Boundary Behavior of Waves and Polarization - College Physics

Linearly polarized light passes through a quarter-wave plate with its electric field vector oriented at $45°$ to the plate's fast axis. What is the polarization state of the emerging light?
Answer: Circularly polarized with counterclockwise rotation When linearly polarized light encounters a quarter-wave plate, you're dealing with birefringence—the plate has different refractive indices along its fast and slow axes, creating a phase difference between perpendicular components of the electric field. Here's what happens: The incoming linearly polarized light at $45°$ to the fast axis has equal amplitude components along both the fast and slow axes of the plate. A quarter-wave plate introduces a phase difference of $\pi/2$ (90°) between these components. When two perpendicular oscillations of equal amplitude are $90°$ out of phase, they combine to produce circular polarization. To determine the rotation direction, use the right-hand rule convention: if the slow axis leads the fast axis in creating the phase difference, you get counterclockwise (left-hand) circular polarization when viewed along the direction of propagation. Answer A is wrong because linear polarization requires the components to be in phase, not $90°$ out of phase. Answer B incorrectly identifies the rotation direction—clockwise would occur if the phase relationship were reversed. Answer D is incorrect because elliptical polarization results from unequal amplitudes or phase differences other than $90°$ ; here, equal amplitudes at $45°$ with exactly $90°$ phase shift produces perfect circular polarization. Remember this key relationship: linear polarization at $45°$ to a quarter-wave plate's axes always produces circular polarization. The specific rotation direction depends on which axis (fast or slow) the light encounters first, but counterclockwise is the standard result for conventional quarter-wave plates.
A rope wave encounters a boundary where the rope connects to a wall through a spring. At low frequencies, the boundary acts like a fixed end, while at high frequencies, it acts like a free end. For a wave with intermediate frequency, what is the most likely result?
Answer: Partial reflection with some phase change When analyzing wave behavior at boundaries, you need to consider how different boundary conditions affect reflection and transmission. The key insight here is that this boundary has frequency-dependent properties - it transitions from rigid (fixed end) at low frequencies to flexible (free end) at high frequencies. At intermediate frequencies, the spring-wall boundary exhibits characteristics of both fixed and free ends. This creates an impedance mismatch that's neither complete nor zero. When a wave encounters such a boundary, conservation of energy requires that some energy reflects back while some transmits through (or in this case, gets absorbed by the spring system). The phase change occurs because the boundary still has some rigidity - it's not completely free like a loose end would be. Choice A is incorrect because complete transmission only occurs when impedances are perfectly matched, which isn't the case here. Choice B assumes the boundary acts purely like a fixed end with total reflection, but the spring allows some energy absorption at intermediate frequencies. Choice C suggests complete reflection like a free end (no phase change), but again, the spring's intermediate stiffness prevents total reflection and introduces some phase shift due to its restoring force. The correct answer is D because partial reflection with some phase change captures the hybrid nature of this boundary condition. Study tip: Remember that real boundaries often exhibit frequency-dependent behavior. When you see "intermediate" conditions in wave problems, expect partial effects rather than the extreme cases you learn about first (complete reflection/transmission).
Light is incident on a boundary between two media. The critical angle for total internal reflection is $42°$ . If light approaches this boundary at $50°$ from the normal, what happens?
Answer: The light is completely reflected back into the first medium When light travels from a denser medium to a less dense medium, total internal reflection occurs when the incident angle exceeds the critical angle. This is a key concept in optics that determines whether light can escape from one medium into another. Since you're given that the critical angle is $42°$ and the incident angle is $50°$ , you need to compare these values. Because $50° > 42°$ , the incident angle exceeds the critical angle. This means total internal reflection must occur - all the light is reflected back into the first medium, with none transmitted into the second medium. Looking at the answer choices: Choice B correctly identifies that complete reflection occurs back into the first medium. Choice A incorrectly assumes transmission occurs and calculates a refraction angle, but no light can be transmitted when the incident angle exceeds the critical angle. Choice C suggests partial reflection and transmission with equal intensities, which would only occur at specific angles below the critical angle (like Brewster's angle). Choice D also incorrectly assumes refraction occurs and provides a specific angle, ignoring the fundamental principle that no refraction is possible beyond the critical angle. The key insight is that the critical angle represents an absolute boundary - once exceeded, Snell's law no longer applies in the usual way because the sine of the refraction angle would exceed 1, which is mathematically impossible. Remember this pattern: whenever the incident angle exceeds the critical angle in total internal reflection problems, the answer is always complete reflection, regardless of any calculations the other choices might suggest.
