Double-Slit Interference - AP Physics 2

1. In a double-slit experiment, the central bright fringe is at the midpoint of the screen. Point $T$ is moved upward along the screen until the path difference becomes exactly one wavelength. Which change in brightness occurs as $T$ reaches that location?
   Answer: It becomes bright because the waves return to being in phase This problem examines double-slit interference as a function of position on the screen. Starting from the central bright fringe (zero path difference), moving point T upward increases the path difference between the two slits. When the path difference reaches exactly one wavelength, the waves arrive in phase again (after one complete cycle offset), producing constructive interference and a bright fringe. Choice B incorrectly associates different amplitudes with darkness, but equal-amplitude waves can produce either bright or dark fringes depending on phase. The strategy is to track path difference: integer wavelengths yield bright fringes.
2. In a stable double-slit pattern, point $Y$ corresponds to a path difference of $3\lambda$. Which outcome occurs at $Y$?
   Answer: A bright fringe because the waves arrive in phase This question tests understanding of double-slit interference. In a stable double-slit pattern, the path difference determines whether waves interfere constructively or destructively at each point. A path difference of 3λ means one wave has traveled exactly three complete wavelengths farther than the other, creating a phase difference of 1080° or equivalently 0° (since 3 × 360° = 1080° = 0° after removing complete cycles). Waves arriving with zero effective phase difference interfere constructively, producing a bright fringe at point Y. Choice A incorrectly claims that odd multiples of λ are destructive, confusing them with odd half-multiples like 1.5λ or 2.5λ. The fundamental rule remains: integer multiples of λ produce constructive interference, while half-integer multiples produce destructive interference.
3. In a double-slit experiment, a point $R$ on the screen is observed to be a dark fringe. Which statement about the path difference $ d$ between the two slits to point $R$ is consistent with this observation (coherent, in-phase slits)?
   Answer: $d$ is a half-integer multiple of the wavelength This problem addresses double-slit interference conditions for dark fringes. Dark fringes occur when waves from the two slits arrive out of phase, causing destructive interference. This happens when the path difference d equals a half-integer multiple of the wavelength: d = (n + 1/2)λ where n = 0, 1, 2, etc. Choice A incorrectly states integer multiples produce dark fringes, but integer multiples actually produce bright fringes. The key insight is that destructive interference requires waves to be half a cycle out of phase, corresponding to half-integer wavelength path differences.
4. A student observes a stable double-slit interference pattern using coherent light. The student increases the screen distance while keeping slit separation and wavelength constant. What happens to the spacing between bright fringes?
   Answer: It increases because the same angular separation corresponds to a larger linear separation farther away This question tests double-slit interference. The linear spacing between bright fringes on the screen is given by Δy = λL/d, where L is the screen distance. When screen distance increases, the fringe spacing increases proportionally because the same angular separation between fringes translates to a larger linear separation at a greater distance. This is analogous to how shadows grow larger when you move a screen farther from the light source. Choice B incorrectly suggests that wave spreading affects fringe spacing, when it's actually the geometric projection that matters. Remember that fringe spacing is directly proportional to screen distance.
5. A student observes a stable double-slit interference pattern using a laser. They then place a thin transparent sheet in front of only one slit, increasing the optical path length through that slit. The overall pattern remains, but the locations of bright fringes shift. Which statement best explains the shift?​
   Answer: The added sheet changes the phase of one wave, changing where path differences produce constructive interference This question tests understanding of double-slit interference. When a thin transparent sheet is placed in front of one slit, it increases the optical path length through that slit because light travels slower through the material than through air. This additional optical path length introduces a phase shift for waves passing through that slit relative to the other slit. Since the phase relationship between waves from the two slits has changed, the locations where constructive and destructive interference occur shift on the screen, causing the entire fringe pattern to move. Choice B incorrectly suggests that the sheet changes intensity distribution—while the sheet may slightly reduce intensity through absorption, this doesn't cause fringe shifts. The key concept is that interference patterns depend on the phase difference between waves, which can be altered by changing the optical path length through one path.
