Physics - Waves and Optics

Nature of Waves

A wave is a disturbance that transfers energy from one point to another without transferring matter. Waves can be classified by the direction of particle motion relative to the direction of wave propagation.

Transverse Waves

In a transverse wave, the particle displacement is perpendicular to the direction of wave propagation.

Examples: light waves, water waves (surface), waves on a string, electromagnetic waves.

Longitudinal Waves

In a longitudinal wave, the particle displacement is parallel to the direction of wave propagation.

Examples: sound waves, compression waves in a spring.

Wave Terminology

Term	Symbol	SI Unit	Definition
Wavelength	$\lambda$	m	Distance between two consecutive points in phase
Frequency	$f$	Hz	Number of complete oscillations per second
Period	$T$	s	Time for one complete oscillation
Amplitude	$A$	m	Maximum displacement from equilibrium
Wave speed	$v$	m/s	Speed at which the wave propagates
Wavefront	—	—	Line joining all points in phase

The Wave Equation

$v = f\lambda$

This is a fundamental relationship connecting wave speed, frequency, and wavelength.

Since $f = \frac{1}{T}$, we can also write:

$v = \frac{\lambda}{T}$

info

The wave equation applies to all types of waves: transverse, longitudinal, mechanical, and electromagnetic.

Worked Example 1

A sound wave has a frequency of $440 \mathrm{ Hz}$ and a wavelength of $0.78 \mathrm{ m}$. Find the speed of sound.

Solution

$v = f\lambda = 440 \times 0.78 = 343.2 \mathrm{ m/s}$

Worked Example 1b

A light wave has a frequency of $5.0 \times 10^{14} \mathrm{ Hz}$. Find its wavelength in air.

Solution

$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{5.0 \times 10^{14}} = 6.0 \times 10^{-7} \mathrm{ m} = 600 \mathrm{ nm}$

Phase and Phase Difference

Two points on a wave are in phase if they have the same displacement and are moving in the same direction. Points separated by a whole number of wavelengths are in phase.

Phase difference is the fraction of a cycle by which one wave leads or lags another:

$\mathrm{Phase difference} = \frac{\Delta x}{\lambda} \times 360^\circ = \frac{\Delta x}{\lambda} \times 2\pi \mathrm{ rad}$

Points that are half a wavelength apart are in antiphase (phase difference $= 180^\circ$ or $\pi \mathrm{ rad}$).

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Wave Properties

Reflection

Waves on a String

Adjust frequency, amplitude, and damping to explore wave behaviour on a virtual string.

Reflection occurs when a wave bounces off a boundary. The law of reflection states:

• The angle of incidence equals the angle of reflection: $\theta_i = \theta_r$
• The incident ray, reflected ray, and normal all lie in the same plane

Refraction

Refraction is the change in direction of a wave as it passes from one medium to another with a different wave speed.

Snell's Law:

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

Where $n$ is the refractive index of the medium and $\theta$ is the angle measured from the normal.

The refractive index is defined as:

$n = \frac{c}{v}$

Where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.

info

info towards the normal. When entering a less dense medium, it speeds up and bends away from the normal.

Total Internal Reflection

Total internal reflection occurs when:

1. Light travels from a denser medium to a less dense medium ($n_1 \gt n_2$)
2. The angle of incidence exceeds the critical angle $\theta_c$

The critical angle is found when the refracted ray travels along the boundary ($\theta_2 = 90^\circ$):

$\sin\theta_c = \frac{n_2}{n_1}$

Worked Example 2

Light travels from glass ($n = 1.5$) to air ($n = 1.0$). Find the critical angle.

Solution

$\sin\theta_c = \frac{1.0}{1.5} = 0.667$

$\theta_c = \sin^{-1}(0.667) = 41.8^\circ$

Diffraction

Diffraction is the spreading of waves as they pass through a gap or around an obstacle. The amount of diffraction depends on the ratio of the wavelength to the gap width.

• Maximum diffraction occurs when the gap width is approximately equal to the wavelength
• When the gap is much wider than the wavelength, diffraction is negligible

Interference

Interference occurs when two or more waves superpose (overlap). The principle of superposition states that the resultant displacement at any point is the sum of the individual displacements.

Constructive interference occurs when waves meet in phase:

$\mathrm{Path difference} = n\lambda \quad (n = 0, 1, 2, \ldots)$

Destructive interference occurs when waves meet in antiphase:

$\mathrm{Path difference} = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, \ldots)$

Worked Example 3

Two coherent sources are $0.5 \mathrm{ mm}$ apart and produce an interference pattern on a screen $1.2 \mathrm{ m}$ away. The fringe spacing is $1.2 \mathrm{ mm}$. Find the wavelength of the light.

Solution

Using the double-slit formula:

$\lambda = \frac{ay}{D} = \frac{0.5 \times 10^{-3} \times 1.2 \times 10^{-3}}{1.2} = 5 \times 10^{-7} \mathrm{ m} = 500 \mathrm{ nm}$

Worked Example 3b

In a double-slit experiment, light of wavelength $550 \mathrm{ nm}$ passes through slits $0.4 \mathrm{ mm}$ apart onto a screen $1.5 \mathrm{ m}$ away. Find the separation between the central bright fringe and the first dark fringe.

Solution

The first dark fringe occurs at path difference $= \lambda/2$:

$y = \frac{(n + 0.5)\lambda D}{a} = \frac{0.5 \times 550 \times 10^{-9} \times 1.5}{0.4 \times 10^{-3}} = \frac{4.125 \times 10^{-7}}{4 \times 10^{-4}} = 1.03 \times 10^{-3} \mathrm{ m} = 1.03 \mathrm{ mm}$

If you get this wrong, revise: Double-slit interference — dark fringes occur at half-integer multiples of $\lambda$.

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Stationary Waves

A stationary (standing) wave is formed when two waves of the same frequency and amplitude travel in opposite directions and superpose.

Properties of Stationary Waves

• Nodes: points of zero amplitude (always at rest)
• Antinodes: points of maximum amplitude
• The distance between adjacent nodes $= \frac{\lambda}{2}$
• The distance between a node and the adjacent antinode $= \frac{\lambda}{4}$
• All points between nodes are in phase
• Points in adjacent loops are in antiphase

Stationary Waves on a String

For a string of length $L$ fixed at both ends:

Mode	Frequency	Wavelength	Description
Fundamental (1st harmonic)	$f_1 = \frac{v}{2L}$	$\lambda_1 = 2L$	One antinode
2nd harmonic	$f_2 = 2f_1$	$\lambda_2 = L$	Two antinodes
3rd harmonic	$f_3 = 3f_1$	$\lambda_3 = \frac{2L}{3}$	Three antinodes
$n$th harmonic	$f_n = nf_1$	$\lambda_n = \frac{2L}{n}$	$n$ antinodes

Worked Example 4

A string of length $0.8 \mathrm{ m}$ has a fundamental frequency of $220 \mathrm{ Hz}$. Find the wave speed and the frequency of the third harmonic.