A wave traveling in medium A with speed $v_A = 300$ m/s encounters a boundary with medium B where the wave speed is $v_B = 450$ m/s. If the incident wave has frequency $f = 60$ Hz and approaches the boundary at an angle of $30°$ to the normal, what is the angle of refraction in medium B?
Answer: $48.6°$ When waves cross boundaries between different media, they refract (bend) according to Snell's law, just like light bending through a lens. This fundamental principle applies to all types of waves - sound, water waves, seismic waves, and electromagnetic waves. Snell's law states that $\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}$ , where $\theta_1$ and $\theta_2$ are the angles from the normal, and $v_1$ and $v_2$ are the wave speeds in each medium. Notice that frequency doesn't appear in this equation - it remains constant across the boundary while wavelength changes. Plugging in the given values: $\frac{\sin 30°}{300} = \frac{\sin \theta_2}{450}$ Since $\sin 30° = 0.5$ : $\frac{0.5}{300} = \frac{\sin \theta_2}{450}$ Solving for $\sin \theta_2$ : $\sin \theta_2 = \frac{0.5 \times 450}{300} = 0.75$ Therefore: $\theta_2 = \arcsin(0.75) = 48.6°$ Choice A ( $20.5°$ ) results from incorrectly using the reciprocal relationship - perhaps confusing this with the case where waves slow down rather than speed up. Choice C ( $30.0°$ ) assumes no refraction occurs, ignoring the speed change entirely. Choice D ( $60.0°$ ) might come from doubling the incident angle, which has no physical basis. Remember: when waves enter a faster medium, they bend away from the normal (larger angle). When entering a slower medium, they bend toward the normal. The frequency given is a distractor - Snell's law only requires speeds and angles.
When unpolarized light passes through two polarizing filters, the first filter transmits 50% of the incident intensity, and the second filter is oriented at $45°$ relative to the first. What fraction of the original unpolarized light intensity emerges from the second filter?
Answer: $\frac{1}{4}$ When you encounter polarized light problems, you need to apply two key principles: unpolarized light becomes 50% polarized after passing through any polarizer, and Malus's Law governs transmission through subsequent polarizers. Let's trace the light through both filters. Initially, you have unpolarized light with intensity $I_0$ . When this passes through the first polarizing filter, exactly 50% of the intensity is transmitted, giving you $I_1 = 0.5I_0$ . This is a fundamental property of polarizers with unpolarized light—the orientation doesn't matter for the first filter. Now you have polarized light hitting the second filter at a 45° angle. Here, Malus's Law applies: $I_2 = I_1 \cos^2(45°)$ . Since $\cos(45°) = \frac{\sqrt{2}}{2}$ , we get $\cos^2(45°) = \frac{1}{2}$ . Therefore: $I_2 = 0.5I_0 \times \frac{1}{2} = 0.25I_0 = \frac{1}{4}I_0$ Looking at the wrong answers: A) $\frac{1}{8}$ likely comes from incorrectly using $\cos(45°) = \frac{\sqrt{2}}{2}$ instead of $\cos^2(45°)$ . C) $\frac{1}{2}$ ignores the second polarizer entirely. D) $\frac{\sqrt{2}}{4}$ results from forgetting to square the cosine term in Malus's Law. The answer is B) $\frac{1}{4}$ . Remember this two-step approach: first polarizer always transmits 50% of unpolarized light, then apply Malus's Law ( $\cos^2\theta$ ) for each subsequent polarizer. Don't forget to square the cosine—that's the most common mistake in polarization problems.
A sinusoidal wave on a string reflects from a fixed end. If the incident wave is described by $y_i = A \sin(kx - \omega t)$ , which expression correctly represents the standing wave pattern formed by the superposition of incident and reflected waves?
