DP Physics SL C.3 Wave Phenomena

DP Physics SL C.3 Wave Phenomena - Full notes

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DP Physics SL | C.3 Wave Phenomena

DP PHYSICS · THEME C: WAVE BEHAVIOUR

C.3

Wave Phenomena

Full notes for SL · IB-style

Contents

1	Reflection and refraction
2	Diffraction and superposition
3	Standing waves
4	The Doppler effect
5	Exam technique
6	Data-based questions and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use the law of reflection and Snell's law
explain total internal reflection, calculate critical angles and describe optical fibres
describe diffraction and superposition
apply path difference and coherence to interference, including Young's double slit
describe standing waves on strings and in pipes and calculate harmonics
describe the Doppler effect and use Δλ/λ ≈ v/c

This booklet covers the SL content of this subtopic. Additional Higher Level material is not included.

Guiding questions

How are observations of wave behaviours at a boundary between media explained? How is the wave model used to explain interference and diffraction? What happens when waves superpose, and when sources or observers move?

How this topic is examined

C.3 appears in Papers 1 and 2 with many calculations: Snell's law, critical angles, double-slit fringe spacing, harmonics on strings and in pipes, and Doppler shifts. HL adds single slits, gratings and Doppler equations.

Draw ray diagrams with normals; measure angles from the normal.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Reflection and refraction

Waves meeting boundaries.

Think about it

Why does a straw look bent in a glass of water, and why do diamonds sparkle?

When a wave meets a boundary between two media, part of it is reflected and part transmitted. The transmitted wave changes speed and, if it arrives at an angle, direction. These ideas explain mirrors, lenses, optical fibres and mirages.

1.1 Reflection

At a boundary, the angle of incidence equals the angle of reflection, both measured from the normal. Speed, frequency and wavelength are unchanged. A wave reflected from a denser medium (or a fixed end of a string) undergoes a phase change of π (it is inverted).

Figure 1. The law of reflection: i = r, measured from the normal.

WORKED EXAMPLE 1 Mirror (basic)

A ray strikes a mirror at 35° to the surface. Find the angle of reflection.

Solution

Angle to normal = 55°M1

Angle of reflection = 55°A1

Common mistake

Measuring angles from the surface instead of the normal.

1.2 Reflection of pulses on strings

A pulse reflected from a fixed end is inverted (phase change of π); a pulse reflected from a free end is not inverted. This phase change determines where nodes form in standing waves.

Figure 2. Reflection of a pulse at a fixed end and at a free end.

Quick check

1. Is there a phase change when light reflects from glass in air?

Answers: 1. yes, π (reflection from a denser medium)

1.3 Huygens' principle

Every point on a wavefront acts as a source of secondary wavelets; the new wavefront is the envelope of these wavelets. This simple construction explains reflection, refraction and diffraction.

Figure 3. Huygens' construction for a plane wavefront.

Quick check

1. How does Huygens' principle explain diffraction at a narrow gap?

Answers: 1. the few points in the gap act as sources whose wavelets spread out in all forward directions

1.4 Refraction and Snell's law

Refraction is the change of direction when a wave crosses a boundary at an angle and its speed changes. The refractive index n = c/v. Snell's law: n₁ sin θ₁ = n₂ sin θ₂. Equivalently sin θ₁/sin θ₂ = v₁/v₂ = λ₁/λ₂. Light bends towards the normal when it slows down.

Figure 4. Refraction of light entering glass: the ray bends towards the normal.

WORKED EXAMPLE 2 Glass block (standard)

Light enters glass (n = 1.50) from air at 40° to the normal. Find the angle of refraction and the speed of light in the glass.

Solution

sin θ₂ = sin 40°/1.50: θ₂ = 25.4°M1 A1

v = 3.0 × 10⁸/1.50 = 2.0 × 10⁸ m s⁻¹A1

WORKED EXAMPLE 3 Water to glass (exam level)

Light travels from water (n = 1.33) into glass (n = 1.50) at 50°. Find the angle of refraction.

