5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Monochromatic light passes through two very narrow slits and falls on a distant screen, producing a pattern of bright and dark bands. Which explanation accounts for this pattern?
Answer and reasoning
ALight travels straight through the slits, and each bright band is an image of a slit. A student who uses the ray model of light picks this. Straight-line paths through two slits would light only two narrow strips; they cannot account for many bands spread over a region much wider than the slits, or for the dark bands between them.
BLight spreads out from each slit, and the waves from the two slits overlap and interfere.Correct Each slit is narrow enough to diffract the light, so waves spread from both slits into the whole region beyond the barrier. Where the two waves overlap they interfere: bright bands where they arrive in step, dark bands where they arrive half a cycle out of step. The pattern needs both effects.
CLight is refracted at the edges of each slit, which bends it into separate bright bands. A student who thinks the bands come from light bending at the slit edges picks this. Refraction occurs where light crosses into a different medium; the bands come from the superposition of the waves spreading from the two slits.
DWaves from the two slits interfere where they meet, so no spreading at the slits is needed. A student who thinks the pattern is interference alone picks this. Without diffraction at each slit the light would not spread into the wide region where the two waves overlap, so there would be nothing to interfere across the screen.
In Young's double-slit experiment, light from a single source passes through two slits and forms bands on a screen. Which claim about light, with its supporting evidence from the pattern, is correct?
Answer and reasoning
ALight is a stream of particles, since it arrives in separate bright bands on the screen. A student who thinks separate bands show separate particles picks this. The key feature is the dark bands: adding light from the second slit makes some points darker, which waves can do and particles traveling along separate paths cannot.
BLight travels only in straight lines, since the brightest band lies straight ahead. A student who uses the ray model picks this. The brightest band does lie straight ahead, but light also reaches many bands far to either side, well outside the straight-line paths through the slits, so the pattern contradicts straight-line travel.
CLight is refracted at the edges of slits, since it reaches points hidden behind the barrier. A student who explains the spreading as refraction at the slit edges picks this. Light does reach points behind the barrier, but because it diffracts and the waves from the two slits interfere; refraction happens where light crosses into a different medium.
DLight has wave properties, since light from two slits can combine to make dark bands.Correct The dark bands are places that receive light from either slit alone but little light when both are open. Light adding to light to give darkness is superposition of waves, which particles traveling along separate paths cannot produce.
Light of wavelength 600 nm passes through two narrow slits onto a screen 2.0 m away. The graph shows the intensity I of the light on the screen against position y, measured from the center of the pattern. What is the separation of the slits?
Answer and reasoning
A0.40 mmCorrect The maxima are 3.0 mm apart, so the first-order maximum is at ymax = 3.0 mm. d = mλL/ymax = (1)(600 × 10⁻⁹ m)(2.0 m)/(3.0 × 10⁻³ m) = 4.0 × 10⁻⁴ m = 0.40 mm.
B0.13 mm A student who uses the outermost maximum shown, at 9.0 mm, as if it were first order picks this. That maximum is third order; with m = 3, d = 3λL/y = 0.40 mm.
C0.20 mm A student who uses the 6.0 mm between the first-order maxima on opposite sides as ymax picks this. ymax for the first-order maximum is measured from the center: 3.0 mm.
D0.80 mm A student who thinks ΔD = λ at a minimum picks this, using the first minimum, at 1.5 mm, as if it were the first-order position. At that minimum ΔD = λ/2; the first-order maximum is at 3.0 mm.
Working From the graph, maxima at y = 0, ±3.0, ±6.0, ±9.0 mm; minima halfway between (±1.5 mm, …). First order: ymax = 3.0 mm. d = mλL/ymax = (600 × 10⁻⁹ m)(2.0 m)/(3.0 × 10⁻³ m) = 4.0 × 10⁻⁴ m = 0.40 mm. Largest angle shown: 9.0 mm/2.0 m ≈ 0.26°, so the small-angle relation applies.
Light of wavelength 600 nm is incident normally on a diffraction grating that has 800 lines per millimeter. At what angle from the normal is the second-order maximum?
Answer and reasoning
A16° A student who writes the grating condition with cos θ picks this: cos θ = 0.96 gives 16°. The path difference between neighboring slits is d sin θ, so sin θ = 0.96.
B29° A student who leaves out the order number picks this: sin θ = λ/d = 0.48 gives 29°, the first-order angle. For the second order, sin θ = 2λ/d = 0.96.
