Cambridge Lower Secondary CheckpointStage 6

Physics: Light and sound

Science Stage 6 Chapter Notes

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Physics: Light and sound
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In this chapter

  1. 1Properties of Light Waves
  2. 2Reflection of Light
  3. 3Refraction of Light
  4. 4Total Internal Reflection
  5. 5Image Formation by Lenses
  6. 6Properties of Sound Waves
  7. 7Echoes and Speed of Sound
  8. 8Pitch, Loudness and Quality of Sound

1. Properties of Light Waves

Light is a form of energy that travels as an electromagnetic wave. It is a transverse wave, meaning the oscillations (vibrations) are perpendicular to the direction of energy transfer. Unlike sound, light does not require a medium and can travel through a vacuum, such as the space between the Sun and Earth. Light travels in straight lines, a property called rectilinear propagation. The speed of light in a vacuum is the fastest speed possible in the universe, approximately 300,000,000 metres per second (3.0 x 10^8 m/s).

c = 3.0 x 10^8 m/s

Key term

Transverse Wave: A wave in which the oscillations are perpendicular (at 90 degrees) to the direction of energy transfer.

Fun fact

It takes about 8 minutes and 20 seconds for light from the Sun to reach Earth, so if the Sun were to suddenly disappear, we wouldn't know for over 8 minutes.

Worked example 13 marks

The Moon is approximately 384,000 km from the Earth. Calculate the time it takes for light reflected from the Moon to reach the Earth. Take the speed of light to be 3.0 x 10^8 m/s.

  1. 1

    Step 1: Convert the distance from km to metres. Distance = 384,000 km = 384,000 × 1000 m = 3.84 x 10^8 m.

  2. 2

    Step 2: State the formula relating speed, distance and time: speed = distance / time. Rearrange for time: time = distance / speed.

  3. 3

    Step 3: Substitute the values into the formula. time = (3.84 x 10^8m) / (3.0 x 10^8 m/s).

  4. 4

    Step 4: Calculate the final answer. time = 1.28 s.

Recap

  • Light is a transverse electromagnetic wave.
  • Light can travel through a vacuum.
  • Light travels in straight lines.
  • The speed of light in a vacuum is 3.0 x 10^8 m/s.

Quick check

  1. What type of wave is a light wave?1 mark
  2. State the value for the speed of light in a vacuum.1 mark

2. Reflection of Light

Reflection is the bouncing of light off a surface. A plane mirror is a flat, smooth reflecting surface. When light hits a plane mirror, it follows two laws of reflection. The first law states that the angle of incidence(i) is equal to the angle of reflection (r). The second law states that the incident ray, the reflected ray, and the normal all lie in the same plane. The 'normal' is an imaginary line drawn perpendicular (at 90°) to the mirror at the point of incidence. The image formed by a plane mirror has specific characteristics: it is virtual (cannot be projected onto a screen), upright, the same size as the object, and laterally inverted (left and right are swapped).

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

Key term

Virtual Image: An image that cannot be formed on a screen, created where rays of light only appear to originate from or converge.

Examiner insight

Examiners award marks for accurately drawn ray diagrams with arrows on all light rays to show the direction of travel.

Common pitfall

A common mistake is measuring the angle of incidence from the surface of the mirror instead of from the normal.

Worked example 12 marks

An object is placed 5 cm in front of a plane mirror. A ray of light leaves the top of the object and strikes the mirror at an angle of incidence of 30°. State the distance of the image from the mirror and the angle of reflection.

  1. 1

    Step 1: Recall the properties of an image in a plane mirror. The image distance is equal to the object distance. Therefore, the image is 5 cm behind the mirror.

  2. 2

    Step 2: Recall the first law of reflection: angle of incidence = angle of reflection.

  3. 3

    Step 3: State the angle of reflection. Since the angle of incidence is 30°, the angle of reflection is also 30°.

Worked example 24 marks

Describe the characteristics of an image formed in a plane mirror.

  1. 1

    The image is virtual (it cannot be formed on a screen).

  2. 2

    The image is upright (the same way up as the object).

  3. 3

    The image is the same size as the object.

  4. 4

    The image is laterally inverted (left and right are swapped).

  5. 5

    The image is as far behind the mirror as the object is in front.

