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Colonel Frank Seely School

Exampro A-level Physics (7407/7408) 3.3.2.3 Refraction at a plane surface

Name:

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Time: 321

Marks: 275

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Q1.Monochromatic light passes from air into water. Which one of the following statements is true?

A The velocity, frequency and wavelength all change

B The velocity and frequency change but not the wavelength

C The velocity and wavelength change but not the frequency

D The frequency and wavelength change but not the velocity

(Total 1 mark)

Q2. The diagram below shows three wavefronts of light directed towards a glass block in the air. The direction of travel of these wavefronts is also shown.

Complete the diagram to show the position of these three wavefronts after partial reflection and refraction at the surface of the glass block.

(Total 3 marks)

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Q3. Figures 1 and 2 each show a ray of light incident on a water-air boundary. A, B, C and D show ray directions at the interface.

Figure 1 Figure 2

(a) Circle the letter below that corresponds to a direction in which a ray cannot occur.

A B C D(1)

(b) Circle the letter below that corresponds to the direction of the faintest ray.

A B C D(1)

(Total 2 marks)

Q4.A small intense light source is 1.5 m below the surface of the water in a large swimming pool, as shown in the diagram.

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(a) Complete the paths of rays from the light source which strike the water surface at X, Y and Z.

(b) Calculate the diameter of the disc through which light emerges from the surface of the water.

speed of light in water = 2.25 × 108 m s–1

speed of light in air = 3.00 × 108 m s–1

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Q5.(a) The diagram shows a 'step index' optical fibre. A ray of monochromatic light, in the plane of the paper, is incident in air on the end face of the optical fibre as shown in the diagram.

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(i) Draw on the diagram the complete path followed by the ray until it emerges at the far end.

(ii) Name the process which occurs as the ray enters the end of the optical fibre.

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(iii) The core has a refractive index of 1.50, clad in a material of refractive index 1.45. Calculate the critical angle of incidence at the core-cladding interface.

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(b) (i) Give one reason why a cladding material is used in an optical fibre.

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(ii) In part (a)(iii), the cladding material has a refractive index of 1.45. Explain why it would be advantageous to use cladding material of refractive index less than 1.45.

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

(c) State one use of optical fibres.

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(Total 11 marks)

Q6.The diagram shows a cross-section of one wall and part of the base of an empty fish tank, viewed from the side. It is made from glass of refractive index 1.5. A ray of light travelling in air is incident on the base at an angle of 35° as shown.

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(a) Calculate the angle θ.

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

(b) (i) Calculate the critical angle for the glass-air interface.

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(ii) Hence, draw on the diagram the continuation of the path of the ray through the glass wall and out into the air. Mark in the values of all angles of incidence, refraction and reflection.

(6)(Total 8 marks)

Q7.Two prisms made from different glass are placed in perfect contact to form a rectangular block surrounded by air as shown. Medium 1 has a smaller refractive index than medium 2.

(a) A ray of light in air is incident normally on medium 1 as shown. At the boundary between medium 1 and medium 2 some light is transmitted and the remainder reflected.

(i) Sketch, without calculation, the path followed by the refracted ray as it enters medium 2 and then emerges into the air.

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(ii) Sketch, without calculation, the path followed by the reflected ray showing it emerging from medium 1 into the air.

(4)

(b) The refractive index of medium 1 is 1.40 and that of medium 2 is 1.60.

(i) Give the angle of incidence at the boundary between medium 1 and medium 2.

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(ii) Calculate the angle of refraction at this boundary.

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(c) Calculate the critical angle for a ray passing from medium 2 into the air.

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(Total 10 marks)

Q8.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.

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(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

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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

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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)

Q9.A glass plate surrounded by air is made up of two parallel sided sheets of glass in perfect contact as shown in the figure. Medium 1, the top sheet of glass, has a smaller refractive index than medium 2.

(a) A ray of light in air is incident on the top sheet of glass and is refracted at an angle of 40° as shown in the figure. At the boundary between medium 1 and medium 2 some light is transmitted and the remainder reflected.

On the figure, sketch without calculation, the following:

(i) the path followed by the transmitted ray showing it entering from the air at the top and emerging into the air at the bottom;

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(ii) the path followed by the reflected ray showing it emerging from medium 1 into the air.

(4)

(b) The refractive index of medium 1 is 1.35 and that of medium 2 is 1.65.

(i) Calculate the angle of incidence where the ray enters medium 1 from the air.

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(ii) Calculate the angle of refraction at the boundary between medium 1 and medium 2.

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(c) Total internal reflection will not occur for any ray incident in medium 1 at the boundary with medium 2.

Explain, without calculation, why this statement is true.

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

(Total 10 marks)

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Q10. The diagram shows a ray of light passing from air into a glass prism at an angle of incidence θi. The light emerges from face BC as shown.refractive index of the glass = 1.55

(a) (i) Mark the critical angle along the path of the ray with the symbol θc.

(ii) Calculate the critical angle, θc.

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(b) For the ray shown calculate the angle of incidence, θi.

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(c) Without further calculations draw the path of another ray of light incident at the same point on the prism but with a smaller angle of incidence. The path should show the ray emerging from the prism into the air.

(3)(Total 8 marks)

Q11. The diagram below shows a liquid droplet placed on a cube of glass. A ray of light from air, incident normally on to the droplet, continues in a straight line and is refracted at the liquid to glass boundary as shown.refractive index of the glass = 1.45

(a) Calculate the speed of light

(i) in the glass,

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(ii) in the liquid droplet.

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(b) Calculate the refractive index of the liquid.

