3A Wave motion I
2
Chapter 2
Refraction of Light
Practice 2.1 (p. 42)
1
Refraction of Light
11
C
By Snell’s law,
sin θa
sin 60 
nX =
=
= 1.51
sin (90   55 )
sin θ X
2
C
3
C
By n =
c
,
v
speed of light in glass
c 3  10 8
=
= 1.82  108 m s1
n
1.65
=
4
12
B
nA sin A = nB sin B
sin r
sin 30 
=
sin 16 .5 sin 19 .5
1.6  sin 60 = 1.4  sin r
r = 25.2
5
B
6
A
7
The light ray does not bend only if it enters
r = 81.8
The angle of refraction is 81.8.
c
(b) By n = ,
v
c
vA
n
1.4 0.875
nA
=
= B =
=
c
n A 1 .6
vB
1
nB
glass along the normal, i.e. only if the angle
of incidence is 0.
8
(a) By Snell’s law,
Angle of incidence = 90  35 = 55
By Snell’s law,
The ratio is 0.875 : 1.
sin θair
nalcohol =
sin θalcohol
sin θair
sin 55 
sin alcohol =
=
nalcohol
1.36
13
alcohol = 37.0
The angle of refraction is 37.0.
9
Refractive index
sin  a
sin 30 
=
=
= 1.46
sin 
sin 20 
10
Light travels at different speeds in different
media. Refraction results from the change in
the speed of light when light crosses a
boundary.
New Senior Secondary Physics at Work
1
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
By Snell’s law,
sin θ a
ng =
sin θ g
sin g =
3
B
By Snell’s law,
sin  a sin 45 
n=
=
= 1.414
sin  sin 30 
1
 1 
C = sin–1   = sin–1 
 = 45
n
 1.414 
sin θa
sin 45 
=
ng
1.50
g = 28.1
The angles that ray A makes with the normal
The critical angle for the liquid-air interface is
at the air-glass interfaces are 28.1, 28.1, and
45.
4
45.
The angles that ray B makes with the normal
at the air-glass interfaces are 0, 0, 0 and 0.
14
Refraction of Light
Critical angle
1
 1 
= sin–1   = sin–1 
 = 43.2
n
 1.46 
5
When he sees the fish above water surface,
the apparent depth of the fish is smaller than
its real depth. Therefore, the Indian should
aim his spear at a point below where the fish
appears to him.
15
(a)
6
By Snell’s law,
nA sin A = nB sin B
If A equals the critical angle, B = 90.
1.7  sin 44.9 = nB  sin 90
nB = 1.20
The refractive index of medium B is 1.20.
7
(b) Stars appear slightly higher.
(a) Critical angle
1
 1 
= sin–1   = sin–1 
 = 37.3
n
 1.65 
 
(b) Maximum value of 
Practice 2.2 (p. 56)
1
D
2
B
= 90 – 37.3 = 52.7
8
(3): Total internal reflection does not occur
when light passes from an optically less dense
medium to an optically denser medium, no
matter how large the angle of incidence is.
New Senior Secondary Physics at Work
2
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3A Wave Motion I
9
Chapter 2
(a) By Snell’s law,
sin θ a
ng =
sin θ g
1.5 =
Critical angle C for water
1
 1 
= sin–1   = sin–1 
 = 48.75
n
 1.33 
sin 60 
sin w
Diameter of the diver’s view
= 2  radius of the cone
w = 35.3
= 2  3  tan C
x = 90  w = 54.7
( = C)
= 2  3  tan 48.75
On side BC,
= 6.84 m
y = x = 54.7
12
On side CD,
Air near the ground is hotter and has a lower
refractive index. With a continuous decrease
angle of incidence
of the refractive index, light from the sky is
= 90  54.7 = 35.3 = w
gradually refracted more towards the
By Snell’s law,
sin θ a
ng =
sin θ g
1.5 =
Refraction of Light
horizontal. When the light meets a layer of air
at an angle beyond the critical angle, total
internal reflection takes place. As a result, a
sin z
sin 35.3
person sees the image of the sky when he
looks down, i.e. a mirage is formed.
z = 60
(b) The angle of emergence of the ray is
Revision exercise 2
equal to the angle of incidence.
10
Multiple-choice (p. 59)
Critical angle for diamond-air interface
1
 1 
= sin–1   = sin–1 
 = 24.4
n
 2.42 
1
By Snell’s law,
sin θa
nm =
sin θm
sin 30 
sin m =
1.4
Critical angle for crystal-air interface
1
 1 
= sin–1   = sin–1 
 = 30
n
 2.00 
The critical angle for a crystal is larger, so a
m = 20.9
smaller amount of light going into a crystal is
internally reflected back. Therefore, a crystal
does not have the same brilliance as a
2
C
3
C
By Snell’s law,
sin i
sin 40
n=
=
= 1.62
sin r sin 90  66.6
diamond.
