An air molecule is situated at point X in the pipe at t = 0. Describe the motion of this air molecule during one complete cycle of the standing wave beginning from t = 0.

The speed of sound c for longitudinal waves in air is given by

$c=\sqrt{\frac{K}{\rho }}$

where ρ is the density of the air and K is a constant.

A student measures f to be 120 Hz when the length of the pipe is 1.4 m. The density of the air in the pipe is 1.3 kg m^–3. Determine, in kg m^–1 s^–2, the value of K for air.

Demonstrate, using a second ray, that the image appears to come from the position indicated.

Outline why the observer detects a series of increases and decreases in the intensity of the received signal as the boat moves along the line XY.

«air molecule» moves to the right and then back to the left ✔

returns to X/original position ✔

wavelength = 2 × 1.4 = «2.8 m» ✔

c = «f λ =» 120 × 2.8 «= 340 m s⁻¹» ✔

K = «ρc² = 1.3 × 340² =» 1.5 × 10⁵ ✔

construction showing formation of image ✔

Another straight line/ray from image through the wall with line/ray from intersection at wall back to transmitter. Reflected ray must intersect boat.

interference pattern is observed

OR

interference/superposition mentioned ✔

maximum when two waves occur in phase/path difference is nλ

OR

minimum when two waves occur 180° out of phase/path difference is (n + ½)λ ✔

A series of dark and bright fringes appears on the screen. Explain how a dark fringe is formed.

The wavelength of the beam as observed on Earth is 633.0 nm. The separation between a dark and a bright fringe on the screen is 4.50 mm. Calculate D.

Calculate the wavelength of the light in water.

State two ways in which the intensity pattern on the screen changes.

superposition of light from each slit / interference of light from both slits

with path/phase difference of any half-odd multiple of wavelength/any odd multiple of $\pi$ (in words or symbols)

producing destructive interference

Ignore any reference to crests and troughs.

[3 marks]

evidence of solving for D «D = $\frac{sd}{\lambda }$»

«$\frac{4.50\times {10}^{-3}\times 0.300\times {10}^{-3}}{633.0\times {10}^{-9}}$ × 2» = 4.27 «m»

Award [1] max for 2.13 m.

[2 marks]

$\frac{633.0}{1.33}$ = 476 «nm»

[1 mark]

distance between peaks decreases

intensity decreases

[2 marks]

Calculate, in m s^–1, the speed for this wave.

Calculate, in Hz, the frequency for this wave.

The graph also shows the displacement of two particles, P and Q, in the medium at t = 0. State and explain which particle has the larger magnitude of acceleration at t = 0.

State the number of all other points on the string that have the same amplitude and phase as X.

The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string.

v = «$\frac{0.05}{0.20\times {10}^{-3}}=$» 250 «m s^–1»✔

λ = 0.30 «m» ✔
$f$ = «$\frac{250}{0.30}=$» 830 «Hz» ✔

NOTE: Allow ECF from (a)(i)
Allow ECF from wrong wavelength for MP2

Q ✔
acceleration is proportional to displacement «and Q has larger displacement» ✔

3 «points» ✔

first harmonic mode drawn ✔

NOTE: Allow if only one curve drawn, either solid or dashed.

Describe two ways in which standing waves differ from travelling waves.

Outline how a standing wave forms in the tube.

The tube is raised until the loudness of the sound reaches a maximum for a second time.

Draw, on the following diagram, the position of the nodes in the tube when the second maximum is heard.

Between the first and second positions of maximum loudness, the tube is raised through 0.37 m. The speed of sound in the air in the tube is 320 m s⁻¹. Determine the frequency of the sound emitted by the loudspeaker.

energy is not propagated by standing waves ✓

amplitude constant for travelling waves OR amplitude varies with position for standing waves OR standing waves have nodes/antinodes ✓

phase varies with position for travelling waves OR phase constant inter-node for standing waves ✓

travelling waves can have any wavelength OR standing waves have discrete wavelengths ✓

OWTTE

«sound» wave «travels down tube and» is reflected ✓

incident and reflected wave superpose/combine/interfere ✓

OWTTE

Do not award MP1 if the reflection is quoted at the walls/container

nodes shown at water surface AND $\frac{2}{3}$way up tube (by eye) ✓

Accept drawing of displacement diagram for correct harmonic without nodes specifically identified.

