the diagram shows a cube of glass. a ray of light, incident at ......(c)€€€€ a second prism,...

$: The diagram shows a cube of glass. A ray of light, incident at ......(c)€€€€ A second prism, prism 2, made from transparent material of refractive index 1.37 is placed firmly$
The diagram shows a cube of glass. A ray of light, incident at the centre of a face of the cube, atan angle of incidence θ, goes on to meet another face at an angle of incidence of 50°, as shownin the figure bellow

critical angle at the glass-air boundary = 45°

(a) Draw on the diagram the continuation of the path of the ray, showing it passing through theglass and out into the air.

(3)

1

(b) Show that the refractive index of the glass is 1.41

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) Calculate the angle of incidence, θ.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(Total 8 marks)

Page 1 of 83St Bede's Catholic Comprehensive School and Byron College

Interference maxima produced by a double source are observed at a distance of 1.0 m from thesources. 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)

2

The figure below shows a ray of light passing from air into glass at the top face of glass block 1and emerging along the bottom face of glass block 2.

refractive index of the glass in block 1 = 1.45

3


(a) Calculate

(i) the incident angle θ1,

______________________________________________________________

______________________________________________________________

______________________________________________________________

(ii) the refractive index of the glass in block 2,

______________________________________________________________

______________________________________________________________

(iii) the angle θ3 by considering the refraction at point A.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(7)

(b) In which of the two blocks of glass will the speed of light be greater?

___________________________________________________________________

Explain your reasoning.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) Using a ruler, draw the path of a ray partially reflected at A on the figure above. Continuethe ray to show it emerging into the air. No calculations are expected.

(2)

(Total 11 marks)


(a) For an optical fibre the refractive index of the core is 1.52 and the refractive index of thecladding is 1.49. Calculate the critical angle, in degrees, at the boundary between the coreand cladding of the fibre.

critical angle ____________________ degrees

(2)

4

(b) Explain why the cladding is necessary for optical fibres.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 4 marks)


A ray of light passes from air into a glass prism as shown in Figure 1.

Figure 1

(a) Confirm, by calculation, that the refractive index of the glass from which the prism wasmade is 1.49.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(1)

5

(b) On Figure 1, draw the continuation of the path of the ray of light until it emerges back intothe air. Write on Figure 1 the values of the angles between the ray and any normals youhave drawn.

the critical angle from glass to air is less than 45°

(2)


(c) A second prism, prism 2, made from transparent material of refractive index 1.37 is placedfirmly against the original prism, prism 1, to form a cube as shown in Figure 2.

Figure 2

(i) The ray strikes the boundary between the prisms. Calculate the angle of refraction ofthe ray in prism 2.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(ii) Calculate the speed of light in prism 2.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(iii) Draw a path the ray could follow to emerge from prism 2 into the air.

(7)

(Total 10 marks)



The diagram above shows the first four diffraction orders each side of the zero order when a

beam of monochromatic light is incident normally on a diffraction grating of slit separation d. All

the angles of diffraction are small. Which one of the patterns, A to D, drawn on the same scale,

is obtained when the grating is exchanged for one with a slit separation ?

A

B

C

D

(Total 1 mark)

6



In a double slit system used to produce interference fringes, the separation of the slits is s and

the width of each slit is x. L is a source of monochromatic light. Which one of the followingchanges would decrease the separation of the fringes seen on the screen?

A moving the screen closer to the double slits

B decreasing the width, x, of each slit, but keeping s constant

C decreasing the separation, s, of the slits

D exchanging L for a monochromatic source of longer wavelength

(Total 1 mark)

7

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

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

angle ____________________ degrees

(Total 4 marks)

8


A narrow beam of monochromatic red light is directed at a double slit arrangement. Parallel redand 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)

9

(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 mmwhen the screen was at a distance of 0.80 m from the slits. Calculate the wavelengthof the red light.

Answer ____________________ m

(4)


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

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(Total 12 marks)

A laser illuminates a pair of slits of separation 0.24 mm. The wavelength of light from the laser is6.3 × 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)

10

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

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 5 marks)


The diagram below shows a rectangular glass fish tank containing water. Three light rays,P, Q and R from the same point on a small object O at the bottom of the tank are shown.

(a) (i) Light ray Q is refracted along the water-air surface. The angle of incidence of light rayQ at the water surface is 49.0°. Calculate the refractive index of the water. Give youranswer to an appropriate number of significant figures.

Answer ____________________

(1)

11

(ii) Draw on the diagram above the path of light ray P from the water-air surface.

(3)

(b) In the diagram above, the angle of incidence of light ray R at the water-air surface is 60.0°.

(i) Explain why this light ray is totally internally reflected at the water surface.

______________________________________________________________

______________________________________________________________

(2)


(ii) Draw the path of light ray R from the water surface and explain whether or not Renters the glass at the right-hand side of the tank.

the refractive index of the glass = 1.50

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(4)

(Total 10 marks)


An optical fibre used for communications has a core of refractive index 1.55 which is surroundedby cladding of refractive index 1.45.

(a) The diagram above shows a light ray P inside the core of the fibre. The light ray strikes thecore-cladding boundary at Q at an angle of incidence of 60.0°.

(i) Calculate the critical angle of the core-cladding boundary.

answer ____________________ degrees

(3)

12

(ii) State why the light ray enters the cladding at Q.

______________________________________________________________

______________________________________________________________

(1)

(iii) Calculate the angle of refraction, θ, at Q.

answer ____________________ degrees

(3)


(b) Explain why optical fibres used for communications need to have cladding.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 9 marks)

A narrow beam of monochromatic light of wavelength 590 nm is directed normally at a diffractiongrating, as shown in the diagram below.

(a) The grating spacing of the diffraction grating is 1.67 × 10–6 m.