C74°Correct The slit spacing is d = 1/(800 mm⁻¹) = 1.25 × 10⁻³ mm = 1.25 × 10⁻⁶ m. For the second-order maximum, d sin θ = 2λ, so sin θ = 2(600 × 10⁻⁹ m)/(1.25 × 10⁻⁶ m) = 0.96 and θ = 74°.
D55° A student who uses the small-angle approximation picks this: θ ≈ 2λ/d = 0.96 rad = 55°. At this angle the approximation fails badly; sin θ = 0.96 gives 74°.
Working d = 1 mm/800 = 1.25 × 10⁻⁶ m. d sin θ = mλ, m = 2: sin θ = 1.2 × 10⁻⁶/1.25 × 10⁻⁶ = 0.96 → θ = 73.7° ≈ 74°. Distractors: m = 1 → sin θ = 0.48 → 28.7°; cos θ = 0.96 → 16.3°; θ = 0.96 rad = 55.0°.
White light is incident on a diffraction grating. In the first-order spectrum on each side of the central maximum, which color appears farthest from the central maximum, and why?
Answer and reasoning
AViolet, since shorter wavelengths are deflected more, as in a prism. A student who carries over the order of a prism spectrum picks this. A prism separates colors by refraction; a grating separates them by interference, and d sin θ = λ puts the shortest wavelength (violet) at the smallest angle.
BRed, since red light travels faster in air than any other color. A student who thinks colors travel at different speeds in air picks this. All colors travel at essentially the same speed in air; red appears farthest because its wavelength is longest.
CNone: the first-order maxima of all colors are at one angle. A student who thinks the slits alone fix where the maxima are picks this. The angle satisfies d sin θ = λ, so each wavelength has its first-order maximum at a different angle, spreading white light into a spectrum.
DRed, since its longer wavelength needs a larger angle to make d sin θ = λ.Correct The first-order maximum of each wavelength is where d sin θ = λ. Red light has the longest visible wavelength, so it needs the largest angle and appears at the outer edge of each first-order spectrum.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
14.8.A.1 Double-slit interference pattern Fix
Double-slit interference pattern
The pattern of bright and dark bands produced on a screen when monochromatic light of wavelength λ passes through two narrow slits a distance d apart. It is caused by a combination of diffraction (the light spreading out from each narrow slit, so that the waves from the two slits overlap) and interference (the superposition of the overlapping waves).
Monochromatic light
Light of a single wavelength (a single color). A double slit lit by monochromatic light gives bands of one color; white light gives colored bands.
Slit separation, d
The distance between the centers of two neighboring slits. Unit: meter (m). It is a different length from the width a of each slit.
Interference-only pattern of a double slit
The pattern predicted when only the interference of the waves from the two slits is considered: bright maxima that are uniformly spaced along the screen (for small angles), all equally bright, with dark minima halfway between them.
Bright and dark bands (maxima and minima)
A bright band is where the waves from the two slits arrive in step and interfere constructively; a dark band is where they arrive half a cycle out of step and interfere destructively. Energy is redistributed across the screen, not destroyed.
Path length difference, ΔD
The difference between the distances traveled by the waves from the two slits to a point on the screen, ΔD = r₂ − r₁. Unit: meter (m). Interference is constructive where ΔD = mλ (m = 0, 1, 2, …) and destructive where ΔD = (m + ½)λ.
Angle θ
The angle between the direction from the slits to a point on the screen and the normal (perpendicular) to the plane of the slits. For light leaving two slits a distance d apart at angle θ, ΔD = d sin θ.
Order of a maximum, m
The whole number that labels a bright maximum by its path length difference, ΔD = mλ. The central bright band is the zeroth-order maximum (m = 0); the next bright band on each side is first order (m = 1), and so on.
Small-angle double-slit relation
For small angles (θ < 10°), sin θ ≈ tan θ = y/L, so the mth-order maximum is at a distance ymax from the middle of the central bright band given by d(ymax/L) ≈ mλ, where L is the distance from the slits to the screen. Neighboring bright bands are therefore λL/d apart.
Distance to the mth-order maximum, ymax
The distance on the screen from the middle of the central bright band to the mth-order bright maximum on one side. Unit: meter (m).
Single-slit diffraction envelope
The broad pattern produced by diffraction at each slit of width a. In a real double-slit pattern the interference maxima lie within this envelope, so their brightness falls away from the center; the envelope is wider for narrower slits.