Solution

$v = f_1 \times \lambda_1 = 220 \times 2(0.8) = 352 \mathrm{ m/s}$

$f_3 = 3f_1 = 3 \times 220 = 660 \mathrm{ Hz}$

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Sound Waves

Properties of Sound

Sound is a longitudinal mechanical wave that requires a medium to propagate. It cannot travel through a vacuum.

Property	Description
Speed in air at $20^\circ\mathrm{C}$	Approximately $343 \mathrm{ m/s}$
Speed in water	Approximately $1480 \mathrm{ m/s}$
Speed in steel	Approximately $5960 \mathrm{ m/s}$
Audible range	$20 \mathrm{ Hz}$ to $20,000 \mathrm{ Hz}$
Infrasonic	Below $20 \mathrm{ Hz}$
Ultrasonic	Above $20,000 \mathrm{ Hz}$

Intensity and Loudness

The intensity of a sound wave is the power per unit area:

$I = \frac{P}{A}$

Intensity is proportional to the square of the amplitude:

$I \propto A^2$

The loudness level is measured in decibels (dB):

$\beta = 10\log_{10}\left(\frac{I}{I_0}\right)$

Where $I_0 = 10^{-12} \mathrm{ W/m}^2$ is the threshold of hearing.

warning

warning physical quantity. A $10 \mathrm{ dB}$ increase corresponds to a $10\times$ increase in intensity, but is perceived as roughly a doubling of loudness.

Worked Example 5

The intensity of one sound is $10^{-4} \mathrm{ W/m}^2$ and another is $10^{-2} \mathrm{ W/m}^2$. Find the difference in loudness level.

Solution

$\Delta\beta = 10\log_{10}\left(\frac{10^{-2}}{10^{-4}}\right) = 10\log_{10}(100) = 10 \times 2 = 20 \mathrm{ dB}$

Ultrasound

Ultrasound has frequencies above $20,000 \mathrm{ Hz}$. Applications include:

• Medical imaging (sonography)
• Non-destructive testing of materials
• Cleaning of precision instruments
• Depth sounding (echo sounding)

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The Electromagnetic Spectrum

Electromagnetic waves are transverse waves that can travel through a vacuum at the speed of light $c = 3 \times 10^8 \mathrm{ m/s}$.

EM Spectrum Summary

Type	Wavelength Range	Frequency Range	Source
Radio waves	$\gt 0.1 \mathrm{ m}$	$\lt 3 \times 10^9 \mathrm{ Hz}$	Oscillating circuits
Microwaves	$0.1 \mathrm{ mm}$ to $0.1 \mathrm{ m}$	$3 \times 10^9$ to $3 \times 10^{11} \mathrm{ Hz}$	Magnetron
Infrared	$700 \mathrm{ nm}$ to $1 \mathrm{ mm}$	$3 \times 10^{11}$ to $4 \times 10^{14} \mathrm{ Hz}$	Hot objects
Visible light	$400 \mathrm{ nm}$ to $700 \mathrm{ nm}$	$4 \times 10^{14}$ to $7.5 \times 10^{14} \mathrm{ Hz}$	Luminous objects
Ultraviolet	$10 \mathrm{ nm}$ to $400 \mathrm{ nm}$	$7.5 \times 10^{14}$ to $3 \times 10^{16} \mathrm{ Hz}$	Hot objects, discharge tubes
X-rays	$0.01 \mathrm{ nm}$ to $10 \mathrm{ nm}$	$3 \times 10^{16}$ to $3 \times 10^{19} \mathrm{ Hz}$	Electron bombardment
Gamma rays	$\lt 0.01 \mathrm{ nm}$	$\gt 3 \times 10^{19} \mathrm{ Hz}$	Radioactive decay

All EM waves:

• Travel at $c = 3 \times 10^8 \mathrm{ m/s}$ in vacuum
• Are transverse waves
• Can be polarised
• Can travel through a vacuum
• Obey the wave equation $c = f\lambda$
• Can be reflected, refracted, and diffracted

tip

For the DSE exam, remember the order of the EM spectrum from longest to shortest wavelength: Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays (RMIVUXG).

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Light

Reflection of Light

Plane mirrors produce virtual images that are:

• The same size as the object
• The same distance behind the mirror as the object is in front
• Laterally inverted (left-right reversed)
• Upright

Refraction of Light

When light passes from one medium to another, it changes speed and direction. The refractive index:

$n = \frac{\sin i}{\sin r} = \frac{c}{v}$

Dispersion

White light is composed of a continuous spectrum of colours. Dispersion occurs because the refractive index of a medium varies with wavelength (shorter wavelengths refract more):

$n_{\mathrm{violet}} \gt n_{\mathrm{blue}} \gt n_{\mathrm{green}} \gt n_{\mathrm{yellow}} \gt n_{\mathrm{orange}} \gt n_{\mathrm{red}}$

A prism separates white light into its constituent colours because each colour has a slightly different refractive index in the glass.

Lenses

Convex (converging) lens: Thicker in the centre, converges parallel rays to the focal point.

Concave (diverging) lens: Thinner in the centre, diverges parallel rays as if from the focal point.

Thin Lens Formula

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Where:

• $f$ = focal length (positive for convex, negative for concave)
• $v$ = image distance (positive for real image, negative for virtual)
• $u$ = object distance (always negative in the real-is-positive convention)

Magnification:

$m = \frac{v}{u} = \frac{h_i}{h_o}$

Where $h_i$ is the image height and $h_o$ is the object height.

Ray Diagram Rules for Convex Lenses

1. A ray parallel to the principal axis refracts through the focal point
2. A ray through the centre of the lens passes straight through undeviated
3. A ray through the focal point refracts parallel to the principal axis

Worked Example 6

An object is placed $30 \mathrm{ cm}$ from a convex lens of focal length $20 \mathrm{ cm}$. Find the image position and magnification.

Solution

$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$

$\frac{1}{v} + \frac{1}{30} = \frac{1}{20}$

$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$

$v = 60 \mathrm{ cm}$

$m = \frac{v}{u} = \frac{60}{-30} = -2$

The image is real, inverted, and twice the size of the object, located $60 \mathrm{ cm}$ on the other side of the lens.

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Optical Fibres

Optical fibres use total internal reflection to transmit light signals over long distances with minimal loss.

Structure

An optical fibre consists of:

• A core with higher refractive index
• A cladding with lower refractive index

Light entering the fibre at an angle less than the acceptance angle undergoes repeated total internal reflection and travels along the fibre.

Applications

• Telecommunications (high-speed data transmission)
• Medical endoscopy
• Decorative lighting

warning

warning acceptance angle. The acceptance angle is the maximum angle at which light can enter the fibre and still undergo total internal reflection.

Worked Example 7

An optical fibre has a core of refractive index $1.50$ and cladding of refractive index $1.45$. Find the critical angle at the core-cladding boundary.

Solution

$\sin\theta_c = \frac{n_{\mathrm{cladding}}}{n_{\mathrm{core}}} = \frac{1.45}{1.50} = 0.967$

$\theta_c = \sin^{-1}(0.967) = 75.2^\circ$

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Polarisation

Polarisation is the process of restricting the vibrations of a transverse wave to one plane.

