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Colonel Frank Seely School

Exampro A-level Physics (7407/7408) 3.3.2.1 Interference

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Time: 264

Marks: 206

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Q1.(a) The diagram below shows schematically an arrangement for producing interference fringes using a double slit.

A dark fringe (minimum intensity) is observed at the point labelled P.

(i) Show clearly on the diagram the distance that is equal to the path difference between the light rays from the two slits to the point P.

(1)

(ii) Explain how the path difference determines that the light intensity at point P is a minimum.

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(iii) Explain briefly the role of diffraction in producing the interference patterns (You may draw a sketch to support your explanation if you wish.)


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(b) In one experiment the separation of the slits is 4.0 × 10–4 m. The distance from the slits to the screen is 0.60 m.

Calculate the distance between the centres of two adjacent dark fringes when light of wavelength 5.5 × 10–7 m is used.

(2)

(c) A student has learned that electrons behave like waves and decides to try demonstrate this using the arrangement in the diagram above. The lamp is replaced by a source of electrons and the system is evacuated.

The student accelerates the electrons to a velocity of 1.4 × 106 m s–1. The beam of electrons is then incident on the double slits. The electrons produce light when incident on the screen.

mass of an electron = 9.1 × 10–31 kg Planck constant = 6.6 × 10–34 J s

(i) Calculate the de Broglie wavelength associated with the electrons.(3)

(ii) Explain briefly, with an appropriate calculation, why the student would be unsuccessful in demonstrating observable interference using the slit separation of 4.0 × 10–4 m.

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(Total 13 marks)

Q2.The diagram represents the experimental arrangement used to produce interference fringes in Young’s double slit experiment.


The spacing of the fringes on the screen will increase if

A the width of the single slit is increased

B the distance XY between the two slits is increased

C a light source of lower frequency is used

D the distance between the single and double slits is decreased

(Total 1 mark)

Q3.A laser emits light of wavelength 6.3 × 10–7 m and is used to illuminate a double slit which has a separation of 2.4 × 10–4 m. Interference fringes are observed 4.2 m from the slits.

(a) Calculate the fringe separation.(2)

(b) The double slit acts as a pair of coherent sources. Explain what is meant by coherent sources.

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(c) The diagram shows the light from the slits, S1 and S2, meeting at P where the first dark fringe is observed.


Explain why a dark fringe is observed at P.

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(Total 7 marks)

Q4.Young’s two slit interference pattern with red light of wavelength 7.0 × 10–7 m gives a fringe separation of 2.0 mm.

What separation, in mm, would be observed at the same place using blue light of wavelength 45 × 10–7 m?

A 0.65

B 1.3

C 2.6

D 3.1

(Total 1 mark)

Q5.Short pulses of sound are reflected from the wall of a building 18 m away from the sound source. The reflected pulses return to the source after 0.11 s.

(a) Calculate the speed of sound.

Speed of sound ........................................(3)

(b) The sound source now emits a continuous tone at a constant frequency. An observer, walking at a constant speed from the source to the wall, hears a regular rise and fall in the intensity of the sound. Explain how the minima of intensity occur.

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(Total 6 marks)

Q6. The diagram below shows a laboratory ultrasound transmitter emitting ultrasonic waves through two slits placed 0.20 m apart. A receiver, moving along line AB, parallel to the line of the slits, detects regular rises and falls in the strength of the signal. A student measures a distance of 0.22 m between the first and the third maxima in the signal when the receiver is 2.5 m from the slits.

(a) (i) Calculate the distance between successive maxima.

Distance between successive maxima .....................(1)

(ii) Calculate the wavelength of the ultrasonic waves.


Wavelength ...............................(2)

(b) One of the slits is now covered. No other changes are made to the experiment.

State the differences between the observations made as the receiver is moved along AB before and after this change. Explain the changes that you mention.

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(Total 6 marks)

Q7. The diagram below shows an arrangement used to demonstrate the interference of sound waves. The two loudspeakers act as coherent sources of sound.


(a) Explain what is meant by the term coherent sources.

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(b) In the diagram, the loudspeakers are separated by 8.5 m and are emitting sound of wavelength 0.77 m. When a sound engineer walks along the line AB, 65 m from the loudspeakers, he observes a regular rise and fall in the sound intensity.

(i) Explain this observation.

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(ii) Calculate the distance moved along AB between two consecutive maxima of sound.

Distance moved .................................................(2)

(Total 6 marks)

Q8. An interference pattern is produced using monochromatic light from two coherent sources. The separation of the two sources is 0.25 mm and the fringe separation is 7.8 mm. The interference pattern is observed on a screen that is 3.5 m from the sources.

(a) Calculate the wavelength of the light used to produce the interference pattern.


wavelength ........................................(3)

(b) The figure below shows light from two coherent sources, S1 and S2, superimposing to create a bright fringe at point Q. Q is equidistant from S1 and S2. The diagram is not to scale.

Explain how the dark fringe at the point P is caused.

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(Total 6 marks)

Q9.A laser is used with a double slit positioned 5.0 m from a white screen. The separation of the slits is 0.25 mm and the wavelength of the laser light is 630 mm.


The screen is removed and a linear air track positioned so that a glider on the air track moves in the same plane that was occupied by the screen.

A light dependent resistor, LDR, is attached to a glider and connected by loosely hanging leads to a datalogger. When the glider moves at a constant speed, the datalogger records the output voltage from a circuit containing the LDR. Output from the datalogger, plotted against time, is shown below.

(a) (i) Explain why the LDR output voltage varies with the position of the glider.

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(ii) Calculate the separation between two adjacent positions of the glider when the


LDR is under maximum illumination.

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(iii) Use your answer to (ii) and the graph to calculate a speed for the glider consistent with these results.

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(b) A potential divider current is used to derive an output from the LDR.

In this experiment the resistance of the LDR is 10 kΩ when under maximum illumination and 100 kΩ when under minimum illumination. The value of R is 50 Ω. Calculate the values of V1 and V2 shown on the graph and state which corresponds to maximum illumination.

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(c) If the experiment were to be repeated how could you ensure that the glider is launched with the same speed each time?

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(Total 10 marks)

Q10.In a Young’s double slits interference arrangement the fringe separation is s when the wavelength of the radiation is λ, the slit separation w and the distance between the slits and the plane of the observed fringes D. In which one of the following cases would the fringe separation also be s?

wavelength slit separationdistance between

slits and fringes

A 2λ 2w 2D

B 2λ 4w 2D

C 2λ 2w 4D

D 4λ 2w 2D

(Total 1 mark)

Q11.The diagram for this question is drawn to scale and 1 mm on the diagram represents an actual distance of 5 mm.