Students often think Light passes straight through each opening as a narrow beam, so a double slit makes one bright band directly behind each slit (and a grating one band behind each of its slits). In fact No. Each slit is narrow enough that light spreads out (diffracts) from it, so light from both slits reaches a wide region of the screen. Where the two waves overlap they interfere, giving many evenly spaced bright bands rather than one band behind each slit, and the central bright band lies opposite the barrier between the slits.
Students often think Light is refracted as it grazes the edges of the slits, and this bending sorts it into separate bright bands. In fact No. The bands come from the superposition of the waves that spread from the two slits: bright where they arrive in step, dark where they arrive half a cycle out of step. Refraction is the bending of light as it crosses into a different medium, and passing an edge of an opaque barrier in air is not such a crossing.
14.8.A.2 Young's double-slit experiment Fix
Young's double-slit experiment
The experiment in which light from one source passes through two slits and forms bands of light and dark on a screen. The dark bands, places where adding light from the second slit reduces the brightness, showed that light has wave properties.
Students often think The separate bright bands of the double-slit pattern show that light arrives as a stream of particles, so Young's experiment is evidence for a particle model of light. In fact No. The dark bands are places that receive light from either slit alone but little light when both slits are open. Light from two slits combining to give darkness is superposition, which particles traveling along separate paths cannot produce, so Young's pattern showed that light has wave properties.
14.8.A.3 Determining properties from a pattern Fix
Determining properties from a pattern
Measurements on a double-slit pattern can be used to find a property of the slits or of the light: for example, the spacing of the bright bands together with λ and L gives d, or together with d and L gives λ.
14.8.A.4 Diffraction grating Fix
Diffraction grating
A collection of many evenly spaced parallel slits or openings. Its interference pattern is the combination of the diffraction patterns of all the slits superimposed; its bright maxima occur at the angles where d sin θ = mλ, with d the spacing between neighboring slits (d = 1/N for N lines per unit length).
Students often think The spread of a grating's pattern is set by the overall width of the grating, so a grating with more slits (a wider grating) has its maxima closer together, like a wider single slit. In fact No. The angles of a grating's bright maxima are set by the spacing d between neighboring slits and by λ: d sin θ = mλ. Adding more slits at the same spacing does not move the maxima.
Students often think The more slits there are, the more the light is spread, so a grating's maxima are farther apart than those of a double slit with the same slit spacing. In fact No. For the same slit spacing d and wavelength, a grating's bright maxima are at the same angles as a double slit's. Gratings usually spread light much more than the double slits used in class because their slits are much closer together (d is often about a micrometer), not because there are more of them.
14.8.A.5 White-light spectrum from a grating Fix
White-light spectrum from a grating
When white light falls on a grating, the central maximum (m = 0) is white because every wavelength has ΔD = 0 there. Each higher-order maximum is spread into a spectrum, with the longest wavelength (red) farthest from the central maximum because it needs the largest angle to satisfy d sin θ = mλ.
Students often think Shorter wavelengths (violet, blue) are deflected more than longer ones (red) in a grating or double slit, just as in a prism. In fact No. In a grating the angle of each maximum is set by d sin θ = mλ, so a longer wavelength gives a larger angle: red light is deflected most and appears farthest from the central maximum. A prism gives the opposite order because it separates colors by refraction, which is greatest for violet light.
Students often think The bright maxima are at positions fixed by the slits alone, so light of every wavelength has its maxima in the same places. In fact No. The maxima occur where d sin θ = mλ, so the same slits put the maxima of different wavelengths at different angles. Only the central maximum (m = 0), at θ = 0, is in the same place for every wavelength.
13 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 13
The graphs numbered 1 to 4 show possible patterns of light intensity I against position y on a screen, for monochromatic light passing through two narrow slits. Which graph shows the pattern predicted when only the interference of the waves from the two slits is considered?
Answer and reasoning
AGraph 1 A student who thinks light travels straight through each slit picks this graph, with one bright peak behind each slit. The waves from the two slits spread and overlap, giving many evenly spaced maxima, not two.
BGraph 2 A student who treats a double-slit pattern as a single-slit pattern picks this graph, with one broad central peak and weak side peaks. That shape comes from diffraction at one slit; interference between the waves from two slits gives evenly spaced maxima.