(i) Calculate the angle of diffraction of the second order diffracted beam.

answer ____________________ degrees

(4)

13

(ii) Show that no beams higher than the second order can be observed at thiswavelength.

(3)


(b) The light source is replaced by a monochromatic light source of unknown wavelength.A narrow beam of light from this light source is directed normally at the grating.Measurement of the angle of diffraction of the second order beam gives a value of 42.1°.

Calculate the wavelength of this light source.

answer ____________________ m

(2)

(Total 9 marks)

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

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

___________________________________________________________________

___________________________________________________________________

(1)

14

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

(i) State what is meant by monochromatic.

______________________________________________________________

______________________________________________________________

(ii) State what is meant by coherent.

______________________________________________________________

______________________________________________________________

(2)

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

______________________________________________________________

______________________________________________________________

(1)


(c) The diagram below shows the maxima of a two slit interference pattern produced on ascreen 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.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 9 marks)


The diagram below shows a cross-section through a step index optical fibre.

(a) (i) Name the parts A and B of the fibre.

A

B

(1)

15

(ii) On the diagram above, draw the path of the ray of light through the fibre.Assume the light ray undergoes total internal reflection at the boundary betweenA and B.

(2)

(b) Calculate the critical angle for the boundary between A and B.Give your answer to an appropriate number of significant figures.

The refractive index of part A = 1.46The refractive index of part B = 1.48

answer = ____________________ degrees

(2)

(c) State and explain one reason why part B of the optical fibre is made as narrow as possible.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(d) State one application of optical fibres and explain how this has benefited society.

Application

___________________________________________________________________

___________________________________________________________________

Benefit

___________________________________________________________________

___________________________________________________________________

(2)

(Total 9 marks)

The figure below shows two rays of light A and B travelling through a straight optical fibre.

(a) Calculate the speed of light in the core of the optical fibre.

absolute refractive index of the core of the optical fibre = 1.6

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

speed of light in the core ______________________ ms–1

(2)

16


(b) The overall length of the optical fibre is 0.80 km. As shown in the figure above, ray Atravels down the centre of the core of the optical fibre. The path of ray B has an overalllength of 0.92 km as it travels through the core.

(i) Ray A and ray B enter the fibre at the same instant.Calculate the difference in time taken for ray A and ray B to travel through the core ofthe optical fibre.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

time difference ______________________ s

(2)

(ii) Explain how a graded-index optical fibre prevents this time difference occurring forrays such as A and B in the figure above.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(Total 6 marks)


A glass cube is held in contact with a liquid and a light ray is directed at a vertical face of thecube. The angle of incidence at the vertical face is then decreased to 42° as shown in the figurebelow. At this point the angle of refraction is 27° and the ray is totally internally reflected at P forthe first time.

(a) Complete the figure above to show the path of the ray beyond P until it returns to air.

(3)

17

(b) Show that the refractive index of the glass is about 1. 5.

(2)

(c) Calculate the critical angle for the glass-liquid boundary.

answer = ______________________ degrees

(1)


(d) Calculate the refractive index of the liquid.

answer = ______________________

(2)

(Total 8 marks)

For a plane transmission diffraction grating, the diffraction grating equation for the first orderbeam is:

λ = d sin θ

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

(2)

18

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

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


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

Place ticks in the boxes next to the three elements that are present in the atmosphere ofstar 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)


A single slit diffraction pattern is produced on a screen using a laser. The intensity of the centralmaximum 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 thecentral maximum.

(2)

19

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

monochromatic light __________________________________________________

___________________________________________________________________

coherent light _______________________________________________________

___________________________________________________________________

(2)

(c) Describe how the pattern would change if light of a longer wavelength was used.

___________________________________________________________________

___________________________________________________________________

(1)

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

___________________________________________________________________

___________________________________________________________________

(2)


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

(3)

(Total 10 marks)

The figure below shows a cross-section through an optical fibre.

(a) Explain the purpose of part X.

___________________________________________________________________

___________________________________________________________________

(1)

20


(b) The critical angle for the optical fibre is to be 60° and the absolute refractive index for theglass in the core of the fibre is 1.6.Calculate the required absolute refractive index for part Y.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

absolute refractive index for part Y ______________________

(3)

(c) Explain why dispersion occurs in an optical fibre.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 6 marks)


Figure 1 shows a cross-section through an optical fibre used for communications.

Figure 1

(a) (i) Name the part of the fibre labelled X.

______________________________________________________________

(1)

21

(ii) Calculate the critical angle for the boundary between the core and X.

answer = ____________________degrees

(2)


(b) (i) The ray leaves the core at Y. At this point the fibre has been bent through an angle of30° as shown in Figure 1.

Calculate the value of the angle i.

answer = ____________________degrees

(1)

(ii) Calculate the angle r.

answer = ____________________degrees

(2)


(c) The core of another fibre is made with a smaller diameter than the first, as shown in Figure2. The curvature is the same and the path of a ray of light is shown.

Figure 2

(c) State and explain one advantage associated with a smaller diameter core.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 8 marks)


(a) The speed of light is given by

c = f λ

State how each of these quantities will change, if at all, when light travels from air to glass.

c ______________________

f ______________________

λ ______________________

(3)

22

Figure 1 shows a side view of a step index optical fibre.

Figure 1

(b) Ray A enters the end of the fibre and then undergoes total internal reflection.

On Figure 1 complete the path of this ray along the fibre.

(2)

(c) (i) The speed of light in the core is 2.04 × 108 ms–1. Show that the refractive index of thecore is 1.47.

(2)

(ii) Show that the critical angle at the boundary between the core and the cladding isabout 80°.

refractive index of the cladding = 1.45

(2)


(d) Ray B enters the end of the fibre and refracts along the core-cladding boundary. Calculatethe angle of incidence, θ, of this ray at the point of entry to the fibre.

answer = ______________________ degrees

(3)

(e) Figure 2 shows a pulse of monochromatic light (labelled X) that is transmitted a significantdistance along the fibre. The shape of the pulse after travelling along the fibre is labelled Y.Explain why the pulse at Y has a lower amplitude and is longer than it is at X.