S1 and S2 are identical coherent transmitters emitting, in phase, microwaves with a wavelength of 25 mm. They are positioned 250 mm apart on a horizontal surface and a detector can be placed anywhere along the line YY' which is in the same plane as the transmitters and parallel to the line containing S1 and S2.

(a) Explain what is meant by coherent.

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(b) By making measurements on the diagram and using the scale, determine the number of wavelengths in the path

(i) S1R,

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(ii) S2R.


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(iii) Use your answers to (i) and (ii) to determine whether or not you expect the signal received by a detector placed at R to be a maximum. Explain your answer.

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(c) Describe how you would expect the signal strength to vary as the detector is moved from R to P via Q.

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(d) Calculate the frequency of the microwaves.

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(Total 10 marks)

Q12.The diagram shows two identical loudspeakers, A and B, placed 0.75 m apart. Each loudspeaker emits sound of frequency 2000 Hz.


Point C is on a line midway between the speakers and 5.0 m away from the line joining the speakers. A listener at C hears a maximum intensity of sound. If the listener then moves from C to E or D, the sound intensity heard decreases to a minimum. Further movement in the same direction results in the repeated increase and decrease in the sound intensity.speed of sound in air = 330 m s–1

(a) Explain why the sound intensity

(i) is a maximum at C,

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(ii) is a minimum at D or E.

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(b) Calculate

(i) the wavelength of the sound,

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(ii) the distance CE.

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(Total 8 marks)

Q13.

A double slit interference experiment is performed using monochromatic light of wavelength λ. The centre of the observed pattern is a bright fringe. What is the path difference between two waves which interfere to give the third dark fringe from the centre?

A 0.5 λ

B 1.5 λ

C 2.5 λ

D 3.5 λ(Total 1 mark)


Q14.(a) A double slit interference experiment is set up in a laboratory using a source of yellow monochromatic light of wavelength 5.86 × 10–7 m. The separation of the two vertical parallel slits is 0.36 mm and the distance from the slits to the plane where the fringes are observed is 1.80 m.

(i) Describe the appearance of the fringes.

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(ii) Calculate the fringe separation, and also the angle between the middle of the central fringe and the middle of the second bright fringe.

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(iii) Explain why more fringes will be seen if each of the slits is made narrower, assuming that no other changes are made.

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(b) Light of wavelength 5.86 × 10–7 m falls at right angles on a diffraction grating which has 400 lines per mm.


(i) Calculate the angle between the straight through image and the first order image.

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(ii) Determine the highest order image which can be seen with this arrangement.

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(c) Give two reasons why the diffraction grating arrangement is more suitable for the accurate measurement of the wavelength of light than the two-slit interference arrangement.

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(Total 15 marks)

Q15.In a double slit interference arrangement the fringe spacing is w when the wavelength of the radiation is λ, the distance between the double slits is s and the distance between the slits and the plane of the observed fringes is D. In which one of the following cases would the fringe spacing also be w?

wave length distance between slits

distance between slits and fringes

A 2λ 2s 2D


B 2λ 4s 2D

C 2λ 2s 4D

D 4λ 2s 2D

(Total 1 mark)

Q16.The diagram shows two closely spaced narrow slits illuminated by light from a single slit in front of a monochromatic light source. A microscope is used to view the pattern of bright and dark fringes formed by light from the two slits.

(a) (i) Explain qualitatively why these fringes are formed.

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(ii) Describe what is observed if one of the narrow slits is covered by an opaque


object.

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(b) The microscope is replaced by a fibre-optic detector linked to a computer. The detector consists of the flat end of many optical fibres fixed together along a line. The other end of each optical fibre is attached to a light-sensitive diode in a circuit connected to a computer. The signal to the computer from each diode is in proportion to the intensity of light incident on the diode. The computer display shows how the intensity of light at the detector varies along the line of the detector when both of the narrow slits are open.

(i) Describe and explain how the pattern on the display would change if the slit separation were increased.

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(ii) Each fibre consists of a core of refractive index 1.50 surrounded by cladding of refractive index 1.32. Calculate the critical angle at the core-cladding boundary.

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(iii) The diagram below shows a light ray entering an optical fibre at point P on the flat end of the fibre. The angle of incidence of this light ray at the core-cladding boundary is equal to the critical angle. On the diagram, sketch the path of another light ray from air, incident at the same point P, which is totally internally reflected at the core-cladding boundary.

(7)

(Total 15 marks)

Q17.


Coherent monochromatic light of wavelength λ emerges from the slits X and Y to form dark fringes at P, Q, R and S in a double slit apparatus. Which one of the following statements is true?

A When the distance D is increased, the separation of the fringes increases.

B When the distance between X and Y is increased, the separation of the fringes increases.

C When the width of the slit T is decreased, the separation of the fringes decreases.

D There is a dark fringe at P because (YP − XP) is 2λ .(Total 1 mark)

Q18.Interference fringes, produced by monochromatic light, are viewed on a screen placed a distance D from a double slit system with slit separation s. The distance between the centres of two adjacent fringes (the fringe separation) is w. If both s and D are doubled, what will be the new fringe separation?

A

B w

C 2w

D 4w(Total 1 mark)

Q19.In a Young’s double slit interference experiment, monochromatic light placed behind a single slit illuminates two narrow slits and the interference pattern is observed on a screen placed some distance away from the slits. Which one of the following decreases the separation of the fringes?

A increasing the width of the single slit

B decreasing the separation of the double slits

C increasing the distance between the double slits and the screen


D using monochromatic light of higher frequency(Total 1 mark)

Q20.

Point sources of sound of the same frequency are placed at S1 and S2. When a sound detector is slowly moved along the line PQ, consecutive maxima of sound intensity are detected at W and Y and consecutive minima at X and Z. Which one of the following is a correct expression for the wavelength of the sound?

A S1X – S1W

B S1Y – S1X

C S1X – S2X

D S1Y – S2Y(Total 1 mark)

Q21.The diagram shows a microwave transmitter T which directs microwaves of wavelength eat two slits S1 and S2 formed by metal plates. The microwaves that pass through the two slits are detected by a receiver.


receiverat O

When the receiver is moved to P from O, which is equidistant from S1 and S2, the signal received decreases from a maximum to a minimum. Which one of the following statements is a correct deduction from this observation?