• Only transverse waves can be polarised
• Sound waves (longitudinal) cannot be polarised
• Polaroid filters allow only vibrations in one plane to pass through
• When two Polaroid filters are perpendicular (crossed), no light passes through

Malus's Law:

$I = I_0 \cos^2\theta$

Where $I_0$ is the intensity of polarised light incident on the analyser and $\theta$ is the angle between the polariser and analyser.

Worked Example 8

Unpolarised light of intensity $200 \mathrm{ W/m}^2$ passes through two Polaroid filters. The second filter's transmission axis is at $60^\circ$ to the first. Find the intensity after the second filter.

Solution

After the first filter (intensity halved): $I_1 = 100 \mathrm{ W/m}^2$

After the second filter: $I_2 = I_1\cos^2 60^\circ = 100 \times 0.25 = 25 \mathrm{ W/m}^2$

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Summary Table

Topic	Key Formula	Key Concept
Wave equation	$v = f\lambda$	Relates speed, frequency, wavelength
Snell's law	$n_1\sin\theta_1 = n_2\sin\theta_2$	Law of refraction
Critical angle	$\sin\theta_c = \frac{n_2}{n_1}$	Total internal reflection threshold
Double-slit	$\lambda = \frac{ay}{D}$	Interference pattern
Stationary waves	$\lambda_n = \frac{2L}{n}$	Harmonics on a string
Decibels	$\beta = 10\log_{10}\left(\frac{I}{I_0}\right)$	Loudness level
Thin lens	$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$	Image formation
Malus's law	$I = I_0\cos^2\theta$	Polarisation

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Exam Tips

• Remember that only transverse waves can be polarised; use this fact to distinguish wave types.
• In refraction problems, always measure angles from the normal, not the surface.
• For double-slit interference, the path difference determines constructive ($n\lambda$) or destructive ($(n + 0.5)\lambda$) interference.
• When drawing ray diagrams, always use a ruler and label the focal points.
• In stationary wave problems, the distance between adjacent nodes is $\lambda/2$, not $\lambda$.
• For the EM spectrum, know the approximate wavelength ranges and one application for each type.

Exam-Style Practice Questions

Question 1: Light of wavelength $600 \mathrm{ nm}$ is incident on a pair of slits separated by $0.2 \mathrm{ mm}$. The screen is $1.5 \mathrm{ m}$ away. Find the distance between the central maximum and the third bright fringe.

Solution

$y = \frac{n\lambda D}{a} = \frac{3 \times 600 \times 10^{-9} \times 1.5}{0.2 \times 10^{-3}} = \frac{2.7 \times 10^{-6}}{2 \times 10^{-4}} = 0.0135 \mathrm{ m} = 13.5 \mathrm{ mm}$

Question 2: A concave lens has a focal length of $15 \mathrm{ cm}$. An object is placed $25 \mathrm{ cm}$ from the lens. Find the image position.

Solution

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{-15}$

$\frac{1}{v} + \frac{1}{25} = -\frac{1}{15}$

$\frac{1}{v} = -\frac{1}{15} - \frac{1}{25} = \frac{-5 - 3}{75} = -\frac{8}{75}$

$v = -9.375 \mathrm{ cm}$

The image is virtual, upright, and $9.375 \mathrm{ cm}$ from the lens on the same side as the object.

Question 3: The critical angle for a glass-air boundary is $42^\circ$. Find the refractive index of the glass.

Solution

$n = \frac{1}{\sin 42^\circ} = \frac{1}{0.669} = 1.49$

Question 4: Unpolarised light of intensity $300 \mathrm{ W/m}^2$ passes through a Polaroid filter. What is the transmitted intensity?

Solution

The first Polaroid reduces the intensity by half: $I = 150 \mathrm{ W/m}^2$.

Question 5: A stationary wave is set up on a string of length $1.2 \mathrm{ m}$. The speed of the waves on the string is $24 \mathrm{ m/s}$. Find the frequency of the second harmonic.

Solution

For the second harmonic: $\lambda_2 = L = 1.2 \mathrm{ m}$

$f_2 = \frac{v}{\lambda_2} = \frac{24}{1.2} = 20 \mathrm{ Hz}$

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Doppler Effect

Definition

The Doppler effect is the change in the observed frequency (or wavelength) of a wave when there is relative motion between the source and the observer.

General Formula

$f' = f \left(\frac{v \pm v_o}{v \mp v_s}\right)$

Where:

• $f'$ = observed frequency
• $f$ = source frequency
• $v$ = speed of the wave in the medium
• $v_o$ = speed of the observer (positive if moving towards the source)
• $v_s$ = speed of the source (positive if moving towards the observer)

Doppler Effect for Sound

When the source moves towards a stationary observer:

$f' = f \left(\frac{v}{v - v_s}\right)$

The observed frequency is higher than the source frequency.

When the source moves away from a stationary observer:

$f' = f \left(\frac{v}{v + v_s}\right)$

The observed frequency is lower than the source frequency.

Applications

• Radar speed guns: Police use Doppler radar to measure vehicle speed
• Medical ultrasound: Doppler ultrasound measures blood flow velocity
• Astronomy: Red shift indicates galaxies moving away from us; blue shift indicates approach
• Weather radar: Doppler radar detects the movement of precipitation

info

Red shift occurs when a light source moves away from the observer. The observed wavelength increases (shifts towards the red end of the spectrum). This is key evidence for the expansion of the universe.

Worked Example 9

A police car sounding a siren of frequency $500 \mathrm{ Hz}$ travels at $30 \mathrm{ m/s}$ towards a stationary observer. The speed of sound is $340 \mathrm{ m/s}$. Find the frequency heard by the observer.

Solution

$f' = f \left(\frac{v}{v - v_s}\right) = 500 \times \frac{340}{340 - 30} = 500 \times \frac{340}{310}$

$f' = 500 \times 1.097 = 548.4 \mathrm{ Hz}$

Worked Example 10

After the police car passes the observer and moves away, what frequency is heard?

Solution

$f' = f \left(\frac{v}{v + v_s}\right) = 500 \times \frac{340}{340 + 30} = 500 \times \frac{340}{370}$

$f' = 500 \times 0.919 = 459.5 \mathrm{ Hz}$

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Sound Intensity and Attenuation

Inverse Square Law for Sound

For a point source emitting sound uniformly in all directions, the intensity decreases with the square of the distance:

$I = \frac{P}{4\pi r^2}$

Where $P$ is the power of the source and $r$ is the distance from the source.

Attenuation

Sound intensity decreases as it travels through a medium due to:

• Spreading of the wavefront (inverse square law)
• Absorption by the medium (energy converted to heat)
• Scattering

The attenuation of sound in a medium is often expressed in decibels per unit distance.

Worked Example 11

A source emits sound with power $0.01 \mathrm{ W}$. Find the intensity at distances of $5 \mathrm{ m}$ and $20 \mathrm{ m}$ from the source.

