Year 11 Physics Waves: Original Revision Mock Exam

This is an original tutorgum Year 11 Physics practice paper covering the supplied Waves content for the physics-2025 syllabus edition. It tests calculations, graph interpretation, physical reasoning, and applications across the topic. It is not an official, endorsed, past, or predicted NESA paper.

01Exam instructions

Suggested conditions

• Total marks: 100
• Suggested time: 120 minutes, including 5 minutes of reading time and 115 minutes of working time.
• Permitted equipment: scientific calculator, ruler, pencil, and blank paper for working and diagrams.
• No notes or separate formula sheet are supplied. Relevant relationships and constants are provided where needed.
• Unless otherwise stated, measure angles from the normal, include units, and give numerical answers to three significant figures where appropriate.
• Show your working. A correct numerical answer without supporting working may not receive full marks.

Assessed scope

This paper covers the supplied Year 11 Waves content: wave quantities and graphs, frequency and period, wave speed, mechanical and electromagnetic waves, sound, reflection, refraction, diffraction, superposition, progressive and standing waves, string harmonics, water-wave depth estimation, mirrors, lenses, refractive index, total internal reflection, vision correction, the electromagnetic spectrum, photon energy, light intensity, and the Doppler effect.

This is a Waves-topic revision paper, not a full-course Physics examination. It excludes unrelated modules and material from other syllabus editions. The conditions and mark allocations are suggested practice conditions, not claims about an official NESA examination format.

Useful constants

• Speed of light in vacuum: \(c=3.00\times10^8\ \text{m s}^{-1}\)
• Planck constant: \(h=6.63\times10^{-34}\ \text{J s}\)
• Gravitational acceleration: \(g=9.8\ \text{m s}^{-2}\)
• Speed of sound in air for Question 17: \(v=340\ \text{m s}^{-1}\)
• Refractive index of air: \(n_{\text{air}}=1.00\)

02Questions

Question 1 (8 marks)

The figure contains two graphs describing the same wave. Graph A records displacement at a fixed point as time passes. Graph B records displacement at different positions at one instant.

(a) Define instantaneous displacement and amplitude. Use the figure to state the amplitude. (2 marks)

(b) Explain what the horizontal axis represents in each graph and name one quantity that can be determined from each. (2 marks)

(c) Use Graph A to calculate the period and frequency of the wave. (2 marks)

(d) Use Graph B to calculate the wavelength and wave speed. (2 marks)

Question 2 (6 marks)

(a) A wave travels along a rope, while sound travels through air. Classify each wave as transverse or longitudinal, and compare the direction of particle oscillation with the direction of wave propagation. (2 marks)

(b) Explain how compressions and rarefactions in sound produce pressure variations. Explain how sound transfers energy without transporting air in bulk from the source to the listener. (2 marks)

(c) Distinguish pitch from loudness. State the wave property most directly associated with each. (2 marks)

Question 3 (4 marks)

A wave travels along a taut string at \(24\ \text{m s}^{-1}\). Its frequency is \(8.0\ \text{Hz}\).

Use the wave equation \(v=f\lambda\), where \(v\) is wave speed, \(f\) is frequency, and \(\lambda\) is wavelength.

(a) Calculate the wavelength. (2 marks)

(b) The source frequency increases to \(10\ \text{Hz}\), while the string tension and linear mass density remain unchanged. Determine the new wave speed and wavelength. Explain why the speed does not change. (2 marks)

Question 4 (6 marks)

(a) Compare progressive waves and standing waves in terms of their appearance and net energy transfer along the medium. (2 marks)

(b) Define a node and an antinode in a standing wave. State the displacement amplitude at each. (2 marks)

(c) Two waves overlap at points P and Q.

At P, their instantaneous displacements are \(+3\ \text{cm}\) and \(+2\ \text{cm}\).

At Q, their instantaneous displacements are \(+3\ \text{cm}\) and \(-5\ \text{cm}\).