A The path difference S1O – S2O = 0.5 λ

B The path difference S1O – S2O = λ

C The path difference S1P – S2P = 0.5 λ

D The path difference S1P – S2P = λ(Total 1 mark)

Q22. (a) State what is meant by coherent sources of light.

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(b)

Figure 1

Young’s fringes are produced on the screen from the monochromatic source by the arrangement shown in Figure 1.

You may be awarded marks for the quality of written communication in your answers.


(i) Explain why slit S should be narrow.

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(ii) Why do slits S1 and S2 act as coherent sources?

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(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 8 marks)


Q23.Interference maxima produced by a double source are observed at a distance of 1.0 m from the sources. In which one of the following cases are the maxima closest together?

A red light of wavelength 700 nm from sources 4.0 mm apart

B sound waves of wavelength 20 mm from sources 50 mm apart

C blue light of wavelength 450 nm from sources 2.0 mm apart

D surface water waves of wavelength 10 mm from sources 200 mm apart(Total 1 mark)

Q24. 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?

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

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(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.

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(Total 12 marks)

Q25. A laser illuminates a pair of slits of separation 0.24 mm. The wavelength of light from the laser is 6.3 × 10–7 m. Interference fringes are observed on a screen 4.3 m from the slits.

(a) Calculate the fringe separation. Give an appropriate unit for your answer.


fringe separation .......................................(3)

(b) State the conditions necessary for two light sources to be coherent.

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(Total 5 marks)

Q26. A diffraction grating has 300 lines per mm. It is illuminated with monochromatic light of wavelength 540 nm.Calculate the angle of the 2nd order maximum, giving your answer to the appropriate number of significant figures.

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angle .......................................... degrees(Total 4 marks)


Q27. Just over two hundred years ago Thomas Young demonstrated the interference of light by illuminating two closely spaced narrow slits with light from a single light source.

(a) What did this suggest to Young about the nature of light?

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(b) The demonstration can be carried out more conveniently with a laser. A laser produces coherent, monochromatic light.

(i) State what is meant by monochromatic.

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(ii) State what is meant by coherent.

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(iii) State one safety precaution that should be taken while using a laser.

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(c) The diagram below shows the maxima of a two slit interference pattern produced on a screen when a laser was used as a monochromatic light source.


The slit spacing = 0.30 mm.The distance from the slits to the screen = 10.0 m.

Use the diagram above to calculate the wavelength of the light that produced the pattern.

answer = ...................................... m(3)

(d) The laser is replaced by another laser emitting visible light with a shorter wavelength.State and explain how this will affect the spacing of the maxima on the screen.

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(Total 9 marks)

Q28. For a plane transmission diffraction grating, the diffraction grating equation for the first order beam is:


λ = d sin θ

(a) The figure below shows two of the slits in the grating. Label the figure below with the distances d and λ.

(2)

(b) State and explain what happens to the value of angle θ for the first order beam if the wavelength of the monochromatic light decreases.

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(c) A diffraction grating was used with a spectrometer to obtain the line spectrum of star X shown in the figure below. Shown are some line spectra for six elements that have been obtained in the laboratory.

Place ticks in the boxes next to the three elements that are present in the atmosphere of star X.


(2)

(d) The diffraction grating used to obtain the spectrum of star X had 300 slits per mm.

(i) Calculate the distance between the centres of two adjacent slits on this grating.

answer = ................................. m(1)

(ii) Calculate the first order angle of diffraction of line P in the figure above.

answer = ........................ degrees(2)

(Total 9 marks)


Q29.The diagram below shows the paths of microwaves from two narrow slits, acting as coherent sources, through a vacuum to a detector.

(a) Explain what is meant by coherent sources.

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(b) (i) The frequency of the microwaves is 9.4 GHz.

Calculate the wavelength of the waves.


wavelength = ................................. m(2)

(ii) Using the diagram above and your answer to part (b)(i), calculate the path difference between the two waves arriving at the detector.

path difference = ................................. m(1)

(c) State and explain whether a maximum or minimum is detected at the position shown in the diagram above.

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(d) The experiment is now rearranged so that the perpendicular distance from the slits to the detector is 0.42 m. The interference fringe spacing changes to 0.11 m.

Calculate the slit separation. Give your answer to an appropriate number of significant figures.

slit separation = ................................. m(3)

(e) With the detector at the position of a maximum, the frequency of the microwaves is now doubled. State and explain what would now be detected by the detector in the same position.


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(Total 14 marks)

Q30.The figure below shows a spectrometer that uses a diffraction grating to split a beam of light into its constituent wavelengths and enables the angles of the diffracted beams to be measured.

(a) Give one possible application of the spectrometer and diffraction grating used in this way.

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(1)

(b) (i) When the spectrometer telescope is rotated from an initial angle of zero degrees, a spectrum is not observed until the angle of diffraction θ is about 50°. State the order of this spectrum.

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(ii) White light is directed into the spectrometer. Light emerges at A and B. State one difference between the light emerging at B compared to that emerging at


A.

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(c) The angle of diffraction θ at the centre of the observed beam B in the image above is 51.0° and the grating has 1480 lines per mm.

Calculate the wavelength of the light observed at the centre of beam B.

wavelength .......................................... m(3)

(d) Determine by calculation whether any more orders could be observed at the wavelength calculated in part (c).

(2)

(Total 8 marks)

Q31.Monochromatic light of wavelength 490 nm falls normally on a diffraction grating that has 6 × 105 lines per metre. Which one of the following is correct?

A The first order is observed at angle of diffraction of 17°.

B The second order is observed at angle of diffraction of 34°.

C The third and higher orders are not produced.

D A grating with more lines per metre could produce more orders.


(Total 1 mark)

Q32.When comparing X-rays with UV radiation, which statement is correct?

A X-rays have a lower frequency.

B X-rays travel faster in a vacuum.

C X-rays do not show diffraction and interference effects.

D Using the same element, photoelectrons emitted usingX-rays have the greater maximum kinetic energy.

(Total 1 mark)

Q33.Read through the following passage and answer the questions that follow it.

Measuring the speed of sound in air

After the wave nature of sound had been identified, many attempts were made tomeasure its speed in air. The earliest known attempt was made by the Frenchscientist Gassendi in the 17th century. The procedure involved timing the intervalbetween seeing the flash of a gun and hearing the bang from some distance away.Gassendi assumed that, compared with the speed of sound, the speed of light is 5infinite. The value he obtained for the speed of sound was 480 m s−1. He alsorealised that the speed of sound does not depend on frequency.A much better value of 350 m s−1 was obtained by the Italian physicists Borelli andViviani using the same procedure. In 1740 another Italian, Bianconi, showed thatsound travels faster when the temperature of the air is greater. 10In 1738 a value of 332 m s−1 was obtained by scientists in Paris. This isremarkably close to the currently accepted value considering the measuringequipment available to the scientists at that time. Since 1986 the accepted valuehas been 331.29 m s−1 at 0 °C.