Solution

At $r = 5 \mathrm{ m}$:

$I_1 = \frac{0.01}{4\pi(5)^2} = \frac{0.01}{314.16} = 3.18 \times 10^{-5} \mathrm{ W/m}^2$

At $r = 20 \mathrm{ m}$:

$I_2 = \frac{0.01}{4\pi(20)^2} = \frac{0.01}{5026.55} = 1.99 \times 10^{-6} \mathrm{ W/m}^2$

Note that doubling the distance reduces the intensity by a factor of 4 (inverse square law).

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Resonance

Definition

Resonance occurs when a system is driven at its natural frequency, resulting in a large amplitude of oscillation.

Conditions for Resonance

• The driving frequency must equal the natural frequency of the system
• Energy is transferred efficiently from the driving force to the system

Examples of Resonance

Example	Description
Swing	Pushing a swing at its natural frequency makes it go higher
Tuning fork	A tuning fork can cause another of the same frequency to vibrate
Musical instruments	Air columns in wind instruments resonate at specific frequencies
Bridge collapse	Tacoma Narrows Bridge collapsed due to resonant vibrations from wind
Microwave oven	Microwaves at the resonant frequency of water molecules heat food

Forced Vibrations vs Natural Vibrations

• Natural frequency: The frequency at which a system oscillates freely when disturbed
• Forced vibrations: Vibrations caused by an external periodic driving force
• Resonance: Maximum amplitude occurs when driving frequency = natural frequency

Damping

Damping reduces the amplitude of oscillations over time by dissipating energy.

• Light damping: Oscillations gradually decrease; sharp resonance peak
• Heavy damping: Oscillations decrease rapidly; broader, lower resonance peak
• Critical damping: System returns to equilibrium in the shortest time without oscillating
• Over-damping: System returns to equilibrium slowly without oscillating

warning

warning excessive vibrations at resonance. Car shock absorbers and building dampers are examples.

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Sound Waves in Air Columns

Closed Pipe (Closed at One End)

A closed pipe supports standing waves with a node at the closed end and an antinode at the open end.

Mode	Frequency	Wavelength
Fundamental (1st harmonic)	$f_1 = \frac{v}{4L}$	$\lambda_1 = 4L$
3rd harmonic	$f_3 = 3f_1$	$\lambda_3 = \frac{4L}{3}$
5th harmonic	$f_5 = 5f_1$	$\lambda_5 = \frac{4L}{5}$

Only odd harmonics are present in a closed pipe.

Open Pipe (Open at Both Ends)

An open pipe supports standing waves with antinodes at both open ends.

Mode	Frequency	Wavelength
Fundamental (1st harmonic)	$f_1 = \frac{v}{2L}$	$\lambda_1 = 2L$
2nd harmonic	$f_2 = 2f_1$	$\lambda_2 = L$
3rd harmonic	$f_3 = 3f_1$	$\lambda_3 = \frac{2L}{3}$

All harmonics are present in an open pipe.

Worked Example 12

A closed pipe of length $0.85 \mathrm{ m}$ produces a fundamental frequency of $100 \mathrm{ Hz}$. Find the speed of sound.

Solution

$v = 4Lf_1 = 4 \times 0.85 \times 100 = 340 \mathrm{ m/s}$

Worked Example 13

Find the frequency of the second harmonic that a closed pipe of length $0.5 \mathrm{ m}$ can support. Take the speed of sound as $340 \mathrm{ m/s}$.

Solution

A closed pipe only supports odd harmonics. The "second harmonic" that exists is actually the 3rd harmonic:

$f_3 = \frac{3v}{4L} = \frac{3 \times 340}{4 \times 0.5} = \frac{1020}{2} = 510 \mathrm{ Hz}$

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Applications of Wave Theory

Ultrasound Imaging

Medical ultrasound uses high-frequency sound waves ($1$ to $15 \mathrm{ MHz}$) to create images of internal body structures.

Principle:

• A piezoelectric transducer emits ultrasound pulses
• Sound waves reflect at boundaries between tissues of different acoustic impedance
• The time delay of returning echoes is used to determine the depth of structures
• The intensity of returning echoes is used to construct a 2D image

Acoustic impedance:

$Z = \rho v$

Where $\rho$ is the density of the medium and $v$ is the speed of sound in the medium.

The fraction of ultrasound reflected at a boundary depends on the difference in acoustic impedance between the two media. A gel is applied between the transducer and the skin to eliminate the air gap (which would cause nearly total reflection due to the large impedance mismatch).

Worked Example 14

Ultrasound pulses are sent into the body and the echo from an organ returns after $40 \mathrm{ \mu s}$. The speed of ultrasound in tissue is $1540 \mathrm{ m/s}$. Find the depth of the organ.

Solution

$d = \frac{vt}{2} = \frac{1540 \times 40 \times 10^{-6}}{2} = \frac{0.0616}{2} = 0.0308 \mathrm{ m} = 3.08 \mathrm{ cm}$

(The factor of 2 accounts for the pulse travelling to the organ and back.)

Seismic Waves

Earthquakes generate several types of waves:

• P-waves (Primary): Longitudinal waves that travel through solids and liquids; fastest
• S-waves (Secondary): Transverse waves that travel through solids only; slower than P-waves
• Surface waves: Travel along the Earth's surface; cause the most damage; slowest

The fact that S-waves cannot travel through liquids provides evidence that the Earth's outer core is liquid.

Worked Example 15

P-waves are detected at a seismograph station $6000 \mathrm{ km}$ from the earthquake epicentre. If the P-wave speed is $8.0 \mathrm{ km/s}$ and the S-wave speed is $4.5 \mathrm{ km/s}$, find the time difference between the arrival of P-waves and S-waves.

Solution

P-wave arrival time: $t_P = \frac{6000}{8.0} = 750 \mathrm{ s} = 12.5 \mathrm{ minutes}$

S-wave arrival time: $t_S = \frac{6000}{4.5} = 1333 \mathrm{ s} = 22.2 \mathrm{ minutes}$

Time difference: $t_S - t_P = 583 \mathrm{ s} = 9.7 \mathrm{ minutes}$

This time difference is used to determine the distance from the epicentre, and data from multiple stations can triangulate the exact location.

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Problem Set

Problem 1: Wave Speed Calculation

A water wave has a wavelength of $3.0 \mathrm{ m}$ and a frequency of $0.25 \mathrm{ Hz}$. Find the wave speed.

Solution

$v = f\lambda = 0.25 \times 3.0 = 0.75 \mathrm{ m/s}$

If you get this wrong, revise: The wave equation $v = f\lambda$.

Problem 2: Double-Slit Interference

Light of wavelength $480 \mathrm{ nm}$ passes through two slits $0.3 \mathrm{ mm}$ apart onto a screen $2.0 \mathrm{ m}$ away. Find the separation of adjacent bright fringes.

Solution

$y = \frac{\lambda D}{a} = \frac{480 \times 10^{-9} \times 2.0}{0.3 \times 10^{-3}} = \frac{9.6 \times 10^{-7}}{3 \times 10^{-4}} = 3.2 \times 10^{-3} \mathrm{ m} = 3.2 \mathrm{ mm}$

If you get this wrong, revise: Double-slit fringe spacing formula $y = \lambda D / a$.

