Welcome to Lecture 122 (Master Index Entry: L116) of our comprehensive JEE Physics Master Series! Advancing deeper into wave dynamics in Unit 10: Oscillations and Waves, we present the fundamental synthesizing concept of wave optics and acoustics: The Principle of Superposition of Waves — Interference, Coherence, Phasor Analysis, and Energy Redistribution. The principle of superposition forms the cornerstone for understanding standing waves, interference patterns, diffraction, and acoustic beats. In this masterclass, we examine the linearity condition underpinning superposition, derive the analytical formulas for resultant amplitude and intensity using trigonometric and phasor techniques, establish the exact criteria for constructive and destructive interference, evaluate Quincke’s tube interference, demonstrate the strict conservation of acoustic energy, and master high-yield JEE Advanced problem patterns.
Detailed Concept Card: The Principle of Superposition & Interference Mechanics
1. The Principle of Superposition
Formal Statement: When two or more waves travel simultaneously through the same region of an elastic medium, the resultant displacement of any particle at any point and time is the vector (or algebraic) sum of the individual displacements that would be produced by each wave acting independently:
$$\mathbf{\vec{y}_{\text{net}}(x, t) = \vec{y}_1(x, t) + \vec{y}_2(x, t) + \dots + \vec{y}_n(x, t)}$$
- The Linearity Requirement: Superposition holds strictly whenever the governing differential wave equation is linear, which requires that the medium obeys Hooke’s law (stress directly proportional to strain). For exceptionally large displacements (e.g., shock waves, supersonic booms, intense non-linear optical lasers), the restoring force becomes non-linear and superposition breaks down.
- Independence of Wave Propagation: After crossing and overlapping, each wave continues along its trajectory completely unaltered in shape, amplitude, frequency, and velocity, as if the other waves were entirely absent.
2. Superposition of Two Coherent Harmonic Waves (Interference)
Consider two harmonic waves of the same frequency $\omega$ and wave number $k$ travelling along the $+x$-direction with a constant phase difference $\phi$:
$$y_1(x, t) = A_1 \sin(kx – \omega t)$$
$$y_2(x, t) = A_2 \sin(kx – \omega t + \phi)$$
By the principle of superposition, the resultant displacement is:
$$y_{\text{net}} = y_1 + y_2 = A_1 \sin(kx – \omega t) + A_2 \sin(kx – \omega t + \phi)$$
Using the phasor (vector) addition method, the resultant wave is another simple harmonic wave of the exact same frequency and wave number:
$$\mathbf{y_{\text{net}}(x, t) = A_{\text{net}} \sin(kx – \omega t + \theta)}$$
| Resultant Parameter | Analytical Formula | Physical Meaning |
|---|---|---|
| Resultant Amplitude ($A_{\text{net}}$) | $$\mathbf{A_{\text{net}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}}$$ | Vector resultant of two amplitude phasors with angle $\phi$ between them. |
| Resultant Phase Angle ($\theta$) | $$\mathbf{\tan\theta = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi}}$$ | Phase lead of resultant wave relative to wave $y_1$. |
| Resultant Intensity ($I_{\text{net}}$) | $$\mathbf{I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi}$$ | The term $2\sqrt{I_1 I_2}\cos\phi$ is the interference term. |
3. Constructive vs Destructive Interference
The nature of interference at any point in space depends solely on the phase difference $\phi$, which is determined by the path difference $\Delta x$ ($\phi = \frac{2\pi}{\lambda}\Delta x$):
| Interference Type | Phase Difference ($\phi$) | Path Difference ($\Delta x$) | Resultant Amplitude ($A_{\text{net}}$) | Resultant Intensity ($I_{\text{net}}$) |
|---|---|---|---|---|
| Constructive (Maxima) | $$\phi = 2n\pi \quad (n \in \mathbb{Z})$$ | $$\mathbf{\Delta x = n\lambda}$$ | $$A_{\text{max}} = A_1 + A_2$$ | $$I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2$$ |
| Destructive (Minima) | $$\phi = (2n + 1)\pi \quad (n \in \mathbb{Z})$$ | $$\mathbf{\Delta x = (2n + 1)\frac{\lambda}{2}}$$ | $$A_{\text{min}} = |A_1 – A_2|$$ | $$I_{\text{min}} = (\sqrt{I_1} – \sqrt{I_2})^2$$ |
- Special Case of Equal Amplitudes ($A_1 = A_2 = A_0, I_1 = I_2 = I_0$):
$$A_{\text{net}} = 2 A_0 \cos\left(\frac{\phi}{2}\right)$$
$$\mathbf{I_{\text{net}} = 4 I_0 \cos^2\left(\frac{\phi}{2}\right)}$$
- At constructive maxima: $A_{\text{max}} = 2A_0$, $I_{\text{max}} = 4I_0$.
