Cambridge Lower Secondary CheckpointStage 9

Physics: Light and sound

Science Stage 9 Chapter Notes

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Physics: Light and sound
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1. Properties of Waves

A wave is a disturbance that transfers energy from one place to another without transferring matter. There are two main types of waves. Transverse waves are where the particles of the medium oscillate perpendicular to the direction of energy transfer; examples include light and all other electromagnetic waves. Longitudinal waves are where the particles oscillate parallel to the direction of energy transfer; sound is a key example. Key properties of a wave include: Amplitude (the maximum displacement from the equilibrium position), Wavelength (λ, the distance between two consecutive identical points on a wave), Frequency (f, the number of complete waves passing a point per second, measured in Hertz, Hz), and Period (T, the time taken for one complete wave to pass a point). The speed of a wave(v) is related to its frequency and wavelength by the wave equation: v = fλ.

v = f × λ

f = 1 / T

Key term

Frequency (f): The number of complete oscillations or waves that pass a given point in one second, measured in Hertz (Hz).

Examiner insight

Examiners expect clear, well-labelled diagrams to illustrate the difference between transverse and longitudinal waves, including labels for wavelength and amplitude.

Common pitfall

Confusing the direction of wave travel with the direction of particle oscillation for transverse and longitudinal waves.

Worked example 13 marks

A water wave has a wavelength of 2.0 m. A floating cork is observed to bob up and down, completing 5 full oscillations in 10 seconds. Calculate the speed of the water wave.

  1. 1

    Step 1: Calculate the frequency (f). Frequency is the number of oscillations per second.

  2. 2

    Number of oscillations = 5

  3. 3

    Time taken = 10 s

  4. 4

    Frequency, f = Number of oscillations / Time taken = 5 / 10 = 0.5 Hz.

  5. 5

    Step 2: Use the wave equation to find the speed (v).

  6. 6

    The wavelength (λ) is given as 2.0 m.

  7. 7

    The wave equation is v = f × λ.

  8. 8

    v = 0.5 Hz × 2.0 m = 1.0 m/s.

  9. 9

    The speed of the water wave is 1.0 m/s.

Recap

  • Waves transfer energy without transferring matter.
  • In transverse waves, oscillations are perpendicular to the direction of energy transfer.
  • In longitudinal waves, oscillations are parallel to the direction of energy transfer.
  • Wave speed is calculated by multiplying frequency by wavelength (v = fλ).
  • Frequency is the number of waves per second, measured in Hertz (Hz).

Quick check

  1. Is a sound wave transverse or longitudinal? Explain why.2 marks
  2. A wave has a frequency of 50 Hz and a wavelength of 3.0 m. What is its speed?1 mark

2. Sound Waves, Pitch and Loudness

Sound is a mechanical wave, which means it needs a medium (like air, water, or solids) to travel through; it cannot travel in a vacuum. It is a longitudinal wave, consisting of a series of compressions (areas of high pressure) and rarefactions (areas of low pressure). The two main properties of a sound we perceive are pitch and loudness. Pitch is determined by the frequency of the sound wave. A high frequency results in a high-pitched sound, while a low frequency results in a low-pitched sound. Loudness is determined by the amplitude of the sound wave. A wave with a large amplitude carries more energy and sounds louder, while a wave with a small amplitude sounds quieter. The typical range of human hearing is from 20 Hz to 20,000 Hz (20 kHz).

Key term

Pitch: The perceived highness or lowness of a sound, which is determined by the frequency of the sound wave.

Common pitfall

Stating that sound waves are transverse or that they can travel through a vacuum.

Fun fact

Dolphins and bats use sound waves with frequencies over 100,000 Hz (ultrasound) for echolocation to navigate and find prey.

Worked example 12 marks

A sound wave has an amplitude of 0.1 mm and a frequency of 800 Hz. The wave changes to have an amplitude of 0.2 mm and a frequency of 400 Hz. Describe two ways the sound has changed.

