A.1.vi then supplies the envelope that dims the outer ones.
Treating a diffraction grating as a different equation. It uses the same $\Delta D = d\sin\theta$ , with $d$ the line spacing. What changes is the sharpness of the maxima, not the condition that locates them.
Converting units at the last step. Slit widths arrive in millimetres or micrometres, wavelengths in nanometres, screen distances in metres. Convert everything to metres before substituting, or the powers of ten will not survive.
Reading $m$ as a mass. The sheet's symbol list says "order or mass". In both slit relations it is an integer order number.

Where they coincide, and why that lulls you

The two experiments overlap in three ways, and each one makes the pair feel like one topic until a question forces them apart.

They coincide in the algebra. Both relations are $\Delta D = m\lambda$ with a geometric expression substituted for $\Delta D$ , so both reduce to the same rearrangement. A student who works only by rearranging symbols will get the same expression twice and have no reason to notice that the two answers mean different things.

They coincide in the apparatus. A real double slit is two single slits, so its pattern contains a single-slit pattern. 14.8.A.1 says the result is caused by a combination of wave diffraction and wave interference. There is no experiment in which one relation applies and the other is irrelevant, only experiments where one effect dominates the appearance.

They coincide in the vocabulary. The equation sheet's conventions box calls both experiments diffraction. Topic 14.8 calls the two-slit case interference. 14.8.A.4 calls a grating's output an interference pattern made from superimposed diffraction patterns. So the word in the question stem is not a reliable guide to which relation is wanted; the number of openings is.

Where they genuinely part company is the centre of the pattern. The double-slit relation has a valid $m = 0$ and the single-slit relation does not, and that difference is not cosmetic. It is the sheet saying that a double slit's central feature is one of the family of maxima the equation locates, while a single slit's central feature is a bright band the equation was never about. Every distance in a single-slit pattern is measured from something the equation does not describe.

If you keep one sentence from this page, keep the subscript: $y_{\text{min}}$ for one slit, $y_{\text{max}}$ for two.

Same numbers, opposite brightness

Light of wavelength $\lambda = 650 \ \mathrm{nm}$ falls on a screen $L = 3.00 \ \mathrm{m}$ away. In experiment A it passes through a single slit of width $a = 0.200 \ \mathrm{mm}$ . In experiment B it passes through two slits separated by $d = 0.200 \ \mathrm{mm}$ . For each, find the distance from the centre of the pattern to the first-order feature the small-angle relation locates, state whether that feature is bright or dark, and confirm the approximation is allowed.

Convert to metres. $\lambda = 650 \ \mathrm{nm} = 6.50 \times 10^{-7} \ \mathrm{m}$ , and $a = d = 0.200 \ \mathrm{mm} = 2.00 \times 10^{-4} \ \mathrm{m}$ .
Experiment A. 14.7.A.4.iv prints $a\left(\dfrac{y_{\text{min}}}{L}\right) \approx m\lambda$ , so $y_{\text{min}} \approx \dfrac{m\lambda L}{a}$ .
With $m = 1$ : $y_{\text{min}} \approx \dfrac{(1)(6.50 \times 10^{-7})(3.00)}{2.00 \times 10^{-4}} = \dfrac{1.95 \times 10^{-6}}{2.00 \times 10^{-4}} = 9.75 \times 10^{-3} \ \mathrm{m} = 9.75 \ \mathrm{mm}$ .
The subscript says minimum, so that point is dark: it is the first-order minimum, and the bright centre of the pattern extends from $-9.75 \ \mathrm{mm}$ to $+9.75 \ \mathrm{mm}$ .
Experiment B. 14.8.A.1.v prints $d\left(\dfrac{y_{\text{max}}}{L}\right) \approx m\lambda$ , so $y_{\text{max}} \approx \dfrac{m\lambda L}{d}$ .
With $m = 1$ and $d$ numerically equal to $a$ , every factor is the same: $y_{\text{max}} \approx 9.75 \times 10^{-3} \ \mathrm{m} = 9.75 \ \mathrm{mm}$ .
The subscript says maximum, so that point is bright: it is the first-order bright fringe, and the next one sits at $19.5 \ \mathrm{mm}$ .
Check the approximation. $\tan\theta = \dfrac{9.75 \times 10^{-3}}{3.00} = 3.25 \times 10^{-3}$ , so $\theta = 0.186$ degrees, well inside the $10$ degrees both 14.7.A.4.iv and 14.8.A.1.v require.
Note what did the work here. Identical inputs, identical arithmetic, and the only thing separating a dark band from a bright one is which subscript the sheet printed on $y$ .

Both come out at $9.75 \ \mathrm{mm}$ from the centre. For the single slit that point is dark, the first-order minimum. For the double slit it is bright, the first-order maximum. The small-angle condition holds at $\theta = 0.186$ degrees.

