14.7 Diffraction

1. ALight changing direction as it passes at an angle from air into a glass of water
   A student who confuses diffraction with refraction picks this. Light changing direction as it crosses from one medium into another is refraction, caused by the change in the speed of light at the boundary.
2. BWhite light separating into a band of colors as it passes through a glass prism
   A student who associates every rainbow of colors with diffraction picks this. A prism separates colors because each color travels at a different speed in glass and is refracted by a different amount; that is dispersion, not diffraction.
3. COcean waves spreading into the calm water behind a narrow harbor entrance Correct
   Diffraction is a wave spreading out as it passes the edge of an obstacle or goes through an opening. The waves passing through the harbor entrance spread out into the sheltered regions on either side behind the harbor wall, in the same medium.
4. DA shout returning as an echo a few seconds later from the face of a distant cliff
   A student who confuses diffraction with reflection picks this. An echo is sound that has bounced back from a surface, which is reflection; diffraction is spreading through an opening or around an edge.

CED 14.7.A.1 · Read this in Fix

1. ASound's wavelengths are comparable to the doorway's width; light's are far smaller. Correct
   Diffraction is most pronounced when the opening is comparable in size to the wavelength. The sounds of speech have wavelengths from roughly 0.1 m to a few meters, comparable to the width of a doorway (about 1 m), so sound spreads widely beyond the doorway. Visible light has wavelengths of about 5 × 10⁻⁷ m, millions of times smaller than the doorway, so its spreading is far too small to notice.
2. BLight is a ray that travels in straight lines, while sound is a wave that spreads.
   A student who thinks light can only travel as straight rays picks this. Light is also a wave and does diffract; through a slit comparable to its wavelength it spreads widely. It hardly spreads through a doorway because the doorway is enormous compared with its wavelength.
3. CSound is carried around the corner by moving air, which does not carry light.
   A student who pictures sound as something carried along by flowing air picks this. The air only vibrates about its position; sound reaches the hallway because the sound wave spreads out as it passes through the doorway.
4. DLight moves too fast to bend around the doorway's edges; sound is slower.
   A student who thinks faster waves travel in straighter lines picks this. The spreading depends on the wavelength compared with the width of the opening, not on the speed of the wave; light spreads widely through a narrow enough slit.

CED 14.7.A.2 · Read this in Fix

1. ALight bounces off the two edges of the slit into a few particular directions, which become the bright bands.
   A student who looks for a ray explanation picks this. Reflection from the thin edges of the slit does not produce the pattern; the bands come from interference of wavefronts from all across the opening.
2. BThe two edges of the slit act as two point sources, like the two slits of a double slit.
   A student who maps the single slit onto a double slit picks this. Every part of the opening sends out wavefronts, not only the edges; at the first dark band the edges' wavefronts are in fact in step, and the canceling happens between pairs of wavefronts across the opening.
3. CLight is refracted as it passes through the slit, bending by different amounts into separate bands.
   A student who confuses diffraction with refraction picks this. Refraction is the bending of light as it crosses into a different medium; passing through a slit in air is not such a crossing. The bands come from interference of wavefronts from all across the opening.
4. DWavefronts from different parts of the slit interfere, adding in some directions and canceling in others. Correct
   Diffraction spreads the light out beyond the slit, and wavefronts from different parts of the opening overlap on the screen. Where they arrive in step they add (bright bands); where they arrive out of step they cancel (dark bands).

CED 14.7.A.3 · Read this in Fix

1. AThe center gets light passing straight through; the dark band is in the edges' shadow.
   A student who uses a ray model picks this. The first dark band lies inside the region lit by the diffracted light, between bright bands; it is caused by interference, not by the edges blocking light traveling in straight lines.
2. BWavefronts arrive mostly in step at the center and cancel in pairs at the dark band. Correct
   At the middle of the central band the wavefronts from all parts of the opening travel nearly equal distances, so they arrive in step, or very nearly so, and add. At the first dark band the path length difference between the edges is one wavelength, so each wavefront from the top half of the opening meets one from the bottom half half a wavelength out of step, and they cancel in pairs.
3. CWavefronts add at the center, and at the dark band their energy is destroyed.
   A student who thinks canceling waves destroy their energy picks this. The wavefronts do cancel at the dark band, but energy is conserved: it is redistributed to the bright bands, which are correspondingly brighter.
4. DThe light reaching the dark band is dimmer before it arrives, so it cannot add up.
   A student who thinks brightness decides whether wavefronts add or cancel picks this. The wavefronts reaching the dark band come from the same opening as those reaching the center and are about as strong; they cancel because they arrive out of step, not because they are dim.