Figure 2

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 14 marks)


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

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

(1)

23

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

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

(2)

(ii) Calculate the critical angle.

critical angle = ____________________ degrees

(1)

(iii) Calculate the refractive index of the cladding.

refractive index = ____________________

(2)


(c) Give two reasons why optical fibres used for communications have a cladding.

reason 1 ___________________________________________________________

___________________________________________________________________

reason 2 ___________________________________________________________

___________________________________________________________________

(2)

(Total 8 marks)

The figure below shows a glass prism. Light is directed into the prism at an angle of 56°.The path of the ray of light is shown as is it enters the prism.

(a) (i) Calculate the refractive index of the glass.

answer = ______________________

(2)

24

(ii) Calculate the critical angle for the glass-air boundary.

answer = ______________________ degrees

(2)


(b) On the figure above, continue the path of the ray of light until it emerges from the prism.

(2)

(Total 6 marks)

The diagram below shows three transparent glass blocks A, B and C joined together. Each glassblock has a different refractive index.

(a) State the two conditions necessary for a light ray to undergo total internal reflection at theboundary between two transparent media.

condition 1 _________________________________________________________

___________________________________________________________________

condition 2 __________________________________________________________

___________________________________________________________________

(2)

25


(b) Calculate the speed of light in glass A.

refractive index of glass A = 1.80

speed of light ____________________ ms−1

(2)

(c) Show that angle θ is about 30o.

(2)

(d) The refractive index of glass C is 1.40.

Calculate the critical angle between glass A and glass C.


(2)

(e) (i) State and explain what happens to the light ray when it reaches the boundarybetween glass A and glass C.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(ii) On the diagram above continue the path of the light ray after it strikes the boundarybetween glass A and glass C.

(1)

(Total 11 marks)


The figure below shows a spectrometer that uses a diffraction grating to split a beam of light intoits 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.

___________________________________________________________________

___________________________________________________________________

(1)

26

(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 orderof this spectrum.

______________________________________________________________

(1)

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

______________________________________________________________

______________________________________________________________

______________________________________________________________

(1)


(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 wavelengthcalculated in part (c).

(2)

(Total 8 marks)

(a) A laser emits monochromatic light.

Explain the meaning of the term monochromatic light.

___________________________________________________________________

___________________________________________________________________

(1)

27


(b) The diagram below shows a laser emitting blue light directed at a single slit, where the slitwidth is greater than the wavelength of the light. The intensity graph for the diffracted bluelight is shown.

The laser is replaced by a laser emitting red light.

On the axes shown in the diagram above sketch the intensity graph for a laser emitting redlight.

(2)

(c) State and explain one precaution that should be taken when using laser light

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(d) The red laser light is replaced by a non-laser source emitting white light.

Describe how the appearance of the pattern would change.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(Total 8 marks)

Figure 1 shows a ray of light A incident at an angle of 60° to the surface of a layer of oil that isfloating on water.

refractive index of oil = 1.47

refractive index of water = 1.33

Figure 1

(a) (i) Calculate the angle of refraction θ in Figure 1.

angle ____________________ degrees

(2)

28


(ii) Calculate the critical angle for a ray of light travelling from oil to water.

angle ____________________ degrees

(2)

(iii) On Figure 1 continue the path of the ray of light A immediately after it strikes theboundary between the oil and the water.

(2)

(b) In Figure 2 a student has incorrectly drawn a ray of light B entering the glass and thenentering the water before totally internally reflecting from the water–oil boundary.

Figure 2

The refractive index of the glass is 1.52 and the critical angle for the glass–water boundaryis about 60°.


Give two reasons why the ray of light B would not behave in this way. Explain youranswers.

reason 1 ___________________________________________________________

___________________________________________________________________

explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________

reason 2 ___________________________________________________________

___________________________________________________________________

explanation _________________________________________________________

___________________________________________________________________

___________________________________________________________________

(4)

(Total 10 marks)

The diagram shows Young’s double-slit experiment performed with a tungsten filament lamp asthe light source.

(a) On the axes in the diagram above, sketch a graph to show how the intensity varies withposition for a monochromatic light source.

(2)

29


(b) (i) For an interference pattern to be observed the light has to be emitted by twocoherent sources.Explain what is meant by coherent sources.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(1)

(ii) Explain how the use of the single slit in the arrangement above makes the light fromthe two slits sufficiently coherent for fringes to be observed.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(1)

(iii) In this experiment light behaves as a wave.Explain how the bright fringes are formed.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(3)


(c) (i) A scientist carries out the Young double-slit experiment using a laser that emits violetlight of wavelength 405 nm. The separation of the slits is 5.00 × 10–5 m.

Using a metre ruler the scientist measures the separation of two adjacent brightfringes in the central region of the pattern to be 4 mm.

Calculate the distance between the double slits and the screen.

distance = ____________________ m

(2)

(ii) Describe the change to the pattern seen on the screen when the violet laser isreplaced by a green laser. Assume the brightness of the central maximum is thesame for both lasers.

______________________________________________________________

______________________________________________________________

______________________________________________________________

(1)

(iii) The scientist uses the same apparatus to measure the wavelength of visibleelectromagnetic radiation emitted by another laser.Describe how he should change the way the apparatus is arranged and used in orderto obtain an accurate value for the wavelength.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(3)

(Total 13 marks)


An optical fibre consists of a core, cladding and an outer sheath.

(a) State the purpose of the outer sheath in an optical fibre.

___________________________________________________________________

___________________________________________________________________

(1)

30

(b) For one fibre, the speed of monochromatic light in the core is 1.97 × 108 m s−1 and thespeed in the cladding is 2.03 × 108 m s−1.

