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EDEXCEL INTERNATIONAL A LEVEL PHYSICS • UNIT 2

PHYSICS 02

Waves. Light. Photons. Electricity. Explore the physics behind wave behaviour, quantum phenomena and electrical systems.

COMPLETE LEARNING SYSTEM • EQUATIONS • PRACTICALS • EXAM PRACTICE

TOPIC 01

Waves and Particle Nature of Light

Unit 2 begins with the behaviour and properties of waves, followed by refraction, diffraction, interference and stationary waves. The unit then develops the idea that electromagnetic radiation can behave both as a wave and as a stream of photons.

Core idea:

A wave transfers energy and information without requiring the overall transfer of matter from one place to another.

Key Wave Quantities

v = fλ

TOPIC 02

Transverse and Longitudinal Waves

Transverse Waves

In a transverse wave, the oscillations are perpendicular to the direction in which the wave transfers energy.

Examples include electromagnetic waves and waves travelling along a stretched string.

Longitudinal Waves

In a longitudinal wave, the oscillations are parallel to the direction of energy transfer.

Sound in air is a longitudinal wave. It consists of compressions and rarefactions caused by variations in pressure and molecular displacement.

Wave Graphs

Transverse waves can be represented by displacement against distance or time. Longitudinal waves can be represented using pressure or molecular displacement.

TOPIC 03

Wavefronts, Phase and Superposition

Wavefront

A wavefront is a line or surface joining points that are at the same phase of oscillation.

Phase

Two points are in phase when they have the same displacement and are moving in the same direction at the same time.

Path Difference

Path difference is the difference between the distances travelled by two waves to reach a particular point.

Superposition

When two or more waves overlap, their resultant displacement is the algebraic sum of their individual displacements.

Interference:

Constructive interference occurs when waves reinforce each other, while destructive interference occurs when they oppose each other.

Phase difference = (path difference / wavelength) × 360°

TOPIC 04

Stationary Waves

A stationary wave is produced when two waves of the same frequency, wavelength and amplitude travel in opposite directions and superpose.

Nodes

A node is a point where the displacement is always zero.

Antinodes

An antinode is a point where the displacement has its maximum amplitude.

Important:

Adjacent nodes are separated by λ/2, while a node and its nearest antinode are separated by λ/4.

Wave on a Stretched String

v = √(T / μ)

Here T is the tension in the string and μ is the mass per unit length.

WORKED EXAMPLE

A string has a tension of 80 N and a mass per unit length of 0.020 kg m⁻¹. Find the wave speed.

v = √(80 / 0.020)

v = √4000

v ≈ 63.2 m s⁻¹

TOPIC 05

Intensity of Radiation

Intensity is the power transferred per unit area perpendicular to the direction of energy transfer.

I = P / A

where I is intensity, P is power and A is area.

Inverse Square Relationship

For radiation spreading uniformly from a point source, intensity decreases as the square of the distance from the source increases.

I ∝ 1 / r²

TOPIC 06

Refraction

Refraction occurs when a wave changes direction because its speed changes when it enters a different medium.

Refractive Index

n = c / v

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

Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

The angles are measured from the normal to the boundary.

Critical Angle

sin C = 1 / n

For light travelling from a material of refractive index n into air, the critical angle C is the angle of incidence for which the refracted ray travels along the boundary.

Total Internal Reflection

Total internal reflection occurs when light travels from a more optically dense medium to a less optically dense medium and the angle of incidence is greater than the critical angle.

CORE PRACTICAL 04

Measuring Refractive Index

The refractive index of a transparent solid or liquid can be determined experimentally by measuring angles of incidence and refraction.

TYPICAL METHOD
  1. Place the transparent material on paper.
  2. Draw a normal at the point where the light enters.
  3. Use a ray box or laser to produce an incident ray.
  4. Measure the angle of incidence and angle of refraction.
  5. Repeat for several different angles.
  6. Use Snell's law or a suitable graph to determine n.

Take care to avoid parallax when measuring angles and repeat measurements to improve reliability.

TOPIC 07

Polarisation

Plane polarisation occurs when the oscillations of a transverse wave are restricted to one direction.

Key point:

Polarisation is evidence that a wave is transverse, because longitudinal waves cannot be plane-polarised.

Electromagnetic waves are transverse and can therefore be polarised.