CGraph 3 A student who thinks the interference of the two waves makes the maxima fade picks this graph. The waves arrive in step at every maximum, so interference alone gives equally bright maxima; the fading comes from diffraction at each slit, which this question sets aside.
DGraph 4Correct Interference alone gives bright maxima wherever the path length difference is a whole number of wavelengths. For small angles these are evenly spaced along the screen, and the waves from the two slits arrive in step at every one of them, so every maximum is equally bright: equal, evenly spaced peaks.
Working Interference only: maxima where ΔD = d sin θ = mλ, evenly spaced (y ≈ mλL/d) and all equally bright, since the two waves arrive in step at each. The graph with equal, evenly spaced peaks is the key; two peaks = ray model; one broad peak = single-slit pattern; fading peaks = interference within the single-slit envelope.
In a double-slit experiment with monochromatic light, point P is at the center of a dark band on the screen. Which statement about the light waves arriving at P from the two slits is correct?
Answer and reasoning
ATheir paths from the two slits differ by exactly one whole wavelength. A student who has exchanged the conditions for bright and dark picks this. A path difference of one whole wavelength brings the waves back in step, crest with crest, which gives a bright band.
BEach has traveled a whole number of wavelengths plus one-half from its slit. A student who counts wavelengths along one path picks this. The distance from either slit on its own does not decide the result; what matters is the difference between the two distances, which is an odd number of half-wavelengths at a dark band.
CThey arrive half a cycle out of step, so their displacements cancel.Correct At the center of a dark band the path length difference is an odd number of half-wavelengths, so a crest from one slit always arrives with a trough from the other. The two displacements are equal and opposite and cancel at every instant.
DThey cancel at P, and the energy that they carried to P is destroyed there. A student who thinks destructive interference destroys energy picks this. The displacements cancel at P, but energy is conserved: it is redistributed to the bright bands, which are brighter than light from either slit alone.
The figure shows the pattern of bright and dark bands on a screen in a double-slit experiment with monochromatic light of wavelength λ. C is the central bright band, and the arrow marks point P at the center of a dark band. What is the path length difference ΔD between the waves reaching P from the two slits?
Answer and reasoning
A1.5λCorrect At C, ΔD = 0. Moving outward, each band passed adds λ/2: the first dark band has ΔD = 0.5λ, the next bright band 1.0λ, and P, the second dark band, 1.5λ.
B0.5λ A student who thinks every dark band has ΔD = λ/2 picks this. That is true only of the first dark band on each side of C. Counting from C, P is the second dark band, so ΔD = 1.5λ.
C2.0λ A student who thinks dark bands occur at whole-number path differences picks this, giving P, the second dark band, ΔD = 2λ. Whole-number path differences give bright bands; dark bands are at odd numbers of half-wavelengths.
D2.5λ A student who numbers the first dark band m = 1 in ΔD = (m + ½)λ picks this, giving P (taken as m = 2) ΔD = 2.5λ. The first dark band has m = 0 and ΔD = 0.5λ, so P, the second, has m = 1 and ΔD = 1.5λ.
Working C: ΔD = 0 (m = 0 maximum). Bright bands at ΔD = mλ, dark bands at ΔD = (m + ½)λ, m = 0, 1, 2, … . P is the second dark band to the right of C, so m = 1 and ΔD = (1 + ½)λ = 1.5λ.
Microwaves of wavelength 3.0 cm pass through two slits. A detector is placed in turn at positions A, B, C and D. The table shows, for each position, the distances r₁ and r₂ from the two slits to the detector and whether the detector reading is high or low. Which claim is supported by the data?
Answer and reasoning
AHigh readings occur where r₁ is a whole number of wavelengths. A student who counts wavelengths along one path picks this. r₁ is 20.5 wavelengths at B, where the reading is high, and 21 wavelengths at C, where it is low, so the data contradict the claim; what matters is the difference r₂ − r₁.
BHigh readings occur where r₂ − r₁ is an odd number of half-wavelengths. A student who has exchanged the conditions for bright and dark picks this. At C, r₂ − r₁ = 4.5 cm, three half-wavelengths, and the reading is low; at B and D, whole numbers of wavelengths, it is high.
CHigh readings occur where r₁ and r₂ are equal, and at no other place. A student who thinks the waves are in step only after equal paths picks this. At B and D the distances differ, by 3.0 cm and 6.0 cm, yet the readings are high, because those differences are whole numbers of wavelengths.
