Master

🌊 Superposition (Cambridge 9702)

Syllabus Overview

This topic covers the behavior of waves when they meet. It is divided into four main sections:

1. Stationary Waves
2. Diffraction
3. Interference
4. The Diffraction Grating

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8.1 Stationary Waves & The Principle of Superposition

The fundamental rule governing all interference and stationary wave phenomena is the Principle of Superposition. You will be asked to define this in almost every exam series.

Definition: Principle of Superposition

When two (or more) waves meet / overlap (at a point), the resultant displacement is the (vector) sum of the individual displacements. (Ref: 9702_s25_ms_21 Q4, 9702_w23_ms_21 Q4)

Formation of Stationary Waves

A stationary (standing) wave is formed when two identical waves travelling in opposite directions superpose.

• Nodes: Points of zero amplitude (destructive interference).
• Antinodes: Points of maximum amplitude (constructive interference).
• Wavelength ($\lambda$): The distance between adjacent nodes (or adjacent antinodes) is exactly $\lambda /2$. The distance between a node and the adjacent antinode is $\lambda /4$.

Model Answer: Formation of a Stationary Wave

Question: An electromagnetic wave reflects at an aluminium sheet. Explain how the stationary wave, including its nodes and antinodes, is formed. (Ref: 9702_w25_qp_24 Q4) Explanation: The incident wave and the reflected wave travel in opposite directions and overlap. According to the principle of superposition, their displacements add together.

• At points where the waves are $180^{\circ }$ out of phase, they cancel out to form nodes (zero amplitude).
• At points where they meet in phase, they reinforce each other to form antinodes (maximum amplitude).

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8.2 Diffraction

Diffraction explains how waves behave when they pass through a gap or around an obstacle.

Definition: Diffraction

The bending / spreading of a wave as it passes an edge or a slit. (Ref: 9702_s12_ms_23 Q6)

Effect of Gap Width

The extent of diffraction depends on the ratio of the wavelength ($\lambda$) to the gap width ($a$).

• Maximum diffraction occurs when the gap width is approximately equal to the wavelength ($a\approx \lambda$).
• If the gap is much larger than the wavelength ($a\gg \lambda$), the wave passes through mostly unchanged, with only slight spreading at the edges.

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8.3 Interference

For stable, observable interference patterns to form, the two sources of waves must be coherent.

Definition: Coherence

Waves are coherent if they have a constant phase difference. (Ref: 9702_m19_ms_22 Q5) Note: Do not just say “same phase” or “same wavelength”. The phrase “constant phase difference” is strictly required by the mark scheme.

Conditions for Observable Interference Fringes

1. The waves must meet/overlap.
2. The sources must be coherent.
3. The waves should have roughly the same amplitude (so dark fringes cancel out effectively).
4. (For transverse waves) They must be unpolarized or polarized in the same plane.

The Double-Slit Equation

For a double-slit experiment (like Young’s slits), the separation between adjacent bright (or dark) fringes is given by:

$x=\frac{\lambda D}{a}$

• $x$ = fringe separation / fringe width (m)
• $\lambda$ = wavelength (m)
• $D$ = distance from slits to screen (m)
• $a$ = distance between the centres of the two slits (m)

Graphing the Relationship

Examiners love to ask you to sketch the relationship between fringe width ($x$) and slit separation ($a$). Since $x\propto \frac{1}{a}$, the graph of $x$ (y-axis) against $a$ (x-axis) is a curved line with a negative gradient of decreasing magnitude (a standard $1/x$ curve). (Ref: 9702_s25_ms_21 Q4)

Amplitude and Intensity

The intensity ($I$) of a wave is directly proportional to the square of its amplitude ($A$). $I\propto A^2$

Model Calculation: Finding Resultant Intensity

Question: Two waves, with intensities $I$ and $4I$, superpose in phase. Determine, in terms of $I$, the maximum possible intensity of the resulting wave. (Ref: 9702_s21_ms_23 Q4) Step 1: Convert intensities to relative amplitudes. $A_1\propto \sqrt{I}$ and $A_2\propto \sqrt{4I}=2\sqrt{I}$ Step 2: Add the amplitudes (because they meet in phase for a maximum). $A_{max}=A_1+A_2=\sqrt{I}+2\sqrt{I}=3\sqrt{I}$ Step 3: Square the resultant amplitude to find resultant intensity. $I_{max}\propto (A_{max}{)}^2=(3\sqrt{I}{)}^2=9I$

Changing the Experimental Conditions (High-Yield 🚨)

Examiners frequently ask what happens to the interference pattern if a parameter is changed. Memorize these standard answers:

Change Made to Experiment	Effect on Fringe Separation ($x$)	Effect on Appearance of Fringes	Reference
Reduce intensity of one slit	No change	Bright fringes become darker; Dark fringes become brighter. (Incomplete cancellation)	9702_s20_ms_21
Increase intensity of both slits	No change	Bright fringes become brighter; Dark fringes are unchanged (Still perfectly cancel out).	9702_w18_ms_21
Use red light instead of green/blue	Increases	Fringes spread further apart because red light has a longer wavelength.	9702_s21_ms_23

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8.4 The Diffraction Grating

A diffraction grating consists of thousands of narrow parallel slits. It produces much sharper and brighter maxima than a double slit.

The Grating Equation

$d\sin \theta =n\lambda$

• $d$ = grating spacing (distance between adjacent slits in meters). If given lines per mm ($N$), $d=\frac{1}{N\times 10^3}$.
• $\theta$ = angle of the maximum from the central normal.
• $n$ = order of the maximum ($0,1,2,...$).
• $\lambda$ = wavelength.

To find the maximum number of orders produced, set $\theta =90^{\circ }$ ($\sin \theta =1$) and solve for $n$, rounding down to the nearest whole number. Total number of bright spots observed = $2n+1$ (including the zero order).

Model Answer: Explaining the Diffraction Grating Pattern

Question: White light is incident on a diffraction grating. Describe how the principle of superposition is used to explain (1) the white light at the zero order, and (2) the difference in position of red and blue light in the first order. (Ref: 9702_w12_ms_21 Q4)

Answer:

1. Zero Order (White Light): At the center, the path difference for all waves from all slits is zero. Therefore, every wavelength meets in phase (constructive interference), producing a maximum. Because all wavelengths/colors mix together at this point, the light appears white.
2. First Order (Colored Spectrum): To obtain a maximum, the path difference must equal exactly $\lambda$ (or phase difference $=360^{\circ }$). Because the wavelength ($\lambda$) of red light is greater than that of blue light, the red and blue waves must travel at different angles to achieve their respective one-wavelength path differences. Hence, maxima occur at different angles, creating a spectrum.

Exam Tip: Red vs. Blue Light

• Red light has the longest visible wavelength ($\approx 700\text{ nm}$) $\to$ it diffracts the most (largest $\theta$).
• Blue/Violet light has the shortest visible wavelength ($\approx 400\text{ nm}$) $\to$ it diffracts the least (smallest $\theta$).
• Because blue light bends less, a grating will show more orders for blue light than for red light. (Ref: 9702_s12_ms_23 Q6)