1
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)
2
The diagram shows a ray of monochromatic light, in the plane of the paper, incident on the end
face of an optical fibre.
(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.
.............................................................................................................
.............................................................................................................
(4)
(b)
(i)
Calculate the critical angle for the optical fibre at the air boundary.
refractive index of the optical fibre glass = 1.57
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
The optical fibre is now surrounded by cladding of refractive index 1.47. Calculate
the critical angle at the core-cladding boundary.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 1 of 89
(iii)
State one advantage of cladding an optical fibre.
.............................................................................................................
.............................................................................................................
(6)
(Total 10 marks)
3
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
(a)
Figure 2
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)
Page 2 of 89
4
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)
5
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.
(a)
Complete the paths of rays from the light source which strike the water surface at X, Y and
Z.
Page 3 of 89
(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
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(Total 7 marks)
6
(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.
(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.
...............................................................................................................
Page 4 of 89
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(7)
(b)
(i)
Give one reason why a cladding material is used in an optical fibre.
...............................................................................................................
...............................................................................................................
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(c)
State one use of optical fibres.
........................................................................................................................
(1)
(Total 11 marks)
7
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.
Page 5 of 89
(a)
Calculate the angle θ.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(b)
(i)
Calculate the critical angle for the glass-air interface.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(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)
Page 6 of 89
8
The diagram shows two closely spaced narrow slits illuminated by light from a single slit in front
of a monochromatic light source. A microscope is used to view the pattern of bright and dark
fringes formed by light from the two slits.
(a)
(i)
Explain qualitatively why these fringes are formed.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Describe what is observed if one of the narrow slits is covered by an opaque object.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(8)
Page 7 of 89
(b)
The microscope is replaced by a fibre-optic detector linked to a computer. The detector
consists of the flat end of many optical fibres fixed together along a line. The other end of
each optical fibre is attached to a light-sensitive diode in a circuit connected to a computer.
The signal to the computer from each diode is in proportion to the intensity of light incident
on the diode. The computer display shows how the intensity of light at the detector varies
along the line of the detector when both of the narrow slits are open.
(i)
Describe and explain how the pattern on the display would change if the slit
separation were increased.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 8 of 89
(iii)
The diagram below shows a light ray entering an optical fibre at point P on the flat
end of the fibre. The angle of incidence of this light ray at the core-cladding boundary
is equal to the critical angle. On the diagram, sketch the path of another light ray from
air, incident at the same point P, which is totally internally reflected at the
core-cladding boundary.
(7)
(Total 15 marks)
9
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.
Page 9 of 89
(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.
(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.
...............................................................................................................
(ii)
Calculate the angle of refraction at this boundary.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(c)
Calculate the critical angle for a ray passing from medium 2 into the air.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 10 marks)
Page 10 of 89
10
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;
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the angle of refraction at the boundary between medium 1 and medium 2.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
Page 11 of 89
(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.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(Total 10 marks)
11
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.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 12 of 89
(b)
For the ray shown calculate the angle of incidence, θi.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(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)
Page 13 of 89
12
The diagram below shows a liquid droplet placed on a cube of glass. A ray of light from air,
incident normally on to the droplet, continues in a straight line and is refracted at the liquid to
glass boundary as shown.
refractive index of the glass = 1.45
(a)
Calculate the speed of light
(i)
in the glass,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
in the liquid droplet.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 14 of 89
(b)
Calculate the refractive index of the liquid.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(c)
On the diagram above, complete the path of the ray showing it emerge from the glass cube
into the air.
No further calculations are required.
(2)
(Total 7 marks)
13
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.
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.
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 15 of 89
(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°.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(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.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(7)
(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.
Page 16 of 89
(ii)
Compare the times taken for pulses of light to travel along the same length of each
fibre.
Give a reason for your answer.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(Total 10 marks)
14
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
(a)
Calculate the speed of light in the glass.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 17 of 89
(b)
Determine
(i)
the angle of incidence, θ,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
the refractive index of the liquid.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(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.
(2)
(Total 9 marks)
15
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?
.............................................................................................................
Page 18 of 89
(ii)
Explain how the path of R1 demonstrates that the refractive index of glass is greater
than the refractive index of water.
.............................................................................................................
.............................................................................................................
(2)
(b)
Using the following information
A = 47.1°
B = 42.9°
C = E = 41.2°
D = F = 48.8°
calculate
(i)
the refractive index of water,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
the ratio,
.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)
(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)
Page 19 of 89
16
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)
(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 = 1.46
The 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)
Page 20 of 89
(d)
State one application of optical fibres and explain how this has benefited society.