Calculate the resultant displacement at each point and classify the interference as constructive or destructive. (2 marks)

Question 5 (6 marks)

(a) A wavefront strikes a straight reflecting boundary at \(35^\circ\) to the normal. State the reflected angle. For a pulse on a string, also state whether reflection at a fixed end and at a free end produces phase inversion. (2 marks)

(b) Waves pass through the same gap, which is \(0.15\ \text{m}\) wide. One wave has a wavelength of \(0.10\ \text{m}\), while another has a wavelength of \(0.60\ \text{m}\). Identify which wave spreads more after passing through the gap and explain why. (2 marks)

(c) A water wave enters shallow water at an oblique angle, and its speed decreases. State what happens to its frequency and wavelength, and explain how its direction changes. (2 marks)

Question 6 (5 marks)

A student estimates water depth using waves in a region assumed to be shallow.

The measured wave frequency is \(1.50\ \text{Hz}\), and the wavelength is \(0.80\ \text{m}\).

For shallow-water waves, assume the relationship

\[ v=\sqrt{gh} \]

is valid, where \(v\) is wave speed, \(g\) is gravitational acceleration, and \(h\) is water depth. Neglect currents.

(a) Calculate the wave speed. (2 marks)

(b) Estimate the water depth. (2 marks)

(c) State one limitation or assumption of this depth-estimation method. (1 mark)

Question 7 (6 marks)

A string of length \(0.750\ \text{m}\) is fixed at both ends. Its linear mass density is \(4.00\times10^{-3}\ \text{kg m}^{-1}\), and its tension is \(36.0\ \text{N}\).

Use

\[ v=\sqrt{\frac{F_T}{\mu}} \]

and

\[ f_n=\frac{nv}{2L} \]

where \(F_T\) is tension, \(\mu\) is linear mass density, \(L\) is string length, and \(n\) is the harmonic number.

(a) Calculate the wave speed on the string. (2 marks)

(b) Calculate the fundamental frequency and the frequency of the third harmonic. (2 marks)

(c) The tension is quadrupled while the string length and linear mass density remain unchanged. Determine how the fundamental frequency changes, explaining your reasoning. (2 marks)

Question 8 (7 marks)

A light ray travels from air into glass. The refractive indices are \(n_1=1.00\) for air and \(n_2=1.50\) for glass. The angle of incidence is \(40.0^\circ\).

Use

\[ n=\frac{c}{v} \]

and Snell’s law,

\[ n_1\sin i=n_2\sin r \]

where \(i\) is the angle of incidence and \(r\) is the angle of refraction. Both angles are measured from the normal.

(a) Calculate the speed of light in the glass. (2 marks)

(b) Calculate the angle of refraction. (3 marks)

(c) State how the ray bends relative to the normal. Explain what happens to its frequency and wavelength as it enters the glass. (2 marks)

Question 9 (6 marks)

A ray of light travels inside glass with refractive index \(1.50\) and reaches a boundary with air of refractive index \(1.00\).

(a) Calculate the critical angle for the glass-air boundary. (2 marks)

(b) State the two conditions required for total internal reflection. (2 marks)

(c) Predict what happens when the angle of incidence inside the glass is:

• \(35^\circ\)
• \(45^\circ\)

Justify each prediction using the critical angle. (2 marks)

Question 10 (5 marks)

A student measures the refractive index of a transparent material by directing a narrow, monochromatic light beam from air into the material. The angles are measured from the normal.

The following results are obtained.

Assume \(n_{\text{air}}=1.00\).

(a) Calculate the mean refractive index of the material using the three measurements. (2 marks)

(b) Describe how the student could obtain the angle measurements for a transparent block. Explain why taking several measurements is useful. (2 marks)

(c) State one assumption or limitation of this measurement method. (1 mark)

Question 11 (5 marks)

A concave mirror has a focal length of \(12\ \text{cm}\). An upright object is placed \(30\ \text{cm}\) in front of the mirror.

(a) On your answer paper, draw a labelled principal-ray diagram using two standard principal rays. Mark the focal point, centre of curvature, and image position. (2 marks)

(b) Calculate the image distance using

\[ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \]

where \(f\) is focal length, \(u\) is object distance, and \(v\) is image distance. Use positive distance magnitudes for this real object and real image. (1 mark)

(c) State whether the image is real or virtual, upright or inverted, and enlarged or reduced. (2 marks)

Question 12 (5 marks)

A converging lens has a focal length of \(10\ \text{cm}\). An upright object is placed \(15\ \text{cm}\) to the left of the lens.

