Apply Wave Phenomena - IB Physics

1. Sound waves of frequency 500 Hz and light waves from a red laser are directed towards an open doorway of width 1.0 m. The speed of sound is 340 m s⁻¹. Diffraction of the sound is easily observed but diffraction of the light is not. What is the reason for this difference?
   Answer: The wavelength of the sound is comparable to the doorway's width, while the light's wavelength is much smaller. Significant diffraction occurs when the wavelength of a wave is on the order of the size of the aperture or obstacle. The wavelength of the sound is $\lambda = v/f = 340/500 = 0.68$ m, which is comparable to the 1.0 m doorway. The wavelength of red light is approximately 700 nm ($7 \times 10^{-7}$ m), which is vastly smaller than the doorway. Therefore, sound diffracts noticeably while light does not.
2. A water wave with amplitude $A = 0.20\text{ m}$ and wavelength $\lambda = 3.0\text{ m}$ travels at speed $v = 1.5\text{ m/s}$. The wave encounters a barrier with a circular opening of diameter $d = 1.5\text{ m}$. What type of wave behavior will be most prominent beyond the opening?
   Answer: Significant diffraction because the opening size is comparable to the wavelength, causing wave spreading When a wave encounters an opening or obstacle, the key factor determining the behavior is the relationship between the opening size and the wavelength. This is a classic diffraction scenario where you need to compare these two measurements. Here, the wavelength is $\lambda = 3.0\text{ m}$ and the opening diameter is $d = 1.5\text{ m}$. Since the opening size is about half the wavelength (comparable in magnitude), significant diffraction will occur. When waves pass through openings similar in size to their wavelength, they spread out considerably beyond the opening rather than continuing in straight lines. Option D correctly identifies this diffraction effect. The opening size being comparable to the wavelength means the waves will bend and spread as they pass through, creating the characteristic semicircular wave pattern beyond the barrier. Option A incorrectly suggests refraction dominates. Refraction occurs when waves change speed between different media, but there's no indication of a medium change here - just a geometric opening in a barrier. Option B misapplies the diffraction principle. While the opening is indeed larger than the wavelength, it's not significantly larger. The rule of thumb is that when the opening is many times larger than the wavelength, you get minimal diffraction. Here, with $d ≈ 0.5\lambda$, diffraction effects are still prominent. Option C incorrectly invokes reflection and impedance mismatch. These concepts apply to boundaries between different media, not to geometric openings of this type. Study tip: Remember that significant diffraction occurs when the obstacle or opening size is comparable to (within a few times) the wavelength. If opening >> wavelength, minimal diffraction; if opening ≈ wavelength, significant diffraction.
3. Light travels from water (n=1.33) into a layer of oil (n=1.47), which rests on top of glass (n=1.52). At which interface(s) is total internal reflection impossible, regardless of the angle of incidence?
   Answer: Both the water-oil and oil-glass interfaces. Total internal reflection can only occur when light travels from a medium of higher refractive index to a medium of lower refractive index. At the water-oil interface, light travels from n=1.33 to n=1.47 (lower to higher), so TIR is impossible. At the oil-glass interface, light travels from n=1.47 to n=1.52 (lower to higher), so TIR is also impossible.
4. Two coherent sources S1 and S2 emit monochromatic waves of wavelength $\lambda$. They produce an interference pattern. At a point P, the path difference (S2P - S1P) is $3.5\lambda$. What is observed at point P?
   Answer: A minimum intensity, as it is a point of destructive interference. Constructive interference (maximum intensity) occurs when the path difference is an integer multiple of the wavelength ($n\lambda$). Destructive interference (minimum intensity) occurs when the path difference is a half-integer multiple of the wavelength ($(n+1/2)\lambda$). Since the path difference is $3.5\lambda$, which is a half-integer multiple, destructive interference occurs, resulting in a minimum intensity.
5. Two coherent sources $S_1$ and $S_2$ are separated by distance $d = 3.0\text{ m}$ and emit waves in phase with wavelength $\lambda = 2.0\text{ m}$. A detector moves along a line parallel to the line joining the sources, at distance $D = 4.0\text{ m}$ from the midpoint between sources. At what distance from the perpendicular bisector will the detector first encounter destructive interference?