Application
......................................................................................................................
......................................................................................................................
Benefit
......................................................................................................................
......................................................................................................................
(2)
(Total 9 marks)
17
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
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 21 of 89
(c)
Calculate the angle of incidence, θ.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
(Total 8 marks)
18
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
Page 22 of 89
(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. Continue
the ray to show it emerging into the air. No calculations are expected.
(2)
(Total 11 marks)
Page 23 of 89
19
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.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
(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)
Page 24 of 89
(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.
Figure 2
(i)
The ray strikes the boundary between the prisms. Calculate the angle of refraction of
the 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)
Page 25 of 89
20
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 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.
.............................................................................................................
.............................................................................................................
(2)
Page 26 of 89
(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
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
(Total 10 marks)
21
(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)
Page 27 of 89
(b)
Explain why the cladding is necessary for optical fibres.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 4 marks)
22
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.
answer ........................... degrees
(3)
Page 28 of 89
(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)
23
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)
Page 29 of 89
24
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)
(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)
Page 30 of 89
(d)
Calculate the refractive index of the liquid.
answer = .....................................
(2)
(Total 8 marks)
25
The figure below shows a cross-section through an optical fibre.
(a)
Explain the purpose of part X.
......................................................................................................................
......................................................................................................................
(1)
Page 31 of 89
(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.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
absolute refractive index for part Y .................................................
(3)
(c)
Explain why dispersion occurs in an optical fibre.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 6 marks)
Page 32 of 89
26
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
= 1.47
refractive index of water
= 1.33
refractive 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.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(1)
Page 33 of 89
(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)
27
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)
Page 34 of 89
(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.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
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.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
(Total 6 marks)
28
(a)
The speed of light is given by
c=fλ
State how each of these quantities will change, if at all, when light travels from air to glass.
c .......................................................
f ........................................................
λ .......................................................
(3)
Page 35 of 89
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
(2)
Page 36 of 89
(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
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 14 marks)
Page 37 of 89
29
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)
Page 38 of 89
30
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)
Page 39 of 89
(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)
Page 40 of 89
(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
(c)
State and explain one advantage associated with a smaller diameter core.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 8 marks)
Page 41 of 89
31
The diagram below shows a section of a typical glass step-index optical fibre used for
communications.
(a)
Show that the refractive index of the core is 1.47.
(1)
(b)
The refracted ray meets the core-cladding boundary at an angle exactly equal to the critical
angle.
(i)
Complete the diagram above to show what happens to the ray after it strikes the
boundary at X.
(2)
(ii)
Calculate the critical angle.
critical angle = .........................degrees
(1)
(iii)
Calculate the refractive index of the cladding.
refractive index = .....................................
(2)
Page 42 of 89
(c)
Give two reasons why optical fibres used for communications have a cladding.
reason 1......................................................................................................
....................................................................................................................
reason 2......................................................................................................
....................................................................................................................
(2)
(Total 8 marks)
Page 43 of 89
32
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.
condition 1 .....................................................................................................
........................................................................................................................
condition 2 .....................................................................................................
........................................................................................................................
(2)
Page 44 of 89
(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.
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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(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)
Page 45 of 89
33
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.
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)
Page 46 of 89
(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 ........................................................................................................
........................................................................................................................
explanation .....................................................................................................
........................................................................................................................
........................................................................................................................
reason 2 ........................................................................................................
........................................................................................................................
explanation .....................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 10 marks)
Page 47 of 89
34
(a)
Tick (✓) the appropriate boxes in the table to indicate how the wavelength, frequency and
speed of light are affected when a ray of light travels from air into glass.
Wavelength
Frequency
Speed
increases
stays the same
decreases
(2)
(b)
Figure 1 shows a right-angled glass prism in contact with a transparent substance on one
of the faces. One of the other angles of the prism is θ.
Figure 1
Page 48 of 89
(i)
A ray A enters perpendicularly to one face of the prism. It is partially refracted and
partially 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 the
surface of the transparent substance.
Determine the angle α.
angle α = ......................................... degree
(2)
(ii)
Determine the angle θ in Figure 1.
angle θ = ......................................... degree
(2)
Page 49 of 89
(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 transmit
information along an optic fibre.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 10 marks)
Page 50 of 89
35
(a)
Describe the structure of a step-index optical fibre outlining the purpose of the core and the
cladding.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
Page 51 of 89
(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).