DHigh readings occur where r₂ − r₁ is a whole number of wavelengths.Correct r₂ − r₁ is 0 at A, 3.0 cm at B, 4.5 cm at C and 6.0 cm at D, that is 0, 1, 1.5 and 2 wavelengths. The readings are high at A, B and D, where the path difference is a whole number of wavelengths, and low at C, where it is an odd number of half-wavelengths.
Working λ = 3.0 cm. ΔD = r₂ − r₁: A 0 (0λ), high; B 3.0 cm (1λ), high; C 4.5 cm (1.5λ), low; D 6.0 cm (2λ), high. r₁/λ: A 20, B 20.5, C 21, D 22 — B (non-integer) is high and C (integer) is low, so r₁ alone does not predict the reading.
Light of wavelength λ passes through two narrow slits a distance d apart. The second-order bright maximum is at an angle θ₂ from the normal to the slits. Which expression gives λ?
Answer and reasoning
Ad sin θ₂/2Correct At the second-order maximum the path length difference is two wavelengths: ΔD = d sin θ₂ = 2λ, so λ = d sin θ₂/2.
Bd sin θ₂ A student who leaves out the order number picks this, treating the maximum as first order. At the second-order maximum ΔD = 2λ, so d sin θ₂ is two wavelengths, not one.
C2d sin θ₂/3 A student who thinks bright maxima occur at odd numbers of half-wavelengths picks this: taking the second maximum as ΔD = 3λ/2 gives λ = 2d sin θ₂/3. Bright maxima occur where ΔD is a whole number of wavelengths.
D2d sin θ₂ A student who moves the order number across the equals sign as a multiplier picks this. From d sin θ₂ = 2λ, dividing by 2 gives λ = d sin θ₂/2.
Working ΔD = d sin θ; maximum of order m: ΔD = mλ. m = 2: d sin θ₂ = 2λ → λ = d sin θ₂/2. Distractors: m omitted → λ = d sin θ₂; (m + ½) with the second maximum as 1.5 → λ = d sin θ₂/1.5 = 2d sin θ₂/3; multiplying by m → 2d sin θ₂.
Microwaves of wavelength 3.0 cm are incident on two narrow slits whose centers are 6.0 cm apart. At what angle from the normal to the slits is the first-order maximum?
Answer and reasoning
A14° A student who thinks the first maximum is where ΔD = λ/2 picks this: sin θ = 1.5 cm/6.0 cm = 0.25 gives 14°. At the first-order maximum ΔD = λ, so sin θ = 0.50.
B30°Correct At the first-order maximum the path length difference is one wavelength: d sin θ = λ, so sin θ = 3.0 cm/6.0 cm = 0.50 and θ = 30°.
C29° A student who uses the small-angle approximation picks this: θ ≈ λ/d = 0.50 rad = 29°. At this large angle sin θ and θ differ noticeably; sin θ = 0.50 gives θ = 30°.
D60° A student who writes the path difference as d cos θ picks this: cos θ = 0.50 gives 60°. The path difference is d sin θ, which is zero at θ = 0, so sin θ = 0.50 and θ = 30°.
Working d sin θ = mλ with m = 1: sin θ = 3.0 cm/6.0 cm = 0.50 → θ = 30°. Distractors: ΔD = λ/2 → sin θ = 0.25 → 14.5° ≈ 14°; θ = λ/d = 0.50 rad = 28.6° ≈ 29°; cos θ = 0.50 → 60°.
Monochromatic light passes through two slits 0.20 mm apart and falls on a screen 1.5 m away. The centers of the two third-order bright maxima, one on each side of the central maximum, are 18.0 mm apart. What is the wavelength of the light?
Answer and reasoning
A8.0 × 10⁻⁷ m A student who uses the whole 18.0 mm as ymax picks this. ymax is measured from the middle of the central bright band to one third-order maximum, so it is 9.0 mm.
B1.2 × 10⁻⁶ m A student who halves the distance correctly but leaves out the order number picks this, treating the band 9.0 mm from the center as first order. It is the third-order maximum, so the result must be divided by 3.
C4.0 × 10⁻⁷ mCorrect The third-order maximum is half of 18.0 mm, 9.0 mm, from the center. d(ymax/L) ≈ mλ gives λ = d ymax/(mL) = (0.20 × 10⁻³ m)(9.0 × 10⁻³ m)/(3 × 1.5 m) = 4.0 × 10⁻⁷ m.