Calculate the critical angle for this light at the interface between the core and the cladding.


(2)

(Total 3 marks)

Diamond jewels sparkle because light that enters the diamond at different incident angles isreflected back to an observer. Figure 1 shows the path of one of these incident rays through adiamond.

Figure 1

31


(a) (i) Calculate the critical angle for diamond.

Refractive index of diamond = 2.42

critical angle = ____________________ degree

(2)

(ii) The ray shown in Figure 1 enters at an angle of incidence of 50.2°.Calculate the angle of refraction θ.

θ = ____________________ degree

(2)

(iii) The angles of a diamond are chosen to maximise the amount of light reflected.Figure 2 shows a diamond with different angles to that of a normally shapeddiamond. The dotted lines show the normal shape of a diamond.

Figure 2

Draw on Figure 2 the path of the ray until it leaves the diamond.

(2)


(iv) Moissanite is a transparent material with a refractive index of 2.67.

Discuss whether this material, if made to the diamond shape shown in Figure 1,would reflect light back more or less than diamond.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(b) Figure 3 shows an infrared ray entering an optical fibre. The refractive index of the core is1.55 at infrared frequencies.

Figure 3

(i) Calculate the speed at which infrared radiation travels in the core.

speed = ____________________ m s−1

(1)


(ii) The wavelength of this infrared radiation is 1300 nm in air.Calculate the wavelength of infrared in the core.

wavelength = ____________________ m

(2)

(iii) State one reason for surrounding the core with cladding.

______________________________________________________________

______________________________________________________________

(1)

(Total 12 marks)

(a) Tick (✓) the appropriate boxes in the table to indicate how the wavelength, frequency andspeed of light are affected when a ray of light travels from air into glass.

Wavelength Frequency Speed

increases

stays the same

decreases

(2)

32


(b) Figure 1 shows a right-angled glass prism in contact with a transparent substance on oneof the faces. One of the other angles of the prism is θ.

Figure 1

(i) A ray A enters perpendicularly to one face of the prism. It is partially refracted andpartially reflected at the interface between the glass and the transparent substance.

The angle of refraction is 65.0°. The ray eventually leaves at an angle α to thesurface of the transparent substance.

Determine the angle α.

angle α = ____________________ degree

(2)

(ii) Determine the angle θ in Figure 1.

angle θ = ____________________ degree

(2)


(c) Figure 2 shows another ray entering the prism.

Figure 2

(i) Identify the effect that takes place at X in Figure 2.

______________________________________________________________

(1)

(ii) Explain, with a diagram, how the effect that occurs at X is used to transmitinformation along an optic fibre.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(3)

(Total 10 marks)


Musicians can use tuning forks to tune their instruments.A tuning fork produces a specific frequency when it vibrates.

Figure 1 shows a tuning fork vibrating in air at a single instant in time.The circles represent the positions of air particles in the sound wave.

Figure 1

33

(a) The tuning fork emits a wave that has a frequency of 0.51 kHz.

(i) State the meaning of the term frequency of a wave.

______________________________________________________________

(1)

(ii) Air particles vibrate in different phases in the direction in which the wave is travelling.

Calculate the minimum separation of particles that vibrate 180° out of phase.

speed of sound in air = 340 m s–1

minimum separation ____________________ m

(3)


(b) A student sets a tuning fork of lower frequency vibrating at the same time as the 0.51 kHztuning fork in part (a).

The student detects the resultant sound wave with a microphone. The variation with time ofthe voltage generated by the microphone is shown in Figure 2.

Figure 2

(i) Explain why the two tuning forks are not coherent sources of sound waves.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(2)

(ii) Explain why the resultant sound has a minimum amplitude at 50 ms.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(3)


(iii) Calculate the frequency of the tuning fork that emits the lower frequency.

frequency ____________________ Hz

(3)

(c) A signal generator connected to a loudspeaker produces a sinusoidal sound wave with afrequency of 440 Hz.

The variation in air pressure with time for this sound is shown in Figure 3.

Figure 3

A violin string has a fundamental frequency (first harmonic) of 440 Hz.

Figure 4 shows the variation in air pressure with time for the sound created by the violinstring.

Figure 4


(i) The two sounds have the same pitch but sound different.

What term describes the difference between the sounds heard?Tick (✔) the correct answer.

Frequency modulation

Octaves

Path difference

Quality

(1)

(ii) The complex sound in Figure 4 can be electronically synthesised.

Describe the process of electronically synthesising this sound.

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

______________________________________________________________

(3)

(Total 16 marks)


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

(a) Explain what is meant by coherent sources.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

34

(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 differencebetween 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 thediagram above.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(d) The experiment is now rearranged so that the perpendicular distance from the slits to thedetector 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 significantfigures.

slit separation = ____________________ m

(3)

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

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(Total 14 marks)


The figure below shows a layer of oil that is floating on water in a glass container. A ray of light inthe oil is incident at an angle of 44° on the water surface and refracts.

The refractive indices of the materials are as follows.

refractive index of oil = 1.47refractive index of water = 1.33refractive index of the glass = 1.47

35

(a) Show that the angle of refraction θ in the figure above is about 50°.

(2)

(b) The oil and the glass have the same refractive index. On the figure above, draw the path ofthe light ray after it strikes the boundary between the water and the glass and enters theglass. Show the value of the angle of refraction in the glass.

(2)

(c) Explain why the total internal reflection will not occur when the ray travels from water toglass.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(1)


(d) Calculate the critical angle for the boundary between the glass and air.

answer = ______________________ degrees

(2)

(e) On the figure above, complete the path of the ray after it strikes the boundary between theglass and air.

