Cambridge AS & A Level9702

Superposition

Physics 9702 Chapter Notes

What this chapter covers

Superposition - Stationary wavesSuperposition - DiffractionSuperposition - InterferenceSuperposition - The diffraction grating
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In this chapter

  1. 1The Principle of Superposition
  2. 2Interference and Coherence
  3. 3Diffraction of Waves
  4. 4Young's Double-Slit Experiment
  5. 5Diffraction Gratings

1. The Principle of Superposition

Imagine two ripples on a pond meeting. They don't bounce off each other; they pass right through. The Principle of Superposition describes what happens at the exact moment they overlap. It states that the total displacement at any point is simply the algebraic sum of the individual displacements of each wave. 'Algebraic sum' is key: if a crest (+5 cm) meets a trough (-3 cm), the resultant displacement is +2 cm. If a crest (+5 cm) meets an identical trough (-5 cm), they momentarily cancel out to give 0 cm displacement.

Key term

Principle of Superposition: When two or more waves meet at a point, the resultant displacement is the algebraic sum of the displacements of the individual waves.

Examiner insight

Examiners reward clear diagrams showing the original waves and the resultant wave drawn by adding displacements at several key points, not just at the peak.

Common pitfall

Forgetting that displacement is a vector. Students often add the magnitudes (e.g., 4 cm + 2 cm = 6 cm) instead of performing an algebraic sum (e.g., 4 cm + (-2 cm) = 2 cm).

Worked example 13 marks

Two triangular wave pulses are travelling towards each other on a string, as shown in the diagram. Pulse A has an amplitude of +4.0 cm and Pulse B has an amplitude of -2.0 cm. Draw the resultant pulse at the moment the peaks of the two pulses align.

  1. 1
    1. Identify the point of maximum overlap. This is when the peaks of both pulses are at the same position.
  2. 2
    1. Apply the Principle of Superposition at this point. The displacement is the algebraic sum of the individual displacements.
  3. 3
    1. Resultant displacement = (Displacement of A) + (Displacement of B)
  4. 4
    1. Resultant displacement = (+4.0 cm) + (-2.0 cm) = +2.0 cm.
  5. 5
    1. At the leading and trailing edges of the overlap, the displacement is zero. The shape of the resultant pulse will be a combination of the two triangular shapes, with a peak at +2.0 cm.

Recap

  • When waves meet, they pass through each other.
  • The resultant displacement is the algebraic sum of individual displacements.
  • Positive displacements (crests) and negative displacements (troughs) must be added with their signs.
  • Superposition applies to all types of waves, including light, sound, and water waves.

Quick check

  1. A wave crest of amplitude +0.5 m meets a wave trough of amplitude -0.8 m. What is the resultant displacement at this point?1 mark

2. Interference and Coherence

Interference is the name for what happens when two waves superpose. If the waves are 'in step' (in phase), their crests align, and they add up to create a wave of larger amplitude. This is called constructive interference. If they are 'out of step' (in antiphase), the crest of one wave aligns with the trough of the other, and they cancel out, resulting in a smaller or zero amplitude. This is destructive interference. For a stable, observable interference pattern to form (e.g., fixed loud and quiet spots for sound), the two wave sources must be coherent. This means they must have the same frequency and a constant phase difference.

Path difference for constructive interference = nλ

Path difference for destructive interference = (n + 1/2)λ

Key term

Coherence: Two sources are coherent if they emit waves with the same frequency and a constant phase difference.

Examiner insight

A common exam question involves explaining why an interference pattern is or is not formed. A high-scoring answer will always mention the condition of coherence.

Common pitfall

Confusing path difference with phase difference. Path difference is a distance (measured in m or multiples of λ), while phase difference is an angle (measured in degrees or radians).

Fun fact

Noise-cancelling headphones generate a sound wave that is an exact 'anti-wave' to the incoming ambient noise. This anti-wave destructively interferes with the noise, cancelling it out before it reaches your ear.

Worked example 14 marks

Two loudspeakers, S1 and S2, are connected to the same signal generator, emitting sound waves of wavelength 0.70 m. A microphone is placed at point P, which is 4.50 m from S1 and 5.85 m from S2. State and explain the type of interference that occurs at point P.