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 7 marks)
36
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)
Page 52 of 89
(b)
For one fibre, the speed of monochromatic light in the core is 1.97 × 108 m s−1 and the
speed 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.
critical angle ....................................... degrees
(2)
(Total 3 marks)
Page 53 of 89
Mark schemes
1
2
C
[1]
(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 and
increases 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)
(iii)
to protect the core surface
[or to prevent cross-over]
6
[10]
3
(a)
A
B1
(b)
D
B1
[2]
Page 54 of 89
4
reflection wavefront direction sensible
B1
refraction wavefront direction sensible
B1
one pair of wavefronts correctly spaced
B1
[3]
5
(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)
sinθwater = 0.75, θwater = 48.6° (1)
radius = 1.5tan48.6° (1) =1.7m, ∴ diameter = 3.4m (1)
[7]
6
(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)
Page 55 of 89
(ii)
greater acceptance angle (1)
enables more light to be collected
(or smaller critical angle makes escape less likely) (1)
(3)
(c)
endoscopy or communications (1)
(1)
[11]
7
(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]
8
(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
Page 56 of 89
(b)
(i)
fringes closer (1)
(because) each fringe must be closer to the centre for the
same path difference
[or correct use of formula as explanation] (1)
(ii)
sin θc
(1) (= 0.88)
θc = 61.6° (1)
(iii)
for second light ray, diagram to show:
smaller angle of incidence at P than first ray (1)
point of incidence at core / cladding boundary to right of first ray (1)
total internal reflection drawn correctly or indicated
at point of incidence to right of right angle (1)
[alternative if ray enters at P from above:
correct refraction at P (1)
TIR at boundary if refraction at P is correct (1)
angle of incidence visibly ≥ critical angle (1)]
7
[15]
9
(a)
Ray diagram to show:
(i)
refraction towards normal at boundary (1)
emerging ray refracted away from normal (1)
(ii)
reflection at boundary with i ≈ r
emerging ray refracted away from normal (1)
4
Page 57 of 89
(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]
10
(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
(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]
Page 58 of 89
11
(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]
12
(a)
(i)
(use of n =
gives)
cglass = ×
= 2.07 × 108 m s–1 (1)
(ii)
use of
cliquid =
(1)
= 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
Page 59 of 89
(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]
13
(a)
(i)
sin c =
c = 42° (1)
(ii)
(iii)
(1)
(41.8°)
1.5 sin i = sin 40 (1)
i = 25° (1) (25.4°)
(use of c = 41.8° gives i = 26.4°)
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]
14
(a)
cg (=
)=
(1)
= 2.0 × 108 m s–1 (1)
2
Page 60 of 89
(b)
(i)
sin q1 (= n sin q2) = 1.5 × sin 15 (1)
q1 = 23° (1) (22.8°)
(ii)
use of
(1)
n2 =
(or equivalent)
(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]
15
(a)
(i)
(angle) F (1)
(ii)
angle D is greater than angle B
[or at the glass-water boundary, ray R1 refracts away
from the normal] (1)
2
(b)
(i)
(use of sin θc =
n = 1.3 (1)
(ii)
use of
gives)
sin 48.8 =
(1)
(1.33)
(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 of 89
16
(a)
(i)
A: cladding + B: core (1)
1
(ii)
refraction towards the normal line (1)
continuous lines + strikes boundary + TIR correct angles by
eye + maximum 2 TIRs (1)
2
(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 at
different times/pulse broadening/merging of adjacent pulses/’smearing’/
poor resolution/lower transmission rate/lower bandwidth/less distance
between regenerators (1)
or to prevent light/data/signal loss (from core or fibre) (1)
(which would cause) signal to get weaker/attenuation/crossover/data
to be less secure (1)
2
(d)
correct application (1) (endoscope, cytoscope, arthroscope etc,
communications etc)
linked significant benefit stated eg improve medical diagnosis/improve
transmission of data/high speed internet (1)
2
[9]
17
(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
Page 62 of 89
(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]
18
(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]
2
Page 63 of 89
(c)
reflection at boundary with i = r (1)
refraction (at bottom surface) bending away from normal (1)
2
[11]
19
(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, refracted
away from normal (1)
emerging ray (r.h. vertical face) refracting away
from normal (1)
7
[10]
20
(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
Page 64 of 89
(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 and
strikes 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]
21
(a)
sin c = nc/nf (1)
B1
78.6° (1)
B1
2
(b)
required for total internal reflection (1)
B1
avoids signal loss (1)
B1
avoids cross-talk (owtte) (1)
B1
max 2
[4]
Page 65 of 89
22
(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 angle
or 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:
keeps signals secure (1)
maintains quality/reduces pulse broadening/smearing (owtte) (1)
it keeps (most) light rays in (the core due to total internal reflection
at the cladding-core boundary) (1)
it prevents scratching of the core (1)
(keeps core away from adjacent fibre cores) so helps to prevent
crossover 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]
23
24
B
[1]
(a)
reflects at correct angle by eye (use top of ‘27’ and bottom