Solution

1.33 sin 50° = 1.50 sin θM1

θ = 42.8°A1

1.5 Apparent depth

Light from an underwater object bends away from the normal as it leaves the water, so the object appears closer to the surface. For viewing from almost directly above, apparent depth = real depth/n.

Figure 5. Refraction makes a coin appear shallower than it is.

WORKED EXAMPLE 4 Pool (standard)

A coin lies 1.2 m below the surface of water (n = 1.33). How deep does it appear from above?

Solution

1.2/1.33M1

= 0.90 mA1

1.6 Total internal reflection

When light travels from a higher to a lower refractive index and the angle of incidence exceeds the critical angle c, no light is refracted: it is all reflected. sin c = n₂/n₁ (= 1/n into air).

Figure 6. Light inside glass meeting the surface below, at and above the critical angle (n = 1.50).

WORKED EXAMPLE 5 Critical angle (standard)

Calculate the critical angle for glass (n = 1.50) to air and for diamond (n = 2.42).

Solution

Glass: sin c = 1/1.50: c = 41.8°A1

Diamond: c = 24.4°: light is trapped more easily, causing sparkleA1

1.7 Optical fibres

An optical fibre has a glass core surrounded by cladding of slightly lower refractive index. Light entering the core hits the boundary above the critical angle and is totally internally reflected along the fibre with very little loss. Fibres carry internet data and are used in endoscopes.

Figure 7. Light guided along an optical fibre by total internal reflection.

WORKED EXAMPLE 6 Core and cladding (exam level)

A fibre has core n = 1.50 and cladding n = 1.46. Find the critical angle at the core–cladding boundary.

Solution

sin c = 1.46/1.50M1

c = 76.7°A1

In the real world

Undersea optical fibre cables carry over 95% of international internet traffic.

1.8 Refraction of water waves

Water waves slow down in shallower water. When plane waves cross a depth boundary at an angle, the part entering the shallow region slows first, so the wavefronts bend and the wavelength decreases. This is why waves approaching a beach turn to become nearly parallel to the shore.

Figure 8. Plane water waves refracting as they enter shallow water (schematic).

WORKED EXAMPLE 7 Water waves (standard)

Waves of frequency 4.0 Hz travel at 0.32 m s⁻¹ in deep water and 0.20 m s⁻¹ in shallow water. Find the wavelengths and the refractive index for this boundary.

Solution

Deep: 8.0 cm; shallow: 5.0 cmA1

n = v₁/v₂ = 1.6A1

1.9 Measuring refractive index

Trace rays through a rectangular glass block for several angles of incidence and measure the angles of refraction. A graph of sin i against sin r is a straight line through the origin with gradient n.

Figure 9. sin i against sin r for a glass block.

WORKED EXAMPLE 8 Gradient (standard)

Find n from the graph.

Solution

Gradient = 0.90/0.60M1

n = 1.50A1

Try this

Use a semicircular glass block and a ray box to find the critical angle directly; compare 1/sin c with your value of n.

1.10 Dispersion

The refractive index of glass is slightly larger for violet light than for red, so a prism separates white light into a spectrum. Rainbows form by refraction, dispersion and internal reflection in raindrops.

Figure 10. Dispersion of white light by a prism (violet deviated most).

WORKED EXAMPLE 9 Prism (exam level)

Glass has n = 1.51 for red and 1.53 for violet. White light enters at 45°. Find the two angles of refraction.

Solution

Red: sin r = 0.707/1.51: r = 27.9°A1

Violet: r = 27.5°A1

2 Diffraction and superposition

Waves spreading and combining.

2.1 Diffraction

Diffraction is the spreading of waves as they pass through a gap or around an obstacle. It is most noticeable when the gap width is similar to the wavelength. Wavelength, frequency and speed are unchanged.

Figure 11. Diffraction of plane waves through a wide and a narrow gap.

WORKED EXAMPLE 10 Sound and light (exam level)

Explain why you can hear someone around a corner but not see them.