  1. 1

    Step 1: Analyse the change in amplitude. The amplitude has increased from 0.1 mm to 0.2 mm.

  2. 2

    Amplitude determines loudness. An increase in amplitude means the sound becomes louder.

  3. 3

    Step 2: Analyse the change in frequency. The frequency has decreased from 800 Hz to 400 Hz.

  4. 4

    Frequency determines pitch. A decrease in frequency means the sound has a lower pitch.

  5. 5

    Conclusion: The sound has become louder and has a lower pitch.

Worked example 22 marks

An oscilloscope displays a sound wave. How would the trace on the screen change if the sound becomes quieter but higher-pitched?

  1. 1

    Step 1: Relate 'quieter' to a wave property. Quieter means a smaller amplitude.

  2. 2

    On an oscilloscope, amplitude is represented by the vertical height of the wave from the centre line. So, the height of the wave on the screen would decrease.

  3. 3

    Step 2: Relate 'higher-pitched' to a wave property. Higher pitch means a higher frequency.

  4. 4

    On an oscilloscope, frequency is represented by how close together the waves are. Higher frequency means more waves would be seen on the screen in the same horizontal distance. The waves would appear more compressed or 'squashed' horizontally.

  5. 5

    Conclusion: The wave's vertical height (amplitude) would decrease, and the waves would become more frequent (closer together horizontally).

Recap

  • Sound is a longitudinal wave that requires a medium to travel.
  • The pitch of a sound is determined by its frequency.
  • The loudness of a sound is determined by its amplitude.
  • High frequency means high pitch; large amplitude means loud sound.
  • Sound travels as a series of compressions and rarefactions.

Quick check

  1. What property of a sound wave must be changed to make it louder?1 mark
  2. Why can't an astronaut on the Moon hear a nearby explosion?1 mark

3. Reflection and Refraction of Light

Reflection occurs when a wave, such as light, bounces off a surface. The Law of Reflection states that the angle of incidence(i) is equal to the angle of reflection (r), and that the incident ray, reflected ray, and the normal all lie in the same plane. The angles are always measured from the normal, a line perpendicular to the surface at the point of incidence. Refraction is the bending of light as it passes from one medium to another, for example from air to glass. This happens because light travels at different speeds in different media. The amount of bending is described by the refractive index(n) of the material. Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media. When light enters a denser medium (e.g., air to glass), it slows down and bends towards the normal. When it enters a less dense medium (e.g., glass to air), it speeds up and bends away from the normal. If the angle of incidence in the denser medium is increased, the angle of refraction increases until it reaches 90°. The angle of incidence that causes this is called the critical angle (c). For any angle of incidence greater than the critical angle, the light does not refract out but is completely reflected back into the denser medium. This is called Total Internal Reflection (TIR).

Angle of incidence (i) = Angle of reflection (r)

n = sin(i) / sin(r)

sin(c) = 1 / n

Key term

Refractive Index (n): A dimensionless number that describes how fast light travels through a material; a higher refractive index means light travels slower.

Examiner insight

Examiners award marks for correctly drawing ray diagrams for refraction, showing the ray bending in the correct direction (towards or away from the normal) with a straight line in each medium.

Common pitfall

Measuring angles of incidence, reflection, and refraction from the surface instead of from the normal.

Worked example 13 marks

A ray of light travels from air into a block of glass with a refractive index of 1.52. If the angle of incidence is 30°, calculate the angle of refraction.

  1. 1

    Step 1: State Snell's Law. For light entering from air (n₁ ≈ 1) into a medium (n₂ = n), the law is n = sin(i) / sin(r).

  2. 2

    Step 2: Rearrange the formula to solve for the angle of refraction (r).

  3. 3

    sin(r) = sin(i) / n.

  4. 4

    Step 3: Substitute the known values into the equation.

  5. 5

    i = 30°

  6. 6

    n = 1.52

  7. 7

    sin(r) = sin(30°) / 1.52

  8. 8

    sin(r) = 0.5 / 1.52 ≈ 0.3289

  9. 9

    Step 4: Find the angle r by taking the inverse sine.