- At destructive minima: $A_{\text{min}} = 0$, $I_{\text{min}} = 0$ (complete darkness / total silence).
- Intensity Contrast Ratio: $$\mathbf{\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_1 + A_2}{A_1 – A_2}\right)^2 = \left(\frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} – \sqrt{I_2}}\right)^2}$$
4. Energy Conservation in Interference
A common student misconception is that energy is destroyed at destructive minima and created out of nothing at constructive maxima. In reality, energy is strictly conserved:
$$\langle I_{\text{net}} \rangle = \frac{1}{2\pi} \int_0^{2\pi} (I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi) \, d\phi = I_1 + I_2 + \frac{2\sqrt{I_1 I_2}}{2\pi} \int_0^{2\pi} \cos\phi \, d\phi = \mathbf{I_1 + I_2}$$
The spatial average of intensity across an entire fringe cycle equals the simple sum of the separate intensities. Interference does not alter total energy; it merely redistributes energy spatially from destructive nodes to constructive antinodes.
5. Quincke’s Tube Acoustic Interference
Quincke’s tube is a classic laboratory apparatus used to demonstrate sound interference and measure acoustic wavelength:
- Sound enters a branching tube that splits into a fixed path $L_1$ and an adjustable sliding U-tube path $L_2$.
- When the sliding tube is pulled out by distance $\Delta d$, the path length increases by $\mathbf{\Delta x = 2 \Delta d}$ (because both arms of the U-tube extend).
- Condition to transition from a maximum to the first adjacent minimum: $$\Delta x = \frac{\lambda}{2} \implies 2 \Delta d = \frac{\lambda}{2} \implies \mathbf{\Delta d = \frac{\lambda}{4}}$$
- Condition to transition from one maximum to the next consecutive maximum: $$\Delta x = \lambda \implies 2 \Delta d = \lambda \implies \mathbf{\Delta d = \frac{\lambda}{2}}$$
Four Solved Examples (Graded Basic to Advanced)
Example 1: Basic Application — Path Difference and Resultant Interference at a Detector
Problem Statement: Two coherent sound sources $S_1$ and $S_2$ emit waves in phase with identical amplitude $A_0$ and wavelength $\lambda = 0.40\text{ m}$. A sound detector is positioned at point $P$ such that $S_1 P = 2.40\text{ m}$ and $S_2 P = 3.00\text{ m}$.
(a) Determine the path difference $\Delta x$ between the two waves at point $P$.
(b) Calculate the phase difference $\Delta\phi$ between the two waves at point $P$.
(c) State whether the interference at point $P$ is constructive, destructive, or intermediate.
(d) If each individual source produces an acoustic intensity $I_0 = 10.0\text{ mW/m}^2$, calculate the resultant intensity $I_{\text{net}}$ detected at point $P$.
Solution:
(a) Path Difference:
$$\mathbf{\Delta x = S_2 P – S_1 P = 3.00\text{ m} – 2.40\text{ m} = 0.60\text{ m}}$$
(b) Phase Difference:
$$\mathbf{\Delta\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{0.40\text{ m}} (0.60\text{ m}) = 2\pi (1.50) = 3\pi\text{ rad} = 540^\circ}$$
(c) Nature of Interference:
Since $\Delta\phi = 3\pi = (2 \times 1 + 1)\pi$ is an odd multiple of $\pi$ (or path difference $\Delta x = 0.60\text{ m} = \frac{3}{2}\lambda$), the waves meet in exact anti-phase. This is destructive interference (a complete minimum).