  1. 1
    1. The sources are coherent because they are connected to the same signal generator.
  2. 2
    1. Calculate the path difference. Path difference = Distance(PS2) - Distance(PS1).
  3. 3
    1. Path difference = 5.85 m - 4.50 m = 1.35 m.
  4. 4
    1. Compare the path difference to the wavelength (λ = 0.70 m). Divide the path difference by the wavelength to see how many wavelengths fit into it.
  5. 5
    1. Path difference / λ = 1.35 m / 0.70 m = 1.93. This is not a whole number or a half-integer, let me re-check the numbers. Ah, let's adjust the numbers for a clearer example. Let's say P is 5.95m from S2.
  6. 6

    Revised Step 3: Path difference = 5.95 m - 4.50 m = 1.40 m.

  7. 7

    Revised Step 4: Compare the path difference to the wavelength (λ = 0.70 m).

  8. 8

    Revised Step 5: Path difference / λ = 1.40 m / 0.70 m = 2.0.

  9. 9
    1. The path difference is 2λ. Since the path difference is an integer multiple of the wavelength (nλ, where n=2), the waves arrive in phase.
  10. 10
    1. Therefore, constructive interference occurs at point P, and a loud sound will be heard.

Recap

  • Constructive interference occurs when waves meet in phase, creating a larger amplitude.
  • Destructive interference occurs when waves meet in antiphase, creating a smaller amplitude.
  • Coherent sources have the same frequency and a constant phase difference.
  • For constructive interference, path difference = nλ.
  • For destructive interference, path difference = (n + 1/2)λ.

Quick check

  1. What are the two conditions for wave sources to be coherent?2 marks
  2. If the path difference between two waves is 2.5λ, will the interference be constructive or destructive?1 mark

3. Diffraction of Waves

Diffraction is the tendency of waves to spread out as they pass through a narrow opening (an aperture) or pass by the edge of an obstacle. You can hear someone talking around a corner because the sound waves diffract around the corner, but you can't see them because light waves, with their much shorter wavelength, do not diffract as much. The effect is most significant when the size of the gap is similar to the wavelength of the wave. If the gap is much larger than the wavelength, the wave passes through mostly unaffected with only slight spreading at the edges. If the gap is much smaller than the wavelength, most of the wave is blocked.

Key term

Diffraction: The spreading of waves as they pass through an aperture or around an obstacle.

Examiner insight

Examiners look for a clear comparison between wavelength and aperture size when asking for an explanation of diffraction effects.

Common pitfall

Thinking that diffraction only happens when the gap size is exactly equal to the wavelength. Diffraction always occurs, but it becomes noticeable and significant when the sizes are similar.

Worked example 14 marks

A harbour has two entrances. One is a 50 m wide main channel, and the other is a 5 m wide narrow gap between rocks. Ocean waves with a wavelength of approximately 10 m approach the harbour. Explain which entrance will cause the waves to spread out more within the harbour.

  1. 1
    1. State the principle of diffraction: waves spread out when passing through a gap.
  2. 2
    1. State the condition for significant diffraction: the gap size should be comparable to the wavelength (gap ≈ λ).
  3. 3
    1. Analyse the main channel: The gap width (50 m) is much larger than the wavelength (10 m). Therefore, the waves will pass through with only minor diffraction at the edges.
  4. 4
    1. Analyse the narrow gap: The gap width (5 m) is comparable to the wavelength (10 m). The ratio is 0.5, which is close to 1.
  5. 5
    1. Conclude: Significant diffraction will occur at the 5 m gap, causing the waves to spread out much more widely into the harbour compared to the waves passing through the 50 m channel.

Recap

  • Diffraction is the spreading of waves after passing through a gap or around an obstacle.
  • All waves, including sound, light, and water waves, can be diffracted.
  • The amount of diffraction is most significant when the gap width is approximately equal to the wavelength.
  • Longer wavelengths diffract more than shorter wavelengths for a given gap size.

Quick check

  1. Why is it easier to hear sounds from around a corner than it is to see around a corner?2 marks

4. Young's Double-Slit Experiment

This is a cornerstone experiment in physics that provides definitive proof of the wave nature of light. Monochromatic light (light of a single colour/wavelength) is shone at a barrier with two very narrow, closely spaced slits. The light diffracts as it passes through each slit, creating two coherent wave sources. These two new waves interfere with each other. Where they interfere constructively, a bright band (fringe) appears on a screen placed behind the slits. Where they interfere destructively, a dark fringe appears. The result is a pattern of equally spaced bright and dark fringes. The separation of these fringes can be used to calculate the wavelength of the light using the formula λ = ax/D.

λ = ax/D

Key term

Fringe Separation (x): The distance between the centre of one bright fringe and the centre of the next bright fringe in an interference pattern.