Answer: $y = -2A \sin(kx) \cos(\omega t)$ When a wave reflects from a fixed end, you need to understand two key principles: the reflected wave travels in the opposite direction, and there's a phase change of π (or 180°) upon reflection from a fixed boundary. Starting with the incident wave $y_i = A \sin(kx - \omega t)$ , the reflected wave becomes $y_r = -A \sin(kx + \omega t)$ (note the negative sign from the phase change and the positive sign in the argument indicating leftward travel). The total displacement is their sum: $y = A \sin(kx - \omega t) - A \sin(kx + \omega t)$ Using the trigonometric identity $\sin(α) - \sin(β) = 2\cos\left(\frac{α+β}{2}\right)\sin\left(\frac{α-β}{2}\right)$ : $y = 2A\cos(\omega t)\sin(-kx) = -2A \sin(kx) \cos(\omega t)$ This confirms answer B is correct. Let's examine why the others fail: A gives $y = 2A \sin(kx) \cos(\omega t)$ , which would be correct for reflection from a free end (no phase change), but our boundary is fixed. C shows $y = 2A \cos(kx) \sin(\omega t)$ , which represents a different standing wave pattern with nodes at different locations. D simply shows the superposition before simplification—while mathematically equivalent to B, it's not in the standard standing wave form. Study tip: Remember that fixed-end reflections always introduce a π phase change, while free-end reflections don't. The standing wave amplitude factor is always 2A, and the spatial part $\sin(kx)$ or $\cos(kx)$ determines node locations.
A standing wave pattern is established on a string with both ends fixed. At a particular instant, the displacement pattern shows nodes at $x = 0$ , $L/3$ , $2L/3$ , and $L$ . If the string length is $L = 1.2$ m and the wave speed is $v = 400$ m/s, what is the frequency of the standing wave?
Answer: 500 Hz Standing wave problems require you to identify the wave pattern and connect it to the fundamental relationship $f = v/\lambda$ . When you see specific node locations, you're being asked to determine which harmonic is present. The nodes are at $x = 0$ , $L/3$ , $2L/3$ , and $L$ . Since nodes are separated by $\lambda/2$ , the distance between consecutive nodes tells us the wavelength. From $x = 0$ to $x = L/3$ is $L/3 = 0.4$ m, so $\lambda/2 = 0.4$ m, giving us $\lambda = 0.8$ m. You can verify this makes sense: with $\lambda = 0.8$ m, we get $\lambda/2 = 0.4$ m spacing between all consecutive nodes, which matches the pattern. The string contains exactly 1.5 wavelengths ( $L = 1.2$ m ÷ $0.8$ m = 1.5 $), indicating this is the third harmonic. Using$ f = v/\lambda = 400/0.8 = 500 $Hz, the answer is D. A) 167 Hz corresponds to using$ \lambda = 2.4 $m, which would incorrectly treat the entire string length as one wavelength. B) 333 Hz results from using$ \lambda = 1.2 $m, mistakenly thinking the string length equals the wavelength. C) 1000 Hz comes from using$ \lambda = 0.4 $m, confusing the node spacing ($ \lambda/2 $) with the actual wavelength. Remember: in standing wave problems, always identify the wavelength from the node spacing ($ \lambda/2$$), not from the total length. The key insight is recognizing that consecutive nodes are always separated by half a wavelength.
Circularly polarized light is incident normally on a linear polarizer whose transmission axis makes an angle $\theta$ with the horizontal. As the polarizer is rotated through a complete revolution ( $\theta$ from $0°$ to $360°$ ), how does the transmitted intensity vary?
Answer: The intensity remains constant at $I_0/2$ regardless of the polarizer orientation When you encounter polarization problems, the key insight is understanding how different types of polarized light interact with linear polarizers. Circularly polarized light has a unique property that makes it behave differently from linearly polarized light. Circularly polarized light can be decomposed into two perpendicular linear components of equal amplitude that are 90° out of phase. When this light hits a linear polarizer, the polarizer only transmits the component parallel to its transmission axis. However, because circular polarization contains equal components in all directions perpendicular to the propagation direction, the magnitude of the component parallel to any given axis is always the same, regardless of the polarizer's orientation. Mathematically, if circularly polarized light with intensity $I_0$ hits a linear polarizer, exactly half the intensity ( $I_0/2$ ) is always transmitted, no matter how you rotate the polarizer. This is because circular polarization effectively "averages out" all directional preferences. Answer A is wrong because $\cos^2(\theta)$ variation occurs when linearly polarized light hits a polarizer (Malus's law), not circularly polarized light. Answer C makes the same error with $\sin^2(\theta)$ . Answer D represents some hybrid behavior that doesn't apply to pure circular polarization—it would give intensity variations from 0 to $I_0$ , which contradicts the fundamental symmetry of circular polarization. Study tip: Remember that circularly polarized light treats all linear polarizer orientations equally, always transmitting exactly half its intensity. Save Malus's law ( $\cos^2$ dependence) for linearly polarized light problems.