   Answer: $1.44\text{ m}$ because the geometry requires considering the full three-dimensional path differences When you encounter two-source interference problems, you need to carefully analyze the geometry to find where constructive and destructive interference occur. The key is calculating the path difference between waves from each source to the detector position. For destructive interference, the path difference must equal an odd multiple of half-wavelengths: $(2n+1)\lambda/2$. For the first minimum, this means $\lambda/2 = 1.0\text{ m}$. Let's set up coordinates with the midpoint between sources as origin. The sources are at $(\pm 1.5, 0)$ and the detector moves along the line $(x, 4.0)$. The path lengths are: - From $S_1$: $\sqrt{(x+1.5)^2 + 16}$ - From $S_2$: $\sqrt{(x-1.5)^2 + 16}$ Setting the path difference equal to $1.0\text{ m}$ and solving (this involves expanding the square roots and using approximations for the geometry), you get $x = 1.44\text{ m}$, confirming answer C. Answer A incorrectly assumes you can ignore the three-dimensional geometry and treat this as a simple linear problem. While the path difference condition $\lambda/2$ is correct, the geometric calculation is oversimplified. Answer B uses the wrong condition for destructive interference—a full wavelength difference would actually give constructive interference, not destructive. Answer D introduces a nonexistent concept about "effective wavelength" changing with detection geometry, which isn't relevant to this interference problem. Study tip: In two-source interference problems, always draw the geometry carefully and use the Pythagorean theorem to find actual path lengths—don't rely on approximations until you understand the full setup.
6. Light of wavelength 600 nm is used in a double-slit experiment, producing fringes with a separation of 3.0 mm. If the light source is replaced with one that produces fringes with a separation of 2.0 mm, without changing the experimental setup, what is the wavelength of the new light source?
   Answer: 400 nm Fringe separation $s$ is directly proportional to wavelength $\lambda$ (from $s = \lambda D/d$). Therefore, the ratio of separations equals the ratio of wavelengths: $s_1 / s_2 = \lambda_1 / \lambda_2$. Plugging in the values: $3.0 / 2.0 = 600 \text{ nm} / \lambda_2$. Solving for $\lambda_2$ gives $\lambda_2 = 600 \text{ nm} \times (2.0 / 3.0) = 400 \text{ nm}$.
7. In a double-slit experiment, the distance between the slits is halved, and the distance from the slits to the screen is doubled. If the original fringe separation was $s$, what is the new fringe separation?
   Answer: $4s$ The fringe separation is given by the formula $s = \lambda D / d$, where $D$ is the distance to the screen and $d$ is the slit separation. The new distance to the screen is $D' = 2D$, and the new slit separation is $d' = d/2$. The new fringe separation $s'$ is $s' = \lambda D' / d' = \lambda (2D) / (d/2) = 4 (\lambda D / d) = 4s$.
8. For a stable interference pattern to be formed by the light from two sources, the sources must be coherent. This means the waves they emit must have
   Answer: the same frequency and a constant phase difference. Coherence is the property that requires the sources to have a constant phase difference and the same frequency. While having the same amplitude provides for maximum contrast in the interference pattern, it is not a requirement for coherence itself. Being exactly in phase (zero phase difference) is a specific case of coherence, but not the general definition.
9. Two wave pulses travel toward each other on a rope. One pulse has a positive displacement of amplitude 3.0 cm. The other has a negative displacement of amplitude 1.5 cm. What is the resultant displacement of the rope at the instant the pulses completely overlap?
   Answer: 1.5 cm According to the principle of superposition, the resultant displacement at any point is the algebraic sum of the individual displacements. At the point of complete overlap, the resultant displacement is $(+3.0 \text{ cm}) + (-1.5 \text{ cm}) = +1.5 \text{ cm}$.
10. A beam of white light travels within a glass block and strikes the glass-air boundary at an angle of incidence that is gradually increased. Due to dispersion, the refractive index of the glass is slightly different for different colors. Which color of light is the first to undergo total internal reflection?
    Answer: Violet, because it has the highest refractive index in glass. The critical angle is given by $\sin \theta_c = 1/n$. Due to dispersion in glass, the refractive index (n) is largest for violet light (shortest wavelength) and smallest for red light (longest wavelength). A larger refractive index results in a smaller critical angle. Since violet light has the smallest critical angle, it will be the first to undergo total internal reflection as the angle of incidence is increased.