The widths of the two patterns

With $\lambda = 650 \ \mathrm{nm}$ and $L = 3.00 \ \mathrm{m}$ : (a) find the width of the central bright band produced by a single slit of width $a = 0.100 \ \mathrm{mm}$ , and the width of the bright bands either side of it; (b) find the fringe spacing produced by a double slit of separation $d = 0.500 \ \mathrm{mm}$ ; (c) say how you would tell the two patterns apart from a photograph with no scale on it.

Convert: $\lambda = 6.50 \times 10^{-7} \ \mathrm{m}$ , $a = 1.00 \times 10^{-4} \ \mathrm{m}$ , $d = 5.00 \times 10^{-4} \ \mathrm{m}$ .
(a) The minima sit at $y_{\text{min}} \approx \dfrac{m\lambda L}{a} = m\dfrac{(6.50 \times 10^{-7})(3.00)}{1.00 \times 10^{-4}} = m(1.95 \times 10^{-2}) \ \mathrm{m}$ , so at $19.5 \ \mathrm{mm}$ , $39.0 \ \mathrm{mm}$ , $58.5 \ \mathrm{mm}$ and so on from the centre, on each side.
The central bright band runs from the first minimum on one side to the first minimum on the other, so its width is $2 \times 19.5 = 39.0 \ \mathrm{mm}$ .
Every other bright band lies between two consecutive minima on the same side, for example between $19.5$ and $39.0 \ \mathrm{mm}$ , so each is $19.5 \ \mathrm{mm}$ wide. The central band is twice as wide as the rest. That factor of two is not printed on the sheet; it comes from the fact that the central band spans two first-order minima while the others span one gap each.
(b) The maxima sit at $y_{\text{max}} \approx \dfrac{m\lambda L}{d} = m\dfrac{(6.50 \times 10^{-7})(3.00)}{5.00 \times 10^{-4}} = m(3.90 \times 10^{-3}) \ \mathrm{m}$ , so the fringe spacing is $3.90 \ \mathrm{mm}$ and it is the same everywhere, which is 14.8.A.1.i's uniformly spaced maxima.
(c) Two tells, both visible without a ruler. First, the middle: a single-slit pattern has a central band about twice the width of its neighbours, while double-slit fringes are all the same width. Second, the brightness: a single-slit pattern fades fast away from the centre, while a double-slit pattern is far more even, fading only as fast as the single-slit envelope of 14.8.A.1.vi allows.
Check the small-angle condition for the largest distance used. At $39.0 \ \mathrm{mm}$ on a $3.00 \ \mathrm{m}$ screen, $\tan\theta = 1.30 \times 10^{-2}$ and $\theta = 0.745$ degrees, still far below $10$ .

(a) The central bright band is $39.0 \ \mathrm{mm}$ wide and every other bright band is $19.5 \ \mathrm{mm}$ wide. (b) The double-slit fringes are evenly spaced at $3.90 \ \mathrm{mm}$ . (c) A single-slit pattern has a double-width central band and fades quickly; a double-slit pattern has equal-width fringes and stays bright further out.

Working backwards, then doubling the frequency

A single slit is lit with $\lambda = 600 \ \mathrm{nm}$ light and the first minimum is measured at $y_{\text{min}} = 12.0 \ \mathrm{mm}$ on a screen $L = 2.50 \ \mathrm{m}$ away. (a) Find the slit width. (b) The frequency of the light is then doubled. Find the new position of the first minimum. (c) Explain in words why the pattern moved the way it did.

(a) Rearrange 14.7.A.4.iv for the width: $a \approx \dfrac{m\lambda L}{y_{\text{min}}}$ .
With $m = 1$ , $\lambda = 6.00 \times 10^{-7} \ \mathrm{m}$ , $L = 2.50 \ \mathrm{m}$ and $y_{\text{min}} = 1.20 \times 10^{-2} \ \mathrm{m}$ : $a \approx \dfrac{(6.00 \times 10^{-7})(2.50)}{1.20 \times 10^{-2}} = \dfrac{1.50 \times 10^{-6}}{1.20 \times 10^{-2}} = 1.25 \times 10^{-4} \ \mathrm{m}$ , which is $0.125 \ \mathrm{mm}$ .
(b) Frequency does not appear in the slit relation, so convert first. The sheet prints $\lambda = v/f$ , and for light in air $v = c$ , so doubling $f$ at fixed $c$ halves $\lambda$ : the new wavelength is $300 \ \mathrm{nm} = 3.00 \times 10^{-7} \ \mathrm{m}$ .
Substitute back with the same slit: $y_{\text{min}} \approx \dfrac{(1)(3.00 \times 10^{-7})(2.50)}{1.25 \times 10^{-4}} = \dfrac{7.50 \times 10^{-7}}{1.25 \times 10^{-4}} = 6.00 \times 10^{-3} \ \mathrm{m} = 6.00 \ \mathrm{mm}$ .
Every distance in the pattern halves, because $y_{\text{min}}$ is directly proportional to $\lambda$ at fixed $a$ and $L$ .
(c) A shorter wavelength needs a smaller angle to build up the same path length difference $\Delta D = a\sin\theta$ , so the whole pattern closes in towards the axis. This is what the Unit 14 overview is pointing at when it says 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: the chain is frequency to wavelength to position, and no single printed equation covers both ends of it.
Check the approximation at both wavelengths. $\tan\theta = 12.0/2500 = 4.80 \times 10^{-3}$ gives $\theta = 0.275$ degrees, and the halved case is smaller still, so 14.7.A.4.iv's $\theta < 10$ degrees holds throughout.
One physical note to keep the answer honest: $300 \ \mathrm{nm}$ is ultraviolet, outside the visible range 14.4.A.3.ii lists, so the new pattern would need a detector rather than an eye. The optics is unchanged.