Problem 3: Snell's Law

Light travels from air into a glass block at an angle of incidence of $50^\circ$. If the refractive index of the glass is $1.52$, find the angle of refraction.

Solution

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

$1.0 \times \sin 50^\circ = 1.52 \times \sin\theta_2$

$\sin\theta_2 = \frac{0.766}{1.52} = 0.504$

$\theta_2 = 30.2^\circ$

If you get this wrong, revise: Snell's law — remember to measure angles from the normal.

Problem 4: Stationary Waves — Harmonics

A string of length $0.6 \mathrm{ m}$ has a wave speed of $240 \mathrm{ m/s}$. Find the frequencies of the first three harmonics.

Solution

Fundamental: $f_1 = \frac{v}{2L} = \frac{240}{2 \times 0.6} = 200 \mathrm{ Hz}$

2nd harmonic: $f_2 = 2f_1 = 400 \mathrm{ Hz}$

3rd harmonic: $f_3 = 3f_1 = 600 \mathrm{ Hz}$

If you get this wrong, revise: Stationary waves on a string and the harmonic series $f_n = nf_1$.

Problem 5: Decibel Calculation

A sound has an intensity of $10^{-6} \mathrm{ W/m}^2$. Find the loudness level in dB.

Solution

$\beta = 10\log_{10}\left(\frac{10^{-6}}{10^{-12}}\right) = 10\log_{10}(10^6) = 10 \times 6 = 60 \mathrm{ dB}$

If you get this wrong, revise: The decibel formula $\beta = 10\log_{10}(I/I_0)$ and $I_0 = 10^{-12} \mathrm{ W/m}^2$.

Problem 6: Convex Lens — Virtual Image

An object is placed $10 \mathrm{ cm}$ from a convex lens of focal length $15 \mathrm{ cm}$. Find the image position and magnification.

Solution

$\frac{1}{v} - \frac{1}{-10} = \frac{1}{15}$

$\frac{1}{v} + \frac{1}{10} = \frac{1}{15}$

$\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}$

$v = -30 \mathrm{ cm}$

$m = \frac{v}{u} = \frac{-30}{-10} = 3$

The image is virtual, upright, and magnified $3\times$, located $30 \mathrm{ cm}$ on the same side as the object.

If you get this wrong, revise: Thin lens formula for convex lens when object is inside $f$.

Problem 7: Doppler Effect — Moving Observer

An observer moves towards a stationary sound source emitting $400 \mathrm{ Hz}$ at $20 \mathrm{ m/s}$. Speed of sound is $340 \mathrm{ m/s}$. Find the observed frequency.

Solution

$f' = f\left(\frac{v + v_o}{v}\right) = 400 \times \frac{340 + 20}{340} = 400 \times \frac{360}{340} = 400 \times 1.059 = 423.5 \mathrm{ Hz}$

If you get this wrong, revise: Doppler effect formula when the observer is moving — use $v + v_o$ in the numerator.

Problem 8: EM Spectrum — Frequency Range

Microwaves have wavelengths from $1 \mathrm{ mm}$ to $0.1 \mathrm{ m}$. Find the corresponding frequency range.

Solution

$f_1 = \frac{c}{\lambda_1} = \frac{3.0 \times 10^8}{0.1} = 3.0 \times 10^9 \mathrm{ Hz}$

$f_2 = \frac{c}{\lambda_2} = \frac{3.0 \times 10^8}{1 \times 10^{-3}} = 3.0 \times 10^{11} \mathrm{ Hz}$

Frequency range: $3.0 \times 10^9$ to $3.0 \times 10^{11} \mathrm{ Hz}$.

If you get this wrong, revise: EM spectrum and the wave equation $c = f\lambda$.

Problem 9: Closed Pipe Harmonics

A closed pipe of length $0.75 \mathrm{ m}$ supports standing waves. The speed of sound is $340 \mathrm{ m/s}$. Find the frequencies of the first two harmonics.

Solution

1st harmonic (fundamental): $f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.75} = \frac{340}{3} = 113.3 \mathrm{ Hz}$

2nd available harmonic (3rd): $f_3 = 3f_1 = 3 \times 113.3 = 340 \mathrm{ Hz}$

Remember: closed pipes only support odd harmonics.

If you get this wrong, revise: Closed pipe harmonics — only odd harmonics ($f_1, f_3, f_5, \ldots$).

Problem 10: Malus's Law — Crossed Polaroids

Unpolarised light of intensity $400 \mathrm{ W/m}^2$ passes through two Polaroid filters with their transmission axes at $90^\circ$ (crossed). What intensity is transmitted?

Solution

After the first filter: $I_1 = 200 \mathrm{ W/m}^2$

After the second (crossed) filter: $I_2 = I_1\cos^2 90^\circ = 200 \times 0 = 0 \mathrm{ W/m}^2$

No light is transmitted through crossed polaroids.

If you get this wrong, revise: Malus's law — $\cos 90^\circ = 0$.

Problem 11: Optical Fibre — Acceptance Angle

An optical fibre has a core refractive index of $1.55$ and cladding refractive index of $1.48$. Find the maximum angle (from the fibre axis) at which light can enter and still undergo total internal reflection.

Solution

Critical angle:

$\sin\theta_c = \frac{n_{\mathrm{cladding}}}{n_{\mathrm{core}}} = \frac{1.48}{1.55} = 0.9548$

$\theta_c = 72.8^\circ$

Maximum angle from the fibre axis:

$\theta_{\mathrm{max}} = 90^\circ - 72.8^\circ = 17.2^\circ$

If you get this wrong, revise: Optical fibres — the relationship between the acceptance angle (from the axis) and the critical angle (from the normal).

Problem 12: Ultrasound Depth

An ultrasound pulse is reflected from a tissue boundary and returns after $26 \ \mu\mathrm{s}$. The speed of ultrasound in the tissue is $1540 \mathrm{ m/s}$. Find the depth of the boundary.

Solution

$d = \frac{vt}{2} = \frac{1540 \times 26 \times 10^{-6}}{2} = \frac{0.04004}{2} = 0.0200 \mathrm{ m} = 2.00 \mathrm{ cm}$

If you get this wrong, revise: Ultrasound depth calculation — remember the factor of 2 for the round trip.

Problem 13: Sound Intensity — Inverse Square Law

At a distance of $10 \mathrm{ m}$ from a sound source, the intensity is $8.0 \times 10^{-5} \mathrm{ W/m}^2$. What is the intensity at $40 \mathrm{ m}$?

Solution

$\frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{10}{40}\right)^2 = \frac{1}{16}$

$I_2 = \frac{8.0 \times 10^{-5}}{16} = 5.0 \times 10^{-6} \mathrm{ W/m}^2$

If you get this wrong, revise: Inverse square law for sound intensity $I \propto 1/r^2$.

Problem 14: Refraction — Water to Glass

Light travels from water ($n = 1.33$) into glass ($n = 1.50$) at an angle of incidence of $45^\circ$. Find the angle of refraction.