11
New Senior Secondary Physics at Work
A
3
4
B
5
B
nA =
3  10 8
c
=
=2
v A 1.5  10 8
nB =
3  10 8
c
=
= 1.5
v B 2  10 8
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
By Snell’s law,
(b) Angle by which the ray is bent
nA sin A = nB sin B
= 41.7 – 30
2  sin 20 = 1.5  sin B
= 11.7
B = 27.1
6
Refraction of Light
2
A
By Snell’s law,
nP sin P = nQ sin Q
sin  P n Q
=
= constant
sin  Q n P
(1A)
(a) Refractive index
1
=
sin C
1
=
sin 24 .4
= 2.42
(1M)
(1A)
7
B
(b) By Snell’s law,
sin θa
nd =
sin θd
sin r
2.42 =
sin (90   80 )
8
D
r = 24.8
(1A)
Total internal reflection will not take place
(Correct path in oil)
(1A)
when light travels from medium X (an
(Correct path in water)
(1A)
Then,
sin C
sin 35 
=
 C = 38
sin 90  sin 70 
Depth of eyes below water surface
0 .8
=
= 0.695 m
 98  
tan

 2 
9
(1M)
3
A
Refractive index =
sin i
= slope of the graph
sin r
Slope of Z > slope of Y > slope of X
 nZ > nY > nX
By Snell’s law,
sin θa
no =
sin θo
sin 50 
1.47 =
sin p
optically less dense medium) to medium Y (an
optically denser medium).
10
C
11
C
12
(HKCEE 2006 Paper II Q32)
13
(HKCEE 2007 Paper II Q13)
p = 31.4
(1A)
q = p = 31.4
(1A)
Then again by Snell’s law,
Conventional (p. 61)
1
(a) By Snell’s law,
sin θa
nw =
sin θ w
no sin o = nw sin w
1.47  sin 31.4 = 1.33  sin r
(1M)
r = 35.2
(1A)
sin a = 1.33  sin 30
a = 41.7
(1A)
The angle of refraction in air is 41.7.
New Senior Secondary Physics at Work
4
 Oxford University Press 2009
3A Wave Motion I
4
Chapter 2
To the goldfish, the girl appears to be smaller
(b) There may be error in measuring angles.
(1A)
(1A)
and further away from it.
5
Refraction of Light
(1A)
If only one pair of data is used for
calculation, the percentage error will be
(a)
sin i
0.174
0.342
0.5
0.643
sin r
0.139
0.259
0.342
0.438
significant.
(c)
(1A)
The first point (0.139, 0.174) and the
second point (0.259, 0.342) seem to be
sin i
0.766
0.866
0.940
0.985
wrong.
(1A)
sin r
0.515
0.588
0.629
0.669
The error may be due to inaccurate
measurement
(1A)
or uneven texture within the Perspex.
(1A)
6
(a)
(Correct drawing of rays)
(1A)
(Correct indication of the image
position)
(1A)
(b) The letters will appear even higher.
(Correct labelled axes)
(1A)
(Correct straight line)
(1A)
(1A)
7
(1A)
(b) Different colours of light travel at
Refractive index of Perspex
= slope of the graph
0.95  0.44
=
0.65  0.3
(1M)
= 1.46
(1A)
New Senior Secondary Physics at Work
(a) Dispersion
different speeds in the prism.
(1A)
Therefore, they are refracted by different
amounts and separated from each other.
(1A)
5
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
Refraction of Light
Repeat with other angles of incidence.
(c)
Record the results in a table.
(1A)
Plot a graph of sin i against sin r. The
red
graph is a straight line passing through
the origin. This means that sin i is
violet
directly proportional to sin r.
(1A)
(b) Direct a ray of light from the ray box to
(Correct drawing of rays)
(1A)
enter the semicircular glass block from
(Correct labels of coloured light) (1A)
8
(a) Refractive index of water
v
= a
vw
=
its curved edge towards its centre. (1A)
Slowly increase the angle of incidence
until the angle of refraction is 90 (1A)
(1M)
At this moment, the angle of incidence is
3  10 8
equal to the critical angle C.
2.26  10 8
= 1.33
10
(1A)
(b) By Snell’s law,
ng sin g = nw sin w
ng = 1.70
(c)
 1 
= sin 1 

 2.42 
(1A)
= 24.4
The refractive index of the glass is 1.70.
v
By ng = a ,
vg
speed of light in the glass
v
= a
ng
9
(1A)
When there is a layer of oil, the critical angle
for the boundary changes.