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

(2)(Total 7 marks)

Q12. Figure 1 shows a cross-section through a rectangular light-emitting diode (LED). When current passes through the LED, light is emitted from the semiconductor material at P and passes through the transparent material and into the air at Q.

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Figure 1

(a) (i) The refractive index of the transparent material of the LED is 1.5. Calculate the critical angle of this material when the LED is in air.

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(ii) Figure 1 shows a light ray PQ incident on the surface at Q. Calculate the angle of incidence of this light ray at Q if the angle of refraction is 40°.

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(iii) Figure 1 also shows a second light ray PR incident at R at an angle of incidence of 45°. Use Figure 1 to explain why this light ray cannot escape into the air.

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(b) The LED in part (a) is used to send pulses of light down two straight optical fibres of the same refractive index as the transparent material of the LED. The fibres are placed end-on with the LED, as shown in Figure 2. Optical fibre 1 is positioned at Q and the other at S directly opposite P.

Figure 2

(i) Continue the path of the light ray PQ into and along the optical fibre.

(ii) Compare the times taken for pulses of light to travel along the same length of each fibre.

Give a reason for your answer.

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(Total 10 marks)

Q13. The diagram shows a ray of monochromatic light, in the plane of the paper, incident on the end face of an optical fibre.

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(a) (i) Draw on the diagram the complete path followed by the incident ray, showing it entering into the fibre and emerging from the fibre at the far end.

(ii) State any changes that occur in the speed of the ray as it follows this path from the source.Calculations are not required.

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(b) (i) Calculate the critical angle for the optical fibre at the air boundary.

refractive index of the optical fibre glass = 1.57

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(ii) The optical fibre is now surrounded by cladding of refractive index 1.47. Calculate the critical angle at the core-cladding boundary.

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(iii) State one advantage of cladding an optical fibre.

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(Total 10 marks)

Q14. The diagram shows a cross-sectional view of the base of a glass tank containing water. A point monochromatic light source is in contact with the base and ray, R1, from the source has been drawn up to the point where it emerges along the surface of the water.

(a) (i) Which angle, A to F, is a critical angle?

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(ii) Explain how the path of R1 demonstrates that the refractive index of glass is greater than the refractive index of water.

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

(b) Using the following information

A = 47.1°

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B = 42.9°

C = E = 41.2°

D = F = 48.8°

calculate

(i) the refractive index of water,

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(ii) the ratio, .

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(c) Ray R2 emerges from the source a few degrees away from ray R1 as shown.Draw on the diagram above the continuation of ray R2.Where possible show the ray being refracted.

(2)(Total 9 marks)

Q15. The diagram, which is not to scale, shows the cross-section of a 45° right angled glass prism supported by a film of liquid on a glass table. A ray of monochromatic light is incident on the prism at an angle of incidence θ and emerges along the glass - liquid boundary as shown.

refractive index of glass = 1.5

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(a) Calculate the speed of light in the glass.

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(b) Determine

(i) the angle of incidence, θ,

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(ii) the refractive index of the liquid.

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(c) The liquid is now changed to one with a lower refractive index. Draw a possible path for the ray beyond the point A and into the air.

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(2)(Total 9 marks)

Q16. The diagram shows a cube of glass. A ray of light, incident at the centre of a face of the cube, at an angle of incidence θ, goes on to meet another face at an angle of incidence of 50°, as shown in 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 the glass and out into the air.

(3)

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

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(c) Calculate the angle of incidence, θ.

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

(Total 8 marks)

Q17. The figure below shows a ray of light passing from air into glass at the top face of glass block 1 and emerging along the bottom face of glass block 2.

refractive index of the glass in block 1 = 1.45

(a) Calculate

(i) the incident angle θ1,

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

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

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

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Explain your reasoning.

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

(2)(Total 11 marks)

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Q18. 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 was made is 1.49.

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(b) On Figure 1, draw the continuation of the path of the ray of light until it emerges back into the air. Write on Figure 1 the values of the angles between the ray and any normals you have 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 placed firmly against the original prism, prism 1, to form a cube as shown in Figure 2.

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Figure 2

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

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

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(iii) Draw a path the ray could follow to emerge from prism 2 into the air.(7)

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(Total 10 marks)

Q19. (a) For an optical fibre the refractive index of the core is 1.52 and the refractive index of the cladding is 1.49. Calculate the critical angle, in degrees, at the boundary between the core and cladding of the fibre.

critical angle ..................... degrees(2)

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

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(Total 4 marks)

Q20. 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.

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

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

(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.

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

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

the refractive index of the glass = 1.50

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(Total 10 marks)

Q21. An optical fibre used for communications has a core of refractive index 1.55 which is surrounded by 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 the core-cladding boundary at Q at an angle of incidence of 60.0°.

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

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answer ........................... degrees(3)

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

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.............................................................................................................(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.

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(Total 9 marks)

Q22. The diagram below shows a cross-section through a step index optical fibre.

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

A

B

(1)

(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 between A 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.

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

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

Application

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Benefit

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(Total 9 marks)

Q23. A glass cube is held in contact with a liquid and a light ray is directed at a vertical face of the cube. The angle of incidence at the vertical face is then decreased to 42° as shown in the figure below. At this point the angle of refraction is 27° and the ray is totally internally reflected at P for the first time.

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

(3)

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(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)

Q24. The figure below shows a cross-section through an optical fibre.

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(a) Explain the purpose of part X.

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

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

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absolute refractive index for part Y .................................................(3)

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

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(Total 6 marks)

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Q25. The figure below shows a layer of oil that is floating on water in a glass container. A ray of light in the 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

(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 of the light ray after it strikes the boundary between the water and the glass and enters the glass. 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 to glass.