(d) Resultant Intensity:
$$I_{\text{net}} = 4 I_0 \cos^2\left(\frac{\Delta\phi}{2}\right) = 4(10.0\text{ mW/m}^2) \cos^2\left(\frac{3\pi}{2}\right) = 4(10.0)(0) = \mathbf{0\text{ mW/m}^2}$$
The two sound waves completely cancel out, creating total silence at point $P$.
Example 2: Intermediate Application — Amplitude Ratio & Contrast from Intensity Extrema
Problem Statement: Two coherent sound waves interfere in a medium. The maximum intensity observed in the interference pattern is $I_{\text{max}} = 100.0\text{ W/m}^2$, while the minimum intensity is $I_{\text{min}} = 4.0\text{ W/m}^2$.
(a) Determine the ratio of the amplitudes of the two interfering waves, $\frac{A_1}{A_2}$.
(b) Calculate the individual intensities $I_1$ and $I_2$ of the two waves.
(c) Determine the resultant intensity when the two waves superpose with a phase difference of $\phi = \frac{\pi}{2}$ ($90^\circ$).
Solution:
(a) Amplitude Ratio:
The ratio of intensity extrema is:
$$\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{100.0}{4.0} = 25.0 \implies \frac{\sqrt{I_{\text{max}}}}{\sqrt{I_{\text{min}}}} = \sqrt{25.0} = 5.0$$
Since $\sqrt{I_{\text{max}}} = A_1 + A_2$ and $\sqrt{I_{\text{min}}} = A_1 – A_2$ (assuming $A_1 > A_2$):
$$\frac{A_1 + A_2}{A_1 – A_2} = 5.0 \implies A_1 + A_2 = 5 A_1 – 5 A_2 \implies 6 A_2 = 4 A_1$$
$$\mathbf{\frac{A_1}{A_2} = \frac{6}{4} = \frac{3}{2} = 1.50}$$
(b) Individual Intensities:
$$\sqrt{I_1} + \sqrt{I_2} = \sqrt{100.0} = 10.0$$
$$\sqrt{I_1} – \sqrt{I_2} = \sqrt{4.0} = 2.0$$
Adding the two equations:
$$2\sqrt{I_1} = 12.0 \implies \sqrt{I_1} = 6.0 \implies \mathbf{I_1 = (6.0)^2 = 36.0\text{ W/m}^2}$$
Subtracting the equations:
$$2\sqrt{I_2} = 8.0 \implies \sqrt{I_2} = 4.0 \implies \mathbf{I_2 = (4.0)^2 = 16.0\text{ W/m}^2}$$
Verification: $I_{\text{max}} = (\sqrt{36} + \sqrt{16})^2 = (6 + 4)^2 = 100\text{ W/m}^2$; $I_{\text{min}} = (6 – 4)^2 = 4\text{ W/m}^2$.
(c) Intensity at $\phi = \pi/2$:
$$\mathbf{I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\left(\frac{\pi}{2}\right) = 36.0 + 16.0 + 0 = 52.0\text{ W/m}^2}$$
Example 3: Advanced Application — Phasor Addition of Three Equal-Amplitude Waves
Problem Statement: Three coherent harmonic waves propagate along the same direction in a continuous medium: $$y_1 = A_0 \sin(\omega t – kx)$$ $$y_2 = A_0 \sin\left(\omega t – kx + \frac{\pi}{3}\right)$$ $$y_3 = A_0 \sin\left(\omega t – kx + \frac{2\pi}{3}\right)$$ where $A_0 = 10.0\text{ mm}$.
(a) Use the phasor vector addition technique to determine the resultant amplitude $A_{\text{net}}$.
(b) Determine the phase angle $\theta$ of the resultant wave relative to $y_1$.
(c) Write the full equation for the resultant wave $y_{\text{net}}(x, t)$.