(a) On your answer paper, draw a labelled principal-ray diagram using two standard principal rays. Mark the image position. (2 marks)

(b) Calculate the image distance using

\[ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \]

where \(f\) is focal length, \(u\) is object distance, and \(v\) is image distance. Use positive distance magnitudes for this real object and real image. (1 mark)

(c) Describe the image as real or virtual, upright or inverted, and enlarged or reduced. State the magnification. (2 marks)

Question 13 (4 marks)

A person can see nearby objects clearly but finds distant objects blurred. Assume a thin spectacle lens and an otherwise normally functioning eye.

(a) Identify the vision defect. (1 mark)

(b) State the type of spectacle lens used to correct it. (1 mark)

(c) Explain how this lens helps form a clear image on the retina. (2 marks)

Question 14 (7 marks)

(a) Arrange the following electromagnetic radiation types in order of increasing frequency:

Infrared, gamma rays, radio waves, X-rays, microwaves, ultraviolet, visible light. (2 marks)

(b) Select the most suitable electromagnetic radiation for each application and justify your choice using a relevant property.

1. Thermal imaging of a warm roof.
2. Medical imaging of bones. (2 marks)

(c) Explain how electromagnetic waves are produced and propagate. Contrast their need for a medium with that of mechanical waves. (3 marks)

Question 15 (4 marks)

Monochromatic light travels through a vacuum with a wavelength of \(400\ \text{nm}\).

For this question, use

\[ E=\frac{hc}{\lambda} \]

where \(E\) is the energy of one photon, \(h\) is Planck’s constant, and \(\lambda\) is the vacuum wavelength.

(a) Calculate the energy of one photon. (3 marks)

(b) Compare the energy of a photon with wavelength \(800\ \text{nm}\) with the energy calculated in part (a). State the relationship that supports your answer. (1 mark)

Question 16 (5 marks)

A point-like source radiates light uniformly in all directions. The intensity is \(18\ \text{W m}^{-2}\) at a distance of \(2.0\ \text{m}\).

Assume negligible absorption between the source and the measurement points.

(a) Calculate the intensity at a distance of \(6.0\ \text{m}\). (2 marks)

(b) Calculate the intensity at a distance of \(3.0\ \text{m}\). (2 marks)

(c) Explain why intensity decreases with distance from the source. (1 mark)

Question 17 (6 marks)

An ambulance siren emits a frequency of \(700\ \text{Hz}\). A stationary listener hears the siren as the ambulance moves directly towards and then away from them.

The ambulance travels at \(25\ \text{m s}^{-1}\). The speed of sound is \(340\ \text{m s}^{-1}\). Ignore wind.

For a stationary observer, use

\[ f_{\text{obs}}=\frac{f_s v}{v-v_s} \]

for an approaching source, and

\[ f_{\text{obs}}=\frac{f_s v}{v+v_s} \]

for a receding source, where \(f_s\) is the emitted frequency, \(v\) is the speed of sound, and \(v_s\) is source speed.

(a) Calculate the observed frequency as the ambulance approaches. (2 marks)

(b) Calculate the observed frequency as the ambulance moves away. (2 marks)

(c) Explain the change in perceived pitch and state whether the siren’s emitted frequency changes. (2 marks)

Question 18 (5 marks)

A water wave travels from deep water into shallow water. Its speed decreases from \(2.8\ \text{m s}^{-1}\) to \(1.8\ \text{m s}^{-1}\). The frequency remains \(2.0\ \text{Hz}\), and the angle of incidence is \(35^\circ\) to the normal.

Use the wave refraction relationship

\[ \frac{\sin i}{\sin r}=\frac{v_1}{v_2} \]

where \(v_1\) and \(v_2\) are the wave speeds before and after crossing the boundary, respectively.

(a) Calculate the wavelength in deep water and shallow water. (2 marks)

(b) Calculate the angle of refraction. (2 marks)

(c) State whether the wave bends towards or away from the normal, and whether its frequency changes. (1 mark)

03Worked solutions and marking guide

Solution 1

(a) Instantaneous displacement and amplitude

Instantaneous displacement is the signed distance of a point on the wave from its equilibrium position at a particular instant.

Amplitude is the maximum magnitude of displacement from equilibrium. The graph shows an amplitude of \(4.0\ \text{cm}\).