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(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 the glass and air.

(2)(Total 9 marks)

Q26. 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

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speed of light in the core ............................. ms–1

(2)

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

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time difference ............................................... s(2)

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

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(Total 6 marks)

Q27. (a) The speed of light is given by

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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)

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 the core is 1.47.

(2)

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

refractive index of the cladding = 1.45

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(2)

(d) Ray B enters the end of the fibre and refracts along the core-cladding boundary. Calculate the 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 significant distance 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

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(Total 14 marks)

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Q28. 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)

(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)

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Q29. 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)

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

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

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(b) (i) The ray leaves the core at Y. At this point the fibre has been bent through an angle of 30° 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 Figure 2. The curvature is the same and the path of a ray of light is shown.

Figure 2

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

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(Total 8 marks)

Q30.The diagram below shows a section of a typical glass step-index optical fibre used for communications.

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

(1)

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

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

(2)

(ii) Calculate the critical angle.

critical angle = .........................degrees(1)

(iii) Calculate the refractive index of the cladding.

refractive index = .....................................(2)

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

reason 1......................................................................................................

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reason 2......................................................................................................

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

(Total 8 marks)

Q31.The diagram below shows three transparent glass blocks A, B and C joined together. Each glass block has a different refractive index.

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

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condition 1 .....................................................................................................

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condition 2 .....................................................................................................

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

(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 is about 30o.

(2)

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

Calculate the critical angle between glass A and glass C.

critical angle ................................. degrees(2)

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

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

Page 47


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

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

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

(1)(Total 11 marks)

Q32.Figure 1 shows a ray of light A incident at an angle of 60° to the surface of a layer of oil that is floating 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)

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

Page 48


angle ................................ degrees(2)

(iii) On Figure 1 continue the path of the ray of light A immediately after it strikes the boundary 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 then entering 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 boundary is about 60°.

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

reason 1 ........................................................................................................

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

Page 49


explanation .....................................................................................................

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

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

reason 2 ........................................................................................................

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

explanation .....................................................................................................

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

........................................................................................................................(4)

(Total 10 marks)

Q33.Monochromatic light may be characterised by its speed, frequency and wavelength. Which of the following quantities change when monochromatic light passes from air into glass?

A Speed only.

B Speed and wavelength only.

C Speed and frequency only.

D Wavelength and frequency only. (Total 1 mark)

Q34.(a) Describe the structure of a step-index optical fibre outlining the purpose of the core and the cladding.

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

Page 50


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

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

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

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

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

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

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

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

(b) A signal is to be transmitted along an optical fibre of length 1200 m. The signal consists of a square pulse of white light and this is transmitted along the centre of a fibre. The maximum and minimum wavelengths of the light are shown in the table below.

Colour Refractive index of fibre Wavelength / nm

Blue 1.467 425

Red 1.459 660

Explain how the difference in refractive index results in a change in the pulse of white light by the time it leaves the fibre.

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

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

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

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

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

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

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

(c) Discuss two changes that could be made to reduce the effect described in part (b).

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

Page 51


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

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

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

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

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

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

(Total 7 marks)

Page 52


M1.C[1]

M2. reflection wavefront direction sensible

B1refraction wavefront direction sensible

B1one pair of wavefronts correctly spaced

B1[3]

M3. (a) A

B1

(b) D

B1[2]

M4.(a) ray straight through at X (1)ray refracted at >30° at Y (1)ray totally internally reflected at Z (1)

(b)

at critical angle sinθair = 1 (1)

Page 53


sinθwater = 0.75, θwater = 48.6° (1)radius = 1.5tan48.6° (1) =1.7m, ∴ diameter = 3.4m (1)

[7]

M5.(a) (i) diagram to show: (1)

ray refracted towards normal (1) total internal reflection at core-cladding interface (1) i = r indicated (1) ray continues whole length of fibre and emerges, without errors(1)

(ii) refraction (1)

(iii) use of 1n2 = (1)

θC (= θ2) = 75.2° (1)

[or sin θC = 1/n gives sinθC = nclad/ncore (1)

sin θC = 1.45/1.50 (1)

θC = 75.2°(1)](max 7)

(b) (i) to protect outer surface of the core (1)

(ii) greater acceptance angle (1)enables more light to be collected(or smaller critical angle makes escape less likely) (1)

(3)

Page 54


(c) endoscopy or communications (1)(1)

[11]

M6.(a)

θ = 22° (1) (22.48°)(2)

(b) (i) (sinθc = 1/n gives) sinθc = (1)

θc = 42° (1) (41.8°)

(ii) ray diagram to show:one total internal reflection (1)with one angle of reflection marked as 68° (1)correct refraction of ray on exit from top surface with 35° marked (1)angle of incidence of 22° marked at point of exit (1)

(6)[8]

M7.(a)

Ray diagram to show:

Page 55


(i) refraction towards normal at boundary (1)emerging ray refracted away from normal (1)

(ii) reflection at boundary with i ≈ remerging ray refracted away from normal (1)

4

(b) (i) 20° (1)

(ii)

θ2 = 17 (.4)° (1)4

(c) (sin θc = 1 / n gives) sin θc = 1 / 1.60 (1)

= θc = 38.7° (1)2

[10]

M8.(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

Page 56


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]

M9.(a)

(i) incident angle > 40° (1) angle of refraction into medium 2 < 40° (1) emergent ray with correct refraction (1)

(ii) reflection at boundary between media with i ≈ r (1) (hence) emergent ray at approximately same angle as incident ray and showing correct refraction (1)

max 4

Page 57


(b) (i) (use of 1n2 = gives) 1.35 = (1)