Solution

$1.33 \times \sin 45^\circ = 1.50 \times \sin\theta_2$

$\sin\theta_2 = \frac{1.33 \times 0.7071}{1.50} = \frac{0.9404}{1.50} = 0.627$

$\theta_2 = 38.8^\circ$

If you get this wrong, revise: Snell's law — check which medium is denser to determine the direction of bending.

Problem 15: Seismic Waves — S-Wave Evidence

P-waves are detected at a seismograph but S-waves are not detected on the opposite side of the Earth. What does this tell us about the Earth's interior?

Solution

S-waves are transverse waves that cannot travel through liquids. The absence of S-waves on the opposite side of the Earth from an earthquake indicates that part of the Earth's interior is liquid. This is evidence for the existence of the liquid outer core.

If you get this wrong, revise: Seismic wave types and their ability to travel through different states of matter.

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tip

tip Ready to test your understanding of Waves and Optics? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Waves and Optics with other physics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

danger

danger

• Confusing the law of reflection with the law of refraction: The law of reflection states that the angle of incidence equals the angle of reflection (both measured from the normal). Refraction follows Snell's law: n1 sin(i) = n2 sin(r), where the angles are measured from the normal. Students often measure angles from the surface instead of the normal in both cases.

• Drawing ray diagrams incorrectly: In ray diagrams for lenses, the three principal rays are: (1) parallel to the principal axis, then through the focal point, (2) through the centre of the lens (undeviated), (3) through the focal point, then parallel to the principal axis. Missing any of these or drawing them incorrectly leads to wrong image positions.

• Forgetting that real images are inverted and virtual images are upright: For a convex (converging) lens, a real image is formed when the object is beyond the focal length -- it is inverted and can be projected. A virtual image (object inside focal length) is upright and magnified but cannot be projected. The sign convention in the lens equation matters.

• Confusing diffraction with interference: Diffraction is the spreading of waves around obstacles or through gaps. Interference is the superposition of waves from two coherent sources to produce a pattern of maxima and minima. They are different phenomena, though both demonstrate the wave nature of light. A diffraction grating produces an interference pattern.

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Derivations

Derivation: Snell's Law from Wave Theory

When a plane wave crosses a boundary between two media at angle $\theta_1$ (in medium 1 with wave speed $v_1$) and $\theta_2$ (in medium 2 with wave speed $v_2$), the wavefront must be continuous across the boundary.

Consider a wavefront AB that just touches the boundary at point A. By the time point B of the wavefront reaches the boundary at point C, the wave at point A has already travelled into medium 2.

The distance BC $= v_1 \Delta t$ and the distance AD $= v_2 \Delta t$, where $\Delta t$ is the time for the wavefront to sweep across the boundary.

The triangles ABC and ACD share the hypotenuse AC. The angle of incidence $\theta_1$ is related to AC and BC:

$\sin\theta_1 = \frac{BC}{AC} = \frac{v_1 \Delta t}{AC}$

Similarly:

$\sin\theta_2 = \frac{AD}{AC} = \frac{v_2 \Delta t}{AC}$

Dividing: $\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2}$

Since the refractive index $n = c/v$:

$\frac{\sin\theta_1}{\sin\theta_2} = \frac{v_1}{v_2} = \frac{c/(n_2)}{c/(n_1)} = \frac{n_2}{n_1}$

$n_1 \sin\theta_1 = n_2 \sin\theta_2$

Derivation: Path Difference for Thin Film Interference

Light of wavelength $\lambda$ in air is incident on a thin film of thickness $t$ and refractive index $n$. Part of the light reflects from the top surface, and part transmits into the film and reflects from the bottom surface.

The optical path difference between the two reflected rays is:

$\Delta = 2nt\cos\theta_r$

where $\theta_r$ is the angle of refraction inside the film.

However, there is an additional phase change of $\pi$ (half wavelength) for the ray reflected from the top surface (reflection from a denser medium). The ray reflected from the bottom surface does not experience this phase change (reflection from a less dense medium).

Constructive interference (bright fringe): $\Delta = 2nt\cos\theta_r = (m + \frac{1}{2})\lambda$ for $m = 0, 1, 2, \ldots$

Destructive interference (dark fringe): $\Delta = 2nt\cos\theta_r = m\lambda$ for $m = 0, 1, 2, \ldots$

For normal incidence ($\theta_r = 0$): $\Delta = 2nt$.

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Experimental Methods

Determining the Wavelength of Light Using Young's Double Slits

Apparatus: A monochromatic light source, a single slit, a double slit, a screen, and a metre rule.

Procedure:

1. Set up the light source, single slit (to ensure coherence), and double slit.
2. Place the screen at distance $D$ from the double slit ($D \geq 1 \mathrm{ m}$ for accuracy).
3. Measure the fringe spacing $\Delta y$ by measuring the distance across several fringes and dividing by the number of spacings.
4. Calculate: $\lambda = \frac{\Delta y \cdot d}{D}$

Precautions:

• Use a monochromatic source for clear fringes.
• Ensure $D \gg d$ for the small-angle approximation to be valid.
• Measure across many fringes (e.g., 10) to reduce the percentage uncertainty in $\Delta y$.

Sources of error:

• Difficulty in measuring the exact positions of the fringe maxima (broad, not sharp).
• Uncertainty in the slit separation $d$ (very small, typically fractions of a mm).
• $D$ may not be measured accurately if the screen is curved.

Measuring the Refractive Index of Air Using a Michelson Interferometer

Principle: In a Michelson interferometer, light is split into two beams that travel different paths before recombining. If one path passes through a vacuum chamber and the other through air, inserting air into the chamber changes the optical path length.

Procedure:

1. Set up the interferometer with white light (for easy identification of zero path difference).
2. Evacuate the chamber and observe the interference pattern.
3. Slowly admit air into the chamber and count the number of fringes $N$ that pass a reference point.
4. The optical path change: $\Delta = 2(n_{\mathrm{air}} - 1)L = N\lambda$
5. Calculate: $n_{\mathrm{air}} = 1 + \frac{N\lambda}{2L}$

where $L$ is the length of the chamber.

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Data Analysis and Uncertainty

Uncertainty in Wavelength from Double Slit

Given $\lambda = \frac{\Delta y \cdot d}{D}$:

$\frac{\Delta\lambda}{\lambda} = \sqrt{\left(\frac{\Delta(\Delta y)}{\Delta y}\right)^2 + \left(\frac{\Delta d}{d}\right)^2 + \left(\frac{\Delta D}{D}\right)^2}$

Example: $\Delta y = (0.80 \pm 0.02) \mathrm{ mm}$, $d = (0.50 \pm 0.01) \mathrm{ mm}$, $D = (1.50 \pm 0.01) \mathrm{ m}$:

$\lambda = \frac{0.80 \times 10^{-3} \times 0.50 \times 10^{-3}}{1.50} = 2.67 \times 10^{-7} \mathrm{ m} = 267 \mathrm{ nm}$

$\frac{\Delta\lambda}{\lambda} = \sqrt{\left(\frac{0.02}{0.80}\right)^2 + \left(\frac{0.01}{0.50}\right)^2 + \left(\frac{0.01}{1.50}\right)^2} = \sqrt{0.000625 + 0.0004 + 0.0000444} = \sqrt{0.001069} = 0.0327 = 3.3\%$

$\Delta\lambda = 0.033 \times 267 = 8.8 \mathrm{ nm}$

$\lambda = (267 \pm 9) \mathrm{ nm}$

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Additional Worked Examples

Worked Example 11

A light ray enters a glass prism of refractive index $1.52$ at an angle of incidence of $50^\circ$. The prism has an apex angle of $60^\circ$. Calculate the angle of emergence.