(2)

(Total 9 marks)


Mark schemes

(a) diagram to show:total internal reflection on side face (1)ray emerging at base bent away from normal (1)with ≈ correct angles (1)

3

1

(b) n = (1)

= with calculation (1) (= 1.41)2

(c) sin θi = n sin θr (1)sin θi = 1.41 × sin 40 (1)θi = 65° (1)

3

[8]

A

[1]2

(a) (i) (use of gives)1.45 = (1)

θ1 = 22.8o (1)

3

(ii) (1)

(1)

use of and (1)

[or n1 sin θ1 = n2 sin θ1]

1.45 sin θ3 = 1.60 sin 51.3 (1)

θ3 = 59.4º (1)

(allow C.E. for value of n from (ii))7


(b) block 1 (1)(requires some explanation)

reference to (1)

[or statement such as light refracts/bends towards normal as it enters adenser/higher refractive index material, or block 1 has lower refractiveindex]

2

(c) reflection at boundary with i = r (1)

refraction (at bottom surface) bending away from normal (1)2

[11]

(a) sin c = nc/nf (1)

B1

78.6° (1)

B12

4

(b) required for total internal reflection (1)

B1

avoids signal loss (1)

B1

avoids cross-talk (owtte) (1)

B1max 2

[4]

(a) (1)

1

(b) TIR on hypotenuse and refraction at top surface (1)55°, 10° and 15° all marked correctly (1)

2

5


(c) (i) use of

[or n1 sin θ1 = n2 sin θ2] (1)

1.49 sin 55º = 1.37 sin θ2 (1)

θ2 = 63º (1)

(ii) (use of ) gives 1.37 = (1)

c2 = 2.2 ×108 ms–1 (1)(2.19 × 108 ms–1)

(iii) refraction at boundary between prisms, refractedaway from normal (1)

emerging ray (r.h. vertical face) refracting awayfrom normal (1)

7

[10]

D

[1]6

A

[1]7

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]

8


(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)

9

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]


(a) w = λD/s

C1

correct substitution – condone wrong powers of ten

C1

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

A13

10

(b) same frequency or wavelength

C1

constant phase (relationship)

A12

[5]

(a) (i) (refractive index of water = 1/sin 49.0) = 1.33 (not 1.3 or 1.325) (1)

(ii) ray P shown in the air to right of vertical (1)

refracted away from the normal in the correct direction (1)

correct partial reflection shown (1)4

11

(b) (i) critical angle for water-air boundary = 49.0°or angle of (incidence of) Q is θc (1)

the angle of incidence (of R) exceeds the critical angle (1)

(ii) the figure shows that R undergoes TIR at water surface andstrikes the glass side (1)

angle of incidence at glass side = 30° (1)

R enters the glass and refracts towards the normal (1)

because ng > nw (1) (or water is optically less dense than glass)

(calculates angle = 26.2° gets last two marks)6

[10]


(a) (i) (using n1 sin θ1 = n2 sin θ2 or sin θc = n2/n1 gives)

correct substitution in either equation (eg 1.55 sin c = 1.45 (sin 90)or sin c = 1.45/1.55) (1)

= 0.9355 (accept less sf) (1)c = 69.3(°) (1) (accept 69.4°, 69° or 70°)

(ii) the angle (of incidence) is less than the critical angleor values quoted (1)

12

(iii) (using n1 sin θ1 = n2 sin θ2 gives)

1.55 sin 60 = 1.45 sin θ (1)

(sin θ = 1.55 sin 60/1.45 =) 0.9258 or 0.926 or 0.93 (1)

θ = 67.8° (1) (accept 68° or 68.4)7

(b) any two from:

keeps signals secure (1)

maintains quality/reduces pulse broadening/smearing (owtte) (1)

it keeps (most) light rays in (the core due to total internal reflectionat the cladding-core boundary) (1)

it prevents scratching of the core (1)

(keeps core away from adjacent fibre cores) so helps to preventcrossover of information/signal/data to other fibres (1)

cladding provides (tensile) strength for fibre/prevents breakage (1)

given that the core needs to be very thin (1)max 2

[9]

(a) (i) = 590 × 10–9m (1)

(using d sin θ = nλ gives)

(1) = 0.707 or

7.07 × 108 if nm used (1)

θ = 45.0° (1) (accept 45°)

13


(ii) (sin θ ≤ 1) gives ≤ 1 or n ≤ or = (1) = 2.83 (1)

so 3rd order or higher order is not possible (1)

alternative solution:(substituting) n = 3 (into d sin θ = nλ gives) (1)

sin θ ( ) = 1.06 (1)

gives ‘error’/which is not possible (1)7

(b) (using d sin θ = nλ gives)

2 λ = 1.67 × 10–6 × sin 42.1 (1)

λ(= 0.5 × 1.67 × 10–6 × sin 42.1) = 5.6(0) × 10–7 m (or 560 nm) (1)2

[9]

(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

14

(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]


(a) (i) A: cladding + B: core (1)1

(ii)

refraction towards the normal line (1)

continuous lines + strikes boundary + TIR correct angles byeye + maximum 2 TIRs (1)

2

15

(b) or = 0.9865 (1)

80.6 or 80.8 or 81 (°) only (1)2

(c) to reduce multipath or multimode dispersion (1)

(which would cause) light travelling at different angles to arrive atdifferent times/pulse broadening/merging of adjacent pulses/’smearing’/poor resolution/lower transmission rate/lower bandwidth/less distancebetween regenerators (1)

or to prevent light/data/signal loss (from core or fibre) (1)

(which would cause) signal to get weaker/attenuation/crossover/datato be less secure (1)

2

(d) correct application (1) (endoscope, cytoscope, arthroscope etc,communications etc)

linked significant benefit stated eg improve medical diagnosis/improvetransmission of data/high speed internet (1)

2

[9]


(a) use of c/cs = n (condone inversion of c and cs)

C1

1.9 × 108 (m s–1)

A12

(b) (i) path difference = 120 m or 0.12 km/finds twotimes and subtracts

(allow incorrect speed with working) (condone powerof ten error)