Award [0] if waveform is shown below the water surface

$\lambda =0.74 «\text{m}»$ ✓

$f=«\frac{c}{\lambda }=\frac{320}{0.74}=» 430 «\text{Hz}»$ ✓

Allow ECF from MP1

Explain the variation in intensity.

Adjacent minima are separated by a distance of 0.12 m. Calculate $f$.

The metal plate is replaced by a wooden plate that reflects a lower intensity sound wave than the metal plate.

State and explain the differences between the sound intensities detected by the same microphone with the metal plate and the wooden plate.

«incident and reflected» waves superpose/interfere/combine ✓

«that leads to» standing waves formed OR nodes and antinodes present ✓

at antinodes / maxima there is maximum intensity / constructive interference / «displacement» addition / louder sound ✓

at nodes / minima there is minimum intensity / destructive interference / «displacement» cancellation / quieter sound ✓

OWTTE

Allow a sketch of a standing wave for MP2

Allow a correct reference to path or phase differences to identify constructive / destructive interference

wavelength = 0.24 «m» ✓

$f$ = «$\frac{340}{0.24}$=» 1.4 «kHz» OR 1400 «Hz» ✓

Allow ECF from MP1

relates intensity to amplitude ✓

antinodes / maximum intensity will be decreased / quieter ✓

nodes / minimum will be increased / louder ✓

difference in intensities will be less ✓

maxima and minima are at the same positions ✓

OWTTE

ai) On most occasions it looked like students knew more than they could successfully communicate. Lots of answers talked about interference between the 2 waves, or standing waves being produced but did not go on to add detail. Candidates should take note of how many marks the question part is worth and attempt a structure of the answer that accounts for that. At SL there were problems recognizing a standard question requiring the typical explanation of how a standing wave is established.

3aii) By far the most common answer was 2800 Hz, not doubling the value given to get the correct wavelength. That might suggest that some students misinterpreted adjacent minima as two troughs, therefore missing to use the information to correctly determine the wavelength as 0.24 m.

b) A question that turned out to be a good high level discriminator. Most candidates went for an answer that generally had everything at a lower intensity and didn't pick up on the relative amount of superposition. Those that did answer it very well, with very clear explanations, succeeded in recognizing that the nodes would be louder and the anti-nodes would be quieter than before.

Particle P in the metal sheet performs simple harmonic oscillations. When the displacement of P is 3.2 μm the magnitude of its acceleration is 7.9 m s^-2. Calculate the magnitude of the acceleration of P when its displacement is 2.3 μm.

The wave is incident at point Q on the metal–air boundary. The wave makes an angle of 54° with the normal at Q. The speed of sound in the metal is 6010 m s^–1 and the speed of sound in air is 340 m s^–1. Calculate the angle between the normal at Q and the direction of the wave in air.

State the frequency of the wave in air.

Determine the wavelength of the wave in air. 

The sound wave in air in (c) enters a pipe that is open at both ends. The diagram shows the displacement, at a particular time T, of the standing wave that is set up in the pipe.

On the diagram, at time T, label with the letter C a point in the pipe that is at the centre of a compression.

Expression or statement showing acceleration is proportional to displacement ✔

so «$7.9\times \frac{2.3}{3.2}»=5.7«\text{m }{\text{s}}^{-2}$»  ✔

$\sin \theta =\frac{340}{6010}\times \sin {54}^{\circ }$   ✔

θ = 2.6° ✔

f = 250 «Hz» OR Same OR Unchanged ✔

$\lambda =«\frac{340}{250}=»1.36\approx 1.4«\text{m}$» ✔

any point labelled C on the vertical line shown below ✔

eg:

This was well answered at both levels.

Many scored full marks on this question. Common errors were using the calculator in radian mode or getting the equation upside down.

Many used a ratio of the speeds to produce a new frequency of 14Hz (340 x 250/6010). It would have helped candidates if they had been aware that the command term ‘state’ means ‘give a specific name, value or other brief answer without explanation or calculation.’

This was answered well at both levels.

Explain, with reference to the light passing through the slits, why a series of voltage peaks occurs.