  10. 10

    r = sin⁻¹(0.3289) ≈ 19.2°.

  11. 11

    The angle of refraction is 19.2°.

Worked example 22 marks

The refractive index of diamond is 2.42. Calculate the critical angle for the diamond-air boundary.

  1. 1

    Step 1: State the formula for the critical angle (c).

  2. 2

    sin(c) = 1 / n

  3. 3

    Step 2: Substitute the refractive index(n) of diamond into the formula.

  4. 4

    n = 2.42

  5. 5

    sin(c) = 1 / 2.42 ≈ 0.4132

  6. 6

    Step 3: Find the angle c by taking the inverse sine.

  7. 7

    c = sin⁻¹(0.4132) ≈ 24.4°.

  8. 8

    The critical angle for diamond is 24.4°.

Recap

  • The angle of incidence equals the angle of reflection.
  • Refraction is the bending of light due to a change in speed when entering a new medium.
  • Light bends towards the normal when entering a denser medium and away from it when entering a less dense one.
  • Total internal reflection (TIR) occurs when light in a denser medium hits a boundary at an angle greater than the critical angle.
  • Snell's Law (n = sin i / sin r) governs refraction.

Quick check

  1. A light ray hits a mirror at an angle of 40° to the mirror surface. What is the angle of reflection?2 marks
  2. What are the two conditions for total internal reflection to occur?2 marks

4. Image Formation by Lenses

Lenses are curved pieces of glass or plastic that refract light to form images. A converging lens (or convex lens) is thicker in the middle and causes parallel rays of light to converge at a point called the principal focus or focal point (F). A diverging lens (or concave lens) is thinner in the middle and causes parallel rays of light to spread out as if they came from a focal point. The distance from the centre of the lens to the focal point is the focal length (f). Ray diagrams are used to determine the position and nature of the image formed by a lens. The nature of an image is described as: real or virtual, inverted or upright, and magnified, diminished or same size. Real images can be projected onto a screen, while virtual images cannot. To draw a ray diagram, three key rays are used: 1) A ray parallel to the principal axis refracts through the focal point (for a converging lens) or appears to come from the focal point (for a diverging lens). 2) A ray passing through the centre of the lens continues undeviated. 3) A ray passing through the focal point before the lens refracts parallel to the principal axis (for a converging lens).

Key term

Focal Length (f): The distance from the optical centre of a lens to its principal focus (the point where parallel rays of light are brought to a focus).

Common pitfall

Drawing rays that bend at the surfaces of the lens instead of at the central axis line in a simplified ray diagram.

Fun fact

The 'burning glass' used by ancient Greeks to start fires was a simple converging lens, focusing the Sun's rays onto a single point.

Worked example 14 marks

An object is placed in front of a converging lens at a distance greater than twice the focal length (beyond 2F). Draw a ray diagram to find the position and nature of the image. State the characteristics of the image.

  1. 1

    Step 1: Draw the principal axis, the converging lens, and mark the focal points F and 2F on both sides of the lens.

  2. 2

    Step 2: Draw the object as an upright arrow to the left of the lens, at a position beyond 2F.

  3. 3

    Step 3: Draw the first ray from the top of the object, parallel to the principal axis. After passing through the lens, this ray refracts and passes through the focal point F on the other side.

  4. 4

    Step 4: Draw the second ray from the top of the object, passing straight through the centre of the lens without changing direction.

  5. 5

    Step 5: The point where these two rays intersect is the location of the top of the image. Draw the image as an arrow from the principal axis to this intersection point.

  6. 6

    Step 6: Describe the image. The image is located between F and 2F on the opposite side of the lens. It is real (formed by the actual intersection of light rays), inverted (upside down), and diminished (smaller than the object).

Recap

  • Converging (convex) lenses are thicker in the middle and bring parallel rays to a focus.
  • Diverging (concave) lenses are thinner in the middle and spread parallel rays out.
  • Ray diagrams can be used to find the position, size, and nature of an image.
  • A real image is formed where light rays actually cross and can be projected onto a screen.
  • A virtual image is formed where rays only appear to diverge from and cannot be projected.