Recap

  • Reflection is the bouncing of light off a surface.
  • The angle of incidence equals the angle of reflection (i = r).
  • The incident ray, reflected ray, and normal are all in the same plane.
  • An image in a plane mirror is virtual, upright, same size, and laterally inverted.

Quick check

  1. If a ray of light hits a mirror at an angle of 40° to the normal, what is the angle of reflection?1 mark
  2. What does the term 'laterally inverted' mean?1 mark

3. Refraction of Light

Refraction is the change in direction of a wave, such as light, when it passes from one medium to another. This bending occurs because the speed of light changes as it enters a new medium. When light enters a more optically dense medium (like from air to glass), it slows down and bends towards the normal. When it enters a less dense medium (like from glass to air), it speeds up and bends away from the normal. The amount of bending is described by the refractive index(n) of the medium. Snell's Law relates the angle of incidence (i), the angle of refraction (r), and the refractive indices of the two media.

n = sin(i) / sin(r)

n = (speed of light in vacuum) / (speed of light in medium)

Key term

Refractive Index (n): A dimensionless number that describes how fast light travels through a material; a measure of its optical density.

Examiner insight

Marks are often awarded for correctly identifying that refraction occurs due to a change in the speed of light.

Common pitfall

Confusing the angle of incidence/refraction with the angle the ray makes with the surface. Always measure angles from the normal.

Worked example 13 marks

A ray of light travelling in air strikes the surface of a glass block at an angle of incidence of 60°. The refractive index of the glass is 1.5. Calculate the angle of refraction.

  1. 1

    Step 1: State Snell's Law: n = sin(i) / sin(r).

  2. 2

    Step 2: Rearrange the formula to make sin(r) the subject: sin(r) = sin(i) / n.

  3. 3

    Step 3: Substitute the known values: sin(r) = sin(60°) / 1.5.

  4. 4

    Step 4: Calculate sin(60°) = 0.866. So, sin(r) = 0.866 / 1.5 = 0.577.

  5. 5

    Step 5: Find the angle r by taking the inverse sine: r = sin⁻¹(0.577) = 35.3°.

Recap

  • Refraction is the bending of light as it changes medium and speed.
  • Light bends towards the normal when entering a denser medium.
  • Light bends away from the normal when entering a less dense medium.
  • Snell's Law, n = sin(i) / sin(r), relates the angles to the refractive index.

Quick check

  1. What happens to the speed of light when it enters a glass block from air?1 mark
  2. A ray of light passes from water into air. Will it bend towards or away from the normal?1 mark

4. Total Internal Reflection

When light travels from a more dense medium to a less dense medium (e.g., glass to air), it bends away from the normal. As the angle of incidence increases, the angle of refraction also increases, getting closer to 90°. The angle of incidence for which the angle of refraction is exactly 90° is called the critical angle (c). If the angle of incidence is greater than the critical angle, the light does not refract out of the medium at all. Instead, it is completely reflected back into the denser medium. This phenomenon is called Total Internal Reflection (TIR). For TIR to occur, two conditions must be met: 1. The light must be travelling from a more optically dense medium to a less optically dense one. 2. The angle of incidence must be greater than the critical angle. This principle is used in optical fibres and periscopes.

sin(c) = 1 / n

Key term

Critical Angle (c): The angle of incidence in a denser medium for which the angle of refraction in the less dense medium is 90°.

Fun fact

The spectacular sparkle of a cut diamond is largely due to total internal reflection. Its high refractive index gives it a small critical angle, so most light entering it gets trapped inside and reflects many times before exiting, creating flashes of colour.

Worked example 13 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 relating critical angle and refractive index: sin(c) = 1 / n.

  2. 2

    Step 2: Substitute the value of n: sin(c) = 1 / 2.42.

  3. 3

    Step 3: Calculate the value of sin(c): sin(c) = 0.413.

  4. 4

    Step 4: Find the angle c by taking the inverse sine: c = sin⁻¹(0.413) = 24.4°.

Worked example 24 marks

Explain, with the aid of a diagram, how an optical fibre transmits light over long distances.

  1. 1

    An optical fibre consists of a very thin core of high refractive index glass, surrounded by cladding of lower refractive index glass.

  2. 2

    Light is shone into one end of the fibre at an angle greater than the critical angle.