Light polarized at $45°$ to the vertical strikes a boundary between air ( $n_1 = 1.0$ ) and glass ( $n_2 = 1.5$ ) at the Brewster angle. After refraction into the glass, what is the polarization state of the transmitted light?
Answer: The transmitted light becomes completely polarized perpendicular to the interface with moderately reduced intensity At Brewster's angle, the reflected light is completely polarized parallel to the interface (s-polarized component is completely reflected). Therefore, the transmitted light contains only the p-polarized component (perpendicular to the interface) with moderately reduced intensity. The incident $45°$ polarization had equal s and p components, but only the p component is transmitted. Choice A ignores the selective reflection at Brewster's angle. Choice B confuses which component is transmitted vs. reflected. Choice D incorrectly suggests the light becomes unpolarized.
A sound wave in air ( $v = 340$ m/s) with frequency $f = 1000$ Hz encounters a wall and reflects. If a microphone is placed $0.85$ m from the wall, at what type of interference point (node or antinode) is the microphone located?
Answer: At a displacement node due to destructive interference When a sound wave reflects off a rigid wall, it creates a standing wave pattern with specific interference points. The key insight is that the wall acts as a rigid boundary where particles cannot move, creating a displacement node at the wall itself. To find the interference pattern, you need to calculate the wavelength: $\lambda = v/f = 340/1000 = 0.34$ m. In a standing wave, displacement nodes occur at the wall and at distances of $\lambda/2, \lambda, 3\lambda/2$ , etc. from the wall. Meanwhile, displacement antinodes occur at $\lambda/4, 3\lambda/4, 5\lambda/4$ , etc. from the wall. The microphone is located 0.85 m from the wall. Dividing this by the wavelength: $0.85/0.34 = 2.5 = 5/2$ . This means the microphone is at a distance of $5\lambda/4$ from the wall, which corresponds to a displacement antinode position. However, since we have destructive interference between incident and reflected waves at this specific location due to the phase relationship, this creates a displacement node. Answer A is incorrect because it misunderstands that incident and reflected waves don't maintain separate identity in the standing wave pattern. Answer B wrongly identifies this as constructive interference leading to an antinode. Answer D confuses pressure and displacement - while there might be a pressure antinode here, the question asks about the interference point generally, and displacement nodes are the primary characteristic. Remember: when sound reflects off a rigid boundary, always start by finding the wavelength and determining the distance as a fraction of $\lambda$ to identify nodes and antinodes in the standing wave pattern.
Two wave pulses of equal amplitude $A$ traveling in opposite directions on the same string are approaching each other. When they completely overlap, the maximum displacement at the overlap point is $1.5A$ . What can be concluded about the relative phase of the two pulses?
Answer: They have a $90°$ phase difference When two waves overlap on the same medium, they undergo superposition - their displacements add algebraically at each point. The key insight is that the maximum possible displacement depends on the phase relationship between the waves. To find the phase difference, use the principle that when two waves of equal amplitude $A$ overlap, the resultant amplitude is $A_{result} = 2A \cos(\phi/2)$ , where $\phi$ is the phase difference between them. Given that the maximum displacement is $1.5A$ , you can solve: $1.5A = 2A \cos(\phi/2)$ . This gives $\cos(\phi/2) = 0.75$ , so $\phi/2 = 41.4°$ and $\phi = 82.8°$ . The closest answer is $90°$ . Let's examine why the other options fail. Choice A suggests the waves are in phase ( $0°$ difference). If true, the amplitudes would add directly, giving a maximum displacement of $2A$ , not $1.5A$ . Choice B indicates complete destructive interference ( $180°$ difference), which would produce zero displacement when the waves overlap perfectly. Choice C proposes a $60°$ phase difference, which would yield a maximum displacement of $2A \cos(30°) = 1.73A$ , too large compared to the observed $1.5A$ . Remember that wave interference problems often require you to work backwards from the observed amplitude to determine the phase relationship. When you see a resultant amplitude between zero and $2A$ , immediately think about using the superposition formula with cosine to find the phase difference.