Solution

Sound wavelength (about 1 m) is similar to doorway widths, so it diffracts stronglyR1

Light wavelength (about 500 nm) is far smaller, so diffraction is negligibleR1

2.2 Diffraction in everyday life

Long-wave radio (wavelengths of hundreds of metres) diffracts around hills, so it can be received in valleys where FM (about 3 m) cannot. Sound diffracts around corners and through doorways. Diffraction limits the detail any telescope or microscope can resolve.

WORKED EXAMPLE 11 Radio (exam level)

Explain why a 200 kHz radio signal is received behind a hill but a 100 MHz signal is not.

Solution

λ = 1500 m vs 3.0 mM1

Long wavelength comparable to the hill's size diffracts stronglyR1

2.3 The principle of superposition

When two waves meet, the resultant displacement at any point is the sum of the individual displacements. The waves pass through each other unchanged.

Figure 12. Superposition of two waves in phase and in antiphase.

WORKED EXAMPLE 12 Pulses (basic)

Pulses of amplitude 3 cm and −2 cm meet. Find the resultant displacement where their peaks overlap.

Solution

3 + (−2)M1

= 1 cmA1

2.4 Coherence and interference

Interference is the superposition of waves from two coherent sources: same frequency and constant phase difference. Where the path difference is nλ (sources in phase), the waves arrive in phase: constructive interference. Where it is (n + ½)λ, they arrive in antiphase: destructive interference.

Key point

Constructive: path difference = nλ. Destructive: (n + ½)λ (for sources in phase).

WORKED EXAMPLE 13 Speakers (exam level)

Two in-phase speakers emit 680 Hz sound (v = 340 m s⁻¹). A listener is 4.0 m from one and 5.25 m from the other. Is the sound loud or quiet?

Solution

λ = 0.50 m; path difference 1.25 m = 2.5λM1

Destructive: quietA1

2.5 Two-source interference patterns

Two dippers vibrating in phase in a ripple tank produce lines of constructive interference (large amplitude) separated by lines of destructive interference (calm water). The same pattern forms with two loudspeakers or two microwave slits.

Figure 13. Calculated interference pattern from two in-phase sources; dark bands are constructive.

WORKED EXAMPLE 14 Microwaves (exam level)

Microwaves of wavelength 3.0 cm pass through two slits. A detector at a point where the path difference is 7.5 cm records a minimum. Explain.

Solution

7.5 cm = 2.5λM1

Waves arrive in antiphase: destructive interferenceR1

Try this

Connect two loudspeakers to the same signal generator (about 1 kHz) and walk slowly across the room in front of them. Listen for loud and quiet positions.

2.6 Applications of interference

Noise-cancelling headphones produce sound in antiphase with incoming noise. Anti-reflection coatings on lenses use thin-film interference to cancel reflected light. Interferometers measure tiny distances (gravitational-wave detectors detect changes smaller than a proton).

Quick check

1. How do noise-cancelling headphones work?

Answers: 1. they produce sound in antiphase so the waves interfere destructively

2.7 Young's double-slit experiment

Light from a single source passes through two narrow slits a distance d apart, which act as coherent sources. Bright and dark fringes form on a screen a distance D away, with fringe spacing s = λD/d. The experiment provides evidence for the wave nature of light.

Figure 14. Young's double-slit arrangement (not to scale).

Figure 15. Intensity pattern of double-slit fringes (ignoring single-slit effects).

WORKED EXAMPLE 15 Wavelength (standard)

Slits 0.50 mm apart produce fringes 2.6 mm apart on a screen 2.0 m away. Find λ.

Solution

λ = sd/D = 2.6 × 10⁻³ × 0.50 × 10⁻³/2.0M1

= 650 nmA1

WORKED EXAMPLE 16 Changing variables (standard)

Describe the effect on the fringes of (a) moving the screen further away; (b) using blue light.

Solution

(a) s increases (s ∝ D)A1

(b) s decreases (shorter λ)A1

2.8 Phase difference and path difference

Phase difference φ = 2π × (path difference)/λ. A path difference of λ/2 gives φ = π (antiphase).