6. Two slits are illuminated by the same laser, producing stable fringes. The screen point $T$ corresponds to a path difference of $3\lambda/2$. Which fringe occurs at $T$?
   Answer: A dark fringe because the waves arrive out of phase This question tests understanding of double-slit interference. When coherent light from two slits creates an interference pattern, the path difference determines whether waves arrive in phase or out of phase at each screen point. A path difference of 3λ/2 equals 1.5λ, which means one wave travels one and a half wavelengths farther than the other, creating a phase difference of 540° or equivalently 180°. Waves arriving 180° out of phase interfere destructively, producing a dark fringe at point T. Choice C incorrectly suggests that laser brightness affects the number of maxima, but fringe positions depend only on wavelength and geometry, not intensity. To identify fringe types, check if the path difference equals nλ (bright) or (n + 0.5)λ (dark) where n is any integer.
7. Two slits are illuminated by the same laser so the waves leaving the slits have a constant phase relationship. On the screen, the central bright fringe occurs where the path difference is zero. Which change increases the spacing between adjacent bright fringes on the screen without requiring any calculation?​
   Answer: Increase the distance from the slits to the screen This question tests understanding of double-slit interference. In double-slit interference, the spacing between adjacent bright fringes on the screen depends on three factors: wavelength (λ), slit separation (d), and distance to the screen (L). The fringe spacing is proportional to λL/d, meaning it increases when wavelength increases, when screen distance increases, or when slit separation decreases. Since the question asks for increased fringe spacing without calculation, increasing the distance to the screen (choice A) will spread the pattern out, making fringes farther apart. Choice B incorrectly suggests that intensity affects fringe spacing—intensity only changes the brightness of the pattern, not the geometric spacing. The key insight is that fringe spacing is a geometric property determined by the ratio of wavelength and slit separation, scaled by the screen distance.
8. A double-slit pattern is produced on a screen. The wavelength is changed from blue light to red light, with slit separation and screen distance unchanged. Which change occurs to the spacing of adjacent bright fringes?
   Answer: It increases because the wavelength is larger This question tests double-slit interference. The spacing between adjacent bright fringes is given by Δy = λL/d, where λ is wavelength, L is screen distance, and d is slit separation. Red light has a longer wavelength than blue light (approximately 700 nm vs 450 nm). Since fringe spacing is directly proportional to wavelength, changing from blue to red light increases the fringe spacing. Choice A incorrectly relates frequency to interference strength rather than recognizing that wavelength directly determines fringe spacing. Remember that fringe spacing increases with wavelength: longer wavelengths produce wider-spaced fringes.
9. In a dark room, monochromatic laser light passes through two narrow slits separated by a small distance and forms alternating bright and dark fringes on a distant screen. At one point on the screen, the two paths from the slits differ by half a wavelength, so the waves arrive out of phase and the intensity is minimal. Which condition produces a bright fringe at that same point on the screen?​
   Answer: The path difference is changed to an integer multiple of the wavelength This question tests understanding of double-slit interference. In double-slit interference, bright fringes occur when waves from both slits arrive in phase (constructive interference), which happens when the path difference equals an integer multiple of the wavelength (0, λ, 2λ, etc.). Dark fringes occur when waves arrive out of phase (destructive interference), which happens when the path difference equals a half-integer multiple of the wavelength (λ/2, 3λ/2, etc.). The question states that the point currently has a path difference of λ/2, creating a dark fringe. To make this point bright, we need to change the path difference to an integer multiple of the wavelength. Choice A incorrectly suggests that changing intensity affects interference patterns—intensity only affects brightness, not the locations of bright and dark fringes. The key strategy is to remember that interference patterns depend on path difference relative to wavelength, not on the absolute intensity of light.
10. In a double-slit experiment, point $U$ is labeled on the screen where the path difference is $0$. Which condition is met at $U$?