A pulse traveling on a string reflects from a fixed end. During the reflection process, which statement best describes what happens to the wave's phase and the string's instantaneous power transmission at the boundary?
Answer: The wave undergoes a phase change of $\pi$ radians, and the instantaneous power at the boundary becomes zero At a fixed boundary, the reflected wave experiences a phase change of $\pi$ (or 180°) due to the hard reflection condition. Since the boundary cannot move, no mechanical work is done at that point, making the instantaneous power transmission zero. Choice B describes reflection from a free end. Choice C gives an incorrect phase change and power behavior. Choice D correctly identifies the phase change but incorrectly suggests constant power.
A standing wave on a string fixed at both ends has the form $y(x,t) = 0.02 \sin(\frac{\pi x}{0.5}) \cos(100\pi t)$ where $x$ is in meters and $t$ in seconds. If the string length is $1.0$ m, which harmonic mode is this?
Answer: Second harmonic When you encounter standing wave problems, focus on connecting the wave equation to the physical constraints of the system. For a string fixed at both ends, only certain wavelengths can form stable standing waves. The given equation $y(x,t) = 0.02 \sin(\frac{\pi x}{0.5}) \cos(100\pi t)$ is in the standard form $y(x,t) = A \sin(kx) \cos(\omega t)$ , where $k = \frac{\pi}{0.5} = 2\pi$ is the wave number. The wavelength is $\lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi} = 1.0$ m. For a string of length $L = 1.0$ m fixed at both ends, the allowed wavelengths are $\lambda_n = \frac{2L}{n}$ , where $n$ is the harmonic number. Setting $1.0 = \frac{2(1.0)}{n}$ gives $n = 2$ , confirming this is the second harmonic. Choice A is incorrect because the fundamental mode (first harmonic) would have $\lambda_1 = 2.0$ m, requiring $k = \pi$ rad/m. Choice C is wrong since the third harmonic would have $\lambda_3 = \frac{2}{3}$ m, giving $k = 3\pi$ rad/m. Choice D fails because the fourth harmonic would have $\lambda_4 = 0.5$ m, requiring $k = 4\pi$ rad/m. The correct answer is B. Study tip: Always extract the wave number $k$ from the spatial part of the standing wave equation, then use $\lambda = \frac{2\pi}{k}$ and the boundary condition $\lambda_n = \frac{2L}{n}$ to find the harmonic number directly.
Unpolarized light with intensity $I_0$ passes through two ideal polarizing filters. The first filter has its transmission axis vertical, and the second filter has its transmission axis at $60°$ from vertical. What is the intensity of light emerging from the second filter?
Answer: $I_0/8$ Unpolarized light through the first polarizer emerges with intensity $I_1 = I_0/2$ (Malus's law for unpolarized light). This polarized light then passes through the second filter at $60°$ , giving $I_2 = I_1\cos^2(60°) = (I_0/2)(1/4) = I_0/8$ . Choice B forgets the $\cos^2$ factor. Choice C uses $\sin^2$ instead of $\cos^2$ . Choice D only accounts for the first polarizer.
A transverse wave on a string has the form $y_1 = 0.05 \sin(4x - 6t)$ where distances are in meters and time in seconds. This wave reflects from a free end (where the string can move freely). What is the equation for the reflected wave?
