Unit 2 Waves Questions -Hinchley Wood School

Unit 2 waves questions -Hinchley Wood School

Q1. The figure below shows a continuous progressive wave on a rope. There is a knot in the rope.

(a) Define the amplitude of a wave.

......

(2)

(b) The wave travels to the right.
Describe how the vertical displacement of the knot varies over the next complete cycle.

......

(3)

(c) A continuous wave of the same amplitude and frequency moves along the rope from the right and passes through the first wave. The knot becomes motionless.
Explain how this could happen.

......

(3)

(Total 8 marks)

Q2.The diagram below shows a section of a typical glass step-index optical fibre used for communications.

(a) Show that the refractive index of the core is 1.47.

(1)

(b) The refracted ray meets the core-cladding boundary at an angle exactly equal to the critical angle.

(i)Complete the diagram above to show what happens to the ray after it strikes the boundary at X.

(2)

(ii)Calculate the critical angle.

critical angle = ...... degrees

(1)

(iii)Calculate the refractive index of the cladding.

refractive index = ......

(2)

reason 1......

......

reason 2......

......

(2)

(Total 8 marks)

Q3. The diagram below shows a liquid droplet placed on a cube of glass. A ray of light from air, incident normally on to the droplet, continues in a straight line and is refracted at the liquid to glass boundary as shown.
refractive index of the glass = 1.45

(a) Calculate the speed of light

(i) in the glass,

......

(ii) in the liquid droplet.

......

(3)

(b) Calculate the refractive index of the liquid.

......

(2)

(c) On the diagram above, complete the path of the ray showing it emerge from the glass cube into the air.
No further calculations are required.

(2)

(Total 7 marks)

Q4.Figure 1 represents a stationary wave formed on a steel string fixed at P and Q when it is plucked at its centre.

Figure 1

(a) Explain why a stationary wave is formed on the string.

......

(3)

(b) (i) The stationary wave in Figure 1 has a frequency of 150 Hz. The string PQ has a length of 1.2 m.
Calculate the wave speed of the waves forming the stationary wave.

Answer ...... m s–1

(2)

(ii) On Figure 2, draw the stationary wave that would be formed on the string at the same tension if it was made to vibrate at a frequency of 450 Hz.

Figure 2

(2)

(Total 7 marks)

Q5. (a) State two requirements for two light sources to be coherent.

......

(2)

(b)Figure 1

Young’s fringes are produced on the screen from the monochromatic source by the arrangement shown in Figure 1.
Explain how this arrangement produces interference fringes on the screen. In your answer, explain why slit S should be narrow and why slits S1 and S2 act as coherent sources.
The quality of your written answer will be assessed in this question.

......

(6)

(c) The pattern on the screen may be represented as a graph of intensity against position on the screen. The central fringe is shown on the graph in Figure 2. Complete this graph to represent the rest of the pattern by drawing on Figure 2.

Figure 2

(2)

(Total 10 marks)

Q6. A single slit diffraction pattern is produced on a screen using a laser. The intensity of the central maximum is plotted on the axes in the figure below.

(a) On the figure above, sketch how the intensity varies across the screen to the right of the central maximum.

(2)

(b) A laser is a source of monochromatic, coherent light. State what is meant by

monochromatic light ......

......

coherent light ......

......

(2)

......

(1)

(d) State two ways in which the appearance of the fringes would change if the slit was made narrower.

......

(2)

(e) The laser is replaced with a lamp that produces a narrow beam of white light. Sketch and label the appearance of the fringes as you would see them on a screen.

(3)

(Total 10 marks)

Q7. The diagram below shows a section of a diffraction grating. Monochromatic light of wavelength λ is incident normally on its surface. Light waves diffracted through angle θ form the second order image after passing through a converging lens (not shown). A, B and C are adjacent slits on thegrating.

(a) (i) State the phase difference between the waves at A and D.

......

(ii) State the path length between C and E in terms of λ.

......

(iii) Use your results to show that, for the second order image,
2λ = d sin θ, where d is the distance between adjacent slits.

......