(a) Suggest an experiment that will demonstrate the wave nature of sound (line 1).

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(b) Using Gassendi’s value for the speed of sound (line 6), calculate the time between seeing the flash of a gun and hearing its bang over a distance of 2.5 km.

time = ........................ s(1)

(c) Explain why it was necessary to assume that ’compared with the speed of sound, the speed of light is infinite’ (line 5).

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(d) Explain one observation that could have led Gassendi to conclude that ’the speed of sound does not depend on frequency’ (line 7).

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(e) Explain how the value obtained by Borelli and Viviani was ’much better’ than that obtained by Gassendi (line 8).

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(f) The speed of sound c in dry air is given by


c = k

where θ is the temperature in °C, and k is a constant.

Calculate a value for k using data from the passage.

k = ........................ m s−1 K−½

(2)

(g) State the steps taken by the scientific community for the value of a quantity to be ’accepted’ (line 13).

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(Total 10 marks)

Q34.A light source emits light which is a mixture of two wavelength, λ1 and λ2. When the light is incident on a diffraction grating it is found that the fifth order of light of wavelength λ1 occurs at the same angle as the fourth order for light of wavelength λ2. If λ1 is 480 nm what is λ2?

A 400 nm

B 480 nm

C 600 nm

D 750 nm

(Total 1 mark)


Q35.A student has a diffraction grating that is marked 3.5 × 103 lines per m.

(a) Calculate the percentage uncertainty in the number of lines per metre suggested by this marking.

percentage uncertainty = ..................................... %(1)

(b) Determine the grating spacing.

grating spacing = ..................................... mm(2)

(c) State the absolute uncertainty in the value of the spacing.

absolute uncertainty = ..................................... mm(1)


(d) The student sets up the apparatus shown in Figure 1 in an experiment to confirm the value marked on the diffraction grating.

Figure 1

The laser has a wavelength of 628 nm. Figure 2 shows part of the interference pattern that appears on the screen. A ruler gives the scale.

Figure 2

Use Figure 2 to determine the spacing between two adjacent maxima in the interference pattern. Show all your working clearly.

spacing = ..................................... mm(1)

(e) Calculate the number of lines per metre on the grating.

number of lines = .....................................(2)


(f) State and explain whether the value for the number of lines per m obtained in part (e) is in agreement with the value stated on the grating.

........................................................................................................................

........................................................................................................................

........................................................................................................................(2)

(g) State one safety precaution that you would take if you were to carry out the experiment that was performed by the student.

........................................................................................................................

........................................................................................................................

........................................................................................................................(1)

(Total 10 marks)


M1.(a) (i) path difference clearly indicated correctlyB1

(1)

(ii) path difference must be (n + ½)λ not just ½λ(accept odd number of wavelengths) (allow diagram)

.

waves arrive at P antiphase / totally out of phase / exactly out of phase / out of phase by by π or 180° (allow diagram)

B1

interfere destructively / cancel / “crests and troughs” at same timeB1

(3)

(iii) idea of waves spreading out at a slit / slits (allow diagram for this mark)

B1

production of overlapping beams from the double slit or illuminating both slits in double slit

B1(2)

(b) fringe spacing = λD / dor correct substitution of data

C1

0.83 mmA1

(2)

(c) (i) momentum of electrons = 1.27 × 10−24 (kg m s−1)or m × v and correct substitution

C1

wavelength = h / pC1

5.2 × 10−10 mA1

or wavelength = h / p and p = mv; or h / mv

Colonel Frank Seely SchoolC1

correct substitution of dataC1

correct answerA1

(3)

(ii) calculation of fringe spacing ( = 7.8 × 10−7 m)allow e.c.f. for d / D confusion in (b)

or ratio λ / d for eachB1

the fringe spacing is too small (to enable separate fringes to be seen)B1

or calculation of D or of λ / d or speed for λ ≈ dB1

speed in apparatus much larger than that needed to make λ ≈ d

B1(2)

[13]

M2.C[1]

M3.(a) λD / d in symbols or numbersC1

1.1 × 10 m–2

A1(2)

(b) same frequency / wavelength.B1

constant phase differenceB1

(2)


(c) Reference to out of phaseC1

phase difference is 180° or π / path difference is λ2 / explanation ofC1

antiphase eg crest meets troughA1

destructive interference / description of cancelling outB1

(3)[7]

M4.B[1]

M5.(a) distance travelled = 2 × 18 mC1

Speed = 36 / 0.11M1

= 327 m / s [164 m / s scores 2]A1

(b) mention of standing waves or superposition or interferenceB1

mention of two waves, opposite directionsB1

because they are permanently out of phase, permanently destructively interfere, permanently in antiphase

B1[6]

M6. (a) (i) fringe spacing = 0.11 m

B11

(ii) λ = (fringe spacing from a(i)) *d/D [ecf] = 0.11 × 0.2/2.5

M1


= 8.8 × 10–3 m

A12

(b) three maximum from

mention of single-slit difraction aftergood diagram of diffraction intensitymention of interference fringes/two-slit beforeinterference/superposition/Young’s slits or alternative beforegood physics description of superpositione.g. interference between overlapping waves from each slitgood physics description of diffractionequally separated fringes beforecentral maximum is twice width of others afterdecrease in intensity for each successive maxima afterchange in spacing afterfainter from middle outwards after

B33

[6]

M7. (a) Constant/zero phase difference/in phase

B1

same frequency/wavelength

B12

(b) (i) mention of interference C1

describes constructive interference OR destructiveinterference OR discusses path difference

A1

(ii) [λD/d]

0.77*65/8.5

C1

=5.9 [5.89] m


A14

[6]

M8. (a) λ = yd/D

C1

substitution correct: condone

powers of 10

C1

5.6 × 107 m

A13

(b) waves arrive out of phase (with each other)

B1

path difference is 1.5λ

B1

idea of cancellation/destructive interference/negativeinterference destruction

B13

[6]

M9.(a) (i) Because the bright fringes and dark fringes (1) produced by double slit interference (1)


(ii) w = (1) = 12.6 (13) mm (1)

(iii) from graph, time to travel between adjacent bright fringes = 0.25 s (1)

use w = 13 mm and υ = (1)

to give υ = = 52 mm s–1 or 5.2 × 10–2 m s–1 (1)(max 5)