The slits are separated by 1.5 mm and the laser light has a wavelength of 6.3 x 10^–7 m. The slits are 5.0 m from the train track. Calculate the separation between two adjacent positions of the train when the output voltage is at a maximum.

Estimate the speed of the train.

In another experiment the student replaces the light sensor with a sound sensor. The train travels away from a loudspeaker that is emitting sound waves of constant amplitude and frequency towards a reflecting barrier.

The sound sensor gives a graph of the variation of output voltage with time along the track that is similar in shape to the graph shown in the resource. Explain how this effect arises.

«light» superposes/interferes

pattern consists of «intensity» maxima and minima
OR
consisting of constructive and destructive «interference»

voltage peaks correspond to interference maxima

«$s=\frac{\lambda D}{d}=\frac{6.3\times {10}^{-7}\times 5.0}{1.5\times {10}^{-3}}=$» 2.1 x 10^–3 «m» 

If no unit assume m.
Correct answer only.

correct read-off from graph of 25 m s

v = «$\frac{x}{t}=\frac{2.1\times {10}^{-3}}{25\times {10}^{-3}}=$» 8.4 x 10^–2 «m s^–1»

Allow ECF from (b)(i)

ALTERNATIVE 1

«reflection at barrier» leads to two waves travelling in opposite directions

mention of formation of standing wave

maximum corresponds to antinode/maximum displacement «of air molecules»
OR
complete cancellation at node position

Explain why intensity maxima are observed at X and Y.

The distance from S1 to Y is 1.243 m and the distance from S2 to Y is 1.181 m.

Determine the frequency of the microwaves.

Outline one reason why the maxima observed at W, X and Y will have different intensities from each other.

two waves superpose/mention of superposition/mention of «constructive» interference ✔

they arrive in phase/there is a path length difference of an integer number of wavelengths ✔

Ignore references to nodes/antinodes.

path difference = 0.062 «m» ✔

so wavelength = 0.031 «m» ✔

frequency = 9.7 × 10⁹ «Hz» ✔

If no unit is given, assume the answer is in Hz. Accept other prefixes (eg 9.7 GHz)

Award [2 max] for 4.8 x 10⁹ Hz.

intensity varies with distance OR points are different distances from the slits ✔

Accept “Intensity is modulated by a single slit diffraction envelope”.

Many candidates were able to discuss the interference that is taking place in this question, but few were able to fully describe the path length difference. That said, the quality of responses on this type of question seems to have improved over the last few examination sessions with very few candidates simply discussing the crests and troughs of waves.

Many candidates struggled with this question. Few were able to calculate a proper path length difference, and then use that to calculate the wavelength and frequency. Many candidates went down blind paths of trying various equations from the data booklet, and some seemed to believe that the wavelength is just the reciprocal of the frequency.

This is one of many questions on this paper where candidates wrote vague answers that did not clearly connect to physics concepts or include key information. There were many overly simplistic answers like “they are farther away” without specifying what they are farther away from. Candidates should be reminded that their responses should go beyond the obvious and include some evidence of deeper understanding.

Calculate the speed of light inside the ice cube.

Show that no light emerges from side AB.

Sketch, on the diagram, the subsequent path of the light ray.

Determine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.

Specific latent heat of fusion of ice  = 330 kJ kg^–1
Specific heat capacity of ice            = 2.1 kJ kg^–1 k^–1
Density of ice                                = 920 kg m^–3

Outline the difference between the molecular structure of a solid and a liquid.

«v = c$\frac{\text{sin }i}{\text{sin }r}$ =» $\frac{3\times {10}^8\times \text{sin}\left(33\right)}{\text{sin}\left(46\right)}$

2.3 x 10⁸ «m s^–1»

light strikes AB at an angle of 57°

critical angle is «sin^–1$\left(\frac{2.3}{3}\right)$ =» 50.1°

49.2° from unrounded value

angle of incidence is greater than critical angle so total internal reflection

OR

light strikes AB at an angle of 57°

calculation showing sin of “refracted angle” = 1.1

statement that since 1.1>1 the angle does not exist and the light does not emerge

[Max 3 marks]

total internal reflection shown

ray emerges at opposite face to incidence

Judge angle of incidence=angle of reflection by eye or accept correctly labelled angles