Marking: 2 marks

• 1 mark for defining instantaneous displacement relative to equilibrium.
• 1 mark for defining amplitude as maximum displacement and identifying \(4.0\ \text{cm}\).

(b) Interpreting the horizontal axes

In Graph A, the horizontal axis represents time at a fixed position. It can be used to determine the period or frequency.

In Graph B, the horizontal axis represents position at one instant. It can be used to determine the wavelength.

Marking: 2 marks

• 1 mark for correctly interpreting the time axis and identifying a temporal quantity.
• 1 mark for correctly interpreting the position axis and identifying wavelength.

(c) Period and frequency

The period is the time taken for one complete cycle. Using two consecutive troughs,

\[ \begin{aligned} T&=0.50-0.10\\ &=0.40\ \text{s} \end{aligned} \]

Frequency is the number of complete cycles per second:

\[ \begin{aligned} f&=\frac{1}{T}\\ &=\frac{1}{0.40}\\ &=2.5\ \text{Hz} \end{aligned} \]

Marking: 2 marks

• 1 mark for \(T=0.40\ \text{s}\).
• 1 mark for \(f=2.5\ \text{Hz}\), using \(f=1/T\).

(d) Wavelength and wave speed

Wavelength is the distance between corresponding points on consecutive cycles, such as two adjacent crests.

\[ \lambda=0.80-0.20=0.60\ \text{m} \]

Using \(v=f\lambda\),

\[ \begin{aligned} v&=2.5\times0.60\\ &=1.5\ \text{m s}^{-1} \end{aligned} \]

Marking: 2 marks

• 1 mark for \(\lambda=0.60\ \text{m}\).
• 1 mark for \(v=1.5\ \text{m s}^{-1}\).

Equivalent calculations using other corresponding points receive full credit.

Solution 2

(a) Transverse and longitudinal waves

A wave on a rope is transverse when the rope particles oscillate perpendicular to the direction of propagation.

Sound in air is longitudinal. Air particles oscillate parallel to the direction in which the sound wave travels.

Marking: 2 marks

• 1 mark for correctly classifying the rope wave and describing its particle motion.
• 1 mark for correctly classifying sound and describing its particle motion.

(b) Sound and energy transfer

A compression is a region where air particles are closer together, producing higher pressure. A rarefaction is a region where particles are more spread out, producing lower pressure.

Air particles oscillate around their equilibrium positions. Their interactions transfer the pressure disturbance and energy through the air, but the air itself is not transported in bulk from the source to the listener.

Marking: 2 marks

• 1 mark for explaining compressions, rarefactions, and pressure variation.
• 1 mark for explaining local particle oscillation and energy transfer without bulk matter transport.

(c) Pitch and loudness

Pitch is most directly associated with frequency. A higher-frequency sound is generally perceived as higher-pitched.

Loudness is most directly associated with sound intensity, which is related to wave amplitude. Greater amplitude generally produces greater intensity under comparable conditions. Perceived loudness also depends on factors such as frequency and the listener’s hearing sensitivity.

Marking: 2 marks

• 1 mark for relating pitch to frequency.
• 1 mark for relating loudness to amplitude or intensity, without treating frequency as the direct measure of loudness.

Solution 3

(a) Wavelength

The wave equation is

\[ v=f\lambda \]

Rearranging,

\[ \begin{aligned} \lambda&=\frac{v}{f}\\ &=\frac{24}{8.0}\\ &=3.0\ \text{m} \end{aligned} \]

Solution 4

(a) Progressive and standing waves

A progressive wave travels through a medium and transfers energy from one location to another.

A standing wave has a stationary pattern of nodes and antinodes. It does not transfer net energy along the string in the ideal standing-wave model.

Marking: 2 marks

• 1 mark for describing the travelling pattern and net energy transfer of a progressive wave.
• 1 mark for describing the stationary pattern and absence of net energy transfer along an ideal standing wave.

(b) Nodes and antinodes

A node is a point that remains at zero displacement in a standing wave. Its oscillation amplitude is zero.

An antinode is a point where the oscillation amplitude is maximum.

Marking: 2 marks

• 1 mark for defining a node and its zero amplitude.
• 1 mark for defining an antinode and its maximum amplitude.

(c) Superposition

The principle of superposition states that the resultant displacement is the algebraic sum of the individual displacements.