Quick check

1. Find the phase difference for a path difference of 0.75λ.

Answers: 1. 1.5π rad

3 Standing waves

Waves that stay put.

Think about it

Why does a guitar string produce a definite note, and why does pressing a fret raise the pitch?

3.1 Formation of standing waves

A standing (stationary) wave forms when two identical waves travelling in opposite directions superpose, usually a wave and its reflection. It has nodes (zero amplitude) and antinodes (maximum amplitude). Adjacent nodes are λ/2 apart. Energy is not transferred along a standing wave.

Feature	Travelling wave	Standing wave
energy transfer	yes	no
amplitude	same at all points	varies from zero (node) to maximum (antinode)
phase	varies continuously along the wave	same between adjacent nodes; antiphase across a node

Figure 16. A standing wave at several instants. N = node, A = antinode.

Quick check

1. What is the distance between a node and the next antinode?

Answers: 1. λ/4

3.2 Standing waves on strings

A string fixed at both ends has nodes at each end. The fundamental (first harmonic) has λ = 2L and f₁ = v/2L. The nth harmonic has λ = 2L/n and fn = nf₁.

Figure 17. The first three harmonics of a string fixed at both ends.

WORKED EXAMPLE 17 Guitar (standard)

A 0.65 m guitar string has wave speed 286 m s⁻¹. Find the fundamental frequency and the third harmonic.

Solution

f₁ = 286/(2 × 0.65) = 220 HzM1 A1

f₃ = 660 HzA1

WORKED EXAMPLE 18 Fret (exam level)

Pressing a fret shortens the vibrating length to 0.49 m. Find the new fundamental.

Solution

f = 286/(2 × 0.49)M1

= 292 HzA1

3.3 Phase in a standing wave

All points between two adjacent nodes oscillate in phase but with different amplitudes. Points on opposite sides of a node are in antiphase. This contrasts with a travelling wave, where phase changes continuously along the wave.

Quick check

1. Two points on a standing wave are 0.3λ apart on either side of a node. What is their phase difference?

Answers: 1. π (antiphase)

3.4 Investigating standing waves on a string

A string is driven by a vibration generator and passes over a pulley to a hanging mass (fixed tension). Increasing the frequency produces the harmonics in turn. A graph of frequency against harmonic number is a straight line through the origin with gradient f₁ = v/2L.

Figure 18. Resonant frequency against harmonic number for a 1.2 m string.

WORKED EXAMPLE 19 Wave speed (exam level)

The gradient is 30 Hz and the string is 1.2 m long. Find the wave speed.

Solution

f₁ = v/2L: v = 2 × 1.2 × 30M1

= 72 m s⁻¹A1

3.5 Standing waves in pipes

In air columns, closed ends are displacement nodes and open ends are displacement antinodes. Pipe open at both ends: λ = 2L/n, all harmonics. Pipe closed at one end: λ = 4L/n with n odd only.

Figure 19. Displacement patterns in open and closed pipes.

WORKED EXAMPLE 20 Closed pipe (exam level)

Find the fundamental frequency of a 0.34 m pipe closed at one end (v = 340 m s⁻¹) and the next possible frequency.

Solution

λ = 4L = 1.36 m; f₁ = 250 HzM1 A1

Next: 3f₁ = 750 HzA1

WORKED EXAMPLE 21 Open pipe (standard)

Find the first two frequencies of a 0.50 m pipe open at both ends.

Solution

f₁ = 340/1.0 = 340 HzA1

f₂ = 680 HzA1

3.6 Musical instruments

Stringed instruments use standing waves on strings; wind instruments use standing waves in air columns. The note's pitch is set by the fundamental; the mixture of harmonics gives each instrument its characteristic tone (timbre).