(3)

(b) A diffraction grating has 4.5 × 105 lines m–1. It is being used to investigate the line spectrum of hydrogen, which contains a visible blue-green line of wavelength 486 nm. Determine the highest order diffracted image that could be produced for this spectral lineby this grating.

......

(2)

(Total 5 marks)

M1. (a)the maximum displacement (of the wave or medium)

from the equilibrium position

accept ‘rest position’, ‘undisturbed position’, ‘mean position’

(b) (vertically) downwards (¼ cycle to maximum negative displacement)

then upwards (¼ cycle to equilibrium position and ¼ cycle to maximumpositive displacement)

down (¼ cycle) to equilibrium position/zero displacement and correctreference to either maximum positive or negative displacement or correctreference to fractions of the cycle

candidate who correctly describes the motion of a knot 180 degrees out ofphase with the one shown can gain maximum two marks(ie knot initially moving upwards)

stationary wave formed

by superposition or interference (of two progressive waves)

knot is at a node

waves (always) cancel where the knot is

allow ‘standing wave’

[8]

M2.(a) (n =)OR 0.2436 / 0.1657 working must be seen

0.24 / 0.17 = 1.41 is not acceptable

AND ( = 1.4699) = 1.47 given correctly to 3 or more significant figures

Watch for:14.1 / 9.54 = 1.478

(b)(i)ray goes along the boundary

Deviation by no more than 1mm by the end of the diagram.

(partial) reflection shown
(allow dotted or solid line. This mark can be awarded if TIR is shown)

Tolerance: 70° to 85° to normal or labelled e.g. θ and θ, etc

(ii)(90 − 9.54 = ) 80.46 or 80.5 (° ) ( allow 80° )

Don’t allow 81 degrees

(iii)(n = nc sin θ)

allow 80 or 81 degrees here

= 1.47 sin 80.46° ecf bii

=1.45 (1.4496)

Correct answer gains both marks

comment on ‘quality’ of signal is not sufficient

•prevent ‘crossover’ of signal / ensure security of data / prevent loss of information / data / signal

don’t allow ‘leakage’ on its own.

•increase the critical angle / reduce pulse broadening / (modal)dispersion / rays with a small angle of incidence will be refracted out of the core

Don’t allow ‘loss of light’

•increase rate of data transfer

Allow ‘leakage of signal’, etc

max two correct (from separate bullet points)

[8]

M3. (a) (i) (use of n =gives)cglass = ×

= 2.07 × 108 m s–1(1)

(ii) use of (1)

cliquid == 2.26 × 108 m s–1 (1)

(allow C.E. for values of cglass from (i))

(b) use of

to give nliquid == 1.33 (1)

(allow C.E. for value of cliquid)

[or use 1n2 = to give correct answer]

(c) diagram to show :
total internal reflection on the vertical surface(1)
refraction at bottom surface with angle in air greater
than that in the liquid (29.2°) (1)

[7]

M4. (a) (progressive waves travel from centre) to ends and reflect (1)

two (progressive) waves travel in opposite directions along the string (1)

waves have the same frequency (or wavelength) (1)

waves have the same (or similar) amplitude (1)

superposition (accept ‘interference’) (1)

max 3

(b) (i) wavelength (= 2 × PQ = 2 × 1.20 m) = 2.4 m (1)

speed (= wavelength × frequency = 2.4 × 150) = 360 m s–1(1)

(answer only gets both marks)

(ii) diagram to show three ‘loops’ (1)and of equal length and
good shape (1) (or loop of one third length (1))

[7]

M5. (a) same wavelength or frequency (1)

same phase or constant phase difference (1)

(b) The marking scheme for this part of the question includes an overall
assessment for the Quality of Written Communication (QWC).
There are no discrete marks for the assessment of QWC but the
candidates’ QWC in this answer will be one of the criteria used
to assign a level and award the marks for this part of the question.