With sensible refraction in correct direction

mass = «volume x density» (0.75)³ x 920 «= 388 kg»

energy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 10⁷ J»

energy required to melt = 388 x 330 x 10³ «= 1.28 x 10⁸ J»

1.4 x 10⁸ «J» OR 1.4 x 10⁵ «kJ»

Accept any consistent units

Award [3 max] for answer which uses density as 1000 kg^–3 (1.5× 10⁸ «J»)

in solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction

in liquid, bonds between molecules can be broken and re-form

OWTTE

Accept converse argument for liquids

[Max 1 Mark]

Outline how a standing wave is produced on the string.

Show that the speed of the wave on the string is about 240 m s⁻¹.

Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position.

«travelling» wave moves along the length of the string and reflects «at fixed end» ✓

superposition/interference of incident and reflected waves ✓

the superposition of the reflections is reinforced only for certain wavelengths ✓ 

$\lambda =2l=2\times 0.62=«1.24 \text{m}»$ ✓

$v=f\lambda =195\times 1.24=242 «{\text{m\hspace{0.05em}s}}^{-1}»$ ✓

Answer must be to 3 or more sf or working shown for MP2.

straight line through origin with negative gradient ✓

Outline how the standing wave is formed.

Draw an arrow on the diagram to represent the direction of motion of the molecule at X.

Label a position N that is a node of the standing wave.

The speed of sound is 340 m s^–1 and the length of the pipe is 0.30 m. Calculate, in Hz, the frequency of the sound.

The speed of sound in air is 340 m s^–1 and in water it is 1500 m s^–1.

The wavefronts make an angle θ with the surface of the water. Determine the maximum angle, θ_max, at which the sound can enter water. Give your answer to the correct number of significant figures.

Draw lines on the diagram to complete wavefronts A and B in water for θ < θ_max.

the incident wave «from the speaker» and the reflected wave «from the closed end»

superpose/combine/interfere

Allow superimpose/add up

Do not allow meet/interact

[1 mark]

Horizontal arrow from X to the right

MP2 is dependent on MP1

Ignore length of arrow

[1 mark]

P at a node

[1 mark]

wavelength is λ = «$\frac{4\times 0.30}{3}$ =» 0.40 «m»

f = «$\frac{340}{0.40}$» 850 «Hz»

Award [2] for a bald correct answer

Allow ECF from MP1

[2 marks]

$\frac{\sin {\theta }_c}{340}=\frac{1}{1500}$

θ_c = 13«°»

Award [2] for a bald correct answer

Award [2] for a bald answer of 13.1

Answer must be to 2/3 significant figures to award MP2

Allow 0.23 radians

[2 marks]

correct orientation

greater separation

Do not penalize the lengths of A and B in the water

Do not penalize a wavefront for C if it is consistent with A and B

MP1 must be awarded for MP2 to be awarded

[2 marks]

Deduce that a minimum intensity of sound is heard at P.

A microphone moves along the line from P to Q. PQ is normal to the line midway between the loudspeakers.

The intensity of sound is detected by the microphone. Predict the variation of detected intensity as the microphone moves from P to Q.

When both loudspeakers are operating, the intensity of sound recorded at Q is $I_0$. Loudspeaker B is now disconnected. Loudspeaker A continues to emit sound with unchanged amplitude and frequency. The intensity of sound recorded at Q changes to $I_{\mathrm{A}}$.

Estimate $\frac{I_{\mathrm{A}}}{I_0}$.

wavelength$=\frac{340}{850}=0.40 «\mathrm{m}»$✓

path difference $=1.8 «\mathrm{m}»$ ✓

$1.8 «\mathrm{m}»=4.5\lambda$ OR $\frac{1.8}{0.20}=9$«half-wavelengths» ✓

waves meet in antiphase «at P»
OR
destructive interference/superposition «at P» ✓

Allow approach where path length is calculated in terms of number of wavelengths; along path A ($\mathit{56}\mathit{.}\mathit{25}$) and
path B ($\mathit{60}\mathit{.}\mathit{75}$) for MP2, hence path difference $\mathit{4}\mathit{.}\mathit{5}$ wavelengths for MP3 

«equally spaced» maxima and minima ✓

a maximum at Q ✓

four «additional» maxima «between P and Q» ✓

the amplitude of sound at Q is halved ✓
«intensity is proportional to amplitude squared hence» $\frac{I_{\mathrm{A}}}{I_0}=\frac{1}{4}$ ✓

Explain why the received intensity varies between maximum and minimum values.