(b) V2 = value corresp to low resist ∴ bright fringe (1)

= 6 × = 2.0 V (1)

V2 = 6 × = 0.29 V (1)(3)

(c) e.g. use spring between glider & track end (1) compress to same length each time, release (1)

(2)[10]

M10.B[1]

M11.(a) constant phase relationship (1) (1)[or same frequency (wavelength) (1) and same phase difference (1)]

Colonel Frank Seely School2

(b) S1R = 15cm on diagram (1) =75cm ∴ 30 waves (1)S2R = 16cm on diagram (1) = 80cm ∴ 32 waves (1)2 whole waves difference so in phase at R (1) maximum (1)

max 5

(c) (falls then rises to) maximum at Q (1)(then falls and rises to) maximum at P (1)

2

(d) = = 1.2 × 1010 Hz (or 12 GHz) (1)1

[10]

M12.(a) (i) superposition (1)between waves in phase (1)gives constructive interference (1)

(ii) at D or E waves out of phase (1)so destructive interference (1)

max 4

(b) (i) λ = = 0.165m (1)

separation between maxima = (1)

= 1.10(m) (1)

distance CE (= × separation)= 0.55 m (1)4

[8]


M13.C[1]

M14.(a) (i) vertical or parallel (1)equally spaced (1)black and yellow [or dark and light] bands (1)

(ii)

= 2.9 × 10–3 m (1)

tan θ = (1) gives θ = 0.18° (1)

(iii) narrower slits give more diffraction (1)more overlap (so more fringes) (1)fringes same width (1)

(max 8)

(b) (i)

× sin θ = 5.86 × 10–7 (1)

θ = 13.6° (1)

(ii) θ = 90° and correctly used (1)

= 4.3 ∴ 4th order (1)

Colonel Frank Seely School(5)

(c) brighter images (1)large angles (1)sharper (or narrower) lines (1)

(max 2)[15]

M15.B[1]

M16.(a) (i) fringes formed when light from the two slits overlap (or diffracts) (1) slits emit waves with a constant phase difference (or coherent) (1) bright fringe formed where waves reinforce (1) dark fringe formed where waves cancel (1) [or if 3rd and 4th not scored, waves interfere (1)] path difference from slits to fringe = whole number of wavelengths for a bright fringe (1) whole number + half a wavelength for a dark fringe (1) [or phase difference is zero (in phase) for a bright fringe (1) and 180° for a dark fringe (1)]

(ii) (interference) fringes disappear (1) single slit diffraction pattern observed [or single slit interference observed] (1) central fringe (of single slit pattern) (1) side fringes narrower than central fringe (1)

max 8

(b) (i) fringes closer (1) (because) each fringe must be closer to the centre for the same path difference [or correct use of formula as explanation] (1)

(ii) sin θc (1) (= 0.88)

θc = 61.6° (1)

(iii) for second light ray, diagram to show: smaller angle of incidence at P than first ray (1) point of incidence at core / cladding boundary to right of first ray (1)


total internal reflection drawn correctly or indicated at point of incidence to right of right angle (1)

[alternative if ray enters at P from above: correct refraction at P (1) TIR at boundary if refraction at P is correct (1) angle of incidence visibly ≥ critical angle (1)]

7[15]

M17.A[1]

M18.B[1]

M19.D[1]

M20.D[1]

M21.C[1]

M22. (a) same wavelength or frequency (1)(same phase or) constant phase difference (1)

2


(b) (i) narrow slit gives wide diffraction (1)(to ensure that) both S1 and S2 are illuminated (1)

(ii) slit S acts as a point source (1)S1 and S2 are illuminated from same source givingmonochromatic/same λ (1)paths to S1 and S2 are of constant length giving constant phasedifference (1)[or SS1 = SS2 so waves are in phase]

Max 4QWC 1

(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)

2[8]

M23.A[1]

M24. (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 twoslits) 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)


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

orcancellation 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)9

(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]

M25. (a) w = λD/s

C1

correct substitution – condone wrong powers of ten


C1

cao 11(.3) mm or equivalent; unit required

A13

(b) same frequency or wavelength

C1

constant phase (relationship)

A12

[5]

M26. sinθ = nλ/d in this form/correct calculations of d/d = 1/300

C1

substitutes correctly – condone powers of 10

C1

18.9

C1

2 or 3 sf only

A1[4]

M27. (a) showed that light was a wave (rather than a particle)/wave nature(of light) (1)

1

(b) (i) single wavelength (or frequency) (1)1


(ii) (waves/source(s) have) constant phase difference (1)1

(iii) any sensible precaution, eg do not look into laser/do not pointthe laser at others/do not let (regular) reflections enter theeye/safety signs/suitable safety goggles (1)

1

(c) (0.16/8) = 0.02(0) (1)

= (1) ecf from calculation of fringe spacing

= 6.0 × 10–7 m (1) (= 600 nm) ecf from calculation of fringe spacing3

(d) maxima closer together (1)

(quotes equation and states that) spacing is proportional to wavelength/D and s are constant therefore as λ decreases so ω decreases (1)

or links smaller wavelength to smaller path difference (1)2

[9]

M28. (a) λ correct (1)

d correct (1) arrow or line needed, both ends extending beyondcentral black line

2

(b) angle θ gets smaller (1)

because path difference gets smaller/d constant, (λ smaller) sosin θ smaller (1)

max 1 for correct explanation for λ increasing2


(c) boxes 1,5,6 (1)(1)

two correct 1 mark

4 ticks max 1

5 or 6 ticks gets 02

(d) (i) 3.3 × 10–6 m (1) (1/300 = 3.33 × 10–3 mm, 3300 nm) DNA 1 sf hereDNA 1/300 000 as answer

accept 3 1/3 × 10–6, 3.33 × 10–6 recurring, etc1

(ii) (sin θ =) (1)

correct wavelength used and seen (545 to 548 × 10–9)

and 9.4 to 9.6 (°) (1) ecf (d) (i), for correct wavelength only(545 to 548 × 10–9)

2[9]

M29.(a) same wavelength / frequency

constant phase relationship allow ‘constant phase difference’ but not ‘in phase’

2

(b) (i) ( λ = )Use of speed of sound gets zero

3.00 × 108 = 9.4 (109) λ OR

= 3.2 × 10−2 (3.19 × 10−2 m) Allow 0.03


(ii) 3.2 × 10−2 (m) ecf from biDon’t allow ‘1 wavelength’, 1λ, etcDo not accept: zero, 2 , 360 °

1

(c) maximum (at position shown) allow constructive superposition. ‘Addition’ is not enough

constructive interference / reinforcement

ecf for ‘minimum’ or for reference to wrong maximum

(the waves meet) ‘in step’ / peak meets peak / trough meets trough / path difference is (n) λ / in phase

3

(d) s = Don’t allow use of the diagram shown as a scale diagram

ecf biDo not penalise s and w symbols wrong way round in working if answer is correct.