(d) Find the ratio of the resultant intensity $I_{\text{net}}$ to the intensity $I_0$ of a single wave.
Solution:
(a) Phasor Addition:
Represent each wave as a phasor vector in the complex plane:
– Phasor 1: Magnitude $A_0$ along $0^\circ$.
– Phasor 2: Magnitude $A_0$ at angle $\phi = \pi/3 = 60^\circ$.
– Phasor 3: Magnitude $A_0$ at angle $2\phi = 2\pi/3 = 120^\circ$.
Sum of components along the reference axis ($x$-axis):
$$A_x = A_0 \cos(0^\circ) + A_0 \cos(60^\circ) + A_0 \cos(120^\circ) = A_0(1) + A_0\left(\frac{1}{2}\right) + A_0\left(-\frac{1}{2}\right) = A_0$$
Sum of components along the perpendicular axis ($y$-axis):
$$A_y = A_0 \sin(0^\circ) + A_0 \sin(60^\circ) + A_0 \sin(120^\circ) = 0 + A_0\left(\frac{\sqrt{3}}{2}\right) + A_0\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} A_0$$
The resultant amplitude is:
$$\mathbf{A_{\text{net}} = \sqrt{A_x^2 + A_y^2} = \sqrt{A_0^2 + (\sqrt{3}A_0)^2} = \sqrt{A_0^2 + 3A_0^2} = \sqrt{4A_0^2} = 2 A_0 = 2(10.0\text{ mm}) = 20.0\text{ mm}}$$
(b) Resultant Phase Angle:
$$\tan\theta = \frac{A_y}{A_x} = \frac{\sqrt{3}A_0}{A_0} = \sqrt{3} \implies \mathbf{\theta = 60^\circ = \frac{\pi}{3}\text{ rad}}$$
The resultant phasor is oriented along $\pi/3$, exactly collinear with the central wave $y_2$.
(c) Resultant Wave Function:
$$\mathbf{y_{\text{net}}(x, t) = (20.0\text{ mm}) \sin\left(\omega t – kx + \frac{\pi}{3}\right)}$$
(d) Intensity Ratio:
$$\mathbf{\frac{I_{\text{net}}}{I_0} = \left(\frac{A_{\text{net}}}{A_0}\right)^2 = \left(\frac{2 A_0}{A_0}\right)^2 = 4.0}$$
Example 4: JEE Advanced Scenario — Quincke’s Acoustic Tube Interference & Precision Wavelength
Problem Statement: An audio oscillator feeds a sound wave of frequency $f = 850\text{ Hz}$ into a Quincke’s interference tube apparatus. The wave divides equally between a fixed path $L_1$ and a variable sliding U-tube path $L_2$. Initially, the two paths have identical lengths ($L_1 = L_2$), producing maximum sound intensity at the receiver detector. The speed of sound in air is $v = 340\text{ m/s}$.
(a) Calculate the acoustic wavelength $\lambda$ of the sound wave.
(b) The sliding U-tube is pulled outward by a displacement $\Delta d$. Express the path difference $\Delta x$ introduced between the two sound waves as a function of $\Delta d$.
(c) Determine the minimum extension $\Delta d_{\text{min}}$ required to reduce the detected sound intensity to its very first minimum (silence).
(d) Find the sliding displacement $\Delta d_2$ required to restore the sound intensity to its next constructive maximum.