θ1 = 60(2)° (1)

(ii)

θ = 31.7° (1)5

(c) (total internal reflection) only occurs when light goes from a higher to a lower refractive index [or goes from a more dense to a less dense medium/material] (1)

1[10]

M10. (a) (i) θc marked (1)

(ii) sin θc = (1)

θc = 40.2° (1)3

(b) n = (1)

(θ2 = 90 – 75.2 = 14.8°)

θ1 (= sin–1{1.55 sin 14.8}) = 23.3° (1)2

(c) Mark scheme not available.3

[8]

Page 58


M11. (a) (i) (use of n = gives) cglass = ×

= 2.07 × 108 m s–1 (1)

(ii) use of (1)

cliquid = = 2.26 × 108 m s–1 (1)

(allow C.E. for values of cglass from (i))3

(b) use of

to give nliquid = = 1.33 (1)

(allow C.E. for value of cliquid)

[or use 1n2 = to give correct answer]2

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

2[7]

Page 59


M12. (a) (i) sin c = (1)

c = 42° (1) (41.8°)

(ii) 1.5 sin i = sin 40 (1)i = 25° (1) (25.4°) (use of c = 41.8° gives i = 26.4°)

(iii) total internal reflection at R (1)further total internal reflection below Q (1)further total internal reflection (1)

7

(b) (i) light ray enters fibre without refraction (1)total internal reflection at fibre/air surface (1)

(ii) pulse in fibre 1 takes longer because it travels across the fibre

as well as along it (1)3

[10]

M13. (a) (i) diagram to show: refraction towards normal on entry (1)total internal reflection shown along fibre (1)refraction away from normal on leaving glass (1)

(ii) speed of light decreases on entry into glass andincreases on leaving (1)

4

(b) (i) (use of sin θc = gives) sin θc = ) (1)

θc = 39.6° (1)

(ii) = (1) (=1.07)

sin θc = (1)

θc = 69.4° (1)

Page 60


(iii) to protect the core surface[or to prevent cross-over]

6[10]

M14. (a) (i) (angle) F (1)

(ii) angle D is greater than angle B[or at the glass-water boundary, ray R1 refracts awayfrom the normal] (1)

2

(b) (i) (use of sin θc = gives) sin 48.8 = (1)

n = 1.3 (1) (1.33)

(ii) use of (1)

(1)

= 1.1 (1) (1.11)5

(c) ray R2 to have greater angle of refraction in water than ray R1 (1)total internal reflection at water-air boundary (1)

2[9]

Page 61


M15. (a) cg (= ) = (1)

= 2.0 × 108 m s–1 (1)2

(b) (i) sin 1 (= n sin 2) = 1.5 × sin 15 (1)1 = 23° (1) (22.8°)

(ii) use of (1) (or equivalent)

n2 = (1)

= 1.3 (1)5

(c) total internal reflection at A (1)correct refraction out of glass at r.h.surface (1) (same angles as l.h. side)

2[9]

M16. (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

(b) n = (1)

= with calculation (1) (= 1.41)2

Page 62


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

3[8]

M17. (a) (i) (use of gives)1.45 = (1)

θ1 = 22.8o (1)

(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 a denser/higher refractive index material, or block 1 has lower refractive index]

Page 63

Colonel Frank Seely School2

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

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

[11]

M18. (a) (1) 1

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

2

(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

Page 64


[10]

M19. (a) sin c = nc/nf (1)

B1

78.6° (1)

B12

(b) required for total internal reflection (1)

B1

avoids signal loss (1)

B1

avoids cross-talk (owtte) (1)

B1max 2

[4]

M20. (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

(b) (i) critical angle for water-air boundary = 49.0°

Page 65


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]

M21. (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)

(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:

Page 66


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]

M22. (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

(b) or = 0.9865 (1)

80.6 or 80.8 or 81 (°) only (1)2

Page 67


(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]

M23. (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

(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

Page 68


(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]

M24. (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

(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

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spreading of pulse/spreading into different wavelengths

B12

[6]

M25. (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 sf

2

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

(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 index only

1

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

θc = 42.86 (= 43.0(°)) (1)2

(e)

Page 70


2[9]

M26. (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)

Page 71


C1

(penalise use of different speeds)

6.4 × 10–7s (ecf from (a))

A12

(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]

M27. (a) decrease

constant

decrease 3

(b)

straight ray (ignore arrow) reflecting to the right

reflected angle = incident angle (accept correct angle labels if reflected angle is outside

Page 72


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]

M28. (a) (i) sin 56 = nglass sin 30

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

Page 73


(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]

M29. (a) (i) cladding 1

(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

Page 74


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]

M30.(a) (n =) OR 0.2436 / 0.1657 working must be seen0.24 / 0.17 = 1.41 is not acceptable

AND ( = 1.4699) = 1.47 given correctly to 3 or more significant figuresWatch for:14.1 / 9.54 = 1.478

1

(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 degrees

1

(iii) (n = nc sin θ)allow 80 or 81 degrees here

Page 75


= 1.47 sin 80.46° ecf bii

=1.45 (1.4496)Correct answer gains both marks

2

(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 of information / data / signaldon’t allow ‘leakage’ on its own.

