Mixed Unit 2 ABP 03-06-14 Hinchley Wood School

Mixed Unit 2 – ABP – 03-06-14 Hinchley Wood School

Q1. (a) In an experiment, a narrow beam of white light from a filament lamp is directed at normal incidence at a diffraction grating. Complete the diagram in the figure below to show the light beams transmitted by the grating, showing the zero-order beam and the first-order beams.

(3)

(b) Light from a star is passed through the grating.

Explain how the appearance of the first-order beam can be used to deduce one piece of information about the gases that make up the outer layers of the star.

......

(2)

(c) In an experiment, a laser is used with a diffraction grating of known number of lines per mm to measure the wavelength of the laser light.

(i) Draw a labelled diagram of a suitable arrangement to carry out this experiment.

(2)

(ii)Describe the necessary procedure in order to obtain an accurate and reliable value for the wavelength of the laser light.
Your answer should include details of all the measurements and necessary calculations.
The quality of your written communication will be assessed in your answer.

......

(6)

(Total 13 marks)

Q2. A narrow beam of monochromatic red light is directed at a double slit arrangement. Parallel red and dark fringes are seen on the screen shown in the diagram above.

(a) (i) Light passing through each slit spreads out. What is the name for this effect?

......

(1)

(ii) Explain the formation of the fringes seen on the screen.

......

(4)

(iii) The slit spacing was 0.56 mm. The distance across 4 fringe spacings was 3.6 mm when the screen was at a distance of 0.80 m from the slits. Calculate the wavelength of the red light.

Answer ...... m

(4)

(b) Describe how the appearance of the fringes would differ if white light had been used instead of red light.

......

(3)

(Total 12 marks)

Q3.Figure 1 shows a stretched string driven by a vibrator. The right-hand end of a string is fixed to a wall. A stationary wave is produced on the string; the string vibrates in two loops.

Figure 1

(a) State the physical conditions that are necessary for a stationary wave to form on the string.

......

(3)

(b) State how you know that the wave on the string is transverse.

......

(1)

Amplitude ......

Phase ......

(2)

(d) The length of the string is 1.2 m and the speed of the transverse wave on the string is 6.2ms–1. Calculate the vibration frequency of the vibrator.

Vibration frequency ......

(4)

(e) The frequency of the vibrator is tripled.

(i) Sketch the new shape of the stationary wave on Figure 2.

Figure 2

(ii) Show on your diagram three points, P, Q and R that oscillate in phase.

(2)

(Total 12 marks)

Q4. 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)

Q5. (a) The graph shows the variation of tensile stress with tensile strain for two wires X and Y, having the same dimensions, but made of different materials. The materials fracture at the points FX and FY respectively.

You may be awarded marks for the quality of written communication provided in your answer to the following questions.

State, with a reason for each, which material, X or Y,

(i) obeys Hooke’s law up to the point of fracture,

......

(ii) is the weaker material,

......

(iii) is ductile,

......

(iv) has the greater elastic strain energy for a given tensile stress.

......

(8)

(b) An elastic cord of unstretched length 160 mm has a cross-sectional area of 0.64 mm2. The cord is stretched to a length of 190 mm. Assume that Hooke’s law is obeyed for this range and that the cross-sectional area remains constant.

the Young modulus for the material of the cord = 2.0 × 107 Pa

(i) Calculate the tension in the cord at this extension.

......

(ii) Calculate the energy stored in the cord at this extension.

......

(5)

(Total 13 marks)

M1. (a)max three from

central maximum shown

two equally spaced first order maxima

central and one first order labelled correctly

central white maximum

indication of spectra/colours in at least one first order beam

at least one first order beam labelled with violet (indigo or blue) closest to the
centre or red furthest

(b) dark/black lines or absorption spectrum or Fraunhofer lines

(reveal the) composition (of the star’s atmosphere)

accept dark ‘bands’

accept atoms or elements in the star

or the peak of intensity

(is related to) the temperature

or Doppler (blue or red) shift

(speed of) rotation or speed of star (relative to Earth)

(c)(i)grating and screen shown with both labelled

laser or laser beam labelled

(ii)The candidate’s writing should be legible and the spelling, punctuation
and grammar should be sufficiently accurate for the meaning to be clear.

The candidate’s answer will be assessed holistically. The answer will be
assigned to one of three levels according to the following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised, logical and
coherent, using appropriate specialist vocabulary correctly. The form and style
of writing is appropriate to answer the question.