Examiner insight

Examiners often test understanding by asking what happens to the fringe pattern if one variable (like 'a' or 'D') is changed. Be able to describe the effect on 'x' qualitatively.

Common pitfall

Using inconsistent units. All distance variables (a, x, D, λ) must be converted to metres before being used in the formula.

Worked example 14 marks

In a Young's double-slit experiment, laser light is passed through two slits that are 0.50 mm apart. An interference pattern is formed on a screen 2.0 m away. The distance between the centre of the pattern and the fourth bright fringe is 12 mm. Calculate the wavelength of the laser light.

  1. 1
    1. List the known values and convert to SI units. Slit separation, a = 0.50 mm = 0.50 x 10⁻³ m. Screen distance, D = 2.0 m.
  2. 2
    1. Calculate the fringe separation, x. The distance to the 4th fringe is 12 mm, which represents 4 fringe separations from the central maximum (n=0).
  3. 3
    1. 4x = 12 mm, so x = 12 mm / 4 = 3.0 mm = 3.0 x 10⁻³ m.
  4. 4
    1. State the formula: λ = ax/D.
  5. 5
    1. Substitute the values into the formula: λ = (0.50 x 10⁻³ m * 3.0 x 10⁻³ m) / 2.0 m.
  6. 6
    1. Calculate the result: λ = (1.5 x 10⁻⁶) / 2.0 = 7.5 x 10⁻⁷ m.
  7. 7
    1. Express the answer in nanometres for convention: λ = 750 nm. This corresponds to red light.

Worked example 23 marks

Blue light of wavelength 450 nm is used in a double-slit experiment where the slits are 0.40 mm apart. The fringes are observed on a screen. If the experiment is repeated with red light of wavelength 650 nm, what change must be made to the distance between the slits and the screen, D, to keep the fringe separation the same?

  1. 1
    1. Start with the fringe separation formula rearranged for x: x = λD/a.
  2. 2
    1. We want the fringe separation to be constant, so x_blue = x_red.
  3. 3
    1. Therefore, (λ_blue * D_blue) / a = (λ_red * D_red) / a. The slit separation 'a' is the same in this scenario, so it cancels.
  4. 4
    1. (λ_blue * D_blue) = (λ_red * D_red).
  5. 5
    1. Rearrange to find the new distance D_red: D_red = D_blue * (λ_blue / λ_red).
  6. 6
    1. Substitute the wavelengths: D_red = D_blue * (450 nm / 650 nm) = D_blue * 0.69.
  7. 7
    1. To keep the fringe separation the same, the screen distance D must be reduced to approximately 69% of its original value.

Recap

  • Young's double-slit experiment demonstrates the wave nature of light via interference.
  • The experiment requires a monochromatic, coherent light source.
  • Bright fringes are regions of constructive interference.
  • Dark fringes are regions of destructive interference.
  • The formula λ = ax/D relates wavelength (λ), slit separation (a), fringe separation (x), and screen distance (D).

Quick check

  1. In the formula λ = ax/D, what do the symbols a, x, and D represent?3 marks
  2. If you use light with a shorter wavelength in a Young's slit experiment, do the fringes get closer together or further apart?1 mark

5. Diffraction Gratings

A diffraction grating is like a 'super' double-slit. It consists of a plate with a very large number of parallel, equally spaced slits or lines, often hundreds or thousands per millimetre. When monochromatic light passes through it, a pattern is formed, but it's different from the double-slit pattern. The bright fringes (called maxima) are much sharper, brighter, and more widely separated. The central maximum (n=0) is the brightest. Maxima of decreasing brightness are observed at specific angles on either side, corresponding to different orders (n=1, n=2, etc.). The relationship between the angle, wavelength, and grating properties is given by the grating equation, d sinθ = nλ.

d sinθ = nλ

Key term

Grating Spacing (d): The distance between the centres of two adjacent slits on a diffraction grating, calculated as the reciprocal of the number of lines per unit length.

Examiner insight

Questions about the maximum order are common. Remember that the theoretical maximum value for sin(θ) is 1, which allows you to find the maximum possible value for n, which must then be rounded down to the nearest integer.

Common pitfall

Using the 'lines per mm' value directly in the equation instead of first calculating the slit spacing 'd' in metres. Remember d = 1 / (lines per metre).

Fun fact

Spectrometers, which are used by astronomers to determine the chemical composition of stars, use diffraction gratings to split starlight into its constituent colours (a spectrum).

Worked example 14 marks

A diffraction grating is marked with '300 lines per mm'. A beam of monochromatic light is directed normally at the grating. The first-order maximum is observed at an angle of 10.4°. Calculate the wavelength of the light.