= 0.12 (0.1218 m) Correct answer gains first two marks.

= any 2sf number Independent sf mark for any 2 sf number

3

(e) a maximum Candidates stating ‘ minimum ’ can get second mark only

(f × 2 results in) λ/2

path difference is an even number of multiples of the new wavelength ( 2n λ new )

allow ‘path difference is nλ’ / any even number of multiples of the new λ quoted e.g. ‘path difference is now 2 λ’

3[14]


M30.(a) one of: (spectral) analysis of light from stars (analyse) composition of stars chemical analysis measuring red shift \ rotation of stars

insufficient answers: ‘observe spectra’, ‘spectroscopy’, ‘view absorption \ emission spectrum’, ‘compare spectra’, ‘look at light from stars’.

Allow : measuring wavelength or frequency from a named source of lightAllow any other legitimate application that specifies the source of light. E.g.absorbtion \ emission spectra in stars, ‘observe spectra of materials’

1

(b) (i) first order beam first order spectrum first order image

Allow ‘n = 1’ , ‘1’ , ‘one’, 1st

1

(ii) the light at A will appear white (and at B there will be a spectrum) OR greater intensity at A

1

(c) ( d = 1 / (lines per mm × 103) = 6.757 × 10−7 (m) OR 6.757 × 10−4 (mm)

( nλ = d sin θ ) = 6.757 × 10−7 × sin 51.0 ecf only for : • incorrect power of ten in otherwise correct calculation of d • use of d = 1480, 1.48, 14.8 (etc) • from incorrect order in bii

= 5.25 × 10−7 (m) ecf only for : • incorrect power of ten in otherwise correct d • from incorrect order in bii

Some working required for full marks. Correct answer only gets 2Power of 10 error in d gets max 2


For use of d in mm, answer = 5.25 × 10−4 gets max 2n = 2 gets max 2 unless ecf from biiuse of d = 1480 yields wavelength of 1150m

3

(d) n = d (sin90) / λ OR n = 6.757 × 10−7 / 5.25 × 10−7 ecf both numbers from c

= 1.29 so no more beams observed or answer consistent with their working

OR

2 = d (sinθ) / λ OR sinθ = 2 × 5.25 × 10−7 / 6.757 × 10−7 ecf both numbers from c

sinθ = 1.55 (so not possible to calculate angle) so no more beams

OR sin−1(2 × (their λ / their d) ) (not possible to calculate) so no more beams ecf

Accept 1.28, 1.3Second line gets both marksConclusion consistent with working

2[8]

M31.A[1]

M32.B[1]

M33.(a) Suitable experiment eg diffraction through a door / out of a pipe 1

(b) Using c = d / t

t = 2 500 / 480 = 5.2 s


(c) (Measured time is difference between time taken by light and time taken by sound)

Calculation assumes that light takes no time to reach observer, ie speed is infinite

Do not allow “could not know speed of light”1

(d) Sound from gun is a mixture of frequencies. Alternative for 1st mark ‘(so speed is independent of frequency) the sound of the gun is similar when close and far away’

1

All the sound reaches observer at the same time, 1

(e) More accurate, as it is closer to the accepted value. 1

(f) When θ = 0 °C c = 331.29 m s–1

1

Therefore

331.29 = k √273.15

k = 20.045 1

(g) The method and value are published 1

other scientists repeat the experiment using the same method 1

[10]

M34.C


[1]

M35.(a) 2.9% Allow 3%

1

(b) seen 1

0.29 mm or 2.9 x 10-4 m must see 2 sf only1

(c) ± 0.01 mm 1

(d) Clear indication that at least 10 spaces have been measured to give a spacing = 5.24 mm

spacing from at least 10 spaces Allow answer within range ±0.05

1

(e) Substitution in d sinθ = nλThe 25 spaces could appear here as n with sin θ as 0.135 / 2.5

1

d = 0.300 x 10-3 m so number of lines = 3.34 x103

Condone error in powers of 10 in substitutionAllow ecf from 1-4 value of spacing

1

(f) Calculates % difference (4.6%) 1


and makes judgement concerning agreement Allow ecf from 1-5 value

1

(g) care not to look directly into the laser beam OR care to avoid possibility of reflected laser beam OR warning signs that laser is in use outside the laboratory ANY ONE

1[10]


E1.(a) (i) Surprisingly few candidates were able to show the path difference clearly. This was considered to be an easy beginning to the question considering that knowledge of this distance is the first basic step in determining whether a signal is maximum or minimum at a given point.

(ii) There were many good thorough answers to this part. Even candidates who could not identify what the path difference was in (i) were able to produce a clear answer in many instances. Some candidates spoiled their answer by being specific and stating that the path difference had to be a half wavelength. The phase difference resulting from the different path lengths was the most commonly overlooked point.

(iii) Candidates usually gained credit for showing the spreading of waves at a slit but fewer explained clearly that this diffraction was necessary to produce two coherent sources or overlapping beams, which then interfered. Many ignored the question and proceeded to describe the conditions for maxima and minima.

(b) This part was usually well done. Some candidates spoiled their attempt by doubling the answer obtained using the correct formula.

(c) (i) This was completed successfully by the majority of the candidates although there was a reluctance to give the unit as m. Many tried to deduce a composite formula and many of these attempts were unsuccessful.

(ii) Many good answers were given in this part. Candidates were able to undertake a variety of relevant calculations but the most important feature of an explanation was that the fringe spacing would be too small to be visible. Some candidates were clearly confused between slit width and slit separation. These candidates often deduced the ratio of λ / d and then wrote about the need to have a slit width of the same order as the wavelength to produce diffraction. Candidates need to be more sceptical about such a statement as in a typical Young's slits experiment fringes are visible although this condition is clearly not met. A few candidates appreciated that for the same slit width less diffraction would take place and therefore there would be less likelihood of the beams overlapping within 0.6 m.


E5.(a) A simple opening question that was well done by many apart from an extremely common significant penalty deduction. Weaker candidates often did not recall that the sound pulse travels to the wall and back again, thus travelling twice the 18m distance quoted in the question.