TOPIC 08

Diffraction

Diffraction is the spreading of waves when they pass through a gap or around an obstacle.

Huygens' Construction

Huygens' construction treats every point on a wavefront as a source of secondary wavelets. The new wavefront is obtained by drawing the envelope of these wavelets.

Diffraction becomes more significant when the size of the gap is comparable to the wavelength.

Diffraction Grating

nλ = d sin θ

Here n is the order of the maximum, λ is wavelength, d is the grating spacing and θ is the diffraction angle.

WORKED EXAMPLE

A diffraction grating has 600 lines per millimetre. Light of wavelength 500 nm produces a first-order maximum. Find the diffraction angle.

Number of lines per metre = 600 × 1000 = 6.00 × 10⁵ m⁻¹

d = 1 / (6.00 × 10⁵)

d = 1.67 × 10⁻⁶ m

sin θ = λ / d

sin θ = (5.00 × 10⁻⁷) / (1.67 × 10⁻⁶) ≈ 0.300

θ ≈ 17.5°

CORE PRACTICAL 05

Finding the Wavelength of Light

A diffraction grating can be used to determine the wavelength of monochromatic light such as laser light.

METHOD
  1. Place the diffraction grating perpendicular to the incident beam.
  2. Observe the central maximum and higher-order maxima.
  3. Measure the angle θ for a known diffraction order.
  4. Use the number of lines per unit length to calculate d.
  5. Use nλ = d sin θ to determine λ.
  6. Repeat measurements for greater reliability.

TOPIC 09

Wave Nature of Electrons

Electrons can behave as waves as well as particles. Electron diffraction experiments provide evidence for the wave nature of matter.

de Broglie Wavelength

λ = h / p

Since momentum p = mv for a non-relativistic particle:

λ = h / mv

The wave nature becomes more significant for particles with small momentum.

TOPIC 10

Pulse-Echo Techniques

A pulse-echo system sends a short pulse of radiation or ultrasound towards an object and detects the reflected echo.

Finding Distance

distance = vt / 2

The factor of 2 is required because the pulse travels to the object and then back to the detector.

Resolution

The amount of information obtained can be limited by the wavelength of the radiation and the duration of the pulse. Shorter wavelengths and shorter pulses can generally provide better resolution.

TOPIC 11

Wave-Particle Duality of Light

Electromagnetic radiation can be described using both a wave model and a photon model.

Wave Model

The wave model explains phenomena such as interference, diffraction, refraction and polarisation.

Photon Model

The photon model treats electromagnetic radiation as packets of energy called photons.

E = hf

Higher-frequency photons have greater energy.

TOPIC 12

The Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency is incident on it.

Threshold Frequency

The threshold frequency is the minimum frequency required for photoemission from a particular metal.

Work Function

The work function is the minimum energy required to remove an electron from the surface of a metal.

hf = φ + ½mv²max

Electronvolt

The electronvolt is a convenient unit for small amounts of energy associated with particles.

1 eV = 1.60 × 10⁻¹⁹ J
Why the photoelectric effect matters:

The instantaneous emission of electrons above a threshold frequency and the dependence of maximum kinetic energy on frequency provide evidence for the particle nature of electromagnetic radiation.

WORKED EXAMPLE

Light of frequency 8.0 × 10¹⁴ Hz falls on a metal with a work function of 2.0 eV. Find the maximum kinetic energy of an emitted electron.

Photon energy: E = hf

E = (6.63 × 10⁻³⁴)(8.0 × 10¹⁴)

E ≈ 5.30 × 10⁻¹⁹ J

Work function: φ = 2.0 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ J

Kinetic energy = 5.30 × 10⁻¹⁹ − 3.20 × 10⁻¹⁹

KEmax ≈ 2.10 × 10⁻¹⁹ J

TOPIC 13

Atomic Line Spectra

Electrons in atoms can occupy only certain discrete energy levels. When an electron moves between energy levels, electromagnetic radiation can be emitted or absorbed.

Emission

When an electron moves from a higher energy level to a lower energy level, a photon is emitted.

Absorption

When an electron absorbs a photon with the correct energy, it can move to a higher energy level.

ΔE = hf

Therefore:

f = ΔE / h
Key idea:

The discrete nature of atomic energy levels produces discrete spectral lines rather than a continuous spectrum.