CED 14.7.A.4.i · Read this in Fix

1. AA row of bright and dark bands, as from a narrow slit
   A student who thinks every opening gives the same pattern picks this. A slit is narrow in only one direction and spreads light into a row of bands; a circular hole is narrow in every direction and produces rings.
2. BA single bright spot the same size and shape as the hole
   A student who uses a ray model, in which the light passes straight through, picks this. A hole small enough to diffract the light spreads it into a disk much larger than the hole, with rings around it.
3. CA bright central disk surrounded by bright and dark rings Correct
   The diffraction pattern depends on the shape of the opening. A small circular hole is narrow in every direction across it, so the light spreads equally in all directions, giving a central bright disk surrounded by alternating dark and bright rings.
4. DA broad, evenly lit disk much larger than the hole, with no rings
   A student who thinks wavefronts from a single opening cannot interfere picks this, expecting the light to spread out without forming bands. Wavefronts from different parts of the hole do interfere, producing the dark and bright rings around the central disk.

CED 14.7.A.5 · Read this in Fix

1. A1.2 × 10⁻² m
   A student who takes the central bright band as an image of the slit picks this, reading the band's full width, 12 mm, as the slit width. The band is the result of diffraction and is far wider than the slit; a = λL/y_min = 1.3 × 10⁻⁴ m.
2. B6.5 × 10⁻⁵ m
   A student who uses the full width of the central band, 12 mm, as y_min picks this: (6.50 × 10⁻⁷)(1.2)/(1.2 × 10⁻²) = 6.5 × 10⁻⁵ m. y_min is measured from the middle of the pattern to the first dark band, 6 mm.
3. C1.3 × 10⁻⁴ m Correct
   The first dark bands (intensity zero) are at y = ±6 mm, so y_min = 6.0 × 10⁻³ m for m = 1. a(y_min/L) ≈ mλ gives a = λL/y_min = (6.50 × 10⁻⁷ m)(1.2 m)/(6.0 × 10⁻³ m) = 1.3 × 10⁻⁴ m.
4. D8.7 × 10⁻⁵ m
   A student who measures to the first side bright band, at about 9 mm, picks this: (6.50 × 10⁻⁷)(1.2)/(9 × 10⁻³) = 8.7 × 10⁻⁵ m. For a single slit, a(y_min/L) ≈ mλ gives the positions of the dark bands, the zeros of intensity.

Working Graph: intensity zeros at y = ±6, ±12, ±18 mm → y_min(1) = 6.0 × 10⁻³ m. a = λL/y_min = (6.50 × 10⁻⁷ m)(1.2 m)/(6.0 × 10⁻³ m) = 1.3 × 10⁻⁴ m. (y_min/L = 0.005, small angle.) Errors: central width as a → 1.2 × 10⁻² m; 12 mm as y_min → 6.5 × 10⁻⁵ m; first side maximum (≈ 9 mm) → 8.7 × 10⁻⁵ m.

