Diffraction - AP Physics 2

1. A speaker emits sound of wavelength $\lambda$ toward a wall with a narrow opening. The opening width is kept fixed while the speaker is adjusted to emit a longer wavelength $2\lambda$ at the same volume. Compared to before, the sound emerging from the opening spreads
   Answer: more, because the wavelength-to-opening ratio increases This question tests understanding of diffraction. Diffraction is the spreading of waves through openings, with the amount of spreading determined by the wavelength-to-opening ratio. When the wavelength increases from λ to 2λ while the opening width stays fixed, the wavelength-to-opening ratio doubles, causing more diffraction and greater spreading of the emerging sound. This occurs because longer wavelengths diffract more through a given opening size. Choice D incorrectly claims larger wavelength reduces diffraction, which contradicts the fundamental principle that diffraction increases with wavelength for a fixed opening. Remember that diffraction is strongest when the opening is comparable to the wavelength.
2. A ripple tank sends plane water waves of wavelength $\lambda$ toward four gaps in a barrier. Each gap is illuminated with the same wave speed. The gap widths are: (1) $0.5\lambda$, (2) $\lambda$, (3) $2\lambda$, (4) $4\lambda$. Which gap produces the greatest diffraction (widest spreading) after the barrier?
   Answer: Gap width $0.5\lambda$ This question tests understanding of diffraction. Diffraction is the spreading of waves as they pass through an opening or around an obstacle, and the amount of spreading depends on the ratio of wavelength to opening size. When the opening is much larger than the wavelength, waves pass through with minimal bending, but when the opening is comparable to or smaller than the wavelength, significant spreading occurs. The gap with width 0.5λ has the smallest opening relative to the wavelength, producing the greatest diffraction. A common misconception (choice A) is that larger openings cause more diffraction, but this confuses the amount of wave energy passing through with the angular spreading of the wave. Remember: diffraction is strongest when the opening size is comparable to or smaller than the wavelength.
3. A laser beam passes through a single narrow slit and forms a diffraction pattern on a screen. In Trial 1, the slit width is $w$; in Trial 2, the slit width is reduced to $w/2$ while wavelength stays the same. Which trial produces a wider central bright region due to diffraction?
   Answer: Trial 2, because decreasing $w$ increases diffraction spreading This question tests understanding of diffraction. Diffraction causes light passing through a slit to spread out, creating a pattern with a central bright region whose width is inversely proportional to the slit width. When the slit width is reduced from w to w/2 while keeping wavelength constant, the ratio λ/w doubles, causing increased diffraction. This increased diffraction results in a wider central bright region on the screen in Trial 2. Choice A incorrectly assumes that wider slits cause more spreading, which is the opposite of what occurs - this represents a fundamental misconception about diffraction. The key principle is that narrower slits (smaller w) produce wider diffraction patterns.
4. A monochromatic laser passes through a single narrow slit. The slit width is much larger than the wavelength $\lambda$, producing only slight spreading. Which change produces the greatest diffraction spreading on a distant screen?
   Answer: Decrease the slit width while keeping $\lambda$ the same This question tests understanding of diffraction. Diffraction is the bending and spreading of waves as they pass through openings, with the effect becoming more pronounced when the opening size approaches the wavelength. Since the slit width is currently much larger than the wavelength (producing only slight spreading), decreasing the slit width will increase the diffraction effect by making the slit-to-wavelength ratio smaller. The spreading angle in single-slit diffraction is approximately λ/a, where a is the slit width, so reducing a increases the spreading. Choice A incorrectly assumes intensity affects diffraction patterns, but diffraction depends only on geometry and wavelength, not on the brightness of the light. Remember that diffraction is strongest when the opening is comparable to the wavelength.
5. A tank produces circular water waves of wavelength $\lambda$ that encounter a small post. In trial 1 the post diameter is $0.5\lambda$; in trial 2 it is $3\lambda$. The wave speed is unchanged. In which trial will the waves bend around the post the most?
   Answer: Trial 1, because the obstacle size is smaller relative to $\lambda$ This question tests understanding of diffraction. Diffraction is the bending of waves around obstacles, with more bending occurring when the obstacle size is comparable to or smaller than the wavelength. In trial 1, the post diameter (0.5λ) is smaller than the wavelength, allowing waves to bend significantly around it. In trial 2, the post diameter (3λ) is much larger than the wavelength, causing minimal bending as waves are mostly blocked. Choice A incorrectly relates obstacle size to wave intensity, but diffraction depends on the size ratio, not intensity. The strategy is: smaller obstacles relative to wavelength produce more diffraction around them.