• increase the critical angle / reduce pulse broadening / (modal)dispersion / rays with a small angle of incidence will be refracted out of the coreDon’t allow ‘loss of light’

• increase rate of data transferAllow ‘leakage of signal’, etc

max two correct (from separate bullet points) 2

[8]

M31.(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

(b)

For second mark, don’t allow 1.6 × 108

Allow 1.66 × 108 or 1.70 × 108

Allow 1.6. × 108

Page 76


(= 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 θc = 51.058 = 51.1 ° (accept 51)

Correct answer on its own gets both marksDon’t accept 50 by itself

2

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 expectedIf c + 22 > d then REFRACTION expected

ORc + 22 c ) 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 line from top of lower arrow.If ei not answered then ecf (d). If ei and d not answered then ecf c

1[11]

Page 77


M32.(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

(ii) sin θ c = 1.33 / 1.47 OR sin θ c = 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 TIR ✓ angle of reflection = angle of incidence ✓

ignore the path of the ray beyond water / glass boundaryApprox. 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 a lower refractive index than oil ✓

Allow ‘ray doesn’t bend towards normal’ (at glass / water)Allow optical densityBoundary in question must be clearly implied

4[10]

M33.B

Page 78


[1]

M34.(a) Core is transmission medium for em waves to progress (by total internal reflection) ✓Allow credit for points scored on a clear labelled diagram.

1

Cladding provides lower refractive index so that total internal reflection takes place ✓

1

And offers protection of boundary from scratching which could lead to light leaving the core. ✓

1

(b) Blue travels slower than red due to the greater refractive index

Red reaches end before blue, leading to material pulse broadening ✓The first mark is for discussion of refractive index or for calculation of time difference.

1

Alternative calculations for first mark

Time for blue = d / v = d / (c / n) = 1200 / (3 × 108 / 1.467) = 5.87 × 10-6 s

Time for red = d / v = d / (c / n) = 1200 / (3 × 108 / 1.459) = 5.84 × 10-6 s

Time difference = 5.87 × 10-6 – 5.84 × 10-6 = 3(.2) × 10-8 s ✓The second mark is for the link to material pulse broadening

1

(c) Discussions to include:

Use of monochromatic source so speed of pulse constant

Use of shorter repeaters so that the pulse is reformed before significant pulse broadening has taken place.

Use of monomode fibre to reduce multipath dispersion ✓ ✓Answer must make clear that candidate understands the distinction between modal and material broadening.

2[7]

Page 79


Page 80


E2. Drawings were again very poor. Examiners awarded marks not only for correct directions of the wavefronts but also for an awareness that wavefront spacings change in refraction and remain unchanged in reflection. Candidates really must take more trouble over these relatively simple drawings if they are not to throw away marks.

E3. Although many candidates scored full marks on this simple opening question, there was a significant number who could not relate the diagram on the paper to the practical situation of light rays moving across the boundary between two media.

E4.Part (a), which should have proved to be a source of easy marks, produced too many answers in which ridiculous rays were drawn.

There were many completely correct answers to part (b), but some candidates got no further than working out a refractive index from the speeds given.

E5.This question also discriminated well. In part (a)(i) almost half the candidates failed to show the path of the ray changing direction due to refraction on entry into the core of the fibre and very few showed refraction on exit. The fact that they were not familiar with refraction was illustrated further by most of the candidates giving T.I.R. as their answer to part (a)(ii). Correct answers were often produced in part (a)(iii) but frequently the working was very sparse or confused.Part (b) worked well and allowed some candidates to show their clarity of thought. The answer to part (c) required only a statement, but even then some candidates were far too economical with their answers. For example, one word answers such as ‘by surgeons’, ‘television’ or ‘phone’ were not awarded a mark, but would probably have gained a mark if these words had been included in a simple sentence with a specific purpose.

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E6.In general the calculations to this question were performed very well, even by the weaker candidates. In part (a) many candidates did have difficulty at the intermediate stage of writing an inverse sine function and the expression θ = sin 35° / 1.5 sin−1 was seen as often as the correct expression θ = sin−1(sin 35° / 1.5).

Completing the ray diagram in part (b) was the main discriminator in this question. Essentially, the ray diagram was that of a simple light guide. Many candidates showed the light refracting out of the side wall, even though they performed the critical angle calculation correctly. Also, many candidates failed to mark the angles of incidence and refraction at the point of emergence from the top surface and several candidates performed an additional calculation to determine the angle of refraction. It was interesting to observe that candidates who used the equation of Snell’s law in the form n = sin i / sin r rather than n = sin θ1 / sin θ2, which is given in the data sheet, often mixed up the angles with the result that the emerging ray was refracted towards the normal

E7.The answers to this question showed that although a large number of candidates had some knowledge of the subject, they had not covered the specification material adequately. This lack of knowledge was apparent in the ray drawings in part (a), where the typical mark was 2 out of 4. There were some surprising aspects of scoring marks in this section, e.g. two marks were available for drawing the paths of the rays emerging into air and being refracted away from the normal. These two marks covered the same idea but a majority of candidates only scored one.

In part (b)(i) only about a quarter of the entry calculated that the angle of incidence at the boundary was 20°, but then managed to score reasonably well through the use of consequential errors. There were some good attempts at part (c) by a minority of candidates. Most candidates were uncertain whether to use a single refractive index or a relative refractive index throughout the calculations in parts (b) and (c).

E8.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.

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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.

E9.Only a minority of candidates tackled this question satisfactorily. In part (a) about 50% of the scripts showed incorrect refraction at the surface, with the angle of incidence being less than the angle of refraction and also did not show the angle of reflection to be equal to the angle of incidence at the boundary between the two media. A worrying number of candidates failed to use a ruler when completing the diagram and a penalty was applied to freehand drawings.