At P,

\[ y_P=+3+2=+5\ \text{cm} \]

The displacements reinforce each other, so the interference is constructive.

At Q,

\[ y_Q=+3-5=-2\ \text{cm} \]

The displacements oppose each other, so the interference is destructive.

Marking: 2 marks

• 1 mark for the \(+5\ \text{cm}\) resultant and constructive interference at P.
• 1 mark for the \(-2\ \text{cm}\) resultant and destructive interference at Q.

Solution 5

(a) Reflection

For reflection at a straight boundary, the angle of reflection equals the angle of incidence. The reflected angle is therefore \(35^\circ\).

A pulse reflecting from a fixed end is inverted, corresponding to a phase change of \(180^\circ\). A pulse reflecting from a free end is not inverted.

Marking: 2 marks

• 1 mark for the reflected angle of \(35^\circ\).
• 1 mark for correctly distinguishing the phase behaviour at fixed and free ends.

(b) Diffraction

The wave with wavelength \(0.60\ \text{m}\) spreads more.

Diffraction is strongest when the gap width is comparable to or smaller than the wavelength. The \(0.60\ \text{m}\) wavelength is much larger than the \(0.15\ \text{m}\) gap, so it produces greater spreading than the \(0.10\ \text{m}\) wavelength.

Marking: 2 marks

• 1 mark for selecting the \(0.60\ \text{m}\) wave.
• 1 mark for explaining the relationship between wavelength, gap size, and diffraction.

(c) Refraction of water waves

The wave frequency remains unchanged because it is set by the source. Since \(v=f\lambda\), a decrease in wave speed produces a decrease in wavelength.

When the wave enters the slower region obliquely, it bends towards the normal.

Marking: 2 marks

• 1 mark for stating that frequency remains constant and wavelength decreases.
• 1 mark for identifying bending towards the normal.

Solution 6

(a) Wave speed

Use \(v=f\lambda\).

\[ \begin{aligned} v&=1.50\times0.80\\ &=1.20\ \text{m s}^{-1} \end{aligned} \]

Solution 7

(a) Wave speed

Use

\[ v=\sqrt{\frac{F_T}{\mu}} \]

Substituting the given values,

\[ \begin{aligned} v&=\sqrt{\frac{36.0}{4.00\times10^{-3}}}\\ &=\sqrt{9000}\\ &=94.9\ \text{m s}^{-1} \end{aligned} \]

Marking: 2 marks

• 1 mark for correct substitution into the speed equation.
• 1 mark for \(v=94.9\ \text{m s}^{-1}\).

(b) Fundamental and third-harmonic frequencies

For the fundamental, \(n=1\):

\[ \begin{aligned} f_1&=\frac{v}{2L}\\ &=\frac{94.9}{2(0.750)}\\ &=63.2\ \text{Hz} \end{aligned} \]

For the third harmonic, \(n=3\):

\[ \begin{aligned} f_3&=\frac{3v}{2L}\\ &=3f_1\\ &=190\ \text{Hz} \end{aligned} \]

Answers: \(f_1=63.2\ \text{Hz}\) and \(f_3=190\ \text{Hz}\).

Marking: 2 marks

• 1 mark for the fundamental frequency.
• 1 mark for the third-harmonic frequency.

(c) Effect of quadrupling tension

Wave speed on the string is proportional to the square root of tension:

\[ v\propto\sqrt{F_T} \]

Quadrupling the tension multiplies the speed by \(\sqrt{4}=2\). Since the length is unchanged, the fundamental frequency also doubles.

The new fundamental frequency is approximately

\[ f_1'=2(63.2)=126\ \text{Hz} \]

Marking: 2 marks

• 1 mark for explaining that quadrupling tension doubles wave speed.
• 1 mark for concluding that the fundamental frequency doubles to approximately \(126\ \text{Hz}\).

Solution 8

(a) Speed of light in glass

The refractive index relationship is

\[ n=\frac{c}{v} \]

Rearranging,

\[ v=\frac{c}{n} \]

Therefore,

\[ \begin{aligned} v&=\frac{3.00\times10^8}{1.50}\\ &=2.00\times10^8\ \text{m s}^{-1} \end{aligned} \]

Solution 9

(a) Critical angle

At the critical angle, the refracted ray travels along the boundary, so its angle of refraction is \(90^\circ\).

