Superposition

• Displacements add with their signs, and the path difference in wavelengths decides the result: a whole number of wavelengths reinforces, an extra half cancels (for sources in phase).
• Displacements add with their signs. Node to node is half a wavelength. Coherent means a constant phase difference. Round the highest order down.

• 8.1.1 I can explain — explain and use the principle of superposition
• 8.1.2 I can — show an understanding of experiments that demonstrate stationary waves using microwaves, stretched strings and air columns (it will be assumed that end corrections are negligible; knowledge of the concept of end corrections is not required)
• 8.1.3 I can explain — explain the formation of a stationary wave using a graphical method, and identify nodes and antinodes
• 8.1.4 I can understand — understand how wavelength may be determined from the positions of nodes or antinodes of a stationary wave
• 8.2.1 I can explain — explain the meaning of the term diffraction
• 8.2.2 I can — show an understanding of experiments that demonstrate diffraction including the qualitative effect of the gap width relative to the wavelength of the wave; for example diffraction of water waves in a ripple tank
• 8.3.1 I can understand — understand the terms interference and coherence
• 8.3.2 I can — show an understanding of experiments that demonstrate two-source interference using water waves in a ripple tank, sound, light and microwaves
• 8.3.3 I can understand — understand the conditions required if two-source interference fringes are to be observed
• 8.3.4 I can recall — recall and use λ = ax / D for double-slit interference using light
• 8.4.1 I can recall — recall and use d sin θ = nλ
• 8.4.2 I can describe — describe the use of a diffraction grating to determine the wavelength of light (the structure and use of the spectrometer are not included)

Units, significant figures and working are part of the physics. Give a calculated answer to the same number of significant figures as the least precise data, or one more; keep full precision in the working and round only at the end; write the unit with every final answer. A fifth of the qualification is experimental: Papers 3 and 5 test AO3 only, and their questions may be set in contexts outside the syllabus content, so the practical work in this chapter is set out as method, recording, graphs and uncertainties rather than as theory.

• “The distance from one node to the next is one wavelength.” Correct Node to node (or antinode to antinode) is \(\lambda/2\). Node to the next antinode is \(\lambda/4\). So \(\lambda\) is twice the node spacing, and six nodes span five spacings, which is \(2.5\lambda\).
• “The open end of a pipe is a node, because the pipe stops there.” Correct An open end is a displacement antinode: the air there is free to move. A closed end is a displacement node: the air cannot move into the solid end. In this syllabus the antinode is taken to be exactly at the open end, so a tube closed at one end has a fundamental of \(\lambda = 4L\).
• “Coherent means in phase.” Correct Coherent sources have a constant phase difference, of any fixed value, which needs the same frequency. Two sources in antiphase are coherent.
• “A path difference of \(\lambda/2\) always gives a minimum.” Correct Only for sources in phase. For sources in antiphase the rules swap: a path difference of \((n + \tfrac{1}{2})\lambda\) gives a maximum and \(n\lambda\) a minimum. Always check the phase of the sources first.
• “Diffraction shortens the wavelength of the wave that gets through.” Correct Diffraction changes only the direction of travel of parts of the wavefront. The wavelength, frequency and speed are unchanged, because the wave is in the same medium on both sides of the gap.
• “For 300 lines per mm, \(d = 300\).” Correct \(d\) is the spacing between lines: \(d = 1/N\) with \(N\) in lines per metre. 300 lines per mm is \(3.00 \times 10^5\) lines per metre, so \(d = 3.33 \times 10^{-6}\,\mathrm{m}\).
• “The second order is at twice the angle of the first.” Correct \(\sin\theta\), not \(\theta\), is proportional to \(n\). In Worked example 6 the first two orders are at 10.2° and 20.7°, not 10.2° and 20.4°, and the gap widens with every order.
• “When two waves cancel, their energy is destroyed.” Repair The displacements add to zero there, but energy is conserved: it is redistributed to the places where the waves reinforce, and each wave continues unchanged after the crossing.
• “Superposition: the amplitudes add.” Repair The displacements add, with their signs, at each point and instant. Amplitudes add only for waves arriving in phase.
• “Node-to-node distance is one wavelength.” Repair It is \(\lambda/2\); node to the adjacent antinode is \(\lambda/4\).
• “A stationary wave carries energy from the vibrator to the pulley.” Repair A stationary wave transfers no net energy along the string. The vibrator only makes up the energy dissipated.
• “All points on a stationary wave have the same amplitude.” Repair The amplitude varies with position, from zero at a node to a maximum at an antinode. It is the progressive wave that has the same amplitude everywhere.
• “Neighbouring points on a stationary wave are out of phase.” Repair All points between adjacent nodes are in phase; points in adjacent loops are in antiphase.
• “The open end of a pipe is a node.” Repair An open end is a displacement antinode; a closed end is a displacement node.
• “Six nodes span three wavelengths.” Repair Six nodes span five spacings: \(5 \times \lambda/2 = 2.5\lambda\).
• “Diffraction shortens the wavelength.” Repair \(\lambda\), \(f\) and \(v\) are unchanged; only the direction of travel of parts of the wavefront changes.
• “The narrower the gap, the better: a gap much narrower than \(\lambda\) gives the clearest diffraction.” Repair The spreading is greatest when the gap is about \(\lambda\). A much narrower gap still spreads the wave, but lets very little energy through.
• “Coherent means in phase.” Repair Coherent means a constant phase difference, of any fixed value, which needs the same frequency.
• “Two identical torches shone through two slits give an interference pattern.” Repair Independent sources are not coherent: their phase difference changes continually. Use one laser to light both slits.
• “A path difference of \(\lambda/2\) always gives a minimum.” Repair Only for sources in phase. For sources in antiphase, \(\lambda/2\) gives a maximum.
• “\(\lambda = ax/D\) is on the formula sheet.” Repair It is recalled, and so is \(d\sin\theta = n\lambda\). Only \(c\) is given for this topic.
• “\(d\) is the number of lines per mm.” Repair \(d = 1/N\), with \(N\) in lines per metre: 300 lines per mm gives \(d = 3.33 \times 10^{-6}\,\mathrm{m}\).
• “The second order is at twice the first-order angle.” Repair \(\sin\theta\) is proportional to \(n\): 10.2° and 20.7° in Worked example 6.
• “\(d/\lambda = 5.66\), so the highest order is 6.” Repair Always round down: order 6 would need \(\sin\theta = 1.06\). The highest order is 5.

• Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.
• Using it. Every question in this chapter is superposition in disguise. Find each wave's displacement (or phase) at the point, then add. A stationary wave (section B) is two travelling waves added; an interference maximum (section G) is two waves arriving in phase; a grating order (section K) is many waves arriving in phase.
• Interleave with the chapters that use this one. When you reach Topic 17 (oscillations), re-answer “why does a string respond strongly only at certain frequencies?” in terms of resonance. When you reach Topic 22 (quantum physics), re-answer “what does a diffraction pattern prove about whatever made it?” Recalling a topic inside a new context is worth more than another pass over this chapter on its own; at A Level, Paper 4 assumes the whole of the AS content, so nothing here is ever finished with.

• Cambridge International AS & A Level Physics 9702 has five components. Topic 8 is AS Level content, so it is examined in Papers 1, 2 and 3. AS Level content: examined in Paper 1 (multiple choice), Paper 2 (AS structured) and, as practical context, Paper 3. Assumed knowledge for Papers 4 and 5. AS Level candidates take Papers 1, 2 and 3; A Level candidates take all five, either staged over two years (Papers 1–3 in year one, Papers 4 and 5 in year two) or together in one series. Examinations are available in the June and November series, and in March in India.
• Across both the AS Level and the A Level the assessment objectives are weighted AO1 40% (knowledge and understanding), AO2 40% (handling, applying and evaluating information) and AO3 20% (experimental skills and investigations). AS candidates are graded a–e; A Level candidates A*–E. The Data and formulas sheet is printed as page 2 of Papers 1 and 2 and as pages 2 and 3 of Paper 4: it gives the constants and a short list of formulas. Every other equation in this chapter is one the syllabus says you must recall, and this chapter says which is which.
• A multiple-choice item on this topic can turn on one number or one condition: nodes counted as spacings, \(\sin\theta\) against \(\theta\), lines per mm against \(d\), sources in antiphase. A structured question asks you to state the principle of superposition, explain how a stationary wave forms, describe one of the named experiments, or state the conditions for observable fringes.
• The calculations are short: \(\lambda\) from node positions or tube lengths, then \(v = f\lambda\); path difference in wavelengths; \(\lambda = ax/D\); \(d\sin\theta = n\lambda\) and the highest order. All of these are recall; only \(c = 3.00 \times 10^8\,\mathrm{m\,s^{-1}}\) is on the Data sheet.
• The syllabus names the grating determination of \(\lambda\) (vary the order, measure \(2y\) and \(D\) with a metre rule, plot \(\sin\theta\) against \(n\)); stationary waves on a string and in a resonance tube are natural Paper 3 contexts too. The largest uncertainty is usually locating a spot centre, a node or a resonance.
• Read the command word before you decide how much to write. This syllabus has fifteen of them: calculate, comment, compare, define, describe, determine, explain, give, identify, justify, predict, show (that), sketch, state and suggest. Define wants a precise meaning — for a physical quantity, usually an equation in words with every quantity named. State and give want a fact and nothing more. Describe wants the points or the features. Explain wants the reasons and the relationships — a describe-level answer to an explain question is incomplete however well written it is. Show (that) gives you the result and asks for the structured evidence that leads to it, so every step must appear — and a final value worked to one more significant figure than the one printed makes it plain that you calculated it rather than copied it. Sketch wants a freehand graph with its key features — intercepts, asymptotes, the shape — correct, but no plotted scale.