Again, in part (b), only about 50% of the candidates could use Snell’s law correctly when calculating the angle of incidence and refraction. Part (c) on the other hand, was well understood and answered correctly by most candidates. Although many candidates encountered difficulties with this question, the examiners were of the opinion that, in general, the answers did show a slight improvement to those in previous years on the optics question.

E10. Not for the first time in the history of this paper the optics question proved to be the most difficult in the paper. Only the best candidates could identify the critical angle. Candidates were a little more successful in calculating the critical angle but the units of degree were frequently omitted. In (b) candidates failed to get the correct answer because they did not calculate the angle in the glass as (90.0° – 75.2°) and often did not use the correct equation. The ray diagram in part (c) was done badly by a vast majority of candidates and the average mark was about one out of possible three.

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E11. The calculations in part (a) and part (b) were performed successfully by about 50 % of the candidates, but many fell at the first hurdle by trying to use the angles given in the question to calculate the speed of light in glass, rather than equate the refractive index to the ratio of the two speeds. A number of these candidates did however redeem themselves by calculating part (b) successfully. The overall impression created by the examinees was that although the relevant equations were known, they did not have the expertise to decide which equation was appropriate to the given calculation.

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

E12. Many candidates scored well in parts (a)(i) and (ii), although some were unable to use the relevant equations correctly. Almost all candidates were aware that the light ray was totally internally reflected at R in part (iii) although a few failed to mention the subsequent internal reflections of the ray.

In part (b) most candidates scored well, giving a correct ray diagram as well as a correct comparison and reason for the different times taken by the light pulses to travel. Some candidates lost a mark by drawing the refraction at Q incorrectly.

E13. As in previous years only a minority of candidates completed the ray diagram correctly, thus gaining full marks. The first error in part (a) occurred at the point where the ray entered the fibre; many candidates drew the refracted ray along the normal. There was a general lack of care in not making the angle of incidence different in value to the angle of reflection. Additionally, many candidates did not show any refraction at the far end of the fibre. Some of the TIR angles along the fibre were very carelessly drawn and the allocated mark was withheld on several occasions. Although most candidates were aware of the change in the speed of light as the ray entered the fibre, many failed to state that there was also a change in speed as the ray left the fibre.

The calculation in part (b) (i) was usually carried out correctly, but in part (ii) the less able candidates did not attempt to use both refractive indices. Many of the statements made in answer to part (iii) were of a very general nature and did not explicitly refer to surface scratches or contamination.

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E14. As in past papers, the optics question gave rise to the biggest problems in the paper. In part (a) (ii) many candidates did not clearly identify the boundary they were considering. In part (b) (i) the correct data was very often not used and a significant number of candidates used angles from both water and glass. Again, in part (b) (ii) candidates did not always use the appropriate data. For example, some used angles between the ray and the boundary itself. Some of the better candidates obtained the ratio 1.11 for the relative refractive index between glass and water but then failed to relate this to the ratio of the speeds of light in the respective media.

The ray diagram in part (c) was slightly easier than in previous papers but still about one third of the candidates failed to score the 2 marks available. Unlike in some previous papers it was rare to see candidates not using a ruler. It should be pointed out that the command ‘draw’ as opposed to ‘sketch’ in the question implies the use of a ruler and examiners are within their rights to apply a penalty.

E15. The optics question again seems to be the Achilles heel of most candidates and the calculation of the speed of light in part (a) was practically the only section that was done well.

The calculation of è in part (b) (i) was incorrect in about a third of cases because Snell’s law equation was inverted when data was substituted. Only a minority of candidates seemed to be able to cope with calculations involving more than one refractive index, as was required in part (b) (ii). It was extremely common to see

candidates attempting to use the equation at the glass-liquid boundary,which was wrong on two levels. Firstly it ignored the refractive index of the glass, and also the critical angle calculated was not the critical angle for the material for which the refractive index was required. When it came to drawing the path of the ray, there were more incorrect solutions than correct ones. Most candidates drew the ray passing down through the liquid and ignored total internal reflection altogether.

E16. As with previous examination papers, the ray diagram in part (a) was attempted with a general lack of care by most candidates. Very few noted that the emergent angle of the ray at the bottom of the cube was θ, which was one way of compensating for lack of accuracy on the diagram.

In part (b), a few candidates did not use the critical angle information and tried to work backwards from part (c). The problem was, however, tackled quite well on the whole. Part (c) caused more problems, because either an angle of 50° instead of 40° was used, or because Snell’s law equation was inverted.

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E17. Answers to this question were relatively good. The typical candidate knew the appropriate equations to use in part (a) and obtained good marks. The less able candidates, however, failed to come up with the correct equations when there were two media involved, or failed to choose the correct data to substitute in the equation.

In part (b), the explanation given by many less able candidates was insufficient to gain the allocated mark. Simply stating ‘it is less dense’ was quite a common but unsatisfactory answer. The diagram in part (c) was completed quite well by most candidates, but many showed the final ray emerging at 90º to the surface.

E18. Almost all candidates calculated the correct answer to part (a). Less able candidates did not show TIR in the diagram in part (b) and only the best candidates included all the angles required.

Only the top 10% obtained the correct answer to part (c) (i). There were many errors in the answers. Many could not determine the correct incident angle (10° was commonly seen); some did not use both refractive indices in the calculation and others inverted the equations when substituting data. Part (c) (ii) was tackled in an indirect manner by many candidates. They would calculate the speed of light in prism 1 and then use their calculated angles at the internal boundary to find the speed of light in prism 2. When drawing the ray in part (c) (iii), the direction in which the ray was refracted at the boundaries was almost a random choice.

E19. Most candidates used the appropriate equation for part (a) but less able candidates inverted the two refractive indices to achieve a ratio for which they could not take the inverse sine. A very significant minority of candidates suffered a penalty for not using two or three significant figures in their answers.