(a) $a = 0.125 \ \mathrm{mm}$ . (b) The first minimum moves in to $6.00 \ \mathrm{mm}$ . (c) Doubling the frequency halves the wavelength, and $y_{\text{min}}$ is proportional to $\lambda$ , so every feature in the pattern moves to half its former distance from the centre.

Frequently asked questions

What is the difference between the single-slit and double-slit equations?

They locate opposite features. AP Physics 2 essential knowledge 14.7.A.4.iv prints the single-slit relation as slit width a times y-min over L is approximately m times lambda, where y-min is 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 prints the double-slit relation as slit separation d times y-max over L is approximately m times lambda, where y-max is the distance to the m-th order of maximum brightness. The two lines have the same shape and the same right-hand side, so the subscript on y is the only thing distinguishing a dark band from a bright one. Both are valid only for angles below 10 degrees.

Is a the slit width or the slit separation?

The width. The AP Physics 2 equation sheet's symbol list defines a as width and d as separation, on consecutive lines. In Topic 14.7, a is the gap across one opening, measured edge to edge. In Topic 14.8, d is the distance from the centre of one slit to the centre of the other. A real double slit has both: the separation d fixes where the fringes fall, and the width a of each individual slit fixes the envelope that decides how bright they are. On real apparatus d is always larger than a, because two openings closer together than their own width would merge into one.

Why is there no m equals 0 for a single slit?

Because the middle of a single-slit pattern is bright, not dark. AP Physics 2 essential knowledge 14.7.A.4.iv defines y-min as the distance from the middle of the central bright fringe to the m-th order of minimum brightness, so the definition of the variable already states that the centre is a bright fringe. Setting m to zero would place a dark band there, contradicting the definition, so the orders start at m equals 1. The double-slit relation is different: m equals 0 gives y-max equals 0, which is the central maximum, a genuine feature and the point every other distance is measured from.

Why is the central bright band of a single-slit pattern twice as wide?

Because it is bounded by two first-order minima rather than one gap. The minima sit at distances of m times lambda L over a from the centre, so the first ones are one such distance out on each side, and the bright region between them is two of those distances across. Every other bright band lies between two consecutive minima on the same side, so it is only one such distance wide. The factor of two is not printed on the AP Physics 2 equation sheet and is not stated in the CED, but it follows in one line from essential knowledge 14.7.A.4.iv, so derive it rather than memorise it.

What happens to a diffraction pattern if you make the slit narrower?

It spreads out. Rearranging essential knowledge 14.7.A.4.iv gives the distance to the m-th minimum as m times lambda times L divided by the slit width, so halving the width doubles every distance in the pattern. Essential knowledge 14.7.A.2 says the same thing qualitatively: diffraction is most pronounced when the size of the opening is comparable to the wavelength of the wave, so narrowing the slit towards the wavelength maximises the spreading. The same algebra applies to a double slit with the separation: reducing d spreads the fringes apart.

How does doubling the frequency change the fringe spacing?

It halves it. Neither slit relation contains frequency, so the route is to convert first with the printed lambda equals v over f. For light in air the speed is fixed at 3.00 times 10 to the 8 metres per second, so doubling the frequency halves the wavelength. Both slit relations make the distance to a feature directly proportional to the wavelength at fixed geometry, so every distance in the pattern halves. The AP Physics 2 Unit 14 overview names this exact question, saying 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.

Does a diffraction grating use the single-slit or the double-slit equation?

The double-slit one, with d as the spacing between neighbouring lines. AP Physics 2 essential knowledge 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 condition for a maximum is still the path length difference d sine theta equal to a whole number of wavelengths. What changes with many slits rather than two is that the maxima become much sharper, not where they are. Essential knowledge 14.8.A.5 adds that in white light the centre maximum is white while higher orders spread into a rainbow, with red appearing farthest from the centre.