  3. 3

    As the light travels down the fibre and hits the core-cladding boundary, it undergoes total internal reflection.

  4. 4

    This process repeats, with the light bouncing off the inside of the core all the way along the fibre, allowing it to be transmitted over long distances with minimal loss of energy.

  5. 5

    (A diagram should show a light ray entering the fibre and reflecting multiple times off the internal surface).

Recap

  • Total Internal Reflection (TIR) occurs when light is completely reflected within a dense medium.
  • Two conditions for TIR: light travels from dense to less dense, and angle of incidence > critical angle.
  • The critical angle is found using sin(c) = 1 / n.
  • TIR is the principle behind optical fibres used in communications.

Quick check

  1. State the two conditions necessary for total internal reflection.2 marks

5. Image Formation by Lenses

A lens is a piece of transparent material, usually glass or plastic, shaped to refract light in a specific way. There are two main types: a converging (convex) lens, which is thicker in the middle and brings parallel rays of light to a focus, and a diverging (concave) lens, which is thinner in the middle and spreads parallel rays of light out. The point where parallel rays converge (or appear to diverge from) is the principal focus or focal point (F). The distance from the centre of the lens to this point is the focal length (f). Ray diagrams can be drawn to determine the position, size, and nature (real or virtual) of an image formed by a lens. A real image is one that can be projected onto a screen, while a virtual image cannot.

1/f = 1/u + 1/v (Lens Equation)

M = hi / ho = v / u (Magnification)

Key term

Focal Length (f): The distance from the optical centre of a lens to its principal focus (focal point).

Common pitfall

Forgetting that for a virtual image in the lens equation, the image distance 'v' is a negative value.

Worked example 15 marks

An object of height 2 cm is placed 30 cm from a converging lens with a focal length of 20 cm. By drawing a scale ray diagram, determine the position, height, and nature of the image.

  1. 1

    Step 1: Draw the principal axis and the lens. Mark the optical centre (O), and the focal points F and 2F on both sides of the lens (F at 20 cm, 2F at 40 cm).

  2. 2

    Step 2: Draw the object as an arrow of height 2 cm at a distance of 30 cm from the lens (between F and 2F).

  3. 3

    Step 3: Draw a ray from the top of the object parallel to the principal axis. This ray refracts through the lens and passes through the focal point F on the other side.

  4. 4

    Step 4: Draw a second ray from the top of the object passing straight through the optical centre O without deviation.

  5. 5

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

  6. 6

    Step 6: Measure the image distance(v) and image height (hi) from the diagram. The image is formed at 60 cm from the lens and is 4 cm tall. It is real (formed by actual intersection of rays) and inverted.

Worked example 24 marks

A magnifying glass uses a converging lens of focal length 10 cm. If it produces a virtual image magnified 3 times, calculate the object distance (u).

  1. 1

    Step 1: State the magnification formula: M = v/u. We are given M=3, so v/u = 3, which means v = 3u. Since the image is virtual, the image distance v is negative, so v = -3u.

  2. 2

    Step 2: State the lens equation: 1/f = 1/u + 1/v.

  3. 3

    Step 3: Substitute the known values and the expression for v: 1/10 = 1/u + 1/(-3u).

  4. 4

    Step 4: Simplify the equation: 1/10 = 1/u - 1/(3u) = (3-1)/(3u) = 2/(3u).

  5. 5

    Step 5: Solve for u: 3u = 2 * 10 = 20. Therefore, u = 20/3 = 6.67 cm.

Recap

  • Converging (convex) lenses focus light; diverging (concave) lenses spread light.
  • A real image can be projected onto a screen; a virtual image cannot.
  • Ray diagrams are used to find the location and nature of an image.
  • Converging lenses are used as magnifying glasses when the object is within the focal length.

Quick check

  1. What type of image is formed by a diverging lens?1 mark
  2. What is a converging lens used for in a camera?1 mark

6. Properties of Sound Waves

Sound is a form of energy produced by vibrating objects. These vibrations are passed through a medium (a solid, liquid, or gas) as a wave. Sound is a longitudinal wave, meaning the particles of the medium vibrate parallel to the direction of energy transfer. This creates areas where the particles are bunched together, called compressions, and areas where they are spread apart, called rarefactions. Sound cannot travel through a vacuum because there are no particles to carry the vibrations. The speed of sound depends on the medium it travels through; it is fastest in solids, slower in liquids, and slowest in gases. This is because the particles in a solid are closer together and more tightly bound, allowing vibrations to pass more quickly.

v = fλ (Wave speed = frequency × wavelength)

Key term

Longitudinal Wave: A wave in which the particles of the medium vibrate parallel to the direction of energy transfer.