Few candidates were awarded two marks for part (b). Too many seemed confused by the properties of reflection and refraction (and to a lesser extent diffraction). Talk about ‘total internal refraction’ was common. Several candidates talked about multipath dispersion or the idea of this, but most thought that the cladding was a graded index and the cladding was there to increase the relative speeds between the waves. Few recognised that total internal reflection can occur without cladding and that the cladding actually increases the critical angle meaning that transmitted rays are more ‘parallel’ than they would otherwise be and that this reduces multipath dispersion. Candidates were credited for referring to

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cross-talk avoidance (or a clear understanding of cross-talk avoidance) between fibres but not for simply saying that light can leak in and out without cladding. Other candidates suggested that they believed that the cladding is the sheath.

E20. Many candidates incorrectly performed the calculation in part (a) (i) and some lost the mark by failing to round to three significant figures.

Most candidates comfortably picked up the first two marks in (a) (ii). The third mark required a correct indication of the partial reflection and very few candidates showed this.

A majority of candidates were able to point out that the angle exceeded the critical angle in part (b) (i). However, some candidates need to be careful not to say ‘gone past’ the critical angle as this does not clearly indicate ‘greater than’. Only a few went on to mention that the critical angle was 49 degrees.

Many candidates picked up the first two marks for a carefully drawn ray reflecting from the surface in (b) (ii) but many then did not correctly show the ray refracting into the glass. Many missed the fact that TIR only occurs when n1>n2 when a ray travels from one to two. Many also went on to calculate the critical angle for the glass-air boundary (62.5 degrees) which only applies to a ray travelling from glass to water. There was also a common misconception that a ray cannot pass into a medium with a higher refractive index. Some struggled to judge angles by eye and the use of a protractor should perhaps be encouraged for these candidates.

E21. It was very pleasing to see how well the calculations to parts (a) (i) and (iii) were done by candidates of all abilities. Part (a) (ii) also presented little difficulty to the vast majority.

The majority of candidates managed to pick up a mark and many the second mark to part (b). This seems to have been universally well learnt by candidates who often referred to ‘preventing crossover’ and the issue of signal security.

E22. In Part (a) (ii) nearly 50% of candidates did not score any marks. Many did not show the ray deviating towards the normal as it entered and many showed it bending away from the normal. It was common to see the reflected ray at a noticeably different angle to the incident. A significant number did not use a ruler.

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For part (b), some candidates rounded to 0.986 before calculating the angle which was penalised and a significant number gave an answer to four significant figures which was also penalised. However, the majority gained both marks here.

About 30% mentioned multimode dispersion or signal loss in part (c), but only a few picked up the second mark for explaining the consequence of this.

Part (d) was very easy and most candidates picked up both marks. Typical answers described the benefits of endoscopy or high speed internet.

E23. Part (a) states that reflection occurs. However, half of all candidates were unable to show the ray of light reflecting from the glass-liquid surface. Those who did do this tended to also get the second mark for showing the ray refracting away from the normal line as it entered the air.

In part (b), most were able to use the angles given to successfully calculate the refractive index of glass. Most of these also remembered to give their answer to three significant figures (1.47).

For part (c), candidates needed to realise the incident angle had just passed the critical angle and therefore the critical angle would be 63° to two significant figures. Some chose 27° instead of 63°. A common incorrect approach was to use 1.0/1.5 = sin θc.

Part (d) was quite a simple question but perhaps, because it was the last question, some candidates may have been short of time. Some may not have realised that they would get full credit for a correct method if they used their answer to part (c).

E24. In part (a) a large number of candidates thought that part X was the cladding.

Part (b) was answered well, with many candidates correctly using the formula and laying out their working in a structured manner. The most common error seen was due to candidates confusing n2 and n1. This resulted in an answer of 1.85 for part Y’s refractive index.

Part (c) was answered poorly, with many candidates being unsure of dispersion. Some candidates tried to describe graded index fibres and how these reduced dispersion rather than answering the question.

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E25. There were very few mistakes on part (a). Most candidates correctly showed their answer to more than two significant figures, which was required here. Where one mark was lost it was usually for only giving the answer as 50 rather than 50.15.

For part (c), many candidates wrongly stated that there was no TIR because the angle was below the critical angle. Candidates had to use the term refractive index or optical density. Use of ‘density’ by itself was not given credit.

Part (d) was done very well by the majority of candidates. The most common error was to calculate sin = 1.33/1.47 for the glass/water boundary rather than the glass/air.

In part (e), the most common answer was to assume that the ray refracts out of the glass and into the air. Even candidates who correctly calculated the critical angle as 43 degrees did not realise that the ray is one degree beyond the critical angle. Most candidates who correctly showed the ray reflecting did not then show the ray continuing into the water.

E26. The majority of candidates were successful in part (a), where errors occurred these involved mixing up of the c and cs terms in the formula.

Part (b)(i) presented more problems to candidates, with many candidates not converting from kilometres to metres. There were many candidates who found the time taken for Ray A and the time taken for Ray B, then subtracted these to find the time difference but many candidates rounded the time values before subtracting.

There were some excellent answers to part (b)(ii) but these were in the minority. Many candidates found it difficult to describe graded-index and often described step-index or monomode instead.

E27. In part (a) relatively few candidates knew that the frequency remains constant when refraction takes place.

Most drew the ray very well in answer to part (b) and the widespread use of a ruler showed a significant improvement over similar questions in previous examinations. However, a large number did not attempt the question. Those who dropped one mark tended to do so because their ray had a reflected angle that was far too big. Though it is not necessary to use a protractor to gain this accuracy mark, it should be encouraged for those who find it hard to make a good approximation by eye.