Solution

At the first face: $1.00 \times \sin 50^\circ = 1.52 \times \sin r_1$

$\sin r_1 = \frac{0.766}{1.52} = 0.504 \implies r_1 = 30.2^\circ$

Angle of incidence at the second face: $r_2 = A - r_1 = 60^\circ - 30.2^\circ = 29.8^\circ$

Check for TIR: critical angle $\theta_c = \sin^{-1}(1/1.52) = 41.1^\circ$. Since $29.8^\circ < 41.1^\circ$, no TIR occurs.

At the second face: $1.52 \times \sin 29.8^\circ = 1.00 \times \sin\theta_e$

$\sin\theta_e = 1.52 \times 0.497 = 0.755 \implies \theta_e = 49.0^\circ$

Worked Example 12

In a single-slit diffraction experiment, light of wavelength $550 \mathrm{ nm}$ passes through a slit of width $0.10 \mathrm{ mm}$. The screen is $2.0 \mathrm{ m}$ away. Calculate the width of the central maximum.

Solution

For the first minimum in single-slit diffraction: $a\sin\theta = \lambda$, where $a$ is the slit width.

$\sin\theta_1 = \frac{\lambda}{a} = \frac{550 \times 10^{-9}}{0.10 \times 10^{-3}} = 5.5 \times 10^{-3}$

For small angles: $\sin\theta \approx \theta \approx \tan\theta$

Width of central maximum $= 2y_1 = 2 \times D\tan\theta_1 \approx 2D\theta_1$

$= 2 \times 2.0 \times 5.5 \times 10^{-3} = 0.022 \mathrm{ m} = 22 \mathrm{ mm}$

Worked Example 13

A plano-convex lens has a refractive index of $1.55$ and one surface with radius of curvature $20 \mathrm{ cm}$. The other surface is flat. Calculate the focal length.

Solution

Using the lensmaker's equation: $\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

For a plano-convex lens with the curved surface facing left ($R_1 = +20 \mathrm{ cm}$) and the flat surface facing right ($R_2 = \infty$):

$\frac{1}{f} = (1.55 - 1)\left(\frac{1}{20} - \frac{1}{\infty}\right) = 0.55 \times \frac{1}{20} = 0.0275$

$f = \frac{1}{0.0275} = 36.4 \mathrm{ cm}$

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Exam-Style Questions

Question 1 (DSE Structured)

A student uses a converging lens to project an image of a illuminated object onto a screen.

(a) When the object is placed $30 \mathrm{ cm}$ from the lens, a sharp image is formed on a screen $120 \mathrm{ cm}$ from the lens. Calculate the focal length of the lens.

(b) The object is now moved to $20 \mathrm{ cm}$ from the lens. Without calculation, describe the nature of the new image and explain your reasoning.

(c) The student wants to use this lens as a simple magnifying glass. What is the maximum magnification achievable if the near point of the user is $25 \mathrm{ cm}$?

Solution

(a) $u = 30 \mathrm{ cm}$, $v = 120 \mathrm{ cm}$

$\frac{1}{f} = \frac{1}{30} + \frac{1}{120} = \frac{4 + 1}{120} = \frac{5}{120} = \frac{1}{24}$

$f = 24 \mathrm{ cm}$

(b) When the object is at $20 \mathrm{ cm}$ (which is less than $f = 24 \mathrm{ cm}$), the object is inside the focal length. No real image can be formed on a screen. The image is virtual, upright, and magnified, appearing on the same side of the lens as the object.

(c) For maximum magnification with a simple magnifying glass, the virtual image should be at the near point ($25 \mathrm{ cm}$). Using $v = -25 \mathrm{ cm}$ (virtual image):

$\frac{1}{24} = \frac{1}{u} + \frac{1}{-25}$

$\frac{1}{u} = \frac{1}{24} + \frac{1}{25} = \frac{25 + 24}{600} = \frac{49}{600}$

$u = \frac{600}{49} = 12.24 \mathrm{ cm}$

$m = \frac{|v|}{u} = \frac{25}{12.24} = 2.04$

Maximum magnification is approximately 2.0.

Question 2 (DSE Structured)

(a) State the conditions necessary for total internal reflection to occur.

(b) An optical fibre has a core of refractive index $1.48$ and cladding of refractive index $1.42$.

(i) Calculate the critical angle at the core-cladding boundary.

(ii) Calculate the maximum angle of incidence at the fibre end face for which light will be totally internally reflected along the fibre (the acceptance angle).

(c) Explain why optical fibres are used for telecommunications instead of copper cables, giving two advantages.

Solution

(a) Two conditions for total internal reflection:

1. Light must travel from a denser medium to a less dense medium ($n_1 > n_2$).
2. The angle of incidence must exceed the critical angle ($\theta_i > \theta_c$).

(b) (i) $\sin\theta_c = \frac{n_{\mathrm{cladding}}}{n_{\mathrm{core}}} = \frac{1.42}{1.48} = 0.9595$

$\theta_c = \sin^{-1}(0.9595) = 73.7^\circ$

(ii) The maximum angle inside the fibre (measured from the fibre axis) for TIR is $90^\circ - 73.7^\circ = 16.3^\circ$.

Using Snell's law at the end face: $n_{\mathrm{air}}\sin\theta_{\max} = n_{\mathrm{core}}\sin(16.3^\circ) = 1.48 \times 0.281 = 0.416$

$\theta_{\max} = \sin^{-1}(0.416) = 24.6^\circ$

(c) Two advantages of optical fibres over copper cables:

1. Much higher bandwidth: Optical fibres can carry far more data per second than copper cables, enabling high-speed internet and telecommunications.
2. Lower signal loss (attenuation): Signals in optical fibres can travel much longer distances without requiring amplification compared to electrical signals in copper.

(Other valid answers: immune to electromagnetic interference, lighter weight, more secure.)

Question 3 (DSE Structured)

(a) Distinguish between a continuous spectrum and a line spectrum.

(b) Light from a discharge tube containing hydrogen is passed through a diffraction grating with $300 \mathrm{ lines/mm}$. A bright line is observed at an angle of $18.5^\circ$ in the first order. Calculate the wavelength of this line.

(c) Explain how the line spectrum of hydrogen provides evidence for the existence of discrete energy levels in atoms.

Solution

(a) A continuous spectrum contains all wavelengths over a continuous range (e.g., the spectrum from a hot filament or the Sun). A line spectrum consists of discrete, well-defined wavelengths (bright lines on a dark background for emission, or dark lines on a continuous background for absorption).