D3.6 × 10⁻⁶ m A student who multiplies by m instead of dividing picks this: λ = m d ymax/L. Rearranging d(ymax/L) ≈ mλ for λ divides by m.
Working ymax = 18.0 mm/2 = 9.0 mm, m = 3. λ = d ymax/(mL) = (0.20 × 10⁻³)(9.0 × 10⁻³)/(3 × 1.5) = 4.0 × 10⁻⁷ m. Angle: 9.0 mm/1.5 m = 0.006 rad ≈ 0.34°, so the small-angle relation applies.
In a double-slit experiment, light of wavelength 400 nm is replaced by light of wavelength 600 nm, and the pair of slits is replaced by a pair whose separation is half as large. The screen is not moved. Assume small angles. By what factor does the distance between neighboring bright bands on the screen change?
Answer and reasoning
A1.33 times as great as before A student who thinks shorter wavelengths are spread more, as in a prism, picks this, multiplying by 400/600 and then by 2. The spacing λL/d is proportional to λ, so the longer wavelength spreads the bands more.
B3.00 times as great as beforeCorrect For small angles neighboring bright bands are λL/d apart. The wavelength is multiplied by 600/400 = 1.5 and d by 1/2, so the spacing is multiplied by 1.5 × 2 = 3.00.
C0.75 times as great as before A student who thinks the band spacing is proportional to the slit separation picks this: 1.5 × 1/2 = 0.75. The spacing is inversely proportional to d, so halving d doubles it.
D1.50 times as great as before A student who thinks the band spacing is set by the width of each slit, not by their separation, picks this, counting only the change in wavelength. Halving d doubles the spacing as well.
In a double-slit experiment, light of wavelength λ passes through two slits a distance d apart onto a screen a distance L away. Assuming small angles, which expression gives the distance y₁ from the middle of the central bright band to the middle of the first dark band on either side of it?
Answer and reasoning
A1.00λL/d A student who thinks dark bands occur where ΔD is a whole wavelength picks this, setting d(y₁/L) = λ. That is the position of the first-order bright maximum; the first dark band is where ΔD = λ/2.
B1.50λL/d A student who numbers the first dark band m = 1 in ΔD = (m + ½)λ picks this, giving it ΔD = 3λ/2. That is the second dark band; the first has m = 0 and ΔD = λ/2.
C0.50λL/dCorrect The first dark band has ΔD = λ/2. For small angles ΔD ≈ d(y/L), so d(y₁/L) = λ/2 and y₁ = 0.50λL/d, halfway between the central maximum and the first-order maximum at λL/d.
D0.25λL/d A student who thinks the distance y in d(y/L) is measured across the pattern, between the two first dark bands, halves the result and picks this. y is measured from the middle of the central bright band, so y₁ = 0.50λL/d.
Working First minimum: ΔD = λ/2 (m = 0 in (m + ½)λ). ΔD = d sin θ ≈ d(y₁/L) for small angles. d(y₁/L) = λ/2 → y₁ = λL/(2d) = 0.50λL/d. Distractors: ΔD = λ → 1.00λL/d; ΔD = 3λ/2 → 1.50λL/d; y measured across the pattern (m13) → half of 0.50 = 0.25λL/d.
The graph shows the intensity I of light on a screen against position y for monochromatic light passing through two identical narrow slits. The dashed curve is the envelope within which the bright peaks lie. One of the slits is then covered. Which statement describes the new pattern?
Answer and reasoning
AThe narrow peaks vanish, leaving the envelope's shape: one broad peak and faint side peaks.Correct With one slit covered there is no second wave to interfere with, so the evenly spaced narrow peaks disappear. What remains is the diffraction pattern of the open slit, which has the shape of the envelope: one broad central peak, as wide as the envelope's central region, with faint peaks beyond it.
BThe same narrow peaks stay in the same places, each one becoming less intense. A student who thinks the second slit only makes the same pattern brighter picks this: covering it should just dim every peak. The narrow peaks come from interference between the waves from the two slits, so they cannot survive when one slit is covered.
CA single narrow peak remains, located directly in line with the open slit. A student who thinks light goes straight through a slit picks this. The open slit is narrow, so its light still spreads out, giving a broad peak as wide as the envelope's central region.
DThe whole pattern shrinks to half its width, as half the opening is now blocked. A student who thinks less opening gives a narrower beam picks this. Covering one slit does not make the remaining slit any narrower, so its light spreads just as much; the broad envelope stays the same width while the narrow peaks disappear.