Part (d) was a little bit more difficult than previous, similar questions in that the angle of refraction had to be calculated (90 – 80.4 = 9.6°) prior to finding the incident angle. This confused a large number of candidates. Among those who did know what to do, a surprisingly common error was to use 100° for a right angle rather than 90°.

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Candidates tended to focus on one cause of loss in answer to part (e), either ‘multipath dispersion’ or ‘attenuation due to energy loss from the pulse’. This meant they accessed only one of the two marks available. Some guesswork was evident in responses to this question. Some candidates explained that the pulse had its amplitude reduced and length increased in order to fit inside the narrow fibre – ‘the fibre is too thin to let the high amplitude through’ was a typical answer. Other common responses were that the wavelength increased when the light entered the glass (presumably the pulse was interpreted as a waveform) or the lower speed of light in glass caused the broadening effect.

E28. Most students got the answer to part (a) (i) correct. However, examiners were looking for correct rounding and some students lost a mark for 1.6 or 1.65. A common incorrect approach was to select the equation with the ratio of speeds and use the two angles instead of speeds.

Most students were also correct in part (a) (ii). These questions always yield high marks.

Part (b) was quite poorly answered. Rays were not drawn carefully enough. It can be difficult to draw angles well without a protractor. A protractor is often useful for PHYA2. Students who cannot judge equal angles approximately by eye should be encouraged to use a protractor. The slanted edge of the prism in this question makes the judgement more difficult than usual. In this question students lose the mark if their line is more than five degrees from the true angle. Many students thought that the ray would refract rather than undergo total internal reflection even though they had calculated the correct critical angle. Many showed the refracted rays bending towards the normal rather than away.

E29. In part (a)(i) a significant number of students did not know ‘cladding’ and in part (a)(ii) the majority got this one correct. However, a significant number had their calculators in radians mode and gained 1 mark for the correct working but got a wrong answer of 1.31. Some rounded prematurely (eg 1.41/1.46 = 0.97 which leads to an answer of 76° rather than 75°). When using the inverse sine function it is important that the value used has not been rounded to less than 4sf.

Quite a few students gave an answer of 85−30 = 55° or 90−30 = 60° for part (b)(i). In part (b)(ii) most students do very well on Snell’s law questions. Those who got the wrong answer for (b)(i) often got full marks here with the error carried forward taken into account. Some did get the refractive indices the wrong way round or omitted the 1.46 – presumably thinking they were calculating a critical angle for a glass / air boundary.

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In part (c) many students thought that rays would refract ‘when the critical angle is exceeded’; perhaps associating a large angle with being ‘too big’. Many thought that a ray will travel further in a wide core. It will actually travel the same distance if the angle is the same.

E30.(a) This was a relatively easy question of a type that usually yields high marks. However, this produced a surprising number of wrong answers. A common error was to misinterpret n = c / cs as a ratio of angles and use it to justify dividing 14.1° by 9.54°.

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

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

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

E31.(a) This was perhaps not as well understood as could be expected. There was some use of ‘dense’ rather than ‘optically dense’ and this was not accepted. There was also confusion with many over whether the incident angle should be greater than or less than the critical angle in order for TIR to take place.

(b) Most candidates were successful on this, but a few truncated to 1.6 × 108.

(c) This was very well done. Some rounded or truncated 0.52836 to 0.52 or 0.53 which led to a rounding error in the answer. Only a few did not put a 2 or 3 significant figure answer.

(d) This was also very well done but again some problems with rounding or truncating, e.g. 1.4 / 1.8 = 0.78 or 0.77. Rounding to 2 sf should only be done for the final answer.

(e) (i) There was quite a lot of confusion on this one. Some candidates correctly calculated the incident angle of 54° but then went on to explain that this is less than the critical angle of 51° that they correctly calculated in (d). A common incorrect answer was: ‘It enters glass C because the angle is greater than the critical angle’.

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Some candidates do not realise that a calculation can be used to answer a question like this and they instead reasoned that ‘glass C has a lower refractive index, therefore the ray will refract away from the normal’.

Many thought the angle of incidence was 31.9° forgetting to add 22° to this. Some gained only 1 mark because they did not quote the incident angle in their answer.

(ii) Again, some careless lines were drawn. Rulers were often not used, and incident and reflected angles were often very different. To get within the examiners tolerance it is a good idea for students to use a protractor.

E32.(a) (i) Most candidates produced excellent answers, but there were a few slips, especially with use of 1.33 rather than 1.47.

(ii) Most candidates gained 2 marks here but a few did not use the refractive index of water (1.33) for n2. It is perhaps the case that some students believe that n2 is always 1 when calculating the critical angle.

(iii) A common mistake seen here was the use of the phrase ‘Total internal refraction’ rather than ‘Total internal reflection’.

It was also extremely common for candidates to say that light would not TIR because it ‘hadn’t reached the critical angle’ for the water-oil boundary. There would be no TIR (and thus no critical angle) because the light is travelling from a lower to a higher refractive index material and under these conditions, the light will refract and there will only be a partial reflection.

(b) In this question candidates sometimes showed TIR despite having successfully calculated the angle of incidence and the critical angle. Candidates received full credit if their answer was consistent with their previous two answers but this was not seen very often. Of those who chose refraction, some unfortunately had the ray bending towards the normal or even, in a few cases, refracting to the left of the normal line.

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· web viewthis question also discriminated well. in part (a)(i) almost half the candidates failed...

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