Fun fact

The study of sound is called acoustics. In the vacuum of space, there is no sound, which is why science fiction movies that have loud explosions in space are scientifically inaccurate!

Worked example 13 marks

A tuning fork produces a sound wave with a frequency of 440 Hz. If the speed of sound in air is 340 m/s, what is the wavelength of the sound wave?

  1. 1

    Step 1: State the wave equation: v = fλ.

  2. 2

    Step 2: Rearrange the formula to find the wavelength (λ): λ = v / f.

  3. 3

    Step 3: Substitute the given values: λ = 340 m/s / 440 Hz.

  4. 4

    Step 4: Calculate the wavelength: λ = 0.77 m.

Worked example 23 marks

Explain why you see a firework explode before you hear the bang.

  1. 1

    Step 1: State that both light and sound are produced by the firework at the same time.

  2. 2

    Step 2: Recall the speeds of light and sound. Light travels much faster (approximately 3.0 x 10^8 m/s) than sound (approximately 340 m/s).

  3. 3

    Step 3: Conclude that because of this large difference in speed, the light from the explosion reaches your eyes almost instantaneously, while the sound takes a noticeable amount of time to travel the same distance to your ears.

Recap

  • Sound is a longitudinal wave caused by vibrations.
  • Sound requires a medium to travel and cannot pass through a vacuum.
  • A sound wave consists of compressions and rarefactions.
  • Sound travels fastest in solids, slower in liquids, and slowest in gases.

Quick check

  1. What is the name for the part of a sound wave where particles are close together?1 mark
  2. Would a bell ringing inside a vacuum jar be audible? Explain why.2 marks

7. Echoes and Speed of Sound

An echo is a reflection of sound. When a sound wave hits a hard, flat surface, like a wall or a cliff, it can bounce back to the listener. If the time between the original sound and hearing the echo is measured, it's possible to calculate the distance to the reflecting surface. The key formula is distance = speed × time. However, it's crucial to remember that the time measured is for the sound to travel to the surface AND back again. Therefore, the distance travelled by the echo is twice the distance to the reflecting surface. This principle, known as echo sounding or sonar (Sound Navigation and Ranging), is used by ships to measure the depth of the sea and by bats for navigation.

distance = speed × time

depth (d) = (speed of sound in water × time for echo) / 2

Key term

Echo: A reflection of a sound wave that is heard after the original sound.

Examiner insight

Examiners look for clear working that shows you understand the sound travels 'there and back', often by seeing the distance explicitly doubled or the final answer halved.

Common pitfall

The most common error is forgetting to divide the total distance travelled by two when finding the distance to a reflector.

Worked example 13 marks

A person stands 85 metres from a large cliff and shouts. The speed of sound in air is 340 m/s. How long will it take for the person to hear the echo?

  1. 1

    Step 1: Calculate the total distance the sound must travel. The sound travels to the cliff and back, so total distance = 2 × 85 m = 170 m.

  2. 2

    Step 2: State the formula: speed = distance / time. Rearrange for time: time = distance / speed.

  3. 3

    Step 3: Substitute the values: time = 170 m / 340 m/s.

  4. 4

    Step 4: Calculate the time: time = 0.5 s.

Worked example 23 marks

A ship's sonar sends a pulse of ultrasound to the seabed. The echo is detected 1.2 seconds later. If the speed of sound in seawater is 1500 m/s, calculate the depth of the sea.

  1. 1

    Step 1: Find the total distance travelled by the sound pulse using distance = speed × time. Total distance = 1500 m/s × 1.2 s = 1800 m.

  2. 2

    Step 2: Recognise that this is the distance to the seabed and back again. The depth of the sea is half of this total distance.

  3. 3

    Step 3: Calculate the depth: Depth = 1800 m / 2 = 900 m.

Recap

  • An echo is a reflected sound wave.
  • The time taken to hear an echo can be used to calculate distance.
  • In echo calculations, the sound travels twice the distance to the reflector.
  • Sonar and ultrasound imaging are practical applications of echoes.