State and explain the wavelength of the sound measured at M.

B is placed at the first minimum. The frequency is then changed until the received intensity is again at a maximum.

Show that the lowest frequency at which the intensity maximum can occur is about 3 kHz.

Speed of sound = 340 m s⁻¹

movement of B means that path distance is different « between BM and AM »
OR
movement of B creates a path difference «between BM and AM» ✓

interference
OR
superposition «of waves» ✓

maximum when waves arrive in phase / path difference = n x lambda
OR
minimum when waves arrive «180° or $\pi$ » out of phase / path difference = (n+½) x lambda ✓

wavelength = 26 cm ✓

peak to peak distance is the path difference which is one wavelength

OR

this is the distance B moves to be back in phase «with A» ✓

Allow 25 − 27 cm for MP1.

«$\frac{\lambda }{2}$» = 13 cm ✓

$f=$«$\frac{c}{\lambda }=\frac{340}{0.13}=$» 2.6 «kHz» ✓

Allow ½ of wavelength from (b) or data from graph.

This was an "explain" questions, so examiners were looking for a clear discussion of the movement of speaker B creating a changing path difference between B and the microphone and A and the microphone. This path difference would lead to interference, and the examiners were looking for a connection between specific phase differences or path differences for maxima or minima. Some candidates were able to discuss basic concepts of interference (e.g. "there is constructive and destructive interference"), but failed to make clear connections between the physical situation and the given graph. A very common mistake candidates made was to think the question was about intensity and to therefore describe the decrease in peak height of the maxima on the graph. Another common mistake was to approach this as a Doppler question and to attempt to answer it based on the frequency difference of B.

Many candidates recognized that the wavelength was 26 cm, but the explanations were lacking the details about what information the graph was actually providing. Examiners were looking for a connection back to path difference, and not simply a description of peak-to-peak distance on the graph. Some candidates did not state a wavelength at all, and instead simply discussed the concept of wavelength or suggested that the wavelength was constant.

This was a "show that" question that had enough information for backwards working. Examiners were looking for evidence of using the wavelength from (b) or information from the graph to determine wavelength followed by a correct substitution and an answer to more significant digits than the given result.

Outline what is meant by the principle of superposition of waves.

Red laser light is incident on a double slit with a slit separation of 0.35 mm.
A double-slit interference pattern is observed on a screen 2.4 m from the slits.
The distance between successive maxima on the screen is 4.7 mm.

Calculate the wavelength of the light. Give your answer to an appropriate number of significant figures.

Explain the change to the appearance of the interference pattern when the red-light laser is replaced by one that emits green light.

One of the slits is now covered.

Describe the appearance of the pattern on the screen.

when 2 waves meet the resultant displacement

is the «vector» sum of their individual displacements

Displacement should be mentioned at least once in MP 1 or 2.

λ = $\frac{4.7\times {10}^{-3}\times 0.35\times {10}^{-3}}{2.4}$

= 6.9 x 10^–7 «m»

answer to 2 SF

Allow missed powers of 10 for MP1.

green wavelength smaller than red

fringe separation / distance between maxima decreases

Allow ECF from MP1.

bright central maximum

subsidiary maxima «on either side»

the width of the central fringe is twice / larger than the width of the subsidiary/secondary fringes/maxima

OR

intensity of pattern is decreased

Allow marks from a suitably labelled intensity graph for single slit diffraction.

Show that the intensity of solar radiation at the orbit of Mars is about 600 W m^–2.

Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.

The atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the greenhouse effect is not significant on Mars.

use of $I\propto \frac{1}{r^2}$ «1.36 × 10³ × $\frac{1}{{1.5}^2}$» ✔

604 «W m^–2» ✔

use of $\frac{600}{4}$ for mean intensity ✔

temperature/K = «$\sqrt[4]{\frac{600}{4\times 5.67\times {10}^{-8}}}=$» 230 ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

OR

low pressure means too little radiation re-radiated back to Mars ✔

Calculate the wavelength of the wave.