Instrument	Standing wave in	Pitch changed by
guitar, violin	string fixed at both ends	length (frets, fingers), tension
flute	pipe open at both ends	opening holes (effective length)
clarinet	pipe closed at one end	keys (effective length); odd harmonics dominate
drum	membrane	tension and size

Quick check

1. Why does a clarinet sound different from a flute playing the same note?

Answers: 1. different mix of harmonics (closed pipe has mainly odd harmonics)

3.7 Measuring the speed of sound with a resonance tube

A vibrating tuning fork is held over a tube whose water level is lowered. Loud resonance occurs first when L₁ = λ/4 and again when L₂ = 3λ/4. Then L₂ − L₁ = λ/2, which avoids the end-correction problem.

Figure 20. Resonance tube: first resonance at L₁ ≈ λ/4.

WORKED EXAMPLE 22 Resonance tube (exam level)

With a 512 Hz fork, resonances occur at 0.16 m and 0.49 m. Find the speed of sound.

Solution

λ/2 = 0.33 m: λ = 0.66 mM1

v = 512 × 0.66 = 338 m s⁻¹A1

3.8 Comparing travelling and standing waves in practice

On a guitar string, waves travel to each end, reflect and superpose. Only frequencies that fit a whole number of half-wavelengths into the string length survive as standing waves; all other frequencies die out quickly. The sound wave that reaches your ear, by contrast, is a travelling wave carrying energy away.

WORKED EXAMPLE 23 Allowed frequencies (exam level)

A 0.80 m string has wave speed 240 m s⁻¹. Which of 150 Hz, 300 Hz, 400 Hz and 450 Hz can form standing waves?

Solution

f₁ = 240/1.6 = 150 Hz; harmonics are multiples of 150 HzM1

150, 300 and 450 Hz (not 400 Hz)A1

3.9 Resonance and standing waves

A system that supports standing waves has natural frequencies (its harmonics). Driving it at one of these frequencies produces resonance: large-amplitude standing waves. This is how musical instruments, microwave ovens and lasers work.

In the real world

Opera singers can shatter a wine glass by singing at its natural frequency, building up a large-amplitude standing wave in the glass.

4 The Doppler effect

Moving sources and observers.

4.1 Qualitative Doppler effect

When a source moves towards an observer, wavefronts bunch up: the observed wavelength is shorter and frequency higher. When it moves away, wavelength is longer and frequency lower. The same happens if the observer moves.

Figure 21. Wavefronts from a moving source.

WORKED EXAMPLE 24 Siren (standard)

Describe what a pedestrian hears as an ambulance passes.

Solution

Higher pitch as it approachesR1

Lower pitch as it moves awayR1

4.2 Doppler effect for light

For light from sources moving at v ≪ c: Δf/f ≈ Δλ/λ ≈ v/c. Galaxies moving away show redshift; approaching sources show blueshift.

WORKED EXAMPLE 25 Galaxy (exam level)

A spectral line at 656.3 nm is observed at 662.9 nm in a galaxy's spectrum. Find the galaxy's recession speed.

Solution

Δλ/λ = 6.6/656.3 = 0.0101M1

v = 3.0 × 10⁶ m s⁻¹ awayA1

WORKED EXAMPLE 26 Speed trap (exam level)

A police radar at 24 GHz reflects from a car moving towards it at 30 m s⁻¹. The frequency shift is about 2v/c × f. Find the shift.

Solution

Δf = 2 × 30/3.0 × 10⁸ × 2.4 × 10¹⁰M1

= 4.8 kHzA1

In the real world

Weather radar uses Doppler shifts of microwaves reflected from raindrops to measure wind speeds inside storms.

4.3 Redshift and the expanding universe

Light from distant galaxies is redshifted, showing they are receding. Hubble found recession speed is proportional to distance (v = H₀d), evidence that the universe is expanding.

Figure 22. Recession speed against distance for galaxies (H₀ ≈ 70 km s⁻¹ Mpc⁻¹).

WORKED EXAMPLE 27 Hubble (exam level)

A galaxy is 200 Mpc away. Estimate its recession speed and the fractional shift in its spectral lines.