(b) This was not well done. Although a substantial number gained 2 out of 3 on a generous mark scheme, only rarely did an answer give any sense that the pattern of maxima and minima was fixed in space. Answers dealt exclusively with the cancellation issues with no real discussion of the permanence of the cancellation of time and space.

E6. (a) (i) Too many were unable to recognise that a first and third fringe have just two (rather than three) fringe spacings between them. A common error was therefore to quote an answer of 0.073 m.

(ii) Despite the possible error in (a)(i), many were able to go forward and calculate the wavelength of the ultrasound correctly (via an error carried forward if necessary). This was generally done very well with a good level of detail in the calculation.

(b) On the other hand, this description and explanation was poor. Candidates had to compare the double slit pattern with the diffraction pattern that appears if one slit is covered. Very few described the diffraction pattern with any accuracy and even fewer were able to say how it arose. Descriptions of the physics of the original interference pattern were better but still not always convincing. Examiners were left with the inevitable conclusion that candidates had not considered the phenomena tested here in sufficient depth to answer the question.

E7. (a) Candidates were usually able to explain the term coherent source in terms of either phase constancy or same frequency etc. Explanations involving both ideas were rare.

(b) (i) The observation of varying sound levels was often described baldly in terms of constructive or destructive interference (a simple mark given the use of the term in the stem) but only rarely were there good and convincing descriptions of either constructive (using ideas of the waves adding in phase) or destructive


(180° out of phase). The term ‘out of phase’ with no angle specified is used frequently and erroneously by candidates.

(ii) The calculation was often well done and clearly demonstrated. Common errors involved the insertion of a factor of two, or the use of a diffraction formula.

E8. (a) This calculation was done well by nearly all candidates. Only a few had difficulty and this was usually with powers often.

(b) Most of the candidates were able to make an acceptable comment about destructive interference and mention of the waves being out of phase was usual. Very few candidates stated that the path difference at the point in question was one and a half wavelengths. In fact, comments about path difference were relatively rare.

E11.In part (a) the phrases “constant phase relationship” or “constant phase difference” were not as well known as they should have been. The majority of candidates were imprecise in their use of language.

A small number of candidates were handicapped in part (b) by the fact that they did not appear to have a ruler available to them. In general, parts (b)(i) and (b)(ii) of the question were done very well by the large majority of candidates. Those candidates getting parts (b)(i) and (b)(ii) wrong could score on part (b)(iii) if their argument was sensible.

Candidates made a sensible attempt at part (c), but many fell into the unintended trap and stated that Q corresponded to a minimum.

Far too many candidates made arithmetical errors in the straightforward calculation in part (d).


E12.Most candidates answered parts (a) and (b) well. Many candidates showed correctly in part (b) that w = 1.1 m but failed to halve it to obtain CE. Too many candidates did part (b)(ii) by inspecting the diagram, ignoring the “not to scale”, and incorrectly halving the 0.75 m to obtain what they thought looked like CE.

E14.Descriptions of the fringes in part (a)(i) were generally poor. In spite of the emphasis on the word vertical, hardly any candidates referred to the vertical lines which result. Furthermore, simple references to dark and bright equally-spaced bands were very rare. Confusion with the single diffraction pattern was common and some candidates contradicted themselves by referring to equally- spaced fringes at the same time as showing a diagram of the single slit pattern. Some of the descriptions made it difficult to believe that the writer had ever seen double slit interference fringes.

The large majority of candidates correctly calculated the fringe separation but failed to determine the angle. The word ‘angle’ frequently triggered the diffraction grating equation and, for those candidates who did attempt a simple geometrical calculation, many calculated the angle only to the first bright fringe and not the second. Such candidates penalised themselves by failing to read the question carefully. Several candidates who had calculated d correctly were quite prepared to offer a maximum order of thousands as their answer to part (b)(ii).

Answers to part (a)(iii) showed a general weak understanding of the relative parts played by interference and diffraction in this part of the question. One group of candidates failed to score any marks because they confused variation of slit width with slit separation and gave their answer in terms of the latter. For those candidates who did consider the narrowing of each slit, many scored only a single mark for recognising that the light at each slit will be diffracted through a larger angle. Although answers were often given in rather vague terms such as “more diffraction”. To gain full credit, it was necessary to refer to the greater area of overlap of the diffracted beams and to recognise that more fringes are seen in this increased area because the fringe width is unchanged. Hardly any candidates made the last of these two points.

Calculations in parts (b)(i) and (b)(ii) were done well and many candidates scored full marks. Common errors in part (b)(i) included incorrect conversion of units from 400 lines per millimetre, poor arithmetic and the often seen confusion between the grating constant, d, and the number of rulings per unit length, N.

Good answers to part (c) were rare and the two systems were not often compared from a


position of secure knowledge. Very few candidates were able to state that the grating is designed to produce fewer lines which are brighter and more widely spaced. lndeed, many candidates stated that the grating produced more fringes. Most answers concentrated on the accuracy of the readings rather than on how the readings were to be obtained. Answers were, in general, very vague.

E16.This was a long question worth 15 marks and it is pleasing to report that almost all candidates were able to gain a reasonable mark, with some gaining high marks.

In part (a)(i) most candidates knew that light was diffracted from each of the pair of narrow slits and that interference fringes were produced and could be seen in the overlap area. Some candidates referred to coherence in terms of waves being emitted in phase rather than with a constant phase difference. Most candidates were able to explain how a bright fringe or a dark fringe was formed and were able to relate their statements correctly to the path difference or phase difference. A significant number of candidates did not make it clear that a phase difference of 180° is necessary for cancellation of two waves, and often just stated that the waves were out of phase.

In part (a)(ii), it was clear that many candidates did not know the meaning of 'opaque' and thought that some light would pass through the opaque object. Few candidates realised that the fringes seen in (a)(i) would no longer be seen, but a small minority knew that single slit diffraction would take place and were thus able to give a satisfactory description.

Most candidates were aware in part (b)(i) that the fringes would be closer but few were able to give an adequate explanation of why this was so and only the best candidates were able to quote and use the appropriate expression to justify their answer. In part (ii) most candidates gave a correct calculation without any difficulty, but some candidates were unable to make any progress because they calculated the critical angle for a boundary with air. In part (iii) many candidates scored all three marks with a clear, correct diagram. The main reason for not scoring full marks was usually a failure to give the correct point of incidence.

