(d) Light and sound

Syllabus
2024
Topic
Level

Learning objectives

3.14Light wavesKnow that light waves are transverse waves and that they can be reflected and refracted3.15Law of reflectionUse the law of reflection (the angle of incidence equals the angle of reflection)3.16Ray diagramsDraw ray diagrams to illustrate reflection and refraction3.17Refraction practicalPractical: investigate the refraction of light, using rectangular blocks, semi-circular blocks and triangular prisms3.18Refractive index equationKnow and use refractive index n = sin i ÷ sin r.3.19Glass refractive index practicalPractical: investigate the refractive index of glass, using a glass block3.20Total internal reflectionDescribe the role of total internal reflection in transmitting information along optical fibres and in prisms3.21Critical angleExplain the meaning of critical angle c3.22Critical angle equationKnow and use sin c = 1/n for critical angle c and refractive index n.3.23Sound wavesKnow that sound waves are longitudinal waves that can be reflected and refracted324P Human hearing rangeKnow that the frequency range for human hearing is 20–20 000 Hz325P Speed of sound practicalPractical: investigate the speed of sound in air326P Oscilloscope and microphoneUnderstand how an oscilloscope and microphone can be used to display a sound wave327P Sound frequency practicalPractical: investigate the frequency of a sound wave using an oscilloscope328P Pitch and frequencyUnderstand how the pitch of a sound relates to the frequency of vibration of the source329P Loudness and amplitudeUnderstand how the loudness of a sound relates to the amplitude of vibration of the source

Describe light as a transverse wave

Light is a transverse electromagnetic wave: its oscillations are perpendicular to the direction in which the wave travels and transfers energy.

Behaviour What happens to light
reflection light returns into the original medium at a boundary
refraction light changes speed and usually changes direction when it crosses into a different medium

Reflection and refraction describe what happens at a boundary; transverse describes the direction of oscillation. These are independent properties, so a reflected or refracted light wave remains transverse.

Apply the law of reflection

For reflection from a surface, the angle of incidence equals the angle of reflection.

i=r

Draw a normal at 90° to the surface where the incident ray meets it. Measure ii between the incident ray and the normal, and rr between the reflected ray and the normal. For example, i=38i=38^\circ gives r=38r=38^\circ.

Both angles are measured from the normal, not from the reflecting surface. A ray making 30° with the surface makes 60° with the normal.

Construct reflection and refraction ray diagrams

A ray diagram uses straight lines with arrowheads to show the direction of light. Every boundary construction begins with a normal drawn at 90° through the point of incidence.

  1. Draw the boundary and normal.
  2. Draw the incident ray ending at the normal and add an arrow towards the boundary.
  3. For reflection, draw the reflected ray on the original side with r=ir=i.
  4. For refraction into a higher-refractive-index medium, draw the refracted ray closer to the normal; into a lower-index medium, draw it farther from the normal.
  5. Add an arrow showing the continuing direction.

At normal incidence, i=0i=0^\circ, so the ray changes speed but not direction. Through a rectangular block with parallel faces, the emerging ray is parallel to the incident ray but laterally displaced.

Do not measure an angle from the surface or draw a curved ray inside one uniform medium. Direction changes at a boundary; each ray segment is straight.

Investigate refraction with three block shapes

Investigate refraction by tracing a narrow light ray through a transparent block and measuring angles from the normal.

  1. Place the block on paper and draw around it.
  2. Direct a single narrow ray at one face and mark two points on the incident ray and two on the emerging ray.
  3. Remove the block, join the marks with a ruler, and draw the normal at each boundary.
  4. Measure incidence and refraction angles with a protractor.
  5. Repeat for a range of incidence angles.
Shape What it lets the investigation show
rectangular block refraction at two parallel faces and lateral displacement
semicircular block a ray aimed through the centre meets the curved face normally, isolating refraction at the flat face
triangular prism successive non-parallel faces produce an overall change in direction

Keep the block fixed while marking each path and use a thin ray; a wide ray or moving outline makes the measured angles uncertain.

Calculate refractive index from two angles

For light entering a material from air, refractive index compares the sine of the incidence angle with the sine of the refraction angle.

n=\frac{\sin i}{\sin r}

nn has no unit; ii and rr are measured from the normal. Use degree mode on the calculator.

If i=45i=45^\circ and r=28r=28^\circ, then n=sin45/sin28=0.7071/0.4695=1.51n=\sin45^\circ/\sin28^\circ=0.7071/0.4695=1.51.

Do not calculate i/ri/r. The equation uses the sine of each angle, and reversing the numerator and denominator gives the wrong refractive index.

Determine the refractive index of glass

A glass block investigation determines refractive index from repeated measurements of incidence and refraction angles.

  1. Draw around a rectangular glass block on paper.
  2. Shine a narrow ray from air into the block and mark the incident and refracted paths.
  3. Remove the block, draw the normal, and measure ii and rr.
  4. Repeat for several incidence angles, avoiding very small angles that give large percentage uncertainty.
  5. Calculate n=sini/sinrn=\sin i/\sin r for each pair and take the mean after checking anomalies.

A stronger analysis plots sini\sin i on the vertical axis against sinr\sin r on the horizontal axis. Since sini=nsinr\sin i=n\sin r, the gradient of a best-fit line through the origin is nn.

Repeat angle pairs, not just the final arithmetic. A mean cannot reveal a systematic error such as measuring from the surface instead of the normal.

Use total internal reflection in fibres and prisms

Total internal reflection occurs when light travels from a higher-refractive-index medium towards a lower-index medium and the incidence angle is greater than the critical angle. No refracted ray then leaves through that boundary.