Quick check

  1. What type of lens is used as a magnifying glass?1 mark
  2. Describe the image always formed by a single diverging lens.2 marks

5. The Electromagnetic Spectrum

The electromagnetic (EM) spectrum is the range of all types of EM radiation. All EM waves are transverse waves that can travel through a vacuum at the same speed: the speed of light, approximately 3.0 x 10⁸ m/s. They differ from each other in their wavelength and frequency. The spectrum is continuous, but is conventionally divided into seven main regions, in order of increasing frequency (and decreasing wavelength): Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, and Gamma rays. A helpful mnemonic is 'Raging Martians Invaded Venus Using X-ray Guns'. As you move from radio waves to gamma rays, the frequency and energy of the waves increase, while the wavelength decreases. Each part of the spectrum has distinct properties, uses, and potential dangers. For example, radio waves are used for broadcasting, microwaves for cooking and communication, infrared for remote controls and thermal imaging, visible light for sight, ultraviolet for sterilisation (but can cause skin cancer), X-rays for medical imaging, and gamma rays for treating cancer.

c = f × λ

Key term

Electromagnetic Spectrum: The continuous spectrum of electromagnetic waves, arranged in order of frequency or wavelength, that travel at the speed of light in a vacuum.

Examiner insight

Questions often require you to recall the order of the EM spectrum. Marks are also frequently awarded for matching a type of radiation with a specific use and its associated danger.

Fun fact

Wi-Fi uses radio waves, typically at frequencies of 2.4 GHz or 5 GHz, to transmit data wirelessly.

Worked example 13 marks

A radio station broadcasts at a frequency of 98.5 MHz. Calculate the wavelength of these radio waves. The speed of light is 3.0 x 10⁸ m/s.

  1. 1

    Step 1: Convert the frequency to Hertz. 'M' stands for Mega, which is 10⁶.

  2. 2

    f = 98.5 MHz = 98.5 × 10⁶ Hz.

  3. 3

    Step 2: State the wave equation for electromagnetic waves.

  4. 4

    c = f × λ

  5. 5

    Step 3: Rearrange the formula to solve for wavelength (λ).

  6. 6

    λ = c / f

  7. 7

    Step 4: Substitute the values and calculate.

  8. 8

    λ = (3.0 × 10⁸ m/s) / (98.5 × 10⁶ Hz)

  9. 9

    λ ≈ 3.046 m.

  10. 10

    The wavelength of the radio waves is approximately 3.05 m.

Recap

  • All electromagnetic waves are transverse and travel at the speed of light in a vacuum.
  • The EM spectrum, from lowest to highest frequency, is Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma ray.
  • As frequency increases, wavelength decreases and the energy of the wave increases.
  • Visible light is the only part of the EM spectrum that is visible to the human eye.
  • Higher frequency EM waves like UV, X-rays, and gamma rays are ionising and can be harmful to living cells.

Quick check

  1. Which type of electromagnetic wave has the longest wavelength?1 mark
  2. State one use and one danger of X-rays.2 marks

6. Constructive and Destructive Interference

When two or more waves meet at a point, they combine. The Principle of Superposition states that the resultant displacement at that point is the vector sum of the individual displacements. This can lead to two main effects: interference. Constructive interference occurs when two waves meet in phase, meaning their crests align with crests and their troughs align with troughs. The amplitudes of the waves add up, resulting in a wave with a larger amplitude. For sound, this results in a louder sound. For light, this results in a brighter spot. Destructive interference occurs when two waves meet in antiphase (or 180° out of phase), meaning the crest of one wave aligns with the trough of the other. The amplitudes cancel each other out. If the original waves had equal amplitudes, the resultant amplitude is zero. For sound, this results in a much quieter sound or silence. For light, this results in a dark spot. This phenomenon is responsible for the loud and quiet spots you can hear when two speakers play the same note, and the bright and dark 'fringes' seen in experiments with light.