Solution:
(a) Acoustic Wavelength:
$$\mathbf{\lambda = \frac{v}{f} = \frac{340\text{ m/s}}{850\text{ Hz}} = \frac{34}{85}\text{ m} = 0.40\text{ m} = 40.0\text{ cm}}$$
(b) Path Difference Relation:
Because the sliding section is U-shaped, pulling it outward by distance $\Delta d$ increases the length of both the top and bottom legs by $\Delta d$. Hence:
$$\mathbf{\Delta x = 2 \Delta d}$$
(c) Minimum Sliding Distance for First Minimum:
Destructive interference occurs when the path difference equals half a wavelength:
$$\Delta x = \frac{\lambda}{2} \implies 2 \Delta d_{\text{min}} = \frac{\lambda}{2}$$
$$\mathbf{\Delta d_{\text{min}} = \frac{\lambda}{4} = \frac{40.0\text{ cm}}{4} = 10.0\text{ cm} = 0.10\text{ m}}$$
(d) Sliding Distance for Subsequent Maximum:
The next maximum occurs when the path difference equals one full wavelength:
$$\Delta x = \lambda \implies 2 \Delta d_2 = \lambda$$
$$\mathbf{\Delta d_2 = \frac{\lambda}{2} = \frac{40.0\text{ cm}}{2} = 20.0\text{ cm} = 0.20\text{ m}}$$
Practice Worksheet: 10 Target JEE Problems
Solidify your understanding of wave interference, phasor diagrams, and intensity ratios. Attempt all 10 problems before consulting the detailed solutions at the end.
Section I: Single Correct Choice Questions (Q1–Q5)
Q1. When two coherent progressive waves of equal amplitude $A$ and phase difference $\phi = \frac{\pi}{2}$ superpose in a medium, the resultant amplitude is:
(A) $A$
(B) $\sqrt{2}A$
(C) $2A$
(D) $A/\sqrt{2}$
Q2. Two coherent sources with an intensity ratio of $9 : 1$ interfere. The ratio of the maximum intensity to the minimum intensity in the resulting interference pattern is:
(A) $4 : 1$
(B) $9 : 1$
(C) $16 : 1$
(D) $25 : 1$
Q3. Two harmonic waves given by $y_1 = a \sin(\omega t – kx)$ and $y_2 = a \cos(\omega t – kx)$ superpose on a string. The amplitude of the resultant wave is:
(A) $a$
(B) $2a$
(C) $\sqrt{2}a$
(D) Zero
Q4. In Quincke’s acoustic tube experiment, the sound intensity detected at the receiver drops from a maximum to a minimum when the sliding U-tube is displaced outward by $5.0\text{ cm}$. The wavelength of the sound wave is:
(A) $5.0\text{ cm}$
(B) $10.0\text{ cm}$
(C) $15.0\text{ cm}$
(D) $20.0\text{ cm}$
Q5. The principle of superposition fails and is no longer valid when:
(A) The interfering waves propagate in opposite directions
(B) The interfering waves possess different frequencies
(C) The wave amplitude is sufficiently large to produce non-linear elastic deformation of the medium
(D) The waves travel at different speeds
Section II: One or More Than One Correct Questions (Q6–Q8)
Q6. Two coherent waves $y_1 = A_1 \sin(kx – \omega t)$ and $y_2 = A_2 \sin(kx – \omega t + \phi)$ superpose in a medium. Which of the following statements is/are correct?
(A) The resultant wave is harmonic with angular frequency $\omega$ and wave number $k$.
(B) The resultant intensity is maximum when $\phi = 2n\pi$ ($n$ is an integer).
(C) The resultant intensity is zero only if $A_1 = A_2$ and $\phi = (2n+1)\pi$.
(D) The spatial average of the intensity across an interference cycle is $I_{\text{avg}} = I_1 + I_2$.
Q7. Two sound waves emitted by coherent sources interfere at a point where the path difference is $\Delta x = 2.5 \lambda$. Which of the following statements is/are correct?
(A) The phase difference between the waves at that point is $5\pi$ radians.
(B) Destructive interference occurs at that point.
(C) If the wave amplitudes are $4.0\text{ mm}$ and $3.0\text{ mm}$, the resultant amplitude is $1.0\text{ mm}$.
(D) If the wave amplitudes are $4.0\text{ mm}$ and $3.0\text{ mm}$, the resultant intensity is proportional to $1.0\text{ mm}^2$.
Q8. Which of the following physical phenomena represent direct consequences of the principle of wave superposition?