Level / Descriptor
an answer will be expected to meet most of the criteria in the level descriptor / Mark range
Good 3 / – answer includes a good attempt at the explanations required
– answer makes good use of physics ideas including knowledge beyond that given in the question
– explanation well structured with minimal repetition or irrelevant points and uses appropriate scientific language
– accurate and logical expression of ideas with only minor/occasional errors of grammar, punctuation and spelling / 5-6
Modest 2 / – answer includes some attempts at the explanations required
– answer makes use of physics ideas referred to in the question but is limited to these
– explanation has some structure but may not be complete
– explanation has reasonable clarity but has a few errors of grammar and/or punctuation and spelling / 3-4
Limited 1 / – answer includes some valid ideas but these are not organised in a logical or clear explanation
– answer lacks structure
– several errors in grammar, punctuation and spelling / 1-2
0 / – incorrect, inappropriate or no response / 0

the explanations expected in a competent answer should include a coherent selection of the following physics ideas:

• narrow single slit gives wide diffraction

• to ensure that both S1 and S2 are illuminated

• slit S acts as a point source

• narrow single slit ensures it provides coherent sources of light at
S1 and S2

• S1 and S2 are illuminated by same source giving same wavelength

• paths to S1 and S2 are of constant length giving constant phase
difference or SS1 and SS2 so waves are in phase

• light is diffracted as it passes through S1 and S2 and the diffracted
waves overlap and interfere

• where the path lengths from S1 and S2 to the screen differ by
whole numbers, n of wavelengths, constructive interference
occurs producing a bright fringe on the screen

• where the path lengths differ by (n + ½) wavelengths, destructive
interference occurs producing a dark fringe on the screen

(c) graph to show: maxima of similar intensity to central maximum (1)
(or some decrease in intensity outwards from centre)

all fringes same width as central fringe (1)

[10]

M6. (a) 3 subsidiary maxima in correct positions (1)

intensity decreasing (1)

(b) a single wavelength (1)

constant phase relationship/difference (1)

(d) wider/increased separation (1)

lower intensity (1)

(e) distinct fringes shown with subsidiary maxima (1)

indication that colours are present within each subsidiary maxima (1)

blue/violet on the inner edge or red outer for at least one subsidiary
maximum (1)

(middle of) central maximum white (1)

[10]

M7. (a) (i) 0, 2π or 4π [or 0, 360° or 720°] (1)

(ii) 4λ(1)

(iii) sin θ = (1)

[or sin θ = ]

CE = 4λ and AC = 2d(1) (hence result)

[or BD = 2λ and AB = d]

max 3

(b) (limiting case is when θ = 90° or sin θ = 1)

(1)(= 4.6)

highest order is 4th (1)

[5]

E1. Many students had learned the correct definition in part (a) but some gave a description, for example ‘the greatest height of the wave from the middle’. This did not gain marks.

Surprisingly in answer to part (b), many students referred to the equilibrium position as the ‘node’ and maximum amplitude as the ‘antinode’ on a progressive wave. Many use fractions of a cycle to describe the position of the knot but some use angles or fractions of a wavelength which are not appropriate. The biggest loss of marks occurred in the first mark where a large number thought that the knot would be travelling upwards initially.

Part (c) was a fairly easy question with students only needing to state that the ‘knot is at a node on a stationary wave which is caused by superposition’ to get three marks. Most students managed to get two of the marking points. Many did not understand how a node is formed, believing it is the sum of a peak and a trough only, or that the whole rope is stationary, or that the rope is only stationary at a node when cancellation occurs between waves that are 180° out of phase. The two waves that form a stationary wave are not always 180° out of phase in order to cancel at the nodes. Nodes are where the wave always cancels but the phase difference between the waves repeatedly varies from zero to 2π. Cancellation everywhere on the stationary wave only occurs when the waves are in antiphase but cancellation always happens at the nodes because the displacements of the waves are always equal and opposite at those points (or displacements are both zero when in phase and in antiphase). This is a complex situation but there are many simulations available on the internet that help to get these ideas across.

E2.(a)This was a relatively easy question of a type that usually yields high marks. However, this produced a surprising number of wrong answers. A common error was to misinterpret n = c / cs as a ratio of angles and use it to justify dividing 14.1° by 9.54°.