of ‘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 marked
and at least one normal line added (1)
3
Page 66 of 89
(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]
25
(a)
(sheath) to protect fibre (and cladding)/to add strength
(to cladding)/prevent loss of signal from scratches
at least sense of protecting fibre/cladding
disallow anything that could infer that it is cladding
[eg prevent signal
loss/protects info]
treat extra as neutral
cladding explanation zero marks
B1
1
(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)
A1
3
Page 67 of 89
(c)
different wavelengths different speeds/different
wavelengths different refractive indices/different
paths/different angles/different distances
B1
spreading of pulse/spreading into different wavelengths
B1
2
[6]
26
(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) is
lower 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
Page 68 of 89
(e)
2
[9]
Page 69 of 89
27
(a)
use of c/cs = n (condone inversion of c and cs)
C1
1.9 × 108 (m s–1)
A1
2
(b)
(i)
path difference = 120 m or 0.12 km/finds two
times and subtracts
(allow incorrect speed with working) (condone power
of ten error)
C1
(penalise use of different speeds)
6.4 × 10–7s (ecf from (a))
A1
2
(ii)
refractive index varies across (graded-index)
core/refractive index maximum at centre of core
B1
ray travels slower in centre/rays travel faster at
edge/ray B travels faster/ray A travels slower
B1
2
[6]
28
(a)
decrease
constant
decrease
3
Page 70 of 89
(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 =
θc = 80.4
or correct substitution in un-rearranged formula
(80.401) (80.3 to 80.54) (≈ 80°) must see 3 sf or more
2
(d)
angle of refraction = 180 – 90 – 80.4 = 9.6°
sinθ = 147(06) sin 9.6
θ = 14 (= 14.194°)
= 0.25 ecf from first mark
ecf from first mark
range 13 to 15° due to use of rounded values
3
(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/non
axial rays take longer time to travel same distance along fibre
as axial rays
2
[14]
Page 71 of 89
29
(a)
(i)
sin 56 = nglass sin 30
(nglass = sin56/sin30) (= 1.658) = 1.7
2
(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 1.7
2
(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]
30
(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
r = 70
ecf bi
(degrees) (69.79) ecf bi
2
Page 72 of 89
(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) angle
large(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]
31
(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
(b)
(i)
ray goes along the boundary
Deviation by no more than 1mm by the end of the diagram.
(partial) reflection shown
(allow dotted or solid line. This mark can be awarded if TIR is shown)
Tolerance: 70° to 85° to normal or labelled e.g. θ and θ, etc
2
(ii)
(90 − 9.54 = ) 80.46 or 80.5
(° ) ( allow 80° )
Don’t allow 81 degrees
1
Page 73 of 89
(iii)
(n = nc sin θ)
allow 80 or 81 degrees here
= 1.47 sin 80.46°
=1.45
ecf bii
(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 / signal
don’t allow ‘leakage’ on its own.
•
increase the critical angle / reduce pulse broadening / (modal)dispersion / rays
with a small angle of incidence will be refracted out of the core
Don’t allow ‘loss of light’
•
increase rate of data transfer
Allow ‘leakage of signal’, etc
max two correct (from separate bullet points)
2
[8]
32
(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 × 10 8
Allow 1.66 × 108 or 1.70 × 108
Allow 1.6. × 108
(= 1.667 × 10 8) = 1.67 × 108 (ms−1)
2
Page 74 of 89
(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 mark
Allow 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 marks
Don’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 expected
If c + 22 > d then REFRACTION expected
OR
ecf from (c) and (d)
c + 22 < their d (θc )
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]
33
(a)
(i)
sin 60 = 1.47sin θ
−1
(sin
OR
sin θ = sin 60 / 1.47 ✓
0.5891) = 36 (°) ✓ (36.0955°) (allow 36.2)
Allow 36.0
2
(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 0.91
2
Page 75 of 89
(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 a
lower refractive index than oil ✓
Allow ‘ray doesn’t bend towards normal’ (at glass / water)
Allow optical density
Boundary in question must be clearly implied
4
[10]
34
(a)
wavelength
frequency
speed
increases
stays the
same
✓
decreases
✓
✓
middle column correct ✓
first and third column correct ✓
2
(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
Page 76 of 89
(ii)
1.09sin65 = 1.70sinx
or sinx = 0.58
or 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 mark
1
(ii)
diagram showing core / cladding and light ray TIR at interface at least once with
another TIR shown on the diagram or suggested in their explanation ✓
labelling is not required and reflections do not have to be accurate
provided they are shown on the correct side of the normal
light fibre consists of core and cladding with lower refractive index / optical
density ✓
light (incident) at angle greater than the critical angle (results in TIR) ✓
3
[10]
35
(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
Page 77 of 89
(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]
36
(a)
Prevents (physical) damage to fibre / strengthen the fibre / protect the fibre
Allow named physical damage e.g. scratching
B1
Prevent crosstalk
1
(b)
(Relative) refractive index = 1.03
or
Use of sinc = n2 / n1
Calculating the refractive indices and rounding before dividing gives
76.8
C1
76.0° or 76.8°
A1
2
[3]
Page 78 of 89
Examiner reports
2
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.