Solution

v = 70 × 200 = 14 000 km s⁻¹M1

Δλ/λ = v/c = 0.047A1

4.4 Doppler ultrasound

Ultrasound reflected from moving blood cells is Doppler-shifted. Measuring the shift gives the blood speed, used to detect blockages and monitor unborn babies' heartbeats.

Quick check

1. Why does reflection from cells moving towards the probe raise the frequency?

Answers: 1. the moving cells act as moving observers and moving sources, compressing the waves

5 Exam technique

Common question types.

5.1 Method checklist

•Draw the normal and measure all angles from it.
•For TIR, check that light is going from higher to lower n.
•For interference, find path difference in wavelengths.
•For harmonics, sketch the node–antinode pattern before choosing λ.
•For Doppler, decide first whether f should go up or down.

5.2 Multi-step problem: pipes and strings

WORKED EXAMPLE 28 Matching notes (exam level)

A string 0.40 m long resonates in its fundamental with a pipe closed at one end of length 0.25 m (vair = 340 m s⁻¹). Find the frequency and the wave speed on the string.

Solution

Pipe: f = 340/(4 × 0.25) = 340 HzM1 A1

String: v = 2Lf = 2 × 0.40 × 340 = 272 m s⁻¹A1

5.3 Multi-step problem: double slit and grating

WORKED EXAMPLE 29 Comparing patterns (exam level)

The same laser (633 nm) is used with double slits 0.30 mm apart (screen 2.0 m) and with a grating of 600 lines per mm. Find the fringe spacing and the first-order grating angle.

Solution

s = 633 × 10⁻⁹ × 2.0/3.0 × 10⁻⁴ = 4.2 mmA1

d = 1.67 μm; sin θ = 0.380: θ = 22.3°M1 A1

6 Data-based questions and further practice

Applying wave phenomena.

6.1 Experimental data: double slit

A student varies D and measures the fringe spacing with d = 0.40 mm.

D / m	1.0	1.5	2.0	2.5	3.0
s / mm	1.3	2.0	2.6	3.3	3.9

Figure 23. Fringe spacing against slit–screen distance.

WORKED EXAMPLE 30 Wavelength from gradient (exam level)

Find λ from the gradient.

Solution

Gradient = 1.3 × 10⁻³ = λ/dM1

λ = 1.3 × 10⁻³ × 4.0 × 10⁻⁴ = 520 nmA1

6.2 Graded practice

WORKED EXAMPLE 31 Pool depth (exam level)

A pool is 2.0 m deep. Viewed from directly above, how deep does it appear (n = 1.33)?

Solution

Apparent depth = real depth/nM1

= 1.5 mA1

WORKED EXAMPLE 32 Violin (standard)

A violin string has fundamental 440 Hz. The tension is increased so the wave speed rises by 2%. Find the new fundamental.

Solution

f ∝ vM1

= 449 HzA1

WORKED EXAMPLE 33 Two-source water waves (exam level)

Two dippers 6.0 cm apart vibrate in phase producing 1.5 cm waves. How many lines of destructive interference cross the line joining them?

Solution

Path differences from −6.0 to +6.0 cm: odd multiples of 0.75 cmM1

±0.75, 2.25, 3.75, 5.25: 8 linesA1

6.3 Graded practice: refraction and interference

WORKED EXAMPLE 34 Critical angle from data (standard)

Light just escapes a liquid at 48° incidence. Find the liquid's refractive index.

Solution

n = 1/sin 48°M1

= 1.35A1

WORKED EXAMPLE 35 Fringe count (exam level)

Double slits 0.20 mm apart illuminated with 550 nm light give fringes on a screen 1.5 m away. How many bright fringes fit in 2.0 cm?

Solution

s = 550 × 10⁻⁹ × 1.5/2.0 × 10⁻⁴ = 4.1 mmM1

2.0/0.41 ≈ 4–5 fringesA1

WORKED EXAMPLE 36 Speed in glass (standard)

Find the wavelength in glass (n = 1.52) of light that has wavelength 589 nm in air.