11. A guitar string vibrates in its third harmonic mode with frequency $f_3 = 660\text{ Hz}$. The string length is $L = 0.65\text{ m}$ and linear mass density is $\mu = 3.2 \times 10^{-4}\text{ kg/m}$. If the tension in the string is suddenly increased by $44\%$, what will be the new frequency of the third harmonic?
    Answer: $792\text{ Hz}$ because wave speed increases with the square root of tension When you encounter guitar string harmonics problems, you're dealing with standing waves on a fixed string, where both wave speed and harmonic frequencies are involved. The key insight is understanding how tension affects wave speed, which then affects frequency. For a guitar string, the wave speed is $v = \sqrt{\frac{T}{\mu}}$, where $T$ is tension and $\mu$ is linear mass density. The third harmonic frequency is $f_3 = \frac{3v}{2L}$. When tension increases by 44%, the new tension becomes $T_{new} = 1.44T$, so the new wave speed is $v_{new} = \sqrt{\frac{1.44T}{\mu}} = 1.2\sqrt{\frac{T}{\mu}} = 1.2v$. Therefore, the new third harmonic frequency is $f_{3,new} = \frac{3(1.2v)}{2L} = 1.2f_3 = 1.2 \times 660 = 792\text{ Hz}$. Option A incorrectly assumes frequency increases proportionally with tension, but frequency actually increases with the square root of tension due to the wave speed relationship. Option B mentions "harmonic number effects," but the harmonic number doesn't change in this problem—we're still looking at the third harmonic. Option D incorrectly claims harmonic frequencies are independent of tension, but since wave speed depends on tension, all harmonic frequencies will change when tension changes. Remember: for string instruments, frequency changes follow the square root relationship with tension. When tension increases by a factor, frequency increases by the square root of that factor—this appears frequently on IB Physics exams.
12. Monochromatic light is incident on a narrow single slit, and a diffraction pattern is observed on a screen. Which change will cause the central maximum of the diffraction pattern to become wider?
    Answer: Decreasing the frequency of the light. The extent of diffraction, and thus the width of the central maximum, is proportional to the ratio $\lambda / b$, where $\lambda$ is the wavelength and $b$ is the slit width. To make the pattern wider, this ratio must increase. Decreasing the frequency of the light (since $c = f\lambda$) will increase its wavelength $\lambda$, thereby increasing the ratio and widening the pattern.
13. In a Young's double-slit experiment, a point on the screen is equidistant from both slits. At this point, there is a bright fringe. A second point on the screen corresponds to the third dark fringe from the center. What is the path difference between the waves arriving at this second point?
    Answer: $2.5 \lambda$ Dark fringes (destructive interference) occur when the path difference is $(m + 1/2)\lambda$, where $m = 0, 1, 2, ...$. The central fringe is at $m=0$ for a bright fringe. The first dark fringe corresponds to $m=0$, with a path difference of $0.5\lambda$. The second dark fringe corresponds to $m=1$, path difference $1.5\lambda$. The third dark fringe corresponds to $m=2$, with a path difference of $(2 + 1/2)\lambda = 2.5\lambda$.
14. Light passes from air (n ≈ 1.00) into a rectangular block of glass (n = 1.50). The light then emerges from the parallel opposite side of the block into water (n = 1.33). If the angle of incidence in the air is 60°, what is the final angle of refraction in the water?
    Answer: 40.6° For parallel interfaces, the intermediate medium does not affect the relationship between the initial and final angles. Applying Snell's Law between the initial medium (air) and the final medium (water): $n_{air} \sin \theta_{air} = n_{water} \sin \theta_{water}$. This gives $1.00 \times \sin(60°) = 1.33 \times \sin \theta_{water}$. Solving for $\theta_{water}$ gives $\arcsin(\frac{\sin(60°)}{1.33}) \approx 40.6°$.
15. The critical angle for light travelling from a certain transparent medium into air is 38.7°. What is the speed of light within this medium? (Speed of light in air ≈ 3.00 × 10⁸ m s⁻¹)
    Answer: 1.88 × 10⁸ m s⁻¹ First, find the refractive index (n) of the medium using the critical angle formula: $\sin \theta_c = n_2 / n_1$. Here, $n_2$ is air (≈ 1) and $\theta_c = 38.7°$. So, $n_1 = 1 / \sin(38.7°) \approx 1.60$. Next, find the speed of light (v) in the medium using $n = c/v$. So, $v = c/n = (3.00 × 10⁸ m s⁻¹) / 1.60 \approx 1.88 × 10⁸ m s⁻¹$.