TOPIC 14

Electric Current, Charge and Potential Difference

Electric Current

Electric current is the rate of flow of electric charge.

I = ΔQ / Δt

Potential Difference

Potential difference is the energy transferred per unit charge when charge moves between two points.

V = W / Q

Resistance

R = V / I

Ohm's law is the special case where current is directly proportional to potential difference at constant temperature.

TOPIC 15

Charge and Energy Conservation in Circuits

Current at a Junction

Charge cannot accumulate indefinitely at a junction. Therefore the total current entering a junction equals the total current leaving it.

ΣI(in) = ΣI(out)

Potential Differences

Energy is conserved around a complete circuit. The total potential difference supplied is equal to the total potential difference across the components.

ΣV = 0 around a complete loop

TOPIC 16

Resistors in Series and Parallel

Series

In series, the same current passes through each resistor.

Rtotal = R₁ + R₂ + R₃ + ...

Parallel

In parallel, the potential difference across each branch is the same.

1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ + ...

WORKED EXAMPLE

Two resistors of 6 Ω and 3 Ω are connected in parallel. Find the combined resistance.

1/R = 1/6 + 1/3

1/R = 1/6 + 2/6 = 3/6 = 1/2

R = 2 Ω

TOPIC 17

Electrical Power and Energy

P = VI
W = VIt

Using V = IR, further useful forms are:

P = I²R
P = V²/R

WORKED EXAMPLE

A 12 V supply drives a current of 2.5 A through a device. Calculate its power.

P = VI

P = 12 × 2.5

P = 30 W

TOPIC 18

Current-Potential Difference Characteristics

Ohmic Conductor

At constant temperature, an ohmic conductor has current directly proportional to potential difference. Its I-V graph is a straight line through the origin.

Filament Bulb

As current increases, the filament becomes hotter. Its resistance increases, so the I-V relationship is non-linear.

Thermistor

A negative temperature coefficient thermistor has decreasing resistance as temperature increases.

Diode

A diode allows current to flow much more easily in one direction than the other and therefore has a strongly non-linear I-V characteristic.

TOPIC 19

Resistivity

Resistivity is a property of a material that describes how strongly it opposes the flow of electric current.

R = ρl / A

Therefore resistance increases with length and decreases with cross-sectional area, for a given material and temperature.

WORKED EXAMPLE

A wire has resistivity 1.7 × 10⁻⁸ Ω m, length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m². Find its resistance.

R = ρl / A

R = (1.7 × 10⁻⁸ × 2.0) / (1.0 × 10⁻⁶)

R = 0.034 Ω

CORE PRACTICAL 06

Determining Electrical Resistivity

TYPICAL METHOD
  1. Measure the length of the wire accurately.
  2. Measure the diameter at several points using a micrometer.
  3. Calculate the cross-sectional area.
  4. Connect the wire in a suitable circuit.
  5. Measure potential difference and current.
  6. Calculate resistance using R = V/I.
  7. Repeat for different lengths.
  8. Use R = ρl/A to determine the resistivity.

Multiple diameter measurements are useful because the cross-sectional area depends on the square of the radius.

TOPIC 20

Current and Charge Carriers

In a metal, current is produced by the movement of conduction electrons through the material.

I = nqvA

Here n is the number density of charge carriers, q is the charge of each carrier, v is the drift velocity and A is the cross-sectional area.

Important:

The drift speed of electrons in a metal is much slower than the speed at which electrical effects propagate through a circuit.

TOPIC 21

Potential Along a Uniform Wire

For a uniform current-carrying wire, the potential changes steadily with distance along the wire when its resistance per unit length is constant.

Key idea:

Equal lengths of a uniform wire have equal resistance and therefore equal potential drops when the same current flows through them.

TOPIC 22

Potential Dividers

A potential divider uses resistors in series to produce a chosen fraction of the supply potential difference.

Vout = Vs × R₂ / (R₁ + R₂)

The resistor connected to the output determines the fraction of the supply voltage appearing across it.

Variable Resistance

A variable resistor, thermistor or light-dependent resistor can be used to make the output voltage depend on an external condition.

Thermistor

For a negative temperature coefficient thermistor, resistance decreases as temperature increases.

LDR

The resistance of a light-dependent resistor decreases as light intensity increases.

WORKED EXAMPLE

A 12 V supply is connected across a 2 kΩ resistor and a 4 kΩ resistor in series. Find the potential difference across the 4 kΩ resistor.