Solution

λ = 589/1.52M1

= 388 nmA1

6.4 Graded practice: standing waves and Doppler

WORKED EXAMPLE 37 Node spacing (standard)

Adjacent nodes on a string are 0.25 m apart at 120 Hz. Find the wave speed.

Solution

λ = 0.50 mM1

v = 60 m s⁻¹A1

WORKED EXAMPLE 38 Pipe length (exam level)

A closed pipe resonates at 200 Hz and next at 600 Hz (v = 340 m s⁻¹). Find its length.

Solution

200 Hz is the fundamental: L = λ/4 = 1.70/4M1

L = 0.43 mA1

WORKED EXAMPLE 39 Star speed (standard)

A line at 500.0 nm is observed at 499.5 nm. Is the star approaching or receding, and how fast?

Solution

Blueshift: approachingA1

v = (0.5/500) × 3.0 × 10⁸ = 3.0 × 10⁵ m s⁻¹A1

6.5 Graded practice: diffraction and resolution

WORKED EXAMPLE 40 Doorway (standard)

Estimate the wavelength of a 300 Hz sound and explain whether it diffracts noticeably through a 0.9 m doorway.

Solution

λ = 340/300 = 1.1 mA1

Comparable to doorway width: strong diffractionR1

WORKED EXAMPLE 41 Harbour (exam level)

Ocean waves of wavelength 20 m reach a harbour entrance 25 m wide. Describe what happens inside the harbour.

Solution

Gap ≈ λM1

Waves spread out considerably inside the harbour (circular wavefronts)R1

6.6 Graded practice: mixed

WORKED EXAMPLE 42 Fibre time delay (exam level)

Light travels 2.0 km along the axis of a fibre with core n = 1.50. Find the travel time.

Solution

v = 2.0 × 10⁸ m s⁻¹M1

t = 2000/2.0 × 10⁸ = 10 μsA1

WORKED EXAMPLE 43 Speaker spacing (exam level)

Two in-phase speakers 1.2 m apart emit 850 Hz. A listener on the centre line moves sideways until the first quiet point. What is the path difference there?

Solution

λ = 340/850 = 0.40 mM1

First minimum: path difference = λ/2 = 0.20 mA1

WORKED EXAMPLE 44 Bat (exam level)

A bat flying towards a wall emits 40 kHz pulses. State and explain whether the echo it hears is higher or lower than 40 kHz.

Solution

HigherA1

The bat is a moving source (wall receives a higher f) and then a moving observer of the reflection (shifted up again)R1

6.7 Self-test

Try these without notes, then check the answers.

Quick check

1. State Snell's law.

2. Find the critical angle for n = 1.33.

3. What two conditions define coherent sources?

4. Write the harmonic frequencies of a closed pipe in terms of f₁.

5. How far apart are adjacent nodes?

6. What happens to the observed frequency as a source moves away?

Answers: 1. n₁ sin θ₁ = n₂ sin θ₂ 2. 48.8° 3. same frequency and constant phase difference 4. f₁, 3f₁, 5f₁, … 5. λ/2 6. it decreases

6.8 Key terms

Term	Meaning
refractive index	n = c/v
critical angle	angle of incidence giving refraction at 90°: sin c = n₂/n₁
diffraction	spreading of waves through gaps or around obstacles
coherent sources	same frequency, constant phase difference
path difference	difference in distances travelled from two sources
node / antinode	points of zero / maximum amplitude in a standing wave
harmonic	standing-wave mode with frequency nf₁
Doppler effect	change in observed frequency due to relative motion

6.9 Common exam errors

•Measuring angles from the surface rather than the normal.
•Saying frequency changes on refraction.
•Using λ = 2L for all harmonics of a string.
•Forgetting that closed pipes have only odd harmonics.
•Stating coherence as 'same frequency' only (constant phase difference is also needed).