CED 14.7.A.6 · Read this in Fix

1. AThey spread out more beyond the wider gap, gap 2.
   A student who thinks a wider opening makes a wave spread more picks this. More of each wavefront passes through gap 2, but it emerges mostly as a straight band; the narrower gap, about one wavelength wide, produces the greater spreading.
2. BThey spread out more beyond gap 1 than beyond gap 2. Correct
   Diffraction is most pronounced when the width of the opening is comparable to the wavelength. Gap 1 is about one wavelength wide, so the waves beyond it spread out widely; gap 2 is about five wavelengths wide, so the waves pass through it largely as a straight band and spread much less.
3. CThey spread out by the same amount beyond both of the gaps.
   A student who thinks the spreading is set by the wave alone picks this. The waves are identical, but the spreading depends on the wavelength compared with the width of the gap, and the two gaps differ in width.
4. DThey do not spread out beyond either of the two gaps.
   A student who thinks a wave diffracts only through a gap narrower than its wavelength picks this. Diffraction occurs at any opening and is most pronounced for a gap comparable to the wavelength, like gap 1; it is small, but still present, for the wider gap 2.

Working Read from the diagram: gap 1 ≈ λ (same as the wavefront spacing), gap 2 ≈ 5λ. Spreading is greatest for a ≈ λ → more spreading beyond gap 1.

CED 14.7.A.2 · Read this in Fix

1. AReplacing the slit with a wider slit
   A student who thinks the central band is an image of the slit picks this. Because a is in the denominator of y_min ≈ λL/a, a wider slit makes the central band narrower.
2. BReplacing the red laser with a blue laser
   A student who thinks blue light spreads more, as it bends more in a prism, picks this. Blue light has a shorter wavelength than red, and y_min ≈ λL/a is proportional to λ, so the central band gets narrower.
3. CReplacing the slit with a narrower slit Correct
   The first dark band is at y_min ≈ λL/a, so the central bright band, 2λL/a wide, gets wider as the slit width a decreases. A narrower opening makes the light spread out more.
4. DUsing a brighter laser of the same red color
   A student who thinks a brighter beam spreads more picks this. The positions of the dark bands depend only on λ, a and L; a brighter laser of the same wavelength makes the bands brighter without changing their width.

Working Central band width = 2y_min(1) ≈ 2λL/a. Wider when a decreases, λ increases or L increases; independent of brightness.

CED 14.7.A.4 · Read this in Fix

1. AThe total distance each one travels from the opening to the screen
   A student who attends to the distances themselves picks this. Two wavefronts that each travel about the same total distance can still add or cancel; what matters is the difference between the distances, compared with the wavelength.
2. BNothing; wavefronts from one opening cannot interfere, as that needs two
   A student who thinks interference needs two separate slits picks this. Wavefronts from different parts of a single opening do interfere, and their interference produces the bright and dark bands of the single-slit pattern.
3. CHow bright each wavefront is when it reaches that point on the screen
   A student who looks for the cause of bright and dark bands in the brightness of each wave picks this. Whether two wavefronts add or cancel depends on whether they arrive in step, which depends on their path length difference; brightness does not decide it.
4. DThe difference in the distances they travel, compared with the wavelength Correct
   The amount of interference depends on the path length difference ΔD. If ΔD is a whole number of wavelengths the wavefronts arrive in step and add; if it is an odd number of half wavelengths they arrive half a cycle out of step and cancel.

CED 14.7.A.4.ii · Read this in Fix

1. A6.0 × 10⁻⁷ m Correct
   T and M are a/2 = 1.0 × 10⁻⁶ m apart. In the right triangle formed by T, M and the foot of the dashed line, the hypotenuse is a/2 and the angle at T equals θ, so the extra path of the wavefront from M is ΔD = (a/2) sin θ = (1.0 × 10⁻⁶ m)(sin 37°) = 6.0 × 10⁻⁷ m.
2. B8.0 × 10⁻⁷ m
   A student who uses the cosine by habit picks this: (1.0 × 10⁻⁶ m)(cos 37°) = 8.0 × 10⁻⁷ m, the length of the dashed side next to θ. The extra path is the side opposite θ: ΔD = (a/2) sin θ.
3. C7.5 × 10⁻⁷ m
   A student who treats tan θ and sin θ as interchangeable picks this: (1.0 × 10⁻⁶ m)(tan 37°) = 7.5 × 10⁻⁷ m. At 37° they differ; the extra path in the right triangle with hypotenuse a/2 is (a/2) sin θ.
4. D1.2 × 10⁻⁶ m
   A student who applies ΔD = a sin θ to any two points of the slit picks this: (2.0 × 10⁻⁶ m)(sin 37°) = 1.2 × 10⁻⁶ m. That is the path length difference between the two edges, a apart; T and M are only a/2 apart, so ΔD = (a/2) sin θ = 6.0 × 10⁻⁷ m.