By Snell’s law,
nd sin d = no sin o
2.42  sin C = 1.40  sin 90
(1M)
C = 35.3
3  10 8
=
1.70
= 1.76  108 m s1
Critical angle for diamond-air interface
1
= sin 1  
n
(1M)
ng sin 50 = 1.33  sin 78.3
diamond.
(1A)
(1A)
As a result, fewer light rays are totally
reflected and the diamond loses brilliance.
semicircular glass block from its straight
(1A)
edge. The ray must hit the centre of the
11
(1A)
Use the full-circle protractor to measure
the angle of incidence i and the angle of
refraction r.
New Senior Secondary Physics at Work
(1A)
The critical angle increases if oil attaches on a
(a) Direct a ray from the ray box to enter the
block.
(1A)
(a) Reflection occurs at X.
Refraction and
(1A)
reflection occur at Y.
(1A)
(b) (i)
(1A)
6
(1A)
Total internal reflection
(1A)
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
(ii)
12
(a) (i)
By ng =
=
totally internally
reflected light ray
3  10 8
1.45
= 2.07  108 m s1
the correct interface.)
(1A)
(Correct ray diagram)
(1A)
nl  sin l = ng  sin g
sin θ g
nl
=
sin θl
ng
(iii) For each object, only one image is
c
vl
c
vg
(1A)
The surface of a ground glass window is
not smooth.
(1A)
vg
When parallel light rays hit the ground
vl
glass, they have different angles of
incidence.
ground
glass
to an
observer
outside the
room
=
=
sin θ g
sin θl
sin θ g
sin θl
sin 29 .2
 2.07  108
sin 26 .6
= 2.26  108 m s1
(1A)
(Correct method)
(1A)
(b) Refractive index of the liquid
c
=
vl
They would be refracted into the glass
and eventually leave the glass and reach
=
an observer outside the room at different
angles.
=
Speed of light in the liquid droplet
sin θl
=
 vg
sin θ g
(1A)
parallel
light rays
(1A)
(ii) By Snell’s law,
(Total internal reflection occurs on
(c)
c
,
vg
speed of light in the glass
c
=
ng
light ray from
an object
formed by a prism.
Refraction of Light
(1A)
(1M)
3  10 8
2.26  10 8
= 1.33
(1A)
Since the refracted light rays leave the
glass in an irregular pattern, the image
formed is blurred.
(1A)
On the other hand, most light from the
outside can pass through the glass and
reach the room. The room is brighter
when ground glass is used instead of
curtain.
New Senior Secondary Physics at Work
(1A)
7
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
Refraction of Light
(b) The speed of light is greater in block 1.
(c)
(1A)
c
By v = , the smaller the refractive
n
index, the greater the speed of light in it.
Since block 1 has a smaller refractive
index than block 2, the speed of light
will be greater in block 1.
(1A)
(c)
(Total internal reflection on the vertical
surface)
(1A)
(Refraction at bottom surface with angle
in air greater than that in glass)
13
(HKCEE 2004 Paper I Q1)
14
(HKCEE 2005 Paper I Q10)
sin  1
(a) (i) By ng1 =
,
sin  2
15
(1A)
(1M)
sin 1 = 1.45  sin 15.5
1 = 22.8
(1A)
(ii) When light travels from glass block
(Reflection at boundary with i = r) (1A)
2 into the air, refraction occurs.
By Snell’s law,
sin θ a
ng2 =
sin θ g 2
=
(Refraction at the bottom surface
bending away from normal)
(1A)
(1M)
Physics in articles (p. 64)
sin 90
sin 38.7
(a)
= 1.60
The refractive index of the glass in
block 2 is 1.60.
(1A)
(iii) By Snell’s law,
ng2  sin g2 = ng1  sin g1(1M)
1.60  sin 51.3 = 1.45  sin 3
(1M)
3 = 59.4
New Senior Secondary Physics at Work
(1A)
8
 Oxford University Press 2009
3A Wave Motion I
Chapter 2
Refraction of Light
(Incident ray from emitter to the glass-air
interface)
(1A)
(Reflected ray from the glass-air interface to
the sensor)
(1A)
(No refracted ray at the glass-air interface)
(1A)
(b)
(Incident ray from emitter to the glass-air
interface)
(1A)
(Reflected ray from the glass-air interface to
(c)
the sensor)
(1A)
(Refracted ray in the water)
(1A)
Critical angle of the glass-air interface
 1 
= sin 1 
(1A)
 = 36.0
 1.7 
By Snell’s law,
nglass  sin glass = nwater  sin water
1.70  sin C = 1.33  sin 90
C = 51.5
(1A)
The critical angle of the glass-water interface
is 51.5.
The angle of incidence the infra-red light is
45. Therefore, total internal reflection would
occur at a glass-air interface but not at a
glass-water interface.
New Senior Secondary Physics at Work
(1A)
9
 Oxford University Press 2009