•correct use of (n)λ = d sin θ

•and measure appropriate angle (eg ‘to first order beam’ is the minimum required)

•and method to measure angle (eg tan θ = x/D, spectrometer, accept protractor)

•and at least one way of improving accuracy/reliability

•for full marks: also explain how d is calculated, eg d = 1/ lines per mm
(× 103)

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well organised and not
fully coherent. There is less use of specialist vocabulary, or specialist
vocabulary may be used incorrectly. The form and style of writing is less
appropriate.

•use of (n)λ = d sin θ

•and measure appropriate angle (eg ‘to first order beam’ is the minimum required)

•and method of measurement of θ (eg tan θ = x/D, spectrometer, accept protractor) or at least one way of improving accuracy/reliability

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised and may not be
relevant or coherent. There is little correct use of specialist vocabulary. The form
and style of writing may be only partly appropriate.

•use of (n)λ = d sin θ

•or measure appropriate angle (eg ‘to first order beam’ is the minimum required)

•or at least one way of improving accuracy/reliability

Incorrect, inappropriate of no response: 0 marks

No answer or answer refers to unrelated, incorrect or inappropriate physics.

The explanation expected in a competent answer should include

Accuracy/reliability points

•measure between more than one order (eg 2 θ)

•measure θ for different orders (for average λnot average angle)

•check or repeat/repeat for different distances (D)

•use of spectrometer

•use large distance to screen (D)

•protractor with 0.5 degree (or less) intervals

•graphical method: plot sin θ against n (gradient = λ/d)

[13]

M2. (a) (i)diffraction (1)

(ii)any 4 points from

interference (fringes formed) (1)

where light from the two slits overlaps (or superposes) (1)

bright (or red) fringes are formed where light (from the two
slits) reinforces (or interfere constructively/crest meets crest) (1)

dark fringes are formed where light (from the two slits)
cancels (or interferes destructively/trough meets crest) (1)

the light (from the two slits) is coherent (1)

either
reinforcement occurs where light waves are in phase
(or path difference = whole number of wavelengths) (1)

or
cancellation occurs where light waves are out of phase of 180°
(in anti-phase)
(or path difference = whole number + 0.5 wavelengths) (1)
(not ‘out of phase’)

(iii)gives λ = (1)

w (= 3.6/4) = 0.9(0) mm (1) (failure to /4 is max 2)

λ(1) = 6.3 × 10–7 m (1)

(b) central (bright) fringe would be white (1)

side fringes are (continuous) spectra (1)

(dark) fringes would be closer together (because λred > average λwhite) (1)

the bright fringes would be blue on the side nearest the centre
(or red on the side away from the centre) (1)

bright fringes merge away from centre (1)

bright fringes wider (or dark fringes narrower) (1)

max 3

[12]

M3. (a) reflection (or 2 waves travelling in opposite directions) (1)

waves have similar amplitudes (1)

waves have similar frequency (1)

reflected wave loses only a little energy at the wall (1)

max 3

(b) displacement perpendicular to rest position of the string (1)

(c)A larger than B (1)

A 180° out of phase with B (1)

(d)λ = 1.2m (1)

c = fλ(1)

f = 6.2/1.2 (1) 5.2Hz (1)

(e) (i) diagram correct: 6 loops (1)

(ii) Q and R correct (1)

[12]

M4. (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]

M5. (a) (i) X (1)
stress (force) strain (extension) for the whole length (1)

(ii) Y (1)
has lower breaking stress (or force/unit area is less) (1)

(iii) Y (1)
exhibits plastic behaviour (1)

(iv) Y (1)
for given stress, Y has greater extension
[or greater area under graph] (1)

QWC 2

(b) (i) (use of E =gives)

=(1)

(1) for data into correct equation, (1) for correct area

= 2.4 N (1)

(allow C.E. for incorrect area conversion)

(ii) (use of energy stored = ½Fe gives) energy = (1)

= 36 × 10–3 J (1)

(allow C.E. for value of F from (i))

[13]

E1. Part (a) was done well by most students. However, many would have benefitted from using a protractor to get the angles between the zero order and the two first-order beams roughly equal.