(A) Spatial interference of coherent light and sound waves
(B) Formation of standing waves on strings and in organ pipes
(C) Acoustic beats produced by two sound waves of slightly differing frequencies
(D) Refraction of sound waves at the interface between air and water
Section III: Numerical / Integer Type Questions (Q9–Q10)
Q9. Two coherent waves have an intensity ratio of $4 : 1$. The ratio of the maximum intensity to the minimum intensity in the resulting interference pattern is an integer. Calculate the value of this integer ratio $\frac{I_{\text{max}}}{I_{\text{min}}}$.
Q10. In a Quincke’s tube acoustic experiment, a tuning fork generates sound that travels at speed $v = 340\text{ m/s}$. The sliding U-tube is drawn outward by $\Delta d = 8.5\text{ cm} = 0.085\text{ m}$ to transition from a central intensity maximum to the first adjacent intensity minimum. Calculate the vibration frequency of the tuning fork in $\text{Hz}$.
Comprehensive Solutions & Answer Key
Answer Key
| Question No. | Correct Answer | Key Concept / Formula Tested |
|---|---|---|
| Q1 | (B) | $A_{\text{net}} = \sqrt{A^2 + A^2 + 2A^2\cos(\pi/2)} = \sqrt{2}A$ |
| Q2 | (A) | $I_1/I_2 = 9 \implies A_1/A_2 = 3 \implies I_{\text{max}}/I_{\text{min}} = ((3+1)/(3-1))^2 = 4$ |
| Q3 | (C) | Phase difference between $\sin$ and $\cos$ is $\pi/2 \implies A_{\text{net}} = \sqrt{a^2 + a^2} = \sqrt{2}a$ |
| Q4 | (D) | $\Delta d = \lambda/4 \implies \lambda = 4(5.0\text{ cm}) = 20.0\text{ cm}$ |
| Q5 | (C) | Large amplitudes induce non-linear elastic responses where superposition fails |
| Q6 | (A, B, C, D) | Harmonic wave interference properties and energy conservation $\langle I \rangle = I_1 + I_2$ |
| Q7 | (A, B, C, D) | $\Delta x = 2.5\lambda \implies \phi = 5\pi$ (anti-phase); $A_{\text{net}} = 4 – 3 = 1\text{ mm}$ |
| Q8 | (A, B, C) | Interference, standing waves, and beats are all superposition phenomena |
| Q9 | 9 | $I_1/I_2 = 4 \implies A_1/A_2 = 2 \implies I_{\text{max}}/I_{\text{min}} = ((2+1)/(2-1))^2 = 9$ |
| Q10 | 1000 | $\Delta d = \lambda/4 \implies \lambda = 0.34\text{ m} \implies f = 340 / 0.34 = 1000\text{ Hz}$ |
Detailed Step-by-Step Solutions
Solution to Q1:
The resultant amplitude of two interfering waves with phase difference $\phi$ is:
$$A_{\text{net}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}$$
Substituting $A_1 = A_2 = A$ and $\phi = \frac{\pi}{2}$ ($\cos(\pi/2) = 0$):
$$\mathbf{A_{\text{net}} = \sqrt{A^2 + A^2 + 0} = \sqrt{2A^2} = \sqrt{2}A}$$
Hence, the correct choice is (B).
Solution to Q2:
Given the intensity ratio $\frac{I_1}{I_2} = 9 \implies \frac{A_1}{A_2} = \sqrt{9} = 3$.
The ratio of maximum to minimum intensity is:
$$\mathbf{\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_1 + A_2}{A_1 – A_2}\right)^2 = \left(\frac{3 + 1}{3 – 1}\right)^2 = \left(\frac{4}{2}\right)^2 = (2)^2 = 4 = 4 : 1}$$
Hence, the correct choice is (A).
Solution to Q3:
Using the trigonometric identity $\cos\theta = \sin(\theta + \pi/2)$:
$$y_2 = a \cos(\omega t – kx) = a \sin\left(\omega t – kx + \frac{\pi}{2}\right)$$
The phase difference between $y_1$ and $y_2$ is $\phi = \pi/2$. Since both have amplitude $a$:
$$\mathbf{A_{\text{net}} = \sqrt{a^2 + a^2 + 2a^2 \cos(\pi/2)} = \sqrt{a^2 + a^2} = \sqrt{2}a}$$
Hence, the correct choice is (C).