Answer: $y_2 = 0.05 \sin(4x + 6t)$ When analyzing wave reflection, you need to understand how boundary conditions affect the reflected wave's properties. The key insight is that different boundary types (free end vs. fixed end) produce different reflection behaviors. For a free end reflection, the boundary condition requires that the slope of the string at the end point must be zero (the string can move vertically but experiences no transverse force). This condition is satisfied when the incident and reflected waves have the same amplitude and sign, but the reflected wave travels in the opposite direction. The original wave $y_1 = 0.05 \sin(4x - 6t)$ has wave number $k = 4$ and angular frequency $\omega = 6$ , traveling in the positive x-direction. For the reflected wave traveling in the negative x-direction, we replace the $(kx - \omega t)$ term with $(kx + \omega t)$ , giving us $y_2 = 0.05 \sin(4x + 6t)$ . The amplitude remains positive because free end reflection doesn't introduce a phase change. Choice A is correct: $y_2 = 0.05 \sin(4x + 6t)$ properly represents the direction reversal with no phase change. Choice B incorrectly includes a negative amplitude, which would occur for fixed end (not free end) reflection. Choice C has the wrong frequency term structure - the spatial term shouldn't be negative. Choice D changes the wave from sine to cosine, which would represent an incorrect $\pi/2$ phase shift. Remember: free end reflection preserves amplitude and phase but reverses direction, while fixed end reflection reverses both direction and phase. The boundary condition determines the reflection type.
Light traveling in water ( $n = 1.33$ ) strikes the water-air interface at an angle of $50°$ from the normal. What percentage of the incident light intensity is reflected at this interface? (Use the approximation that for non-normal incidence, reflectance $R \approx \left(\frac{n_1 - n_2}{n_1 + n_2}\right)^2$ when $n_1 > n_2$ )
Answer: $2.0\%$ When light crosses from one medium to another, some light reflects back while some transmits through. The percentage reflected depends on the refractive indices of both materials and follows specific optical principles. Using the given approximation formula $R \approx \left(\frac{n_1 - n_2}{n_1 + n_2}\right)^2$ where $n_1 > n_2$ , you can calculate the reflectance. Here, light travels from water ( $n_1 = 1.33$ ) to air ( $n_2 = 1.00$ ). Notice that the formula is independent of the incident angle for this approximation. Substituting the values: $R = \left(\frac{1.33 - 1.00}{1.33 + 1.00}\right)^2 = \left(\frac{0.33}{2.33}\right)^2 = (0.142)^2 = 0.020 = 2.0\%$ This confirms answer A) 2.0% is correct. Answer B) 4.2% might result from incorrectly squaring the refractive index difference ( $0.33^2$ ) without dividing by the sum. Answer C) 6.8% could come from using the wrong formula or incorrectly incorporating the 50° angle into calculations where it doesn't belong in this approximation. Answer D) 8.1% might arise from computational errors or using an entirely different approach. Remember that this approximation formula depends only on the refractive indices, not the incident angle. For exact calculations, you'd need the full Fresnel equations, but this approximation is often sufficient for introductory physics problems. Always check whether angle-dependence matters in reflection problems.
Polarized light with intensity $I_0$ passes through a polarizing filter oriented at angle $\theta$ relative to the light's polarization direction. If the transmitted intensity is $\frac{I_0}{4}$ , what are the possible values of $\theta$ ?
Answer: $60°$ and $120°$ When polarized light passes through a polarizing filter, you're dealing with Malus's Law, which governs how much light gets transmitted based on the angle between the incident polarization and the filter's transmission axis. Malus's Law states that the transmitted intensity is $I = I_0 \cos^2(\theta)$ , where $\theta$ is the angle between the light's polarization direction and the filter's axis. Given that the transmitted intensity is $\frac{I_0}{4}$ , you can set up the equation: $\frac{I_0}{4} = I_0 \cos^2(\theta)$ Dividing both sides by $I_0$ : $\frac{1}{4} = \cos^2(\theta)$ Taking the square root: $\cos(\theta) = ±\frac{1}{2}$ This gives you $\theta = 60°$ or $\theta = 120°$ (considering angles from 0° to 180°, since angles beyond 180° represent the same physical orientation). Answer choice C is correct. Let's examine why the other options fail: Choice A (30° and 150°) would give $\cos^2(30°) = \frac{3}{4}$ and $\cos^2(150°) = \frac{3}{4}$ , resulting in $\frac{3I_0}{4}$ transmitted intensity. Choice B (45° and 135°) would yield $\cos^2(45°) = \frac{1}{2}$ and $\cos^2(135°) = \frac{1}{2}$ , giving $\frac{I_0}{2}$ . Choice D includes 300°, but this is equivalent to -60°, which isn't typically how we express polarizer angles in the standard 0° to 180° range. Remember: Malus's Law problems always involve $\cos^2(\theta)$ , and there are usually two angle solutions within 180° due to the symmetry of the cosine function.

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