State the phase difference between the two waves.

Identify a time at which the displacement of P is zero.

Estimate the amplitude of the resultant wave.

Calculate the length of the tube.

A particle in the tube has its equilibrium position at the open end of the tube.
State and explain the direction of the velocity of this particle at time $t=\frac{T}{8}$.

Draw on the diagram the standing wave at time $t=\frac{T}{4}$.

$T=4\times {10}^{-3}$«s» or $f=250$«Hz» ✓

$\lambda =340\times 4.0\times {10}^{-3}=1.36\approx 1.4$«m» ✓

Allow ECF from MP1.
Award [2] for a bald correct answer.

«±» $\frac{\mathrm{\pi }}{2}/90°$  OR  $\frac{3\mathrm{\pi }}{2}/270°$ ✓

1.5 «ms» ✓

8.0 OR 8.5 «μm» ✓

From the graph on the paper, value is 8.0. From the calculated correct trig functions, value is 8.49.

L = «$\frac{3}{4}\lambda =$» 0.90 «m» ✓

to the right ✓

displacement is getting less negative

OR

change of displacement is positive ✓

horizontal line drawn at the equilibrium position ✓

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻²

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻²

Show that the equilibrium surface temperature of Titan is about 90 K.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

Two microwave transmitters, X and Y, are placed 12 cm apart and are connected to the same source. A single receiver is placed 54 cm away and moves along a line AB that is parallel to the line joining X and Y.

Maxima and minima of intensity are detected at several points along AB.

(i) Explain the formation of the intensity minima.

(ii) The distance between the central maximum and the first minimum is 7.2 cm. Calculate the wavelength of the microwaves.

Radio waves are emitted by a straight conducting rod antenna (aerial). The plane of polarization of these waves is parallel to the transmitting antenna.

An identical antenna is used for reception. Suggest why the receiving antenna needs to be be parallel to the transmitting antenna.

The receiving antenna becomes misaligned by 30° to its original position.

The power of the received signal in this new position is 12 μW.

(i) Calculate the power that was received in the original position.

(ii) Calculate the minimum time between the wave leaving the transmitting antenna and its reception.

i
minima = destructive interference
Allow “crest meets trough”, but not “waves cancel”.
Allow “destructive superposition” but not bald “superposition”.

at minima waves meet 180° or π out of phase
Allow similar argument in terms of effective path difference of $\frac{\lambda }{2}$.
Allow “antiphase”, allow “completely out of phase”
Do not allow “out of phase” without angle. Do not allow $\frac{n\lambda }{2}$ unless qualified to odd integers but accept $\left(n+\frac{1}{2}\right)\lambda$

ii
$\lambda =\frac{sd}{D}$ or $\lambda =\frac{12\times 2\times 7.2}{54}=$ or $\lambda =\frac{12\times 7.2}{54}=$ seen
Award [2] for a bald correct answer.

$\lambda =$ «$\frac{12\times 2\times 7.2}{54}=$» 3.2 «cm»
Award [1 max] for 1.6 «cm»
Award [2 max] to a trigonometric solution in which candidate works out individual path lengths and equates to $\frac{\lambda }{2}$.

ALTERNATIVE 1

the component of the polarized signal in the direction of the receiving antenna

is a maximum «when both are parallel»

ALTERNATIVE 2:

receiving antenna must be parallel to plane of polarisation
for power/intensity to be maximum

Do not accept “receiving antenna must be parallel to transmitting antenna”

ALTERNATIVE 3:

refers to Malus’ law or  I = I₀ cos²θ

explains that I is max when θ = 0

ALTERNATIVE 4:

an electric current is established in the receiving antenna which is proportional to the electric field

maximum current in receiving antenna requires maximum field «and so must be parallel»

i
$I_0=\frac{I}{{\cos }^2\theta }$ or $\frac{12}{{\cos }^{2}30}$ seen
Award [2] for bald correct answer.
Award [1 max] for MP1 if 9 x 10^-6W is the final answer (I and I₀ reversed).
Award [1 max] if cos not squared (14 μW).

1.6 × 10^-5«W»
Units not required but if absent assume W.

ii
1.9 × 10^–4 «s»