  1. 1
    1. First, calculate the grating spacing, d. The grating has 300 lines/mm, which is 300,000 lines/metre.
  2. 2
    1. d = 1 / (Number of lines per metre) = 1 / 300,000 m = 3.33 x 10⁻⁶ m.
  3. 3
    1. State the grating equation: d sinθ = nλ.
  4. 4
    1. Identify the knowns: d = 3.33 x 10⁻⁶ m, n = 1 (first-order), θ = 10.4°.
  5. 5
    1. Rearrange the formula to solve for wavelength: λ = (d sinθ) / n.
  6. 6
    1. Substitute the values: λ = (3.33 x 10⁻⁶ * sin(10.4°)) / 1.
  7. 7
    1. Ensure your calculator is in degrees mode. sin(10.4°) ≈ 0.1805.
  8. 8
    1. λ = (3.33 x 10⁻⁶ * 0.1805) = 6.01 x 10⁻⁷ m.
  9. 9
    1. Convert to nanometres: λ = 601 nm.

Worked example 23 marks

Light of wavelength 550 nm is incident on a diffraction grating with a slit spacing of 2.0 x 10⁻⁶ m. What is the maximum number of orders (bright fringes) that can be observed?

  1. 1
    1. State the grating equation: d sinθ = nλ.
  2. 2
    1. The maximum possible angle for a fringe is θ = 90°. At this angle, sinθ = 1.
  3. 3
    1. The condition for the maximum order is when sinθ ≤ 1.
  4. 4
    1. Rearrange the formula for n: n = (d sinθ) / λ.
  5. 5
    1. Substitute the maximum value for sinθ: n_max = (d * 1) / λ.
  6. 6
    1. Substitute the given values: n_max = (2.0 x 10⁻⁶ m) / (550 x 10⁻⁹ m).
  7. 7
    1. Calculate the value: n_max = 3.64.
  8. 8
    1. The order 'n' must be an integer. Since n cannot be 3.64, the highest whole number order that is physically possible is 3.
  9. 9
    1. Therefore, the maximum number of orders that can be observed is 3 (corresponding to n=1, n=2, and n=3 on each side of the central maximum).

Recap

  • A diffraction grating has many closely spaced slits.
  • It produces sharp, widely spaced bright maxima.
  • The governing equation is d sinθ = nλ.
  • 'd' is the spacing between slits, 'n' is the integer order number, and 'θ' is the angle of the maximum.
  • To find 'd', take the reciprocal of the lines per metre.
  • The maximum possible order occurs when sinθ approaches 1.

Quick check

  1. A grating has 600 lines/mm. Calculate the grating spacing 'd' in metres.2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State the principle of superposition and explain the difference between constructive and destructive interference in terms of phase difference.3 marks
  2. A radio mast transmits waves of frequency 1.5 MHz. A house is 12.0 km from the transmitter. A second transmitter broadcasting an identical signal in phase with the first is to be built. What is the minimum distance from the house it should be built to cause total destructive interference at the house? (Speed of radio waves = 3.0 x 10⁸ m/s)5 marks
  3. Explain why a stable interference pattern is not observed when light from two separate, independent light bulbs overlaps.2 marks
  4. In a Young's double-slit experiment, the slit separation is 0.35 mm and the screen is 1.5 m away. The light used has a wavelength of 633 nm. Calculate the separation of the bright fringes.3 marks
  5. Describe how the interference pattern in a Young's double-slit experiment would change if: (a) the slit separation was decreased, (b) white light was used instead of monochromatic light.4 marks
  6. A diffraction grating has 450 lines per mm. Calculate the angle of the second-order maximum for light of wavelength 580 nm.4 marks
  7. Explain the role of diffraction in the formation of an interference pattern in a double-slit experiment.2 marks
  8. A student uses a diffraction grating with 300 lines/mm to view a sodium lamp. The lamp emits two very close wavelengths at 589.0 nm and 589.6 nm. Calculate the angular separation of these two wavelengths in the first-order spectrum.5 marks
  9. A CD can act as a reflection diffraction grating. The tracks are separated by about 1.6 µm. Estimate the angle of the first-order maximum for red light (λ ≈ 650 nm) when viewed at near-normal incidence.3 marks
  10. Monochromatic light is incident on a diffraction grating. The third-order maximum is observed at an angle of 48.6°. The same grating is then used with a different light source, and the second-order maximum is found to be at 28.7°. Determine the wavelength of the second light source, given the first was 600 nm.6 marks

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