6. A microwave of wavelength $\lambda$ passes through a rectangular opening of width $w$. In Setup A, $w=\lambda/3$; in Setup B, $w=4\lambda$. After the opening, the wavefront spreads. Which setup produces the greatest diffraction?
   Answer: Setup A, because $w$ is much smaller than $\lambda$ This question tests understanding of diffraction. Diffraction is the bending and spreading of waves as they pass through an opening, with the effect being most pronounced when the opening width w is small compared to the wavelength λ. In Setup A, w = λ/3 means the opening is much smaller than the wavelength, causing significant diffraction and spreading. In Setup B, w = 4λ means the opening is much larger than the wavelength, resulting in minimal diffraction with waves continuing mostly straight. Choice A represents a common misconception that larger openings cause more bending, when the opposite is true - smaller openings relative to wavelength cause greater diffraction. The key principle is that diffraction is strongest when the opening is comparable to or smaller than the wavelength.
7. Two identical ripple tanks produce plane waves that pass through the same slit of width $w$. Tank 1 produces wavelength $\lambda_1$ and Tank 2 produces wavelength $\lambda_2$, where $\lambda_2>\lambda_1$. The waves spread after the slit. Which tank shows the greater diffraction?
   Answer: Tank 2, because the ratio $\lambda/w$ is larger This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through an opening, with the amount of spreading determined by the ratio λ/w, where λ is wavelength and w is slit width. Since Tank 2 has a longer wavelength (λ₂ > λ₁) and both tanks use the same slit width w, Tank 2 has a larger λ/w ratio. This larger ratio means Tank 2 produces greater diffraction and more spreading of the waves after the slit. Choice A incorrectly relates frequency to diffraction strength, failing to recognize that longer wavelengths (lower frequencies) actually diffract more for a given opening size. Remember that diffraction increases as wavelength increases relative to the opening size.
8. A loudspeaker emits sound of wavelength $\lambda$ toward a doorway. In Trial 1 the doorway width is $w=0.8\lambda$; in Trial 2 it is widened to $w=2.5\lambda$. The sound spreads into the hallway after passing through the doorway. Which trial shows the greatest diffraction into the hallway?
   Answer: Trial 1, because $w$ is closer to (or smaller than) $\lambda$ This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through an opening, with the effect being strongest when the opening width w is comparable to or smaller than the wavelength λ. In Trial 1, w = 0.8λ means the doorway width is slightly smaller than the wavelength, producing significant diffraction. In Trial 2, w = 2.5λ means the doorway is much wider than the wavelength, resulting in minimal diffraction and the sound continuing mostly straight through. Choice A incorrectly focuses on transmitted intensity rather than the geometric relationship between opening size and wavelength - this misconception confuses diffraction with transmission. Remember that maximum diffraction occurs when the opening is about the same size as or smaller than the wavelength.
9. Two waves approach identical narrow openings. Wave 1 has wavelength $2\,\text{cm}$; wave 2 has wavelength $6\,\text{cm}$. Both travel at the same speed in the medium. Which wave diffracts more after passing through the opening?
   Answer: Wave 2, because its wavelength is larger relative to the opening This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through an opening, with the amount of spreading determined by the ratio of wavelength to opening size. Wave 2 has a wavelength of 6 cm, which is three times larger than wave 1's wavelength of 2 cm. Since both pass through identical openings, wave 2 has a larger wavelength-to-opening ratio and therefore diffracts more. Choice A incorrectly relates frequency to diffraction, but higher frequency means shorter wavelength and less diffraction for a given opening. The key principle: for identical openings, longer wavelength produces more diffraction.
10. Water waves with wavelength $\lambda$ pass through a single slit of fixed width $w$. The student repeats the trial using waves of wavelength $2\lambda$ while keeping the slit width and water depth the same. In both trials, the wave spreads after passing through the slit. Compared to the $\lambda$ trial, how does the spreading change for the $2\lambda$ trial?
    Answer: It is greater, because the ratio $\lambda/w$ is larger This question tests understanding of diffraction. Diffraction occurs when waves encounter an opening or obstacle, with the amount of spreading determined by the ratio λ/w, where λ is wavelength and w is the opening width. When wavelength increases while the slit width remains constant, the ratio λ/w increases, resulting in greater diffraction and more spreading of the waves. In this case, doubling the wavelength from λ to 2λ while keeping w constant doubles the ratio λ/w, causing increased spreading. Choice C incorrectly assumes that diffraction depends only on slit width, ignoring the crucial role of wavelength - this represents a misconception about what controls diffraction. The transferable strategy is that diffraction increases when wavelength increases relative to the opening size.