3
4
5
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.
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.
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.
6
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.
Page 79 of 89
7
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
8
This was a long question worth 15 marks and it is pleasing to report that almost all candidates
were able to gain a reasonable mark, with some gaining high marks.
In part (a)(i) most candidates knew that light was diffracted from each of the pair of narrow slits
and that interference fringes were produced and could be seen in the overlap area. Some
candidates referred to coherence in terms of waves being emitted in phase rather than with a
constant phase difference. Most candidates were able to explain how a bright fringe or a dark
fringe was formed and were able to relate their statements correctly to the path difference or
phase difference. A significant number of candidates did not make it clear that a phase difference
of 180° is necessary for cancellation of two waves, and often just stated that the waves were out
of phase.
In part (a)(ii), it was clear that many candidates did not know the meaning of 'opaque' and
thought that some light would pass through the opaque object. Few candidates realised that the
fringes seen in (a)(i) would no longer be seen, but a small minority knew that single slit diffraction
would take place and were thus able to give a satisfactory description.
Most candidates were aware in part (b)(i) that the fringes would be closer but few were able to
give an adequate explanation of why this was so and only the best candidates were able to quote
and use the appropriate expression to justify their answer. In part (ii) most candidates gave a
correct calculation without any difficulty, but some candidates were unable to make any progress
because they calculated the critical angle for a boundary with air. In part (iii) many candidates
scored all three marks with a clear, correct diagram. The main reason for not scoring full marks
was usually a failure to give the correct point of incidence.
9
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).
Page 80 of 89
10
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.
11
12
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.
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.
13
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.
Page 81 of 89
14
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.
15
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.
16
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.
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.
Page 82 of 89
17
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.
18
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.
19
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.
Page 83 of 89
20
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.
21
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 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.
22
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.
Page 84 of 89
24
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).
25
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.
26
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.
Page 85 of 89
27
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.
28
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°.
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.
Page 86 of 89
29
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.
30
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.
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.
Page 87 of 89
31
32
(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.
(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’ .
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.
Page 88 of 89
33
(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)
34
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.
This question showed up a lack of geometrical knowledge in some students but strong students
sailed through. In (b)(i) most students knew Snell’s Law as applied to the boundary between two
media. Unfortunately, many did not choose to use the correct refractive indices or use the correct
angles. The use of 25° in place of the correct 65° featured prominently. The students who had
problems with (b)(i) also had problems in (b)(ii). There was the potential for an error carried
forward mark for students who presented a value for the angle x as in the mark scheme.
However, only a minority of these students correctly found the answer to this section by
subtracting x from 90°.
Although (c)(ii) asks for a diagram, weaker students sometimes chose not to give one or when
they did it showed an optical fibre without cladding. In this way they failed to gain the first mark.
The other two marks were only obtained by stronger students. More students chose to give an
answer that involved the critical angle than an answer involving the refractive indices. Only the
strongest students scored all three marks.
36
(a)
Most candidates gave a sensible response usually referring to improved strength or
protection for the optical fibre. Those who discussed light leaving or entering the fibre rarely
referred to the consequences for the signal in the fibre or the effect on other signals in
nearby fibres.
(b)
Most completed this straightforward calculation successfully. Some candidates gave
answers that suggested that their calculators were set in radian-mode. Others rounded off
too early in the calculation when finding the refractive index of the core and cladding first
which is inadvisable in any calculation as it can produce answers outside an accepted
range.
Page 89 of 89