7 Common misconceptions and links

Students often think…	Actually…
frequency changes on refraction	frequency is constant; speed and wavelength change
diffraction changes wavelength	wavelength is unchanged
nodes are points of maximum vibration	nodes do not vibrate
standing waves transfer energy along the medium	they do not
Doppler shift depends on the distance to the source	it depends on relative velocity

How C.3 connects to the rest of the course

Topic	Connection
C.2 Wave model	v = fλ and wave graphs
C.1 SHM	resonance
E.2 Quantum physics (HL)	electron diffraction and wave–particle duality
E.5 Fusion and stars	redshift and stellar spectra

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Monochromatic light passes through two slits 0.25 mm apart. Bright fringes are observed on a screen 1.8 m away with a spacing of 4.3 mm.

(a) State the conditions for observing interference fringes. [2]

(b) Calculate the wavelength of the light. [2]

(d) A string of length 0.80 m vibrates in its third harmonic at 360 Hz. Calculate the wave speed. [2]

(e) Light from a star shows a line at 486.1 nm shifted to 486.6 nm. Calculate the star's radial speed. [2]

Markscheme

Part	Answer	Marks
(a)	coherent sources (constant phase difference, same frequency)	A1
	similar amplitude / slit separation small	A1
(b)	λ = sd/D = 4.3 × 10⁻³ × 0.25 × 10⁻³/1.8	M1
	6.0 × 10⁻⁷ m	A1
(c)	fringes closer together	A1
	shorter wavelength; s ∝ λ	R1
(d)	λ = 2L/3 = 0.533 m	M1
	v = 192 m s⁻¹	A1
(e)	v = (0.5/486.1) × 3.0 × 10⁸	M1
	3.1 × 10⁵ m s⁻¹ away	A1

Examiner's comments. In (a), 'same frequency' alone is not enough: coherence requires a constant phase difference. In (d), many used λ = 2L for the third harmonic.

9 Check your understanding

Section A: multiple choice

1. When light passes from air into glass its

A. speed decreases and frequency stays the same B. frequency increases C. wavelength increases D. speed increases

2. Total internal reflection requires light to travel

A. into a denser medium B. at normal incidence C. into a medium of lower refractive index D. at less than the critical angle

3. Coherent sources have

A. the same amplitude only B. constant phase difference and the same frequency C. different frequencies D. random phase

4. Fringe spacing in Young's experiment increases if

A. d increases B. λ decreases C. the slits are wider D. D increases

5. The fundamental of a string fixed at both ends has wavelength

A. L B. 2L C. 4L D. L/2

6. A pipe closed at one end produces

A. odd harmonics only B. all harmonics C. even harmonics only D. no harmonics

7. A receding galaxy shows

A. blueshift B. no shift C. higher frequencies D. redshift

8. Destructive interference occurs for a path difference of

A. λ B. 2λ C. 1.5λ D. 0

Section B: structured

9. Light enters water (n = 1.33) at 50° from air. Calculate the angle of refraction and the speed in water. [3]

10. Explain how a standing wave forms on a string. [3]

11. A pipe open at both ends is 0.50 m long. Calculate its first two harmonic frequencies (v = 340 m s⁻¹). [2]

12. Explain why the sound of a passing car changes pitch. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. sin r = sin 50°/1.33: r = 35.2°; v = 2.26 × 10⁸ m s⁻¹.

10. Wave reflects at fixed end; incident and reflected waves superpose; nodes and antinodes at fixed positions at certain frequencies.

11. 340 Hz and 680 Hz.

12. Approaching: wavefronts compressed, higher f; receding: stretched, lower f.

10 Summary and topic map

Idea	Remember
Reflection	i = r; phase change at fixed end
Refraction	n₁ sin θ₁ = n₂ sin θ₂; sin c = n₂/n₁
Diffraction	significant when gap ≈ λ
Interference	path diff nλ or (n + ½)λ; s = λD/d
Standing waves	nodes λ/2 apart; harmonics
Doppler	Δλ/λ ≈ v/c

Before the exam, make sure you…

•measure angles from the normal
•check coherence conditions
•draw node/antinode patterns for strings and pipes
•identify approaching vs receding for Doppler

Wave Phenomena on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 24. Topic map for Wave Phenomena.

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