Vout = 12 × 4/(2 + 4)

Vout = 12 × 4/6

Vout = 8.0 V

TOPIC 23

e.m.f. and Internal Resistance

Electromotive Force

The electromotive force of a source is the energy supplied by the source per unit charge passing through it.

Internal Resistance

Real cells have internal resistance. Some of the energy supplied by the cell is dissipated inside the cell itself.

ε = I(R + r)

Equivalently, the terminal potential difference is:

V = ε − Ir

Here ε is the e.m.f., I is current, R is external resistance and r is internal resistance.

Difference:

e.m.f. refers to the energy supplied per unit charge by the source, while terminal p.d. is the energy transferred per unit charge in the external circuit.

CORE PRACTICAL 07

Determining e.m.f. and Internal Resistance

METHOD
  1. Connect the cell to a variable external resistance.
  2. Measure the current I and terminal potential difference V.
  3. Repeat for several different values of external resistance.
  4. Plot V against I.
  5. Use the relationship V = ε − Ir.
  6. The y-intercept gives ε.
  7. The magnitude of the gradient gives r.
V = ε − Ir

TOPIC 24

Resistance and Temperature

Metallic Conductors

As temperature increases, the lattice ions vibrate more strongly. Electrons experience more frequent collisions, so the resistance of a metal generally increases.

Negative Temperature Coefficient Thermistor

In an NTC thermistor, resistance decreases as temperature increases. This can be explained by an increase in the number of charge carriers available for conduction.

TOPIC 25

Light-Dependent Resistors

An LDR changes its resistance depending on the intensity of incident light.

Increasing illumination increases the number of charge carriers available for conduction, reducing resistance.

Application:

LDRs can be incorporated into potential-divider circuits to produce automatic responses to changes in light level.

PRACTICE

Unit 2 Mixed Practice

1. A wave has a frequency of 500 Hz and wavelength 0.68 m. Calculate its speed.
Show answer

v = fλ = 500 × 0.68

v = 340 m s⁻¹

2. Explain why a stationary wave has nodes and antinodes.
Show answer

A stationary wave is formed by superposition of two identical waves travelling in opposite directions. Destructive interference produces nodes and constructive interference produces antinodes.

3. State one condition required for total internal reflection.
Show answer

The wave must travel from a medium of higher refractive index to a medium of lower refractive index, and the angle of incidence must exceed the critical angle.

4. A diffraction grating has 500 lines per millimetre. Calculate its grating spacing.
Show answer

500 lines/mm = 5.00 × 10⁵ lines/m.

d = 1/(5.00 × 10⁵)

d = 2.00 × 10⁻⁶ m

5. Calculate the energy of a photon with frequency 6.0 × 10¹⁴ Hz.
Show answer

E = hf

E = (6.63 × 10⁻³⁴)(6.0 × 10¹⁴)

E ≈ 3.98 × 10⁻¹⁹ J

6. A current of 3 A flows for 20 s. Calculate the charge transferred.
Show answer

Q = It = 3 × 20

Q = 60 C

7. Two resistors of 4 Ω and 8 Ω are connected in series. Find their combined resistance.
Show answer

R = 4 + 8

R = 12 Ω

8. A 6 Ω resistor carries a current of 2 A. Calculate the power dissipated.
Show answer

P = I²R

P = 2² × 6

P = 24 W

9. Explain why the resistance of an NTC thermistor decreases when temperature increases.
Show answer

Increasing temperature increases the number of charge carriers available for conduction, causing the resistance to decrease.

10. A cell has e.m.f. 1.50 V and internal resistance 0.20 Ω. A current of 2.0 A flows. Find the terminal potential difference.
Show answer

V = ε − Ir

V = 1.50 − (2.0 × 0.20)

V = 1.10 V

QUICK REVIEW

Unit 2 Key Equations

v = fλ
v = √(T / μ)
I = P / A
n = c / v
n₁ sin θ₁ = n₂ sin θ₂
sin C = 1 / n
nλ = d sin θ
λ = h / p
E = hf
hf = φ + ½mv²max
I = ΔQ / Δt
V = W / Q
R = V / I
P = VI
W = VIt
R = ρl / A
I = nqvA
Vout = Vs × R₂ / (R₁ + R₂)
V = ε − Ir