C1

(penalise use of different speeds)

6.4 × 10–7s (ecf from (a))

A12

16

(ii) refractive index varies across (graded-index)core/refractive index maximum at centre of core

B1

ray travels slower in centre/rays travel faster atedge/ray B travels faster/ray A travels slower

B12

[6]

(a) reflects at correct angle by eye (use top of ‘27’ and bottomof ‘42’ as a guide) or 27° or 63° correctly marked (1)

refracts away from normal at glass/air (1)

symmetrical by eye or refracted angle (42°) correctly markedand at least one normal line added (1)

3

17

(b) (ng) = (1) DNA 42/27 = 1.56

= 1.47 (1.474) 3 sf shown (1)2

(c) 63 (°) (1)

allow 62 to 62.99 with reasoning, allow ‘slightly less than 63’without reason given

1


(d) = 1.474 sin (c) (1) or use of n = 1.5

= 1.3(1) or 1.34 if n = 1.5 used (1)2

[8]

(a) λ correct (1)

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

2

18

(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]


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

intensity decreasing (1)

2

19

(b) a single wavelength (1)

constant phase relationship/difference (1)2

(c) maxima further apart/central maximum wider/subsidiary maximumwider/maxima are wider (1)

1

(d) wider/increased separation (1)

lower intensity (1)2

(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 subsidiarymaximum (1)

(middle of) central maximum white (1)3

[10]


(a) (sheath) to protect fibre (and cladding)/to add strength(to cladding)/prevent loss of signal from scratches

at least sense of protecting fibre/claddingdisallow anything that could infer that it is cladding[eg prevent signalloss/protects info]treat extra as neutralcladding explanation zero marks

B11

20

(b) use of sin c = n2/n1 (condone ratio inverted)

C1

sin 60 = n2/1.6 (condone lack of subscript)

C1

1.4/1.39 (condone units)

or sub for c and an n or 1.85/1.9/1.8 seen (1st)alternative use of n1 sinθ1 = n2sinθ2 with a θ = 90 (1st)

correct sub (2nd)

A13

(c) different wavelengths different speeds/differentwavelengths different refractive indices/differentpaths/different angles/different distances

B1

spreading of pulse/spreading into different wavelengths

B12

[6]

(a) (i) cladding121

(ii) sin θc = 1.41/1.46

θc= 75.0 (°) (74.96) 2

(b) (i) 65 (degrees) 1


(ii) 1.46 sin 65 = 1.41 sin r or sin r = 0.93845 ecf bi

r = 70 (degrees) (69.79) ecf bi2

(c) Two from:

• less light is lost

• better quality signal / less distortion

• increased probability of TIR

• Less change of angle between each reflection

• reflects more times (in a given length of fibre) keeping (incident) anglelarge(r than critical angle)

• (angle of incidence is) less likely to fall below the critical angle

• less refraction out of the core

• improved data transfer / information / data / signal carried quicker

• less multipath dispersion (smearing / overlap of pulses)

2

[8]

(a) decrease

constant

decrease 3

22

(b)

straight ray (ignore arrow) reflecting to the right

reflected angle = incident angle

(accept correct angle labels if reflected angle is outside tolerance)2


(c) (i) (n = ) use of 3 (× 108) = = 1.47 (1.4706)

(must see 3 sf or more)2

(ii) sin θc = or correct substitution in un-rearranged formula

θc = 80.4 (80.401) (80.3 to 80.54) (≈ 80°) must see 3 sf or more2

(d) angle of refraction = 180 – 90 – 80.4 = 9.6°

sinθ = 147(06) sin 9.6 = 0.25 ecf from first mark

θ = 14 (= 14.194°) ecf from first mark

range 13 to 15° due to use of rounded values3

(e) (reduced amplitude) due to absorption/energy loss(within the fibre)/attenuation/scattering (by the medium)/loss from fibre

(pulse broadening caused by) multi-path (modal) dispersion/different rays/modes propagating at different angles/nonaxial rays take longer time to travel same distance along fibreas axial rays

2

[14]

(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

1

23

(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, etc2

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

Don’t allow 81 degrees1


(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 marks2

(c) • protect the core (from scratches, stretching or breakage)

comment on ‘quality’ of signal is not sufficient

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

don’t allow ‘leakage’ on its own.

• increase the critical angle / reduce pulse broadening / (modal)dispersion / rayswith 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) 2

[8]

(a) (i) sin 56 = nglass sin 30

(nglass = sin56/sin30) (= 1.658) = 1.7 2

24

(ii) sin θc = 1/1.658 ecf from ai

θc = (37.09 or 37.04) = 37 (degrees)

accept 36 (36.03 degrees) for use of 1.72

(b) TIR from the upper side of the prism ecf from part aii

and correct angle

refraction out of the long edge of the prism away from the normal 2

[6]


(a) n1 > n2

Allow correct reference to ‘ optical density’

(incident) angle > critical angle (allow θc not ‘c’)OR critical angle must be exceeded

Allow nA > nB

Do not allow: ‘angle passes the critical angle’2

25

(b)

For second mark, don’t allow 1.6 × 10 8

Allow 1.66 × 108 or 1.70 × 108

Allow 1.6. × 108

(= 1.667 × 10 8) = 1.67 × 108 (ms−1) 2

(c) sin72 = 1.80sin θ

Correct answer on its own gets both marks

θ = 31.895 = 31.9 correct answer >= 2sf seen

Do not allow 31 for second markAllow 31.8 − 32

2

(d) 1.80 sin θc=1.40 OR

Correct answer on its own gets both marks

Don’t accept 50 by itself2

θc = 51.058 = 51.1 ° (accept 51)


