1. Topic 4.4
Wave behaviour

2. Wave Behaviour
Reflection in one
dimension
When a wave moves through a
medium, the velocity and shape of
that wave remains constant. This
is so no matter what the medium.

3. This diagram shows a pulse
travelling along a string

4. This diagram shows the pulse
after it has been reflected

5. Notice
The pulse keeps its shape
It is inverted
It has undergone a 180o
phase
change
Or π change in phase

6. This is because the instant the pulse hits the
fixed end, the rope attempts to move the fixed
end upwards
It exerts an upwards force on the fixed end
By Newton’s third law, the wall will exert an
equal but opposite force on the rope
This means that a disturbance will be created
in the rope which, however is downwards and
will start moving to the left

7. If the end of the rope is not fixed but free to
move the situation is different
Most of the pulse would carry on in the
same direction, some would be reflected
but the reflected pulse is in the same phase
as the original pulse
There is a change of direction, but no
inversion here

8. Similar situations occur in
springs and columns of air
It is also why metals are shiny.
The light incident on the
surface is reflected back.

9. Slinky Investigation
Use a slinky or combination to investigate:
1. Reflection of wave at fixed boundary (ie hold
the slinky firm at one end)
2. Reflection of wave at free boundary (do not
hold other end of slinky)
3. Transfer of energy between a heavy and a
light slinky
Fill in notes

10. Wave Behaviour
Reflection in two
dimensions

11. The direction of incident and reflected waves is best described
by straight lines called rays. The incident ray and the
reflected ray make equal angles with the normal. The angle
between the incident ray and the normal is called the angle
of incidence and the angle between the reflected ray and
the normal is called the angle of reflection.
Image from aplusphysics

12. Light is shone from above a ripple
tank onto a piece of white card
beneath
The bright areas represents the
crests
The dark areas represent the troughs

13. These wavefronts can be used to
show reflection (and refraction and
diffraction and interference) of water
waves
Image from agilegeoscience

15. Normal
Angle of
incidence
Angle of
reflection=

16. Image from agilegeoscience

17. Candle in front of mirror
The rays diverge(spread apart) from the tip of
the flame and continue to diverge upon
reflection. These rays appear to originate from
a point located behind the mirror.

18. Candle in front of mirror
This is called a virtual image because the light
does not actually pass through the image but
behaves as though it virtually did. The image
appears as far behind the mirror as the object
is in front of it and the object and the image is
the same.
It is different with a curved mirror.

19. Light and sound reflections
Image from P Hewitt, conceptual physics

20. The Law for Reflection
The angle of incidence is equal to the angle of
reflection
Also - The incident ray, the reflected ray and
the normal lie on the same plane
Use this rule for any ray or wave diagram
involving reflection from any surface

21. For circular waves hitting a flat reflector, the
reflected waves appear to come from a source,
which is the same distance behind the reflector
as the real source is in front of it
Also a line joining these 2 sources is
perpendicular to the reflecting surface

22. O I

23. If a plane wave is incident on a circular
reflector then the waves are reflected so that
they
–Converge on a focus if the surface is
concave
–Appear to come from a focus if the
surface is convex

26. Echos
In the case of sound, a source of sound can be
directed at a plane, solid surface and the
reflected sound can be picked up by a
microphone connected to an oscilloscope.
The microphone is moved until a position of
maximum reading on the oscilloscope is
achieved.
When the position is recorded it is found that
again the angle of incidence equals the angle
of reflection.

27. Wave Behaviour
Refraction

28. Examples
Place a pencil in a beaker of water – what
happens?
Place a coin in a mug, move your head so that
you cant see the coin, pour in water – what
happens?
Why is it hard to catch fish with your hands?

29. The speed of a wave depends only on the
nature and properties of the medium through
which it travels.
This gives rise to the phenomenon of refraction
Refraction is the change of direction of travel
of a wave resulting from a change in speed of
the wave when it enters the other medium at
an angle other than a right angle.

30. Refraction for light
Partial reflection
Incident ray
Incident ray
Refracted ray
Refracted ray
Partial reflection

32. Interpreting diagrams involving refraction
It may help to imagine the ranks of a marching band.
Obviously, the cadence does not change.
Thus the period and the frequency do not change.
But the speed and the
wavelength do change. CONCRETE
DEEP MUD

33. Refraction of Waves in 1 & 2 Dimensions
Light bends towards the normal when;
it enters a more optically dense medium.
Light bends away from the normal;
when it enters a less optically dense medium.
The amount the incident ray is deviated;
depends on the nature of the transparent material

34. Refraction & Huygen’s Principal
Consider a wavefront advancing through
medium 1;
– travelling at velocity v1
– falling at an angle on to a medium 2,
– travelling at velocity is v2
.

35. Refraction & Huygen’s Principal
The wavefront with the points;
A, B, C, D are a source of secondary wavelets.
After a time t, the secondary source wavelet from D;
–has moved a distance s1
= v1
t while,
–wavelet from A has moved,
–smaller distance s2
= v2
t,
–in the denser medium 2,
–where the velocity is less.

36. Refraction & Huygen’s Principal
The time for the wavelet to travel;
from B to B2
is identical,
for the wavelet to travel from C to C2
.
The wavelet from C;
spends a longer time in,
less dense medium 1 in travelling,
from C to C1
than for,
B to travel from B to B1
.

37. Refraction & Huygen’s Principal
Thus there is less time for the wave;
to travel C1
to C2
in the denser medium 2,
than for the wavelet to travel from B1
to B2
.
The distance CC1
is thus less than BB1
.

38. Refraction & Huygen’s Principal
The new wavefront at the end of this time;
envelope of tangents to,
the wavelet wavefronts at A1
B2
C2
D1
.
The direction of movement of the wavefront
has changed;
refracted towards the normal at point A.

39. Deriving Snell’s Law
Since the wave;
travels in a direction,
perpendicular to its wavefront,
∠IAD = 90o
and
∠AA1
D1
= 90o
.

40. Deriving Snell’s Law
From ∠IAD,
i + θ = 90o
,
so i = 90o
- θ.
From ∠N’AD1
,
R + γ = 90o
,
so R = 90o
- γ.
But from ∠D1AN,
α + θ = 90o
,

41. Deriving Snell’s Law
so α = 90o
- θ,
so α = i.
In triangle D1
AA1
,
β + γ + 90o
= 180o
.
⇒β = 90o
- γ,
so β = R

42. Deriving Snell’s Law
In triangle ADD1
,
sin i = v1
t/AD1
In triangle AA1
D1
,
sin R = v2
t/AD1
Dividing

43. Deriving Snell’s Law
• 1n2 = a constantsin
sin
i
R
v t
AD
v t
AD
v t
v t
v
v
n= ÷ = = =1
1
2
1
1
2
1
2
1 2

44. Snell’s Law
When a wave is incident from a vacuum;
– on to a medium M,
– the refractive index is written nM
– is called the absolute refractive index
– of medium M.

45. Snell’s Law
Snell’s law states that the ratio of the sine of
the angle of incidence to the sine of the
refraction is constant and equals the ratio of
the velocity of the wave in the incident medium
to the velocity of the refracting medium.
sin
sin
i
R
v
v
n= =1
2
1 2

46. Snell’s Law

47. This law enