Ch 2 · 8 Superposition · A Level Physics 9702
Topic 2 of 2 in Waves and Superposition — A Level Physics 9702 — 58 lessons.
Introduction
When two waves meet, they simply add. That one rule, the principle of superposition, explains the notes of a guitar string, the bright and dark bands of laser light through two slits, and how a diffraction grating measures the wavelength of light. (Syllabus topic 8: sections 8.1–8.4.)
Key ideas — 8.1 (a) The principle of superposition
Principle of superposition: when two or more waves meet at a point, the resultant displacement is the sum of the individual displacements of the waves at that point (taking direction into account).
🧑🏫 Teacher's explanation
Imagine two people standing on a large, flat sheet of thin rubber. If both people push up at the exact same spot at the same time, the sheet moves up much higher than if only one person pushed. This is because their individual movements combine. Conversely, if one person pushes up while the other pushes down with the same force, the sheet stays flat at that spot.
📖 Definitions
🔍 Worked example — Easy
Two waves are traveling through a medium. At a specific point P, Wave A has a displacement of +0.5 mm and Wave B has a displacement of +0.3 mm. Calculate the resultant displacement at point P.
🔍 Worked example — Standard
Two waves meet at a point. Wave 1 provides a displacement of +4.0 mm. Wave 2 provides a displacement of -2.5 mm. Determine the resultant displacement.
🔍 Worked example — Exam-style
Two identical waves with an amplitude of 5.0 mm are traveling towards each other. At a certain point, they are in phase such that both waves have a displacement of +5.0 mm. At another point, they are out of phase such that one has a displacement of +5.0 mm and the other has -5.0 mm. Calculate the resultant displacement at both points.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 8.1 (b) Stationary waves
A stationary (standing) wave forms when two progressive waves of the same frequency (and similar amplitude), travelling in opposite directions, superpose. Usually a wave reflects back on itself.
🧑🏫 Teacher's explanation
Imagine two people standing on opposite sides of a long, thin trampoline, each shaking it up and down at the same rhythm. If they shake it perfectly in sync but in opposite directions, the trampoline won't look like a traveling wave moving from one person to the other. Instead, certain spots will stay perfectly still (the nodes), while the spots in the middle will bounce up and down violently (the antinodes). This is a stationary wave.
📖 Definitions
🔍 Worked example — Easy
A string is vibrating in a stationary wave pattern. The distance between two adjacent nodes is 0.15 m. Calculate the wavelength λ of the wave.
🔍 Worked example — Standard
A stretched string of length 0.80 m forms a stationary wave with three loops (three antinodes). Calculate the wavelength of the wave.
🔍 Worked example — Exam-style
A microwave experiment uses a transmitter and a metal plate. A detector finds a node at x = 0.20 m and the next antinode at x = 0.35 m. Calculate the frequency of the microwaves if the speed of light c = 3.0 × 10^8 m/s.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — 8.2 Diffraction
Diffraction is the spreading of a wave as it passes through a gap or around the edge of an obstacle.
🧑🏫 Teacher's explanation
Imagine you are walking through a doorway. If the doorway is very wide, you walk straight through without changing direction. However, if the doorway is very narrow, you tend to 'fan out' as you exit, spreading across the room. This is exactly what waves do. When a wave encounters an obstacle or a gap, it doesn't just travel in a straight line; it bends around the edges.
📖 Definitions
🔍 Worked example — Easy
A water wave with a wavelength of 0.05 m passes through a gap in a barrier. Describe the diffraction pattern if the gap width is 0.005 m compared to if the gap width is 0.5 m.
🔍 Worked example — Standard
In a double-slit experiment, the distance between the slits is 0.20 mm and the distance to the screen is 1.5 m. If the first-order bright fringe is observed at 1.2 cm from the center, calculate the wavelength λ of the light.
🔍 Worked example — Exam-style
A diffraction grating has 500 lines per centimeter. Calculate the wavelength of light that produces a maximum at an angle of 30.0^°.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 8.3 Interference
Interference is the superposition of waves from two (or more) coherent sources, which produces a stable pattern of maxima and minima.
🧑🏫 Teacher's explanation
When two waves meet, they don't just bounce off each other like billiard balls. Instead, they pass through each other, and for a brief moment, their displacements add up. This is the principle of superposition: the total displacement at any point is the algebraic sum of the individual displacements of the waves. If the peaks line up, you get a bigger peak (constructive interference); if a peak meets a trough, they cancel out (destructive interference).
📖 Definitions
🔍 Worked example — Easy
A laser with wavelength λ = 633 nm shines through two slits separated by a = 0.20 mm. The screen is located at a distance D = 2.5 m from the slits. Calculate the separation x between adjacent bright fringes.
🔍 Worked example — Standard
In a double-slit experiment, the distance between the slits is 0.15 mm and the screen is 1.2 m away. The distance between the 2nd bright fringe and the 5th bright fringe is measured to be 1.2 mm. Calculate the wavelength λ of the light.
⚠️ Common errors
🧠 Quick check 1
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🎯 Exam question
Key ideas — 8.4 The diffraction grating
A diffraction grating has a very large number of parallel, equally spaced slits. The light from all the slits interferes, giving very sharp, bright maxima at angles given by
🧑🏫 Teacher's explanation
Imagine you are standing in a massive crowd of people, and everyone is trying to clap at the exact same time. If everyone claps perfectly in sync, you hear one massive, deafening 'CRACK' at specific moments. If they are slightly out of sync, you hear a messy, continuous noise. A diffraction grating is like a massive crowd of tiny slits. When light passes through them, each slit acts as a new source of waves. Because the slits are so close together and so numerous, the waves from all these slits overlap.
📖 Definitions
🔍 Worked example — Easy
A diffraction grating has 500 lines per mm. Calculate the slit spacing d in meters.
🔍 Worked example — Standard
A laser with wavelength λ = 633 nm is shone through a grating with d = 2.5 × 10^-6 m. Calculate the angle θ for the first-order maximum (n=1).
🔍 Worked example — Exam-style
A diffraction grating with 400 lines per mm is used to observe the second-order maximum (n=2) of a light source. The angle of diffraction is measured to be 30.0^°. Determine the wavelength λ of the light.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Example 1: stationary wave. On a string, adjacent nodes are 0.25 m apart and the frequency is 120 Hz. λ = 0.50 m, so the speed of the progressive waves is v = fλ = 60 m s⁻¹.
🔍 Worked example — Easy
A string is vibrating in a stationary wave pattern. The distance between two adjacent nodes is 0.25 m. If the frequency of the vibration is 120 Hz, calculate the speed of the progressive wave.
🔍 Worked example — Standard
In a double-slit experiment, the slit separation a is 0.50 mm and the screen distance D is 2.0 m. The distance between ten consecutive bright fringes is 24 mm. Calculate the wavelength λ of the light.
🔍 Worked example — Exam-style
A diffraction grating has 500 lines per mm. Light of wavelength 600 nm is shone through it. Calculate the angle θ of the first-order maximum (n=1).
⚠️ Common errors
Common mistakes
Practice
Answers: (1) When waves meet, the resultant displacement at a point is the sum of the individual displacements. (2) Any three: stationary waves have nodes and antinodes (amplitude varies with position); they transfer no energy; points between adjacent nodes are in phase (in a progressive wave, phase varies continuously along the wave). (3) Sources with a constant phase difference. (4) It doubles (x = λ D/a). (5) d = 3.33 × 10^-6 m; λ = dsinθ/2 = 3.33 × 10^-6 × 0.379/2 = 6.3 × 10^-7 m.