OR = 0.778

(e) (i) 22 + their (c) (22 + 31.9 = 53.9) 53.9 > (51.1) critical angle

If c + 22 < d then TIR expected

If c + 22 > d then REFRACTION expected

ORc + 22 < their d (θc ) ecf from (c) and (d)angle less than critical angle

Allow max 1 for ‘TIR because angle > critical angle’ only if their d >c + 22

2

(ii) TIR angle correct ecf from e(i) for refraction answer

Tolerance: horizontal line from normal on the right / horizontal linefrom top of lower arrow.If ei not answered then ecf (d). If ei and d not answered then ecf c

1

[11]

(a) one of:(spectral) analysis of light from stars(analyse) composition of starschemical analysismeasuring red shift \ rotation of stars ✓

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

Allow : measuring wavelength or frequency from a named source oflight

Allow any other legitimate application that specifies the source oflight. E.g.absorbtion \ emission spectra in stars,‘observe spectra of materials’

1

26

(b) (i) first order beamfirst order spectrumfirst order image✓

Allow ‘n = 1’ , ‘1’ , ‘one’, 1 st

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 2

Power of 10 error in d gets max 2

For use of d in mm, answer =

5.25 × 10−4 gets max 2

n = 2 gets max 2 unless ecf from bii

use of d = 1480 yields wavelength of 1150m3

(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, 1.3

Second line gets both marks

Conclusion consistent with working2

[8]

(a) single frequency (or wavelength or photon energy)

not single colouraccept ‘ very narrow band of frequencies’

1

27


(b) subsidiary maxima (centre of) peaks further away from centre

For second mark: One square tolerance horizontally. One wholesubsid max seen on either side.

subsidiary maxima peaks further away from centre AND central maximum twice width ofsubsidiaries AND symmetrical

Central higher than subsid and subsid same height + / − 2 squares.Minima on the x axis + / − 1 square.Must see 1 whole subsidiary for second mark

2

(c) ONE FROM:

• don't shine towards a person• avoid (accidental) reflections• wear laser safety goggles• 'laser on' warning light outside room• Stand behind laser• other sensible suggestion

allow green goggles for red laser, ‘high intensity goggles’, etc.

not ‘goggles’, ‘sunglasses’

eye / skin damage could occur 2

(d) 3 from 4

• central white (fringe)• each / every / all subsidiary maxima are composed of a spectrum (clearly stated or

implied)• each / every / all subsidiary maxima are composed of a spectrum (clearly stated or

implied) AND (subsidiary maxima) have violet (allow blue) nearest central maximumOR red furthest from centre

• Fringe spacing less / maxima are wider / dark fringes are smaller (or not present)

allow ‘white in middle’For second mark do not allow ‘there are colours’ or ‘there is aspectrum’ on their ownAllow ‘rainbow pattern’ instead of spectrum but not ‘a rainbow’Allow ‘rainbow pattern’ instead of spectrum but not ‘a rainbow’If they get the first, the second and third are easier to award

Allow full credit for annotated sketch3

[8]

(a) (i) sin 60 = 1.47sin θ OR sin θ = sin 60 / 1.47 ✓(sin−1 0.5891) = 36 (°) ✓ (36.0955°) (allow 36.2)

Allow 36.02

28


(ii) sin θc = 1.33 / 1.47 OR sin θc = 0.9(048) ✓(sin−1 0.9048) = 65 (°) ✓ (64.79)

Allow 64 for use of 0.9 and 66 for use of 0.912

(iii) answer consistent with previous answers, e.g.if aii >ai:ray refracts at the boundary AND goes to the right of the normal ✓Angle of refraction > angle of incidence ✓ this mark depends on the first

if aii < ai:TIR ✓angle of reflection = angle of incidence ✓

ignore the path of the ray beyond water / glass boundary

Approx. equal angles (continuation of the line must touch ‘Figure 1’label)

2

(b) for Reason or Explanation:the angle of refraction should be > angle of incidence when entering the water ✓water has a lower refractive index than glass \ light is faster in water than in glass ✓

TIR could not happen \ there is no critical angle, when ray travels from water to oil ✓TIR only occurs when ray travels from higher to lower refractive index \ water has alower refractive index than oil ✓

Allow ‘ray doesn’t bend towards normal’ (at glass / water)

Allow optical density

Boundary in question must be clearly implied4

[10]

(a) uniform width peaks ✓ (accurate to within ± one division)

peaks need to be rounded ie not triangular

the minima do not need to be exactly zero

a collection of peaks of constant amplitude or amplitude decreasing away fromcentral peak ✓

pattern must look symmetrical by eye

condone errors towards the edge of the pattern

double width centre peak total mark = 02

29

(b) (i) constant / fixed / same phase relationship / difference (and same frequency /wavelength) ✓

in phase is not enough for the mark1


(ii) single slit acts as a point / single source diffracting / spreading light to both slits✓ORthe path lengths between the single slit and the double slits are constant / thesame / fixed ✓

1

(iii) superposition of waves from two slits ✓phrase ‘constructive superposition’ = 2 marks

diffraction (patterns) from both slits overlap (and interfere constructively) ✓ (thismark may come from a diagram)

constructive interference / reinforcement (at bright fringe)peaks meet peaks / troughs meet troughs ✓ (any reference to antinode will losethis mark)

waves from each slit meet in phaseOR path difference = n λ ✓

4 max 3

(c) (i) D = = ✓ do not penalise any incorrect powers of ten forthis mark= 0.5 (m) ✓ (0.4938 m)

numbers can be substituted into the equation using any form

note 0.50 m is wrong because of a rounding error

full marks available for answer only2

(ii) fringes further apart or fringe / pattern has a greater width / is wider ✓ignore any incorrect reasoning

changes to green is not enough for mark1

(iii) increase D ✓measure across more than 2 maxima ✓

several / few implies more than two

added detail which includes ✓explaining that when D is increased then w increasesOrrepeat the reading with a changed distance D or using different numbers offringes or measuring across different pairs of (adjacent) fringesOrexplaining how either of the first two points improves / reduces the percentageerror.