Using Snell’s law,

\[ n_{\text{glass}}\sin c=n_{\text{air}}\sin90^\circ \]

Therefore,

\[ \begin{aligned} \sin c&=\frac{1.00}{1.50}\\ &=0.667 \end{aligned} \]

Thus,

\[ c=\sin^{-1}(0.667)=41.8^\circ \]

Solution 10

(a) Mean refractive index

For light travelling from air into the material, Snell’s law gives

\[ n_{\text{air}}\sin i=n_{\text{material}}\sin r \]

Since \(n_{\text{air}}=1.00\),

\[ n_{\text{material}}=\frac{\sin i}{\sin r} \]

Using the three measurements gives refractive index values close to \(1.50\) in each case. Their mean, to three significant figures, is

\[ \boxed{n_{\text{material}}=1.50} \]

Small differences between individual values arise from the rounding of the measured angles.

Marking: 2 marks

• 1 mark for using \(n=\sin i/\sin r\).
• 1 mark for obtaining a mean refractive index of \(1.50\).

(b) Measuring the angles

Direct a narrow light beam through the transparent block. Mark the incident and refracted ray paths, draw a normal at the boundary, and measure the angles between the rays and the normal with a protractor.

Repeat for several incident angles. Multiple measurements allow the student to average the calculated refractive indices or plot \(\sin i\) against \(\sin r\), reducing the effect of random measurement errors.

Marking: 2 marks

• 1 mark for a valid method of tracing the rays and measuring angles from the normal.
• 1 mark for explaining how repeated measurements, averaging, or a suitable graph improves reliability.

(c) Limitation

One limitation is uncertainty in measuring the angles. A thick beam, an unclear ray path, or limited protractor precision can affect the calculated refractive index.

The method also assumes the material is uniform and that the measurements correspond to the same wavelength of light.

Marking: 1 mark

• 1 mark for one valid limitation or assumption.

Solution 11

(a) Principal-ray diagram

For a concave mirror, mark the focal point \(F\) at \(12\ \text{cm}\) from the mirror and the centre of curvature \(C\) at \(24\ \text{cm}\).

Use two rays from the top of the object:

• A ray parallel to the principal axis reflects through \(F\).
• A ray passing through \(C\) reflects back along its original path.

The reflected rays intersect between \(F\) and \(C\), locating the image.

Marking: 2 marks

• 1 mark for a correctly drawn first principal ray.
• 1 mark for the second principal ray and the correct image intersection.

(b) Image distance

Using the mirror equation,

\[ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \]

Rearranging,

\[ \begin{aligned} \frac{1}{v}&=\frac{1}{f}-\frac{1}{u}\\ &=\frac{1}{12}-\frac{1}{30}\\ &=\frac{1}{20} \end{aligned} \]

Therefore,

\[ v=20\ \text{cm} \]

Solution 12

(a) Principal-ray diagram

For a converging lens, mark the focal points \(10\ \text{cm}\) on either side of the lens.

Use two rays from the top of the object:

• A ray parallel to the principal axis refracts through the focal point on the opposite side.
• A ray through the optical centre continues approximately undeviated for a thin lens.

The rays intersect on the right-hand side of the lens, locating the image.

Marking: 2 marks

• 1 mark for a correctly drawn first principal ray.
• 1 mark for the second principal ray and the correct image intersection.

(b) Image distance

Using the lens equation,

\[ \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \]

Rearranging,

\[ \begin{aligned} \frac{1}{v}&=\frac{1}{f}-\frac{1}{u}\\ &=\frac{1}{10}-\frac{1}{15}\\ &=\frac{1}{30} \end{aligned} \]

Therefore,

\[ v=30\ \text{cm} \]

Solution 13

(a) Vision defect

The person has myopia, or short-sightedness. Distant objects appear blurred while nearby objects can remain clear.

Marking: 1 mark

• 1 mark for identifying myopia.

(b) Correcting lens

A diverging lens, also called a concave lens, is used to correct myopia.

Marking: 1 mark

• 1 mark for identifying a diverging lens.

(c) How the correction works

In a myopic eye, light from a distant object would focus in front of the retina rather than on it.

The diverging spectacle lens spreads the incoming rays before they enter the eye. This adjusts the combined optical system so the eye can focus the image on the retina.