Device Role of total internal reflection
optical fibre repeated internal reflections keep light pulses inside the core so encoded information travels along the fibre
prism internal reflection redirects a beam through a chosen angle in devices such as binoculars and periscopes

Both conditions are required: travel from higher to lower refractive index, and i>ci>c. If either condition fails, some light is refracted through the boundary.

Define the critical angle

The critical angle cc is the incidence angle in the higher-refractive-index medium for which the refracted ray in the lower-index medium is at 9090^\circ to the normal and travels along the boundary.

Incidence angle in the higher-index medium Outcome
i<ci<c light refracts out, bending away from the normal
i=ci=c refracted ray travels along the boundary
i>ci>c total internal reflection

The critical angle is not the first angle at which any reflection occurs: partial reflection can occur below cc. It is the threshold that separates refraction out from total internal reflection.

Calculate critical angle and refractive index

For a material-to-air boundary, critical angle and refractive index are related by:

\sin c=\frac{1}{n}

To find the angle, use c=sin1(1/n)c=\sin^{-1}(1/n). To find refractive index, use n=1/sincn=1/\sin c. The angle cc is in degrees and nn has no unit.

For glass with n=1.50n=1.50, c=sin1(1/1.50)=41.8c=\sin^{-1}(1/1.50)=41.8^\circ. A larger refractive index gives a smaller critical angle.

Use inverse sine when finding an angle; c=1/nc=1/n is incorrect. This syllabus equation applies to the material-air boundary described here.

Describe sound as a longitudinal wave

Sound is a longitudinal mechanical wave. Particles of the medium oscillate parallel to the direction of wave travel, forming compressions and rarefactions.

Behaviour What happens to sound Example
reflection sound returns from a boundary an echo
refraction sound changes speed and direction across regions where its speed differs sound bending through air at different temperatures

Sound requires particles to pass on the vibration, so it cannot travel through a vacuum. The particles oscillate locally; they do not travel from the source to the listener.

Use the human hearing range

The syllabus frequency range for human hearing is 20 Hz to 20 000 Hz, which is also 20 Hz to 20 kHz.

Frequency Classification relative to human hearing
below 20 Hz below the hearing range (infrasound)
20 Hz to 20 000 Hz inclusive within the stated hearing range
above 20 000 Hz above the hearing range (ultrasound)

20 kHz=20000 Hz20\text{ kHz}=20 000\text{ Hz}, not 20 Hz. A vibrating source can produce sound outside the human hearing range even though it is still oscillating.

Measure the speed of sound in air

An electronic two-microphone method measures sound travel time over a known distance without relying on human reaction time.

  1. Place two microphones a measured distance dd apart and connect them to an oscilloscope or data logger.
  2. Make a sharp sound close to the first microphone.
  3. Measure the delay Δt\Delta t between the two recorded signals.
  4. Calculate v=d/Δtv=d/\Delta t.
  5. Repeat at several large separations, identify anomalies, and average consistent values or find the gradient of a distance-time graph.

v=\frac{d}{\Delta t}

Measure distance between the microphone positions and keep air temperature as constant as practical because sound speed changes with temperature. A larger separation makes the delay a larger fraction of the measured time and reduces percentage uncertainty.

For an echo method, sound travels to the wall and back, so the distance is 2d2d. A handheld stopwatch is unsuitable for short separations because the travel time is comparable with reaction time.

Display sound with a microphone and oscilloscope

A microphone converts sound-pressure variations into a changing electrical signal; an oscilloscope displays that signal as voltage against time.

Screen direction Represents Setting
horizontal (xx) time timebase: time per division
vertical (yy) signal voltage, related to sound-wave amplitude gain: volts per division

Connect the microphone, produce the sound, then adjust the timebase until at least one complete cycle is visible and the trace is steady. Adjust vertical gain so the trace is large enough to measure without leaving the screen.

The screen is not a picture of air particles moving through space. Its horizontal axis is time, and its vertical displacement represents the microphone's electrical signal.

Measure sound frequency with an oscilloscope

Find sound frequency by measuring the period of a steady oscilloscope trace and using f=1/Tf=1/T.

  1. Connect a microphone and obtain a steady trace with several complete cycles.
  2. Count the horizontal divisions across NN complete cycles.
  3. Multiply divisions by the timebase to find their total time.
  4. Divide by NN to obtain one period TT in seconds.
  5. Calculate f=1/Tf=1/T and repeat the measurement across another set of cycles.

T=\frac{(\text{divisions})(\text{time per division})}{N},\qquad f=\frac{1}{T}

If 8 divisions contain 4 cycles and the timebase is 0.50 ms/division, total time is 4.0 ms. Thus T=4.0/4=1.0 ms=1.0×103 sT=4.0/4=1.0\text{ ms}=1.0\times10^{-3}\text{ s} and f=1000 Hzf=1000\text{ Hz}.

Use the horizontal timebase, not the vertical gain. Convert milliseconds or microseconds to seconds before using f=1/Tf=1/T.

Relate pitch to source frequency

Pitch is the perception linked to the frequency at which the source vibrates: increasing source frequency produces a higher-pitch sound, while decreasing it produces a lower-pitch sound.

Source change Wave change Heard change
vibrates more times each second higher frequency, shorter period higher pitch
vibrates fewer times each second lower frequency, longer period lower pitch

Amplitude does not determine pitch. Two sounds can have the same frequency and pitch but different amplitudes and loudnesses.

Relate loudness to source amplitude

For the same source and listening conditions, a larger vibration amplitude produces a larger-amplitude sound wave and a louder sound; a smaller amplitude produces a quieter sound.

Source vibration Oscilloscope trace Heard sound
larger amplitude taller trace from equilibrium louder
smaller amplitude shorter trace from equilibrium quieter

Changing amplitude does not change frequency or pitch. Loudness at a listener can also change with distance, so the amplitude relationship must compare otherwise similar conditions.