Superposition, Stationary Waves on Strings, Young's Double-Slit & Diffraction Gratings

1

Principle of Superposition, Phase Difference & Coherence

Principle of Superposition: when two or more waves meet at a point, the resultant displacement equals the vector sum of the individual displacements of the waves. Two wave sources are coherent if they have the same frequency (and wavelength) and a constant phase difference. Constructive interference occurs when path difference = nλ (phase difference = 2nπ rad or 360n°, in phase); Destructive interference occurs when path difference = (n + ½)λ (phase difference = (2n + 1)π rad or 180°, antiphase).

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2

Stationary (Standing) Waves vs Progressive Waves

A stationary wave is formed by the superposition of two progressive waves of the same frequency, wavelength, and amplitude travelling in opposite directions (e.g. an incident wave and its reflection on a stretched string or in a microwave/pipe cavity). Nodes (N): points of zero amplitude where waves always meet in antiphase (π rad); distance between adjacent nodes = ½λ. Antinodes (A): points of maximum amplitude where waves meet in phase (0 rad). Unlike progressive waves, stationary waves transfer zero net energy, have amplitude that varies from 0 at nodes to max at antinodes, and all points between two adjacent nodes oscillate IN phase with one another (Points on opposite sides of a single node are in antiphase by π rad).

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3

Stationary Waves on a Stretched String: f₁ = (1 / 2L) √(T / μ)

For a string of length L fixed at both ends (node at each end): First harmonic (fundamental, N-A-N) has L = ½λ₁, so λ₁ = 2L and f₁ = c / 2L = (1 / 2L) √(T / μ), where T is string tension (in N) and μ = m / L is mass per unit length (in kg m⁻¹). The nth harmonic has n antinodes (loops), wavelength λ_n = 2L / n, and frequency f_n = n f₁.

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4

Young's Double-Slit Interference: w = λD / s

Illuminating two narrow, parallel slits separated by distance s with coherent monochromatic light of wavelength λ produces equally spaced bright and dark interference fringes on a screen at distance D (valid when D ≫ s, small-angle approximation): Fringe spacing w = λD / s, where w is the distance between centres of adjacent bright (or adjacent dark) fringes. If white light is used: the central fringe (zero path difference for all λ) is white, and outer fringes are continuous spectra with violet (shortest λ) closest to the centre and red (longest λ) furthest out.

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5

Diffraction Gratings: d sin θ = nλ & Maximum Order (n_max)

A diffraction grating has N slits per metre (or per mm), so the slit spacing is d = 1 / N (e.g. 500 lines mm⁻¹ gives d = 10⁻³ m / 500 = 2.0 × 10⁻⁶ m). Constructive interference maxima occur at discrete angles θ where the path difference between adjacent slits equals nλ: d sin θ = nλ, where n is the integer order of maximum (n = 0, 1, 2.). Compared to a double slit, a diffraction grating produces much sharper, brighter, and more widely spaced maxima, allowing much more accurate measurement of λ. Since sin θ cannot exceed 1 (θ < 90°), the highest visible order is n_max = ⌊d / λ⌋ (always round down to the nearest whole integer; total number of bright spots = 2n_max + 1).

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6

Refraction (n₁ sin θ₁ = n₂ sin θ₂), Critical Angle & Optical Fibres

Refractive index n = c / c_s (ratio of speed of light in vacuum to speed in material; frequency f stays constant while wavelength λ_s = λ / n decreases). Snell's Law: n₁ sin θ₁ = n₂ sin θ₂. Total Internal Reflection (TIR) occurs when light travels from a higher refractive index core (n₁) toward a lower refractive index cladding (n₂ < n₁) and the angle of incidence θ₁ exceeds the critical angle θ_c, where sin θ_c = n₂ / n₁. Step-index optical fibres have a narrow high-n glass core surrounded by a lower-n glass cladding (which protects the core surface from scratches, prevents crossover of signals between adjacent fibres, and increases θ_c so only rays near the axis undergo TIR, reducing modal/multipath dispersion). Material (chromatic) dispersion (violet travels slower than red in glass) is eliminated by using monochromatic laser light; both dispersions cause pulse broadening that limits maximum bit-rate.