11. A ripple tank produces waves of wavelength $\lambda$ that pass through a gap and then spread. The motor is adjusted to double the amplitude while keeping $\lambda$ and the gap width constant. Compared with before, the diffraction (angular spreading) is
    Answer: unchanged, because diffraction depends on $\lambda$ relative to gap width This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through an opening, and the angular spreading depends only on the ratio of wavelength to gap width, not on wave amplitude or intensity. Doubling the amplitude increases the energy and intensity of the waves but does not change the wavelength or gap width, so the diffraction angle remains unchanged. Choice A incorrectly assumes amplitude affects diffraction, but this confuses wave energy with the geometric spreading pattern. Remember: diffraction depends only on the wavelength-to-opening ratio, not on wave amplitude or intensity.
12. Microwaves of wavelength $3.0\,\text{cm}$ pass through a square opening of side $6.0\,\text{cm}$. In a second setup, radio waves of wavelength $30\,\text{cm}$ pass through the same opening. Both have the same power. Which setup shows greater diffraction beyond the opening?
    Answer: Radio waves, because $\lambda$ is larger compared with the opening This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through an opening, with the amount of spreading determined by the ratio of wavelength to opening size. The radio waves (λ = 30 cm) have a ratio of 30/6 = 5, while the microwaves (λ = 3 cm) have a ratio of 3/6 = 0.5. Since the radio waves have a much larger wavelength relative to the opening, they experience greater diffraction. Choice C incorrectly suggests that wave speed varies between different electromagnetic waves in air, but all EM waves travel at speed c in air. Remember: larger wavelength relative to opening size means more diffraction.
13. Ocean waves of wavelength about $20\ \text{m}$ approach a breakwater with a gap. In case 1 the gap is $5\ \text{m}$ wide; in case 2 the gap is $40\ \text{m}$ wide. Wave height and speed are the same. Which case produces the greatest bending into the sheltered region?
    Answer: Case 1, because the gap is much smaller than the wavelength This question tests understanding of diffraction. Diffraction is the bending and spreading of waves as they pass through openings, with maximum spreading when the opening is comparable to or smaller than the wavelength. The ocean waves have wavelength 20 m, while Case 1 has a 5 m gap (much smaller than wavelength) and Case 2 has a 40 m gap (larger than wavelength). Case 1 produces greater diffraction because the gap (5 m) is much smaller than the wavelength (20 m), causing waves to spread dramatically into the sheltered region. Choice A incorrectly claims wider gaps increase diffraction, but actually narrower gaps relative to wavelength increase diffraction. The principle is that diffraction is strongest when the opening is comparable to or smaller than the wavelength.
14. Two identical slits each of width $a$ are used separately with different light. Setup R uses red light ($\lambda_R$) and setup V uses violet light ($\lambda_V$), with $\lambda_R>\lambda_V$ and equal brightness. Which setup produces the greater diffraction (wider central maximum) on a far screen?
    Answer: Setup R, because larger wavelength relative to $a$ increases diffraction This question tests understanding of diffraction. Diffraction is the spreading of light as it passes through slits, with the angular width of the central maximum proportional to λ/a (wavelength divided by slit width). Since red light has a longer wavelength than violet light (λR > λV) and both pass through identical slits of width a, red light has a larger λ/a ratio and produces a wider central maximum. Choice A incorrectly claims higher frequency increases diffraction, but higher frequency means shorter wavelength, which actually decreases diffraction for a fixed slit width. The key principle is that diffraction increases as wavelength increases relative to the opening size.
15. Plane light waves pass a razor blade edge. The wavelength $\lambda$ is not negligible compared with the edge thickness, so light bends into the shadow region. The diffraction into the shadow becomes greatest when
    Answer: the wavelength is increased while the edge is unchanged This question tests understanding of diffraction. Diffraction is the bending of waves around edges or obstacles, with the effect becoming more pronounced as the wavelength increases relative to the obstacle size. When light passes a razor blade edge, some light bends into the geometrical shadow region due to diffraction. Increasing the wavelength while keeping the edge unchanged makes the wavelength larger relative to the edge features, causing more bending into the shadow. Choice D incorrectly suggests that decreasing wavelength increases diffraction, but shorter wavelengths actually bend less around obstacles and create sharper shadows. The key principle is that diffraction is strongest when the opening or obstacle is comparable to the wavelength.