Quick check

  1. A bat sends out a squeak and receives an echo 0.1 s later. How far away is the prey? (Speed of sound = 340 m/s)2 marks

8. Pitch, Loudness and Quality of Sound

Sounds can be described by three subjective qualities: pitch, loudness, and quality (or timbre). Each of these corresponds to a physical property of the sound wave. Pitch describes how high or low a sound is and is determined by the wave's frequency. A high frequency produces a high pitch, while a low frequency produces a low pitch. Loudness is our perception of the sound's intensity and is determined by the wave's amplitude. A wave with a large amplitude is loud, while a wave with a small amplitude is quiet. Quality, or timbre, is what distinguishes two sounds of the same pitch and loudness, for example, a violin and a piano playing the same note. It is determined by the waveform, or the shape of the wave, which is created by the combination of the fundamental frequency and other higher frequencies called overtones. An oscilloscope can be used to display sound waves visually, showing their amplitude, frequency, and waveform.

Key term

Frequency: The number of complete oscillations or cycles of a wave that pass a point per second, measured in Hertz (Hz).

Common pitfall

Confusing the terms 'pitch' and 'loudness' or 'frequency' and 'amplitude'. Remember: Pitch links to Frequency, Loudness links to Amplitude.

Worked example 14 marks

The diagrams show the traces of two sound waves, A and B, on an oscilloscope. Compare the two sounds in terms of their pitch and loudness. Justify your answers.

  1. 1

    Pitch: Sound A has a higher pitch than sound B. This is because its trace on the oscilloscope shows more waves in the same time interval, meaning it has a higher frequency.

  2. 2

    Loudness: Sound A and sound B have the same loudness. This is because their traces have the same height, meaning they have the same amplitude.

Worked example 24 marks

How could you change a sound to make it quieter and lower in pitch? Describe the changes you would see on an oscilloscope screen.

  1. 1

    To make the sound quieter, you must decrease its amplitude. On the oscilloscope, this would be seen as the height of the wave trace decreasing.

  2. 2

    To make the sound lower in pitch, you must decrease its frequency. On the oscilloscope, this would be seen as the waves becoming more spread out, with fewer waves visible on the screen in the same time period.

Recap

  • The pitch of a sound is determined by its frequency.
  • The loudness of a sound is determined by its amplitude.
  • The quality (timbre) of a sound is determined by its waveform.
  • An oscilloscope displays the shape of a sound wave.

Quick check

  1. If you increase the frequency of a sound wave, what happens to its pitch?1 mark
  2. Which property of a sound wave must be increased to make the sound louder?1 mark

End-of-chapter exercise

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

  1. Define a transverse wave and give an example of one.2 marks
  2. A ray of light passes from water (n=1.33) into air at an angle of incidence of 35°. Calculate the angle of refraction. (n for air = 1.00)4 marks
  3. Explain, using the terms 'frequency' and 'amplitude', how a quiet, high-pitched sound differs from a loud, low-pitched sound.4 marks
  4. A boat uses sonar to find a shipwreck. It sends a sound pulse and receives the echo 0.24 seconds later. If the speed of sound in seawater is 1500 m/s, calculate the depth of the shipwreck.3 marks
  5. An object is placed 15 cm from a converging lens of focal length 10 cm. Using the lens equation (1/f = 1/u + 1/v), calculate the image distance (v) and state whether the image is real or virtual.4 marks
  6. Describe the 'bell-in-a-jar' experiment that demonstrates that sound requires a medium to travel. State the observation and conclusion.3 marks
  7. Explain with the aid of a simple diagram how total internal reflection is used to guide light along an optical fibre. You must state the two conditions for TIR to occur.5 marks
  8. A sound wave has a frequency of 2 kHz and a wavelength of 75 cm. (a) Calculate the speed of the sound wave in m/s. (b) How long would it take for this sound to travel 2 km?5 marks
  9. Draw a ray diagram to show the formation of a virtual, upright and magnified image by a converging lens. Label the object, image, focal points (F) and the lens.4 marks
  10. A ray of light is incident on a plane mirror at an angle of 50° to the surface of the mirror. What is (a) the angle of incidence, (b) the angle of reflection, and (c) the angle between the incident and reflected rays?3 marks

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