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1. Q1

   A steel guitar string of length 0.640 m and mass 1.80 × 10⁻³ kg is under a tension of 115 N. Calculate the frequency of its first harmonic (fundamental) and the wavelength and frequency of the third harmonic on the string, and explain how a stationary wave is formed when the string is plucked. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Calculate mass per unit length μ = m ÷ L = (1.80 × 10⁻³ kg) ÷ 0.640 m = 2.813 × 10⁻³ kg m⁻¹ [1 mark]. 2. Substitute into the first harmonic equation: f₁ = (1 ÷ (2 × 0.640)) × √(115 ÷ 2.813 × 10⁻³) = 0.78125 × √(40,889) = 0.78125 × 202.2 = 158 Hz [1 mark]. 3. For the third harmonic (n = 3): wavelength λ₃ = 2L ÷ 3 = (2 × 0.640) ÷ 3 = 0.427 m; frequency f₃ = 3 × f₁ = 3 × 158 = 474 Hz [1 mark]. 4. When plucked, progressive transverse waves travel in both directions along the string and reflect off the fixed ends [1 mark]. 5. The two progressive waves of equal frequency, wavelength, and amplitude travelling in opposite directions superpose, producing fixed nodes at points of destructive interference (antiphase) and antinodes at points of constructive interference (in phase) [1 mark].

2. Q2

   In a Young's double-slit experiment, a laser of wavelength 633 nm illuminates two slits separated by 0.45 mm. Interference fringes are observed on a screen 2.80 m away. Calculate the fringe spacing w, explain how a student should measure w accurately to minimise percentage uncertainty, and describe and explain what happens to the fringe pattern if one of the slits is covered with a thin neutral-density filter that reduces the transmitted amplitude by half. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Calculate fringe spacing: w = λD ÷ s = (6.33 × 10⁻⁷ m × 2.80 m) ÷ (0.45 × 10⁻³ m) = 3.94 × 10⁻³ m = 3.94 mm [1 mark]. 2. To minimise percentage uncertainty, measure the total distance across a large number of fringe spacings (e.g. 10 fringe spaces, from the centre of the 1st bright fringe to the centre of the 11th bright fringe) using a vernier caliper or travelling microscope, then divide by 10 [1 mark]. 3. If one slit transmits half the amplitude (A/2 vs A), the fringe spacing w remains completely unchanged because λ, D, and s are unchanged [1 mark]. 4. At the bright fringes (constructive interference), resultant amplitude decreases from 2A to 1.5A, so the bright fringes become dimmer (lower intensity, since I ∝ A²) [1 mark]. 5. At the dark fringes (destructive interference), the two waves in antiphase no longer cancel completely (A - 0.5A = 0.5A ≠ 0), so the dark fringes are no longer completely black (reduced contrast) [1 mark].

3. Q3

   Monochromatic light of wavelength 589 nm from a sodium vapour lamp is incident normally on a diffraction grating with 5.00 × 10⁵ lines per metre. Calculate the angle θ between the two second-order maxima, and calculate the maximum number of orders visible on each side of the central maximum. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Calculate the grating spacing d: d = 1 ÷ N = 1 ÷ (5.00 × 10⁵ m⁻¹) = 2.00 × 10⁻⁶ m [1 mark]. 2. For the second-order maximum (n = 2), use d sin θ₂ = 2λ: sin θ₂ = (2 × 5.89 × 10⁻⁷) ÷ (2.00 × 10⁻⁶) = 0.589 → θ₂ = sin⁻¹(0.589) = 36.1° [1 mark]. 3. Therefore, the total angle between the two second-order maxima on either side of the central zero-order maximum = 2 × 36.1° = 72.2° [1 mark]. 4. For maximum order, sin θ < 1, so n_max = d ÷ λ = (2.00 × 10⁻⁶) ÷ (5.89 × 10⁻⁷) = 3.396 → round down to integer n_max = 3 orders on each side [1 mark].

4. Q4

   A step-index optical fibre has a glass core of refractive index 1.52 surrounded by a glass cladding of refractive index 1.46. Calculate the speed of light in the core and the critical angle at the core-cladding boundary, and explain what happens to a high-frequency digital signal if pulse broadening becomes too large. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Speed of light in the core: c_s = c ÷ n₁ = (3.00 × 10⁸ m s⁻¹) ÷ 1.52 = 1.97 × 10⁸ m s⁻¹ [1 mark]. 2. Critical angle at the core-cladding interface: sin θ_c = n₂ ÷ n₁ = 1.46 ÷ 1.52 = 0.9605 → θ_c = sin⁻¹(0.9605) = 73.9° [1 mark]. 3. Modal and material dispersion cause each short pulse of light to spread out in time (pulse broadening) as it travels along a long fibre [1 mark]. 4. If pulse broadening is too large, adjacent pulses overlap and merge into one another at the receiver, so the detector cannot distinguish '1's from '0's (causing loss of information / signal degradation) [1 mark].

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