Working Right triangle T–M–F (F = foot of the perpendicular from T on M's ray): hypotenuse TM = a/2 = 1.0 × 10⁻⁶ m, angle at T = θ. ΔD = MF = (a/2) sin θ = (1.0 × 10⁻⁶ m)(sin 37°) = 6.0 × 10⁻⁷ m. Errors: rote a sin θ → 1.2 × 10⁻⁶ m; cos → 8.0 × 10⁻⁷ m; tan → 7.5 × 10⁻⁷ m.

CED 14.7.A.4.iii · Read this in Fix

1. A7.5 × 10⁻⁷ m
   A student who uses the cosine picks this: (6.00 × 10⁻⁷ m)/0.80 = 7.5 × 10⁻⁷ m. With θ measured from the normal, the path length difference is a sin θ, not a cos θ.
2. B1.0 × 10⁻⁶ m Correct
   At the first dark band the path length difference between wavefronts from the two edges of the slit is one wavelength: ΔD = a sin θ = λ. So a = λ/sin θ = (6.00 × 10⁻⁷ m)/0.60 = 1.0 × 10⁻⁶ m. (The angle is large, so the small-angle form a(y_min/L) ≈ mλ does not apply; ΔD = a sin θ does.)
3. C5.0 × 10⁻⁷ m
   A student who takes the edges' path length difference at a dark band to be half a wavelength, a sin θ = λ/2, picks this: (3.0 × 10⁻⁷ m)/0.60 = 5.0 × 10⁻⁷ m. For a single slit the first dark band is where a sin θ = λ, so that the two halves of the slit cancel each other.
4. D3.6 × 10⁻⁷ m
   A student who multiplies by sin θ instead of dividing picks this: (6.00 × 10⁻⁷ m)(0.60) = 3.6 × 10⁻⁷ m. From a sin θ = λ, a = λ/sin θ, which must be larger than λ.

Working First dark band: a sin θ = λ → a = λ/sin θ = 6.00 × 10⁻⁷ m / 0.60 = 1.0 × 10⁻⁶ m. Errors: cos → 7.5 × 10⁻⁷ m; λ/2 condition → 5.0 × 10⁻⁷ m; λ sin θ → 3.6 × 10⁻⁷ m.

CED 14.7.A.4.iii · Read this in Fix

1. A1.0 × 10⁻² m Correct
   For small angles, a(y_min/L) ≈ mλ with m = 1: y_min = λL/a = (5.0 × 10⁻⁷ m)(2.0 m)/(1.0 × 10⁻⁴ m) = 1.0 × 10⁻² m, that is, 1.0 cm. (y_min/L = 0.005, so the small-angle approximation is valid.)
2. B5.0 × 10⁻³ m
   A student who takes the first dark band to be where a(y/L) = λ/2 picks this: λL/(2a) = 5.0 × 10⁻³ m. For a single slit the first dark band is where a(y_min/L) ≈ λ.
3. C5.0 × 10⁻⁵ m
   A student who thinks the central bright band is a straight-through image of the slit picks this, placing the first dark band at the edge of that image, a/2 = 0.050 mm from the center. Diffraction spreads the light far beyond the slit's width: y_min = λL/a = 1.0 cm.
4. D1.0 × 10⁻⁵ m
   A student who substitutes the slit width as 0.10 without converting millimeters to meters picks this: (5.0 × 10⁻⁷)(2.0)/0.10 = 1.0 × 10⁻⁵ m. The slit width is 0.10 mm = 1.0 × 10⁻⁴ m.