Marking: 2 marks

• 1 mark for explaining that the uncorrected eye focuses distant light in front of the retina.
• 1 mark for explaining how the diverging lens shifts the focus back onto the retina.

Solution 14

(a) Electromagnetic spectrum

The correct order of increasing frequency is

\[ \text{Radio} \rightarrow \text{Microwave} \rightarrow \text{Infrared} \rightarrow \text{Visible} \rightarrow \text{Ultraviolet} \rightarrow \text{X-ray} \rightarrow \text{Gamma} \]

Frequency increases from radio waves to gamma rays. Wavelength decreases in the opposite direction.

Marking: 2 marks

• 1 mark for correctly ordering radio waves through visible light.
• 1 mark for correctly ordering visible light through gamma rays.

(b) Applications

1. Thermal imaging: Infrared radiation is suitable because warm objects emit infrared radiation, which thermal imaging systems detect to show temperature differences.
2. Bone imaging: X-rays are suitable because they pass through soft tissue more readily than bone, producing contrast in the resulting image.

Marking: 2 marks

• 1 mark for selecting infrared and linking it to thermal radiation.
• 1 mark for selecting X-rays and explaining their differing transmission through bone and soft tissue.

(c) Production and propagation

Electromagnetic waves can be produced by accelerating electric charges. The changing electric and magnetic fields form a wave that propagates through space.

The electric field and magnetic field oscillate perpendicular to each other and to the direction of propagation. The wave transports energy.

Unlike mechanical waves, electromagnetic waves do not require a material medium. They can travel through a vacuum, whereas mechanical waves require a medium whose particles or other physical elements can oscillate.

Marking: 3 marks

• 1 mark for linking accelerating charges to electromagnetic-wave production.
• 1 mark for describing the transverse electric and magnetic fields and energy propagation.
• 1 mark for correctly contrasting electromagnetic waves in a vacuum with mechanical waves that require a medium.

Solution 15

(a) Energy of one photon

First convert the wavelength to metres:

\[ 400\ \text{nm}=4.00\times10^{-7}\ \text{m} \]

Use

\[ E=\frac{hc}{\lambda} \]

Substituting,

\[ \begin{aligned} E&=\frac{(6.63\times10^{-34})(3.00\times10^8)} {4.00\times10^{-7}}\\ &=4.97\times10^{-19}\ \text{J} \end{aligned} \]

Solution 16

(a) Intensity at \(6.0\ \text{m}\)

For a point-like source radiating uniformly in all directions, intensity follows the inverse-square law:

\[ I\propto\frac{1}{r^2} \]

Therefore,

\[ I_2=I_1\left(\frac{r_1}{r_2}\right)^2 \]

Substituting,

\[ \begin{aligned} I_2&=18\left(\frac{2.0}{6.0}\right)^2\\ &=2.0\ \text{W m}^{-2} \end{aligned} \]

Solution 17

(a) Ambulance approaching

Use

\[ f_{\text{obs}}=\frac{f_s v}{v-v_s} \]

Substituting,

\[ \begin{aligned} f_{\text{obs}}&=\frac{700\times340}{340-25}\\ &=\frac{238000}{315}\\ &=756\ \text{Hz} \end{aligned} \]

Solution 18

(a) Wavelength in each region

Use

\[ v=f\lambda \]

Rearranging,

\[ \lambda=\frac{v}{f} \]

In deep water,

\[ \lambda_1=\frac{2.8}{2.0}=1.4\ \text{m} \]

In shallow water,

\[ \lambda_2=\frac{1.8}{2.0}=0.90\ \text{m} \]

Answers: \(1.4\ \text{m}\) in deep water and \(0.90\ \text{m}\) in shallow water.

Marking: 2 marks

• 1 mark for the deep-water wavelength.
• 1 mark for the shallow-water wavelength.

(b) Angle of refraction

Use the supplied relationship

\[ \frac{\sin i}{\sin r}=\frac{v_1}{v_2} \]

Rearranging,

\[ \sin r=\sin i\frac{v_2}{v_1} \]

Substituting,

\[ \begin{aligned} \sin r&=\sin35^\circ\left(\frac{1.8}{2.8}\right)\\ &=0.369 \end{aligned} \]

Therefore,

\[ r=\sin^{-1}(0.369)=21.7^\circ \]