Key term

Principle of Superposition: When two or more waves overlap, the total displacement at any point is the sum of the individual displacements of each wave at that point.

Common pitfall

Thinking that destructive interference 'destroys' energy. Energy is conserved; it is just redistributed to the points of constructive interference.

Worked example 12 marks

Two identical sound waves from two speakers arrive at a point P. The crest of Wave 1 arrives at the same time as the crest of Wave 2. What type of interference occurs at P and what is the effect on the sound heard?

  1. 1

    Step 1: Identify the phase relationship. The waves arrive 'crest to crest', which means they are in phase.

  2. 2

    Step 2: Apply the principle of superposition. When waves are in phase, their amplitudes add together.

  3. 3

    Step 3: State the type of interference. The addition of amplitudes is called constructive interference.

  4. 4

    Step 4: Describe the effect on the sound. Constructive interference of sound waves results in a larger amplitude, which is perceived as a louder sound.

Worked example 23 marks

A musician notices that in a certain spot in a concert hall, a particular note from the stage sounds very quiet. Explain this observation using the concepts of interference and reflection.

  1. 1

    Step 1: Identify the source of multiple waves. The sound can travel directly from the stage to the musician, and also indirectly by reflecting off a wall (e.g., the back wall).

  2. 2

    Step 2: Consider the path difference. The reflected sound travels a longer path than the direct sound. This path difference can cause the two waves to be out of phase when they arrive at the musician's ear.

  3. 3

    Step 3: Apply the concept of interference. If the path difference is such that the waves arrive in antiphase (crest of one meets trough of the other), destructive interference occurs.

  4. 4

    Step 4: Link interference to the observation. Destructive interference causes the amplitudes to cancel, resulting in a much smaller resultant amplitude. This is perceived as a very quiet sound or a 'dead spot'.

Recap

  • The Principle of Superposition describes how waves combine when they meet.
  • Constructive interference occurs when waves meet in phase, resulting in a larger amplitude (louder/brighter).
  • Destructive interference occurs when waves meet in antiphase, resulting in a smaller amplitude (quieter/darker).
  • Interference requires coherent waves, which have a constant phase difference and the same frequency.
  • Interference patterns of loud/quiet sounds or bright/dark fringes are evidence of the wave nature of sound and light.

Quick check

  1. What is the name for the effect when two waves meet and reinforce each other to create a larger wave?1 mark
  2. For destructive interference to cause complete cancellation, what must be true about the two interfering waves?1 mark

End-of-chapter exercise

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

  1. Define a longitudinal wave and give an example of one.2 marks
  2. List the colours of the visible spectrum in order of increasing wavelength.2 marks
  3. A sound wave has a frequency of 500 Hz and a wavelength of 0.7 m in air. Calculate the speed of the sound wave.2 marks
  4. Describe how the perceived pitch and loudness of a sound would change if its frequency is halved and its amplitude is doubled.2 marks
  5. A ray of light enters a glass block from air at an angle of incidence of 45°. The refractive index of the glass is 1.5. Calculate the angle of refraction.3 marks
  6. Using a ruler, draw a ray diagram to show how a converging lens forms a real, inverted, and diminished image of an object placed beyond 2F. Label the principal axis, focal points (F), the object, and the image.4 marks
  7. Explain the two conditions required for total internal reflection to occur and state one practical application of this phenomenon.3 marks
  8. Two loudspeakers are placed side-by-side and connected to the same signal generator, producing sound waves of the same frequency. A person walks along a line parallel to and in front of the speakers. Explain why they hear the sound alternating between loud and quiet.4 marks
  9. Compare the properties of radio waves and gamma rays in terms of their wavelength, frequency, and the energy they carry.3 marks
  10. An oscilloscope displays a sound wave. Describe how the trace on the screen would change if a second sound, which is quieter and has a higher pitch, is displayed instead. Be specific about amplitude and frequency on the screen.3 marks

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