    Answer: A bright fringe because the waves arrive in phase This question tests understanding of double-slit interference. At the center of a double-slit pattern, where the path difference equals zero, waves from both slits travel exactly the same distance to reach point U. With zero path difference, the waves arrive perfectly in phase (0° phase difference) and interfere constructively, creating the central bright fringe. This is always the brightest point in the pattern because the waves add with maximum amplitude. Choice A incorrectly claims that equal paths cause cancellation, but waves only cancel when they arrive 180° out of phase. The fundamental principle is that zero path difference always produces maximum constructive interference, regardless of wavelength or slit separation.
11. A stable double-slit interference pattern is observed. One slit is covered so only the other slit remains open, with the same light source and screen. What happens to the bright and dark fringes?
    Answer: They vanish because interference requires two coherent paths This question tests understanding of double-slit interference fundamentals. Interference patterns require waves from two coherent sources (the slits) to overlap and interfere. When one slit is covered, only single-slit diffraction remains, producing a broad central maximum with much weaker side maxima—not the characteristic alternating bright and dark fringes of double-slit interference. Choice A incorrectly suggests each slit produces its own interference pattern, but interference requires interaction between waves from multiple sources. The key concept is that double-slit interference patterns vanish when reduced to a single slit.
12. Two coherent slits produce an interference pattern on a screen. At a certain point $Q$, the path difference from the two slits is half a wavelength. Which condition occurs at point $Q$?
    Answer: A dark fringe forms because the waves arrive out of phase at $Q$ This problem involves double-slit interference with a specific path difference condition. A path difference of half a wavelength (λ/2) causes the waves to arrive exactly out of phase—one wave's crest meets the other's trough. This produces destructive interference, resulting in a dark fringe at point Q. Choice B incorrectly states that a half-cycle phase shift produces a bright fringe, when it actually causes cancellation. The key principle is that half-integer wavelength path differences (λ/2, 3λ/2, 5λ/2, etc.) always produce dark fringes due to destructive interference.
13. A ripple tank produces water waves that pass through two narrow openings, creating an interference pattern on the far side. At a location where the crests from both openings arrive together, the water oscillation amplitude is large. Which statement best describes the path difference at that location?​
    Answer: It is an integer multiple of the wavelength This question tests understanding of double-slit interference. In double-slit interference with water waves, constructive interference occurs when wave crests from both sources arrive together, creating large amplitude oscillations. This happens when the path difference between the two sources equals an integer multiple of the wavelength (0, λ, 2λ, etc.). When crests arrive together, they are in phase, which only occurs at these specific path differences. Choice B incorrectly describes the condition for destructive interference, where crests meet troughs. The misconception in choice C is that interference patterns depend on wave properties and geometry, not just wave speed. The key principle is that constructive interference requires waves to arrive in phase, which happens when path differences are integer multiples of the wavelength.
14. Two coherent slits are illuminated by monochromatic light. At point $S$ on the screen, the path difference is $2.5 \lambda$. Which fringe occurs at $S$?
    Answer: A dark fringe occurs because the path difference is a half-integer multiple of $\lambda$ This problem involves double-slit interference with a specific path difference value. A path difference of 2.5λ equals 2λ + λ/2, which is a half-integer multiple of the wavelength. This causes the waves to arrive out of phase (one is shifted by half a cycle relative to the other), producing destructive interference and a dark fringe. Choice A incorrectly assumes any path difference greater than 2λ produces a bright fringe, missing that the fractional part determines the interference type. The principle is that path differences of (n + 1/2)λ always yield dark fringes regardless of the integer n.
15. Monochromatic light passes through two slits and forms fringes on a screen. At point $R$, the path difference equals $0.5\lambda$. Which fringe is observed at $R$?
    Answer: A dark fringe because the waves arrive out of phase This question tests understanding of double-slit interference. When monochromatic light passes through two slits, the interference pattern depends on the path difference between waves reaching each point on the screen. A path difference of 0.5λ means one wave travels half a wavelength farther than the other, creating a phase difference of 180°. Waves that are 180° out of phase interfere destructively, producing a dark fringe at point R. Choice B incorrectly suggests that intensity affects fringe positions, but the pattern depends only on path geometry and wavelength. Remember that half-integer multiples of the wavelength [(n + 0.5)λ] always produce destructive interference, while integer multiples produce constructive interference.