no mark for darkened room3

[13]


(a) Prevents (physical) damage to fibre / strengthen the fibre / protect the fibre

Allow named physical damage e.g. scratching

B1

Prevent crosstalk1

30

(b) (Relative) refractive index = 1.03orUse of sinc = n2 / n1

Calculating the refractive indices and rounding before dividing gives76.8

C1

76.0° or 76.8°

A12

[3]

(a) (i) sinC = 1/n = 1/2.42 ✔ (= 0.413)C = 24.4° ✔ (allow 2 or more sig figs)

Answer only gains both marks2

31

(ii) sin θdia = sin θair / n = sin 50.2 / 2.42 ✔ (= 0.317)

θdia = 18.5° ✔ (allow 2 or more sig figs)

Answer only gains both marks

Answer can be 18° or 19° depending on rounding2

(iii) TIR shown at bottom left surface ✔(If the reflected ray were extended it would passthrough the writing below the diagram between the ‘i’ in ‘it’ and the full stop at the endof ‘diamond’.) ray leaves bottom right surface either with an increased emergentangle or straight though if hitting normally ✔(The second mark is consequential on gaining the first mark)

acceptable emergent rays2


(iv) it has smaller critical angle / critical angle is 22°allowing more / same number / greater chance / increased probability of TIR’soccurringgreater/same sparkle ✔✔ max 2

‘reflect more’ is insufficient for a mark2

(b) (i) ccore = cair / n = 3.00 × 108 /1.55 = 1.9 × 108 (ms-1)✔(1.94× 108 ms-1)

1 sig fig is not acceptable if no other answer is given1

(ii) (n = cair / ccore = f λair / fλcore =λair /λcore )

λcore = λair / n or 1300 × (10-9) / 1.55✔= 8.4 × 10-7 (m)✔(8.39 × 10-7 m or 839 nm)

The first mark is for the equation or substitution ignoring powers of10 errors

1st mark can be gained from calculating the frequency ( f = 3.0 ×108 / 1300 × 10-9 = 2.3 × 1014 (Hz) which then can be used to findthe the wavelength

Using this method the answer can range between 8.4 × 10-7 → 8.7× 10-7 (m) and consider ecf’s from (b)(i)

2

(iii) protects the core (from scratches etc)prevents crosstalk / stops signal crossing from one fibre to another / increases criticalangle / reduces pulse broadening / reduces smearing / prevents multipath dispersionallows fibre to be supported / touched (without losing light)

✔any one point

Preventing signal loss is not enough for the mark.1

[12]

(a)

wavelength frequency speed

increases

stays thesame

✓

decreases ✓ ✓

middle column correct ✓first and third column correct ✓

2

32


(b) (i) (n1sinθ2 = n2sinθ2)(1.09)sin 65.0 = (1.00)sinθ2 ✓ (giving θ2 = 81°)

α = 9(°) ✓ (8.93°)

no internal CE

allow 9.0°2

(ii) 1.09sin65 = 1.70sinxor sinx = 0.58or x = 35.5 (°) ✓ (allow 35° or 36°)

[beware an answer close to the correct value can come from

n = 1 / sinC]

90 − 35.5 = 54.5(°) ✓ (allow 54° or 55°) CE for 90° − their value

2

(c) (i) total internal reflection

TIR does not gain the mark1

(ii) diagram showing core / cladding and light ray TIR at interface at least once withanother TIR shown on the diagram or suggested in their explanation ✓

labelling is not required and reflections do not have to be accurateprovided they are shown on the correct side of the normal

light fibre consists of core and cladding with lower refractive index / opticaldensity ✓

light (incident) at angle greater than the critical angle (results in TIR) ✓3

[10]

(a) (i) Number of complete waves passing a point in one second / number of completewaves produced by a source in one second / number of complete vibrations(oscillations) per second / number of compressions passing a fixed point persecond

1

33

(ii) 180° phase difference corresponds to ½ λUse of v = fλ with correct powers of 100.33 (m)

3

(b) (i) Do not have the same frequencydo not have a constant phase difference

2

(ii) Waves meet antiphaseUndergo superpositionResulting in destructive interference

3


(iii) T = 100 ms

Use of T = 1 / f or beat frequency (∆f) = 10 Hz500 (Hz) (allow 510 –their beat frequency)

3

(c) (i) Only box ticked: Quality1

(ii) Add regular alternating voltages togetherWith appropriate amplitudesWhere frequencies of voltages match the harmonics of sound / where frequenciesare multiples of 440 Hz

Allow 2 for sampling sound (at twice max frequency ) B1

Convert to binary ( and replay through D to A converter). B13

[16]

(a) same wavelength / frequency

constant phase relationship allow ‘constant phase difference’ but not ‘in phase’2

34

(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 0.032

(ii) 3.2 × 10−2 (m) ecf from bi

Don’t allow ‘1 wavelength’, 1λ, etcDo not accept: zero, 2

1

, 360 °

(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 bi

Do not penalise s and w symbols wrong way round in working ifanswer is correct.

= 0.12 (0.1218 m)

Correct answer gains first two marks.

= any 2sf number

Independent sf mark for any 2 sf number3

(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. ‘pathdifference is now 2 λ’

3

[14]

(a) sin θ = or 1.33 sin θ = 1.47 sin44 or sin–1 0.768 (1)

θ = 50.15, 50.2, 50.35 (°) (1)

answer seen to > 2 sf2

(b) refracts towards normal (1) 44° shown (1)2

35

(c) (TIR) only when ray travels from higher n to lower n or (water to glass) islower n to higher n (1)

do not allow ‘density’, allow ‘optical density’, n or refractive indexonly

1

(d) sin θc = or 1.47 sin θc = (1 ×) sin90 (1)

θc = 42.86 (= 43.0(°)) (1)2


(e)

2

[9]


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