(b)(i)Very few candidates showed the partial reflection and there was a mark available for this.

(ii)In this question many candidates wrongly used critical angle = sin 1 / 1.47 (= 42.84). Perhaps because they didn’t have the n for the cladding they used n = 1. However, this would have given the critical angle for the air / core. Another common error was to think that the critical angle was 9.54°.

(iii)Again, many candidates chose the first equation they saw (n = c / cs) and substituted angles instead of speeds.

E3. The calculations in part (a) and part (b) were performed successfully by about 50 % of the candidates, but many fell at the first hurdle by trying to use the angles given in the question to calculate the speed of light in glass, rather than equate the refractive index to the ratio of the two speeds. A number of these candidates did however redeem themselves by calculating part (b) successfully. The overall impression created by the examinees was that although the relevant equations were known, they did not have the expertise to decide which equation was appropriate to the given calculation.

It is difficult to understand why so many candidates find ray drawing so demanding. Only about 10% of the candidates were awarded full marks in part (c). The errors which occurred were not drawing equal angles for internal reflection and refracting the emergent ray towards the normal.

E4. A large number of candidates struggled with part (a). This was mainly due to a lack of understanding of the fact that two waves must be travelling in opposite directions in order for a standing wave to form. They seemed to be describing one wave reflecting back and forth. Those who understood how the stationary wave formed and added further detail went on to score two or three marks fairly easily.

Some candidates in part (b) (i) did not multiply by two and only scored one mark out of the two available.

A majority gained two marks in part (b) (ii). A few candidates knew what to do but their sketch lacked acceptable accuracy, for example, the ‘loops’ were not of similar length. Only a quarter of candidates got the wavelength wrong.

E6. Most candidates gained at least one mark in part (a) for showing that the intensity of peaks reduced with distance from the centre. However, many did not recall the key difference between the pattern for single and double slits – the single slit pattern has a central maximum which is double the width of the subsidiary maxima.

There were many correct definitions of monochromatic and coherent in part (b). A few stated ‘same colour’ for monochromatic and ‘in phase’ for coherent. Neither of these were accepted.

In part (c), many candidates incorrectly used the equation for two slits to show that the maxima were further apart. This was not penalised since an explanation was not asked for.

Many candidates got part (d) the wrong way around, saying that the fringes would be more closely spaced and more intense. There seemed to be some guess work evident here. Candidates need to be able to describe the appearance of the single slit pattern and be aware of how it will change for different wavelengths, slit widths and for monochromatic and white light. Some teachers introduce the equation for the single slit although it is not in the specification. This is not necessary but can certainly help the more mathematically minded students. To illustrate the change in the pattern, a simple demonstration can be carried out with a red and a green laser shone through the same slit onto a screen.

A pleasing number of candidates produced very detailed and high quality answers to part (e), with many gaining all three marks. Some drew a graph of intensity, which did not gain a mark on its own.

E7. Knowledge of the derivation of d sin θ = nλ for the diffraction grating is required by section 13.1.7 of the Specification. Fundamental to this derivation, is familiarity with the concepts of phase and path difference. Part (a) proved to be an effective test of candidates’ understanding in these areas, and the question seemed to strike many candidates with apprehension: blank spaces were fairly common and ridiculous answers very frequent. Phase difference was particularly badly known, with many answers to part (i) expressed in terms of λ A correct answer of 4λ in part (ii) became almost a prerequisite for a successful approach to part (iii). Clearly 2λ = d sinθ can be shown by inserting n = 2 into the standard formula, but this was not the target of part (iii) and no marks could be awarded for such a trivial response.

Several recent questions about the diffraction grating in the Unit 4 Section A papers have covered areas similar in content to part (b), and candidates answers to this part were usually much more satisfactory than those in part (a). There was some confusion between the number of lines per metre (4.5 × 105) and the grating spacing (2.2 × 10–6 m). A small number of candidates took the numbers from their calculations too literally, quoting their final answers for the order as 4.57, whilst others failed to comprehend that this meant that the highest order would be the fourth rather than the fifth.

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