16. Coherent monochromatic light passes through two slits and forms an interference pattern on a screen. At a certain point, the two paths differ by one wavelength. Which statement best describes the interference at that point?
    Answer: A bright fringe occurs because the waves arrive in phase This question tests double-slit interference. When coherent light passes through two slits, the resulting interference pattern depends on the path difference between waves from each slit. A path difference of one wavelength means the waves arrive with the same phase (360° phase difference equals 0° effectively), producing constructive interference and a bright fringe. This is because one wave has completed exactly one more full cycle than the other, bringing them back into phase. Choice A incorrectly claims that equal amplitudes always cancel, ignoring the crucial role of phase. The fundamental principle is that constructive interference occurs when path difference equals an integer multiple of the wavelength.
17. Coherent light passes through two slits, producing bright fringes where the waves arrive in phase. At point $Q$, the path difference is $\tfrac{3}{2}\lambda$. What is observed at $Q$?
    Answer: A dark fringe because a half-integer multiple of $\lambda$ produces destructive interference This question tests double-slit interference. The interference pattern depends on the path difference between waves from the two slits. When path difference equals a half-integer multiple of the wavelength (like 3λ/2 = 1.5λ), the waves arrive with opposite phases (180° out of phase) and interfere destructively, producing a dark fringe. This occurs because one wave has completed 1.5 cycles more than the other, putting them in opposite phases. Choice A incorrectly claims that any multiple of λ/2 produces constructive interference, confusing integer and half-integer multiples. The key rule is: integer multiples of λ give bright fringes, half-integer multiples give dark fringes.
18. A double-slit setup uses coherent light. At point $S$ on the screen, the path difference is $\lambda/4$. Which statement best describes the intensity at $S$?
    Answer: An intermediate intensity because the waves are not fully in or out of phase This question tests understanding of double-slit interference. In double-slit interference, the path difference determines the phase relationship between waves from the two slits. A path difference of λ/4 corresponds to a phase difference of 90°, meaning the waves are neither completely in phase nor completely out of phase. This partial phase difference results in partial interference - neither fully constructive nor fully destructive - producing an intermediate intensity between the bright and dark fringes. Choice A incorrectly assumes any nonzero path difference causes complete cancellation, which only occurs at specific values like λ/2. The general rule is that only path differences of nλ produce maximum brightness and (n + 0.5)λ produce complete darkness; all other values yield intermediate intensities.
19. A double-slit pattern is produced by monochromatic coherent light. At point $W$, the path difference equals $4\lambda$. Which fringe appears at $W$?
    Answer: A bright fringe because the waves arrive in phase This question tests understanding of double-slit interference. When monochromatic coherent light passes through two slits, the path difference at each screen point determines the interference condition. A path difference of 4λ means one wave has traveled exactly four complete wavelengths more than the other, corresponding to a phase difference of 1440° or equivalently 0° (since 4 × 360° = 1440° = 0° after removing complete cycles). Waves arriving with zero effective phase difference interfere constructively, producing a bright fringe at point W. Choice A incorrectly claims that even multiples of λ cause cancellation, but only odd half-integer multiples [(2n+1)λ/2] produce destructive interference. The key insight is that any integer multiple of λ yields constructive interference.
20. In a two-slit experiment, point $P$ on the screen is a dark fringe. Without changing wavelength, which adjustment can make $P$ become a bright fringe?
    Answer: Change the path difference at $P$ from a half-integer multiple of $\lambda$ to an integer multiple of $\lambda$ This question tests double-slit interference. A dark fringe occurs when the path difference is a half-integer multiple of the wavelength (λ/2, 3λ/2, etc.), causing destructive interference. To convert this dark fringe to a bright fringe, the path difference must become an integer multiple of the wavelength (0, λ, 2λ, etc.). This can be achieved by changing the geometry to alter the path difference at point P. Choice B incorrectly assumes that increasing intensity can overcome destructive interference, when phase relationship is what matters. The strategy is to recognize that bright/dark fringes depend on whether path difference is an integer or half-integer multiple of wavelength.

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