(b) $d = \frac{1}{300 \times 10^3} = 3.33 \times 10^{-6} \mathrm{ m}$

For first order ($n = 1$): $d\sin\theta = \lambda$

$\lambda = 3.33 \times 10^{-6} \times \sin 18.5^\circ = 3.33 \times 10^{-6} \times 0.317 = 1.06 \times 10^{-6} \mathrm{ m} = 1060 \mathrm{ nm}$

(Note: This is in the infrared region.)

(c) The line spectrum of hydrogen shows that atoms only emit (or absorb) light at specific, discrete wavelengths. By $E = hf = hc/\lambda$, this means photons of only specific energies are emitted. This is explained by electrons existing in discrete energy levels: when an electron transitions from a higher level to a lower level, it emits a photon with energy equal to the difference between the two levels: $\Delta E = hf$. The discrete wavelengths correspond to specific energy level differences, providing evidence for quantised energy levels.

Question 4 (DSE Structured)

(a) What is meant by the term "coherent" when applied to light sources?

(b) In a Young's double-slit experiment using light of wavelength $580 \mathrm{ nm}$, the slit separation is $0.40 \mathrm{ mm}$ and the screen is $2.5 \mathrm{ m}$ from the slits.

(i) Calculate the fringe spacing.

(ii) Calculate the distance from the central maximum to the third bright fringe.

(iii) If the slit separation is reduced to $0.20 \mathrm{ mm}$, describe and explain the effect on the fringe pattern.

(c) Explain why a single slit is placed before the double slit in this experiment.

Solution

(a) Two sources are coherent if they emit waves with a constant phase relationship (i.e., the phase difference between them does not change with time). For light, this means the waves must have the same frequency and a constant phase difference.

(b) (i) $\Delta y = \frac{\lambda D}{d} = \frac{580 \times 10^{-9} \times 2.5}{0.40 \times 10^{-3}} = \frac{1.45 \times 10^{-6}}{4.0 \times 10^{-4}} = 3.63 \times 10^{-3} \mathrm{ m} = 3.63 \mathrm{ mm}$

(ii) Third bright fringe ($n = 3$): $y_3 = 3\Delta y = 3 \times 3.63 = 10.9 \mathrm{ mm}$

(iii) If the slit separation is halved, the fringe spacing doubles ($\Delta y \propto 1/d$). The fringes become wider and more spread out. However, the fringes also become less bright because less light passes through each narrower slit.

(c) The single slit acts as a point source of light. Light from an extended source (e.g., a lamp) would create multiple sets of interference patterns that overlap and wash out. The single slit ensures that the light reaching the double slit is coherent (comes from a single point). This is necessary for a stable, observable interference pattern.

Question 5 (DSE Structured)

(a) Explain what is meant by polarisation of light.

(b) Unpolarised light of intensity $I_0$ is incident on two polarising filters. The first has its transmission axis vertical, and the second has its transmission axis at $60^\circ$ to the vertical. Calculate the intensity of light transmitted through both filters.

(c) A third polarising filter is now inserted between the first two, with its transmission axis at $30^\circ$ to the vertical. Calculate the new intensity of the transmitted light and compare it with the intensity in part (b).

(d) State one application of polarised light and explain how it works.

Solution

(a) Polarisation is the process of restricting the oscillations of a transverse wave to a single plane. For light, this means the electric field oscillates in only one direction. Unpolarised light has oscillations in all directions perpendicular to the direction of propagation.

(b) After the first filter (vertical): $I_1 = \frac{1}{2}I_0$ (half the light passes, as unpolarised light has equal components in all directions).

After the second filter (at $60^\circ$ to vertical): Using Malus's law:

$I_2 = I_1\cos^2 60^\circ = \frac{1}{2}I_0 \times (0.5)^2 = \frac{1}{2}I_0 \times 0.25 = 0.125I_0 = \frac{I_0}{8}$

(c) After the first filter: $I_1 = \frac{1}{2}I_0$

After the second filter (at $30^\circ$): $I_2 = I_1\cos^2 30^\circ = \frac{1}{2}I_0 \times \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2}I_0 \times 0.75 = 0.375I_0$

After the third filter (at $60^\circ$ to vertical, which is $30^\circ$ from the second filter's axis): $I_3 = I_2\cos^2 30^\circ = 0.375I_0 \times 0.75 = 0.281I_0$

This is greater than $0.125I_0$ (from part b). Inserting an intermediate polariser actually increases the transmitted intensity. This is because the intermediate filter rotates the polarisation direction in two $30^\circ$ steps rather than one $60^\circ$ step, and $\cos^2 30^\circ \times \cos^2 30^\circ > \cos^2 60^\circ$.

(d) Application: Polaroid sunglasses. Reflected light from horizontal surfaces (roads, water) is partially polarised in the horizontal plane. Polaroid sunglasses have their transmission axis vertical, so they block the horizontally polarised reflected glare while allowing other light to pass. This reduces glare and improves visual comfort.

(Other valid applications: liquid crystal displays (LCDs), stress analysis in engineering using photoelasticity, optical activity in sugar solutions.)

Extended Analysis: Resolution of Diffraction Grating

The resolving power $R$ of a diffraction grating is its ability to distinguish between two closely spaced wavelengths:

$R = \frac{\lambda}{\Delta\lambda} = nN$

where $n$ is the order number and $N$ is the total number of illuminated slits.

Example: A grating with $500 \mathrm{ lines/mm}$ is illuminated over a width of $2.0 \mathrm{ cm}$ in the second order.

$N = 500 \times 10^3 \times 0.02 = 10000 \mathrm{ slits}$

$R = nN = 2 \times 10000 = 20000$

Minimum resolvable wavelength difference at $\lambda = 550 \mathrm{ nm}$:

$\Delta\lambda = \frac{\lambda}{R} = \frac{550}{20000} = 0.0275 \mathrm{ nm}$

Extended Worked Example: Single Slit Width

In a single-slit diffraction experiment, the first minimum is observed at an angle of $3.0^\circ$ when light of wavelength $620 \mathrm{ nm}$ is used. Calculate the slit width.

Solution

$a\sin\theta = \lambda$

$a = \frac{\lambda}{\sin\theta} = \frac{620 \times 10^{-9}}{\sin 3.0^\circ} = \frac{620 \times 10^{-9}}{0.0523} = 1.19 \times 10^{-5} \mathrm{ m} = 0.0119 \mathrm{ mm}$

Extended Worked Example: Dispersion in a Prism

A prism has apex angle $60^\circ$ and refractive index $1.52$ for sodium light ($\lambda = 589 \mathrm{ nm}$). Calculate the angle of minimum deviation.

Solution

At minimum deviation, the ray passes symmetrically through the prism:

$n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$1.52 = \frac{\sin\left(\frac{60 + D_m}{2}\right)}{\sin 30^\circ} = \frac{\sin\left(\frac{60 + D_m}{2}\right)}{0.5}$

$\sin\left(\frac{60 + D_m}{2}\right) = 1.52 \times 0.5 = 0.760$

$\frac{60 + D_m}{2} = \sin^{-1}(0.760) = 49.5^\circ$

$60 + D_m = 99.0^\circ \implies D_m = 39.0^\circ$