Working y_min = mλL/a = (1)(5.0 × 10⁻⁷ m)(2.0 m)/(1.0 × 10⁻⁴ m) = 1.0 × 10⁻² m. Errors: λ/2 condition → 5.0 × 10⁻³ m; geometric image edge a/2 → 5.0 × 10⁻⁵ m; a left in mm (0.10) → 1.0 × 10⁻⁵ m.

CED 14.7.A.4.iv · Read this in Fix

1. A1.00 mm
   A student who thinks a narrower slit gives a narrower pattern picks this, treating y_min as proportional to both λ and a: (1/2)(1/2) × 4.00 mm = 1.00 mm. The slit width is in the denominator of y_min ≈ mλL/a, so halving a doubles y_min.
2. B16.0 mm
   A student who thinks shorter-wavelength light spreads more picks this, treating y_min as inversely proportional to both λ and a: 2 × 2 × 4.00 mm = 16.0 mm. y_min is proportional to λ, so halving the wavelength halves y_min.
3. C2.00 mm
   A student who thinks the spreading depends only on the light, not on the slit, picks this, halving y_min for the halved wavelength and ignoring the narrower slit. Halving a doubles y_min, which cancels the effect of halving λ.
4. D4.00 mm Correct
   y_min ≈ mλL/a. Halving λ halves y_min, and halving a doubles it, so the two changes cancel: y_min stays 4.00 mm.

Working y_min ∝ λ/a (L fixed). New/old = (1/2)/(1/2) = 1 → 4.00 mm. Errors: ∝ λa → ×1/4 = 1.00 mm; ∝ 1/(λa) → ×4 = 16.0 mm; ∝ λ only → ×1/2 = 2.00 mm.

CED 14.7.A.4.iv · Read this in Fix

1. AHorizontally, across the narrow width of the slit Correct
   The diffraction pattern depends on the shape of the opening, and the light spreads most across the opening's narrow dimension. The slit is narrow horizontally, so the pattern is a horizontal row of bright and dark bands.
2. BVertically, along the long length of the slit
   A student who thinks the pattern is an image of the opening picks this, expecting a tall stripe shaped like the slit. The spreading is greatest across the slit's narrow width, which is horizontal.
3. CEqually in all directions, as a set of rings
   A student who thinks every opening produces the same kind of pattern picks this. Rings come from a small circular hole, which is narrow in every direction; a slit is narrow in one direction only.
4. DNot at all; the beam stays a small, bright spot
   A student who thinks light always passes through openings in straight lines picks this. Light is a wave, and a slit narrow enough to be comparable to its wavelength spreads it out widely.

CED 14.7.A.5 · Read this in Fix

1. AThe opening for X is wider than the opening used for Y.
   A student who thinks the central band is an image of the opening picks this. The wider central peak comes from the narrower opening: a is in the denominator of y_min ≈ λL/a.
2. BThe openings are equally wide, as the same laser was used.
   A student who thinks the pattern is set by the light alone picks this. With the same wavelength and screen distance, the only way to get central peaks of different widths is openings of different widths.
3. CPattern Y came from two slits, as its maxima are closer.
   A student who takes closely spaced bands as the sign of two slits picks this. Pattern Y has the same shape as X, a central peak twice as wide as its much weaker side peaks, which is the pattern of a single opening. Its bands are closer together only because its opening is wider.
4. DThe opening for X is narrower than the opening for Y. Correct
   Both patterns are single-slit patterns (one wide central peak with much weaker side peaks). The first zeros of X are at ±6 mm and those of Y at ±3 mm. With λ and L the same, y_min ≈ λL/a, so X's opening is narrower, half as wide as Y's.

Working X: zeros at ±6 mm; Y: zeros at ±3 mm. y_min ∝ 1/a (same λ, L) → a_X = a_Y/2: X's opening is narrower.