16. A ripple tank sends plane water waves toward a barrier with a single slit. The slit width is about the same as the wavelength $\lambda$, and the wave speed is unchanged. Which situation shows the greatest diffraction after the slit?
    Answer: Use a narrower slit while keeping $\lambda$ the same This question tests understanding of diffraction. Diffraction is the spreading of waves when they pass through an opening or around an obstacle, and the amount of spreading depends on the ratio of wavelength to opening size. When the slit width is comparable to the wavelength, significant diffraction occurs, with maximum spreading when the slit becomes narrower relative to the wavelength. The wave spreads out more with a narrower slit because the opening constrains the wavefront more severely, causing greater bending. Choice B incorrectly suggests amplitude affects diffraction, which is a common misconception - diffraction depends only on the wavelength-to-opening ratio, not on wave intensity. To maximize diffraction, make the opening size comparable to or smaller than the wavelength.
17. A monochromatic sound wave passes through a doorway into a hallway. The doorway width is fixed. Compared to a higher-frequency sound, a lower-frequency sound has a larger wavelength. Which sound shows the greatest diffraction (spreading) into the hallway?
    Answer: The lower-frequency sound because its wavelength is larger relative to the doorway This question tests understanding of diffraction. Diffraction occurs when waves bend around obstacles or spread out after passing through openings, with the amount of spreading determined by the ratio of wavelength to opening size. Lower frequency sounds have longer wavelengths (since v = fλ and v is constant in air), making their wavelength larger relative to the fixed doorway width. This larger λ/width ratio causes greater diffraction and more spreading into the hallway. Choice A incorrectly assumes wave speed depends on frequency, but all sound waves travel at the same speed in air regardless of frequency. The key strategy is to compare wavelength to opening size: larger wavelength relative to opening means more diffraction.
18. Light of wavelength $500\ \text{nm}$ passes through a slit of width $a$ and forms a diffraction pattern on a far screen. The light source intensity is doubled while $\lambda$ and $a$ stay the same. Compared to before, the angular width of the central maximum is
    Answer: unchanged, because it depends on $\lambda$ relative to $a$ This question tests understanding of diffraction. Diffraction patterns depend on the ratio of wavelength to slit width (λ/a), which determines the angular width of features like the central maximum. When only the light intensity is changed while keeping wavelength and slit width constant, the λ/a ratio remains unchanged, so the angular width of the central maximum stays the same. The pattern becomes brighter but maintains the same angular dimensions. Choice A incorrectly suggests intensity affects diffraction angle, but intensity only affects brightness, not the geometric spreading of the pattern. The key principle is that diffraction geometry depends only on the wavelength-to-opening ratio, not on wave intensity.
19. A student compares two ripple-tank setups with identical barriers and slit widths. Setup 1 uses waves with longer wavelength than setup 2. Both have the same amplitude. Which setup shows greater diffraction after the slit?
    Answer: Setup 1, because the wavelength is larger relative to the slit width This question tests understanding of diffraction. Diffraction is the spreading of waves through openings, with the amount of spreading depending on the ratio of wavelength to opening size. Setup 1, with longer wavelength waves passing through the same slit width, will show greater diffraction because the wavelength is larger relative to the slit width. The diffraction angle is proportional to λ/a, where λ is wavelength and a is slit width, so larger wavelengths produce more spreading. Choice B represents the misconception that smaller wavelengths diffract more, but actually smaller wavelengths relative to the opening result in less spreading. Remember that diffraction is strongest when the opening is comparable to the wavelength.
20. Water waves of wavelength $\lambda$ encounter two openings in a barrier. Opening X has width about $\lambda/4$ and opening Y has width about $2\lambda$. The incoming wave amplitude is the same for both. Which opening produces the greatest diffraction after the barrier?
    Answer: Opening X, because the opening is much smaller than $\lambda$ This question tests understanding of diffraction. Diffraction is the spreading of waves after passing through openings, with the amount of spreading inversely related to the ratio of opening size to wavelength. Opening X has width λ/4 (much smaller than the wavelength), while opening Y has width 2λ (much larger than the wavelength). When an opening is much smaller than the wavelength, waves spread out dramatically in all directions after passing through. Choice A incorrectly assumes more energy through a wider opening means more diffraction, but diffraction is about wave spreading, not energy transmission. The key principle is that diffraction is strongest when the opening is comparable to or smaller than the wavelength.

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