A lamp emits only blue light of wavelength 450 nm and red light of wavelength 650 nm. The light passes through a diffraction grating. The figure shows the central maximum C and the three bright lines nearest to it on one side of the pattern, J, K and L. Which line is the first-order maximum of the red light?
Answer and reasoning
ALine J, the nearest to C A student who thinks shorter wavelengths are deflected more picks this, expecting red to be nearest. d sin θ = λ puts the 450 nm first-order line at the smallest angle, so J is blue.
BLine K, between J and LCorrect The lines appear in order of increasing mλ: 450 nm (blue, first order) at J, 650 nm (red, first order) at K, and 2 × 450 nm = 900 nm (blue, second order) at L. The red first-order line is K.
CLine L, the farthest from C A student who treats every line as first order picks this, expecting red to be farthest. L is the second-order blue line, where mλ = 900 nm, which is more than the 650 nm of the red first-order line.
DC, the central maximum A student who counts the central maximum as the first-order maximum picks this. C is the zeroth-order maximum (m = 0), where every wavelength arrives in step; the first-order red maximum is where d sin θ = 650 nm.
Working Lines in order of mλ: 450 nm (m = 1, blue) → J; 650 nm (m = 1, red) → K; 900 nm (m = 2, blue) → L; next 1300 nm (m = 2, red), not shown. Figure drawn for d = 2.0 μm, L = 1.0 m: y = 0.23 m, 0.34 m, 0.50 m.
A double slit and a diffraction grating have the same spacing d between neighboring slits. Each is lit in turn by the same monochromatic light. How does the grating's pattern of bright maxima compare with the double slit's?
Answer and reasoning
AIts maxima are at smaller angles, since the grating is much wider. A student who thinks the overall width of the grating sets the spread picks this, treating the grating like one wide opening. The maxima depend on the spacing between neighboring slits, which is the same here.
BIts maxima are at larger angles, since more slits spread the light more. A student who credits the wide spread of real gratings to their number of slits picks this. Real gratings spread light widely because their slit spacing is very small; with the same d, the angles are the same.
CIt has many more maxima, one behind each of its many slits. A student who thinks light passes straight through each slit picks this. The waves from all the slits spread and overlap; bright maxima occur only where d sin θ = mλ, at the same angles as for the double slit.
DIts maxima are at the same angles, since slit spacing sets them for both.Correct At a bright maximum of a grating, the waves from every pair of neighboring slits differ in path by a whole number of wavelengths, d sin θ = mλ, the same condition as for a double slit with the same d. The extra slits do not move the maxima.
Light containing two wavelengths, λ₁ and λ₂, passes through the same pair of narrow slits and falls on a screen. The third-order bright maximum of λ₁ is at the same position on the screen as the second-order bright maximum of λ₂. Which expression gives λ₂?
Answer and reasoning
A0.67λ₁ A student who rearranges d sin θ = mλ as λ = m d sin θ writes λ₁ = 3d sin θ and λ₂ = 2d sin θ, and picks this. The order divides: λ = d sin θ/m, so λ₂ = (3/2)λ₁.
B2.00λ₁ A student who counts the central maximum as the first-order maximum gives the 'third-order' maximum of λ₁ a path difference of 2λ₁ and the 'second-order' maximum of λ₂ a path difference of λ₂, and picks this. The central maximum is order zero, so the orders are 3 and 2.
C1.67λ₁ A student who puts bright maxima where the path difference is an odd number of half-wavelengths places the third bright maximum at 5λ₁/2 and the second at 3λ₂/2, giving λ₂ = 5λ₁/3, and picks this. Bright maxima have whole-wavelength path differences, mλ.
D1.50λ₁Correct Both maxima are at the same angle, so they have the same path length difference d sin θ. Bright maxima have ΔD = mλ, so 3λ₁ = 2λ₂ and λ₂ = (3/2)λ₁ = 1.50λ₁.
Working Same position on the screen means the same angle θ, so the same path length difference ΔD = d sin θ. Bright maxima: ΔD = mλ. Third order of λ₁: d sin θ = 3λ₁; second order of λ₂: d sin θ = 2λ₂. So 2λ₂ = 3λ₁ and λ₂ = (3/2)λ₁ = 1.50λ₁. No small-angle approximation is needed.
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