CED 14.7.A.6 · Read this in Fix

1. AnλL/a
   A student who takes the wavelength in water to be nλ picks this. Light is slower in water at the same frequency, so its wavelength is λ/n, shorter than in air, and the pattern narrows rather than widens.
2. BλL/(na) Correct
   In water the frequency is unchanged and the speed is c/n, so the wavelength is λ/n. The first dark band satisfies a(y_min/L) ≈ λ/n, giving y_min ≈ λL/(na): the pattern is narrower in water.
3. CλL/a
   A student who assumes the light keeps its air wavelength in water, because its color is unchanged, picks this. The frequency is unchanged, but the speed falls to c/n, so the wavelength and the pattern shrink by the factor n.
4. DλL/(2na)
   A student who correctly uses the wavelength λ/n but puts the first dark band where the path difference between the slit's edges is half a wavelength picks this. For a single slit the first dark band is where that difference is a whole wavelength, a(y_min/L) ≈ λ/n.

Working Frequency unchanged, speed in water c/n, so wavelength in water λ_w = λ/n. First dark band (m = 1): a(y_min/L) ≈ 1·λ_w, so y_min ≈ λ_w L/a = λL/(na). The pattern shrinks by the factor n compared with air.

CED 14.7.A.4.iv · Read this in Fix

1. A0.50λ₁ Correct
   With λ₁ the first dark bands are at λ₁L/a on each side of the middle, so W = 2λ₁L/a. With λ₂ the fourth-order dark band is at 4λ₂L/a. Setting 4λ₂L/a = 2λ₁L/a gives λ₂ = 0.50λ₁.
2. B0.25λ₁
   A student who takes the full width of the central bright band as the distance y_min to the first dark band sets W = λ₁L/a, gets 4λ₂ = λ₁, and picks this. The first dark bands are on both sides of the middle, each λ₁L/a away, so the central band's width is 2λ₁L/a.
3. C0.29λ₁
   A student who puts the dark bands where the path length difference is (m − ½)λ gets W = λ₁L/a and places the fourth-order dark band at (7/2)λ₂L/a, so λ₂ = (2/7)λ₁ ≈ 0.29λ₁. For a single slit the dark bands are where a sin θ = mλ, so y_min = mλL/a.
4. D2.00λ₁
   A student who thinks shorter wavelengths spread out more treats the band positions as inversely proportional to the wavelength, sets 4/λ₂ = 2/λ₁, and picks this. The positions are proportional to the wavelength, y_min = mλL/a, so the fourth-order band reaches W only when λ₂ is half of λ₁.

Working Small angles: a(y_min/L) ≈ mλ, so y_min = mλL/a. With λ₁ the first-order dark bands lie at ±λ₁L/a, so W = 2λ₁L/a. With λ₂ the fourth-order dark band is at 4λ₂L/a. Setting 4λ₂L/a = 2λ₁L/a gives λ₂ = λ₁/2 = 0.50λ₁.

CED 14.7.A.4.iv · Read this in Fix

1. A6.0 mm Correct
   For small angles y_min = mλL/a. With λ and L unchanged, y_min is proportional to m/a: the second-order band is twice as far out as the first, and halving the slit width doubles every distance again, so y = 4 × 1.5 mm = 6.0 mm.
2. B1.5 mm
   A student who treats the central band as an image of the slit expects a narrower slit to shrink the pattern, halves the distances, doubles for the second order, and picks this. Diffraction spreads the light more through a narrower opening: y_min is inversely proportional to a, so halving a doubles the distances.
3. C9.0 mm
   A student who places the dark bands where the path length difference is (m − ½)λ puts the first at ½λL/a and the second at (3/2)λL/a, three times as far out, then doubles for the narrower slit: 1.5 mm × 3 × 2 = 9.0 mm. The dark bands are where a sin θ = mλ, so the second-order band is twice as far out as the first, not three times.
4. D3.0 mm
   A student who thinks how much light spreads depends on its wavelength alone ignores the change of slit, doubles 1.5 mm for the second order, and picks this. The pattern depends on the slit width too: y_min = mλL/a, so the slit half as wide doubles every distance.

Working Small angles: y_min = mλL/a, so with λ and L fixed, y_min ∝ m/a. Going from m = 1 to m = 2 doubles the distance, and halving a doubles it again: y = (1.5 mm)(2)(2) = 6.0 mm.

CED 14.7.A.4.iv · Read this in Fix