16. A monochromatic light wave travels from a medium with refractive index $n_1$ to a medium with refractive index $n_2$, where $n_1 > n_2$. What are the changes in the wave's speed and wavelength?
    Answer: Speed increases and wavelength increases. The frequency of the wave remains constant as it crosses the boundary. The speed of light in a medium is given by $v = c/n$. Since $n_1 > n_2$, the speed $v$ increases as the light enters the second medium. Because $v = f\lambda$ and $f$ is constant, if $v$ increases, the wavelength $\lambda$ must also increase.
17. A sound wave with frequency $f = 440\text{ Hz}$ travels from air into a medium where its speed increases to $1.5$ times the speed in air. A second sound wave with the same frequency travels from air into a different medium where its wavelength becomes $0.8$ times the wavelength in air. When these two transmitted waves meet at a boundary, what is the ratio of their wavelengths?
    Answer: $1.875$ because wavelength scales with speed in the first medium but changes differently in the second For the first wave: v₁ = 1.5v_air, and since f is constant, λ₁ = v₁/f = 1.5(v_air/f) = 1.5λ_air. For the second wave: λ₂ = 0.8λ_air is given directly. Therefore, λ₁/λ₂ = (1.5λ_air)/(0.8λ_air) = 1.5/0.8 = 1.875.
18. A Doppler radar system operating at frequency $f_0 = 10.0\text{ GHz}$ detects a reflected signal from a moving vehicle. The received frequency is $f_r = 10.000134\text{ GHz}$. If the radar wave travels at $c = 3.00 \times 10^8\text{ m/s}$, what is the speed of the vehicle?
    Answer: $20.0\text{ m/s}$ toward the radar because the wave experiences two Doppler shifts in the round trip When you encounter Doppler radar problems, remember that the wave makes a round trip—from radar to vehicle and back—experiencing two Doppler shifts. Let's work through this systematically. The frequency increased from 10.0 GHz to 10.000134 GHz, indicating the vehicle is approaching the radar. For radar systems, you must use the double Doppler shift formula: $f_r = f_0 \left(1 + \frac{2v}{c}\right)$ when the vehicle approaches, where the factor of 2 accounts for the round trip. Solving for velocity: $10.000134 = 10.0 \left(1 + \frac{2v}{3.00 \times 10^8}\right)$ The frequency ratio is 1.0000134, so: $\frac{2v}{3.00 \times 10^8} = 0.0000134$ Therefore: $v = \frac{0.0000134 \times 3.00 \times 10^8}{2} = 20.0\text{ m/s}$ Answer A incorrectly applies single Doppler shift, giving half the correct speed. Answer B has the wrong direction—frequency increases indicate approach, not recession. Answer C gets the speed right but wrong direction, likely from misinterpreting the frequency change. Answer D correctly recognizes both the direction (toward the radar due to frequency increase) and accounts for the double Doppler effect inherent in radar systems. Study tip: For radar problems, always remember the "double Doppler" rule—the wave bounces off the moving object, so multiply your standard Doppler calculation by 2. Frequency increase means approach, frequency decrease means recession.
19. A wave pulse travels along a rope and reflects from a fixed end. Which statement correctly describes the reflected pulse?
    Answer: It is inverted but has the same speed. Reflection from a fixed end causes a phase inversion of 180 degrees (an upright pulse reflects as an inverted pulse). The speed of the wave is determined by the properties of the medium (the rope), which do not change, so the speed remains the same. Some energy may be absorbed at the boundary, reducing amplitude, but the speed being constant is a more fundamental principle.
20. A thin film of oil (refractive index $n_{oil} = 1.45$) floats on water (refractive index $n_{water} = 1.33$) and appears bright green ($\lambda = 520\text{ nm}$) when viewed from above in air under normal incidence. What is the minimum possible thickness of the oil film?
    Answer: $179\text{ nm}$ because there is a phase change at the top surface only, requiring odd multiples of λ/4 For thin film interference, we need to consider phase changes at interfaces. At air-oil interface (n_air < n_oil): phase change occurs. At oil-water interface (n_oil > n_water): no phase change. With one phase change, constructive interference requires 2nt = (m + 1/2)λ. For minimum thickness (m=0): 2(1.45)t = (1/2)(520nm), so t = 179nm.

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