In part (b), the basic requirement is that students know that dark lines (absorption lines) are seen on the spectra from stars and that these reveal elements present in the outer layers of the star. The mark scheme also credited other uses of a stars spectrum. Many students had the idea that spectral lines revealed elements but few knew about absorption lines. This is an area where students who have taken PHYA1 first may have an advantage since they have studied atomic energy levels and may have seen absorption spectra.

Labelling a laser, diffraction grating and some sort of screen or suitable detector was all that was required for the two marks in part (c) (i). Many students missed out the screen. Some had double slits instead of a grating.

Part (c) (ii) was, in general, poorly answered. Many students did not seem to be familiar with this practical and instead described a two-slit approach to measuring wavelength. Those who seemed familiar with the procedure tended not to fully answer the question which asked for details of all the measurements and necessary calculations. The candidate who leaves out these details is unlikely to be able to score more than two marks out of six even if they have given a reasonable general description of the experiment. For example, they must include details of how the angle is to be measured eg by measuring the distance between the zero order and the first-order beam (using a ruler) and the distance between the screen and the grating. They must then use tan θ = O/A to calculate the angle. Where students knew which equation to use, they tended to know insufficient detail to score more than a few marks. Of those students who did describe the use of a grating, many did not know the meanings of the symbols in the equation eg, d was often thought to be the distance between grating and screen and n, the number of lines per mm or even the refractive index of air. Many described measuring the grating spacing with micrometers or metre rules, forgetting that the question stated that the lines per mm are known.

In short, many of the students who took this exam seemed poorly prepared for this type of question. They were, in some cases, able to produce an answer from a past paper for a closely related, but significantly different, question. Many seemed unaware of the style and quality of answer expected.

Most answers were vague, the literacy level was generally poor and there was a lack of detail regarding the measurements and what should be done with them. This is often the case in the January examination, but it is possible to improve the necessary skills even in the short preparation time available. A few structured lessons on answering this type of question can to be incorporated into schemes of work, allowing students to be fully aware of the expectations.

E2. Part (a) (ii) was answered well by many who knew the terminology very well; most gained three or four marks. The majority of the candidates who did not gain any marks had misinterpreted the words ‘describe the formation’ to mean ‘describe the appearance’ rather than ‘how and why are they formed’.

Most candidates correctly rearranged the double slit formula in (a) (iii). It was then surprising that very few candidates realised they had to divide 3.6 by 4 to get the fringe spacing and this limited them to a maximum of two marks. Again many candidates who understood how to answer the question then failed to get to grips with the powers of ten and dropped marks.

Most candidates did not gain any marks in part (b) and only very few gained full marks. Part of the problem was that many believed that a single continuous spectrum would appear or that each fringe would be a different colour. A useful exercise to overcome candidate’s difficulties with descriptive answers could be to show interference phenomena and ask students to write a detailed description as they are observing the pattern.

E4. 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.

E5. Examiners were pleased to find that part (a) was answered satisfactorily and that candidates not only chose the correct wire but were very often able to provide the correct reason for doing so. Many candidates gained full marks, while a large number only lost one or two marks.
Part (i) was usually correct, although reasons such as ‘the graph is a straight line’ were not accepted. A ‘constant gradient’ was accepted but few candidates gave this as a reason, most givingthe proportionality of the quantities involved. In part (ii) answers such as ‘Y broke before X’ was not accepted. Examiners were looking for a reason in terms of lower breaking stress.

Answers to part (iii) were not so good and candidates who did not know the correct answer attempted an answer in terms of the gradients of the curves or the bending of curve Y as the tensile strain increased. Part (iv) gave the most trouble. Many candidates again tried an explanation in terms of the gradient, but a significant number followed the correct track and gave a reason in terms of the area under such a graph. Unfortunately the majority of these candidates referred to the area under the whole curve, whereas it should have been the area under the curve at a given tensile stress. Surprisingly, many candidates, even when using a given stress, gave the area under X as being greater than that under Y.

The final calculation in part (b) did not cause too much difficulty and, provided the initial equation for the Young modulus was correct, candidates produced a correct answer with correct units. One common error which again arose from not reading the question thoroughly, was using the extended length of the elastic cord as the extension. Converting the cross-sectional area of the cord from mm2 to m2 caused some problems, but this error was carried forward after the initial penalty had been imposed. The calculation in part (ii) was also done well by those who knew the expression for the energy stored, or were aware that it was given in the data sheet. Some answers, resulting from a carry forward of an incorrect force in part (i) gave energies amounting to several million joules. This attracted no comment.

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