E22. Whilst it was generally recognised in part (a) that coherent sources provide waves of the same wavelength (or frequency), the requirement about phase was less well understood. The common answer was that the waves must be ‘in phase’, whilst the accepted answer was that there has to be a constant phase relation between them. Although monochromatic sources that are in phase will be coherent, coherence does not require the sources to be in phase. In part (b), the single monochromatic source is the reason for fulfilling the same A criterion; this was correctly quoted by most. Satisfactory explanations of how the phase criterion is satisfied were very rare indeed, with few references to the paths SS1 and SS2.


Had part (c) required candidates to sketch Young’s fringes, there can be no doubt that the responses would have been much more rewarding. Most candidates were unable to translate their knowledge of the appearance of a familiar phenomenon into the required intensity/position graph. Near the centre of the pattern, the fringes are all of very similar intensity and all should have been drawn with the same width as the central fringe. The majority of wrong answers showed either the single slit diffraction pattern, or fringes having the same width as the central one but with much lower intensity.

E24. 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.

E25. In part (a), most candidates selected the correct equation for the double slit but there was often confusion between the slit separation and the slit-screen separation. A sizeable number of candidates had trouble with calculating the fringe separation to the correct power of 10 and single figure answers and lack of units (each of which was penalised) were not uncommon.

Few candidates knew the definition for coherence in part (b); most interpreted the question as asking for the conditions for a clear interference pattern and so equal or nearly equal amplitudes were frequently mentioned. Many candidates simply expressed the need for constant frequencies, wavelengths, phase, amplitude and velocity. The idea of a constant phase relationship was not generally understood with many candidates incorrectly equating constant phase with constant phase relationship.


E26. Many candidates answered this question well. The most common error was to use the value for the number of lines per mm as the grating spacing. Otherwise, a few candidates had their calculators set in radians rather than degrees. Significant figures were coped with well.

E27. Part (b) (i) was the definition of monochromatic. Most had no problem with this but a significant number simply said ‘one colour’ and this was not enough.

In part (b) (ii) ‘constant phase relationship’ or ‘difference’ was expected but many candidates said ‘in phase’ which was not given credit.

80% picked up the mark for a sensible suggestion in part (b) (iii) such as ‘never point the laser at someone’. The other 20% suggested ‘goggles’, ‘safety goggles’, ‘tinted goggles’ which was not enough. A few candidates said ‘specialised goggles’ or ‘goggles designed for use with lasers’ which was given credit.

Part (c) was a calculation using the two slit formula. 35% scored full marks. Common errors included converting 0.30 mm to 3 × 10–3 m, using 0.16 as w, or using w = 0.16/9 rather than 0.16/8 due to counting dots rather than gaps and incorrectly rearranging the formula.

In part (d), the majority of candidates scored the first mark but were unable to explain why in a convincing manner.

E28. Many candidates did not seem at all familiar with the use of this diagram in the derivation of the grating equation in part (a) and the placing of the labels was often completely random. A large number did not attempt to label the diagram and half of all candidates did not score any marks.

Many who scored one mark had labelled the wavelength correctly but did not accurately indicate the ‘line spacing’ with a suitable arrow or line.

Most candidates gained the first mark in part (b) for realising that sin θ decreased so θ would decrease. Many candidates failed to gain the second mark by not stating that d remained constant. Very few candidates attempted to explain in terms of path difference.

The majority of candidates had no problem matching up the spectral lines in part (c).

In part (d) (i), about half of all candidates were unable to convert lines per mm to line spacing and there was considerable confusion with powers of ten. Many candidates did not convert to metres and many also rounded to one significant figure.


In part (d) (ii) it was expected that the candidate would read an accurate value off the scale.

However, many chose a value to the nearest 10 nm, typically 550 nm. In this situation, it is always best to interpolate when reading off the scale. The uncertainty in this reading can then be expressed by giving the final answer to two significant figures. Line P is somewhere between 545 and 548 nm.

E29.(a) The explanation of coherent should include ‘constant phase relationship’ and ‘same frequency’. Many only picked up one mark for stating one of these. Many said that the waves were ‘in phase’ which was not enough for the mark.

(b) (i) Some did not realise that the speed of light was given on the data sheet. There were many mistakes with the powers of ten; ‘M’ was often interpreted as 106 , 1012 or even 1015.

(ii) Many did not attempt this. Some may have run out of time but many did not understand the concept of path difference.

(c) Some felt that although the waves arrived in phase, the point where they meet is ‘zero displacement’ for both of them and therefore gives a zero reading.

Nodes and antinodes were often referred to as the peaks / troughs and points of zero displacement on a progressive wave, e.g. two troughs meet to give destructive interference. Nodes and antinodes were often mixed up – ‘anti’ wrongly associated with ‘minimum’ perhaps?

(d) Quite a few did not express their answer to 2sf, believing that 3sf was appropriate or that 2 d.p. was required.

To help students remember to address the significant figure issue, one could advise them to draw a line between the instruction: ‘...appropriate number of significant figures’ and the answer line as soon as they read it in the question.

If the student is uncertain about which symbol represents slit separation and which represents fringe spacing, they may still get the right answer and this was not penalised. However, it is essential that s and D are not interchanged and this mistake led to many candidates failing to access 2 of the marks.

E30.(a) There were some rather vague answers here such as ‘To calculate the wavelength of a light’ or ‘to look at the light from stars’. There needed to be a little more than this to get the mark, i.e. a specific example such as ‘analyse the elements present in the atmosphere of a star’ or explain that the composition of a material or gas can be determined.


(b) The candidates who knew this often lacked detail in their answer, e.g. ‘it would be dimmer’. Some thought there would only be one colour at B rather than a spectrum.

Quite a few thought that the wavelength at B would be different from A due to the increased angle.

Some candidates thought that the light at B would be composed of different wavelengths and the white light at A would be a single wavelength.

(c) This was a fairly standard exam question but surprisingly there were few correct answers. Students seemed to be poorly prepared for this question and confusion reigned regarding the meaning of the terms in the grating equation. Use of the lines per mm as the line spacing (d = 1480) was very common.

There was also confusion between line spacing, d, and order, n. Some used 1480 for d and for n.

Candidates often used 1 / 1480 and then failed to convert this into metres.

(d) There were a surprising number of candidates who did not attempt this question.Even if they felt they had the wrong numbers for wavelength and line spacing in part (c), candidates simply needed to divide their d by their λ, and if greater than 1, conclude that no further orders are possible.

There was also some confusion over the method required, e.g. some used the angle given in part (c), (51°), and calculated a new wavelength that would give a second order at that angle.

· web view2021. 3. 14. · (b) the microscope is replaced by a fibre-optic detector linked to a...

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