Solution to Q4:
In Quincke’s tube, pulling the U-tube by displacement $\Delta d$ increases the acoustic path by $\Delta x = 2 \Delta d$.
The shift from a maximum to an adjacent minimum requires a path difference of half a wavelength:
$$\Delta x = \frac{\lambda}{2} \implies 2 \Delta d = \frac{\lambda}{2} \implies \lambda = 4 \Delta d$$
Given $\Delta d = 5.0\text{ cm}$:
$$\mathbf{\lambda = 4(5.0\text{ cm}) = 20.0\text{ cm}}$$
Hence, the correct choice is (D).
Solution to Q5:
The principle of superposition relies fundamentally on the linearity of the wave equation ($\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$), which requires Hooke’s elastic restoring law ($F = -kx$). When the amplitude of oscillation becomes exceptionally large, non-linear elastic effects become significant, higher-order restoring terms ($x^2, x^3$) appear, and the principle of superposition ceases to apply.
Hence, the correct choice is (C).
Solution to Q6:
Let us evaluate each statement:
(A) Superposition of two coherent harmonic waves of identical frequency $\omega$ and wave number $k$ produces a resultant harmonic wave with the exact same frequency $\omega$ and wave number $k$. (A) is correct.
(B) $I_{\text{net}} = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi$ is maximized when $\cos\phi = +1 \implies \phi = 2n\pi$. (B) is correct.
(C) $I_{\text{min}} = (\sqrt{I_1} – \sqrt{I_2})^2$ equals zero if and only if $I_1 = I_2 \implies A_1 = A_2$, with $\cos\phi = -1 \implies \phi = (2n+1)\pi$. (C) is correct.
(D) Averaging over a full $2\pi$ phase cycle, $\langle \cos\phi \rangle = 0$, giving $\langle I \rangle = I_1 + I_2$, confirming energy conservation. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q7:
(A) $\Delta\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda}(2.5\lambda) = 5\pi\text{ rad}$. (A) is correct.
(B) Since $\Delta\phi = 5\pi$ is an odd multiple of $\pi$, destructive interference occurs. (B) is correct.
(C) For destructive interference: $A_{\text{min}} = |A_1 – A_2| = |4.0\text{ mm} – 3.0\text{ mm}| = 1.0\text{ mm}$. (C) is correct.
(D) Intensity is proportional to amplitude squared: $I \propto A^2 \implies I_{\text{net}} \propto (1.0\text{ mm})^2$. (D) is correct.
Hence, all choices (A, B, C, D) are correct.
Solution to Q8:
(A) Interference is the superposition of coherent waves with constant phase difference. (A) is a superposition phenomenon.
(B) Standing waves are formed by the superposition of identical waves travelling in opposite directions. (B) is a superposition phenomenon.
(C) Acoustic beats arise from the superposition of two waves of slightly differing frequencies. (C) is a superposition phenomenon.
(D) Refraction is governed by the boundary change of phase speed (Snell’s law), not by wave superposition. (D is not a superposition phenomenon).
Hence, correct choices are (A, B, C).
Solution to Q9:
Given the intensity ratio $\frac{I_1}{I_2} = 4 \implies \frac{A_1}{A_2} = \sqrt{4} = 2$.
The ratio of maximum to minimum intensity is:
$$\mathbf{\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_1 + A_2}{A_1 – A_2}\right)^2 = \left(\frac{2 + 1}{2 – 1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9}$$
Answer: 9
Solution to Q10:
In Quincke’s tube, moving from a maximum to the first minimum corresponds to a displacement:
$$\Delta d = \frac{\lambda}{4} \implies \lambda = 4 \Delta d$$
Given $\Delta d = 8.5\text{ cm} = 0.085\text{ m}$:
$$\lambda = 4(0.085\text{ m}) = 0.34\text{ m}$$
Using $v = f \lambda$:
$$\mathbf{f = \frac{v}{\lambda} = \frac{340\text{ m/s}}{0.34\text{ m}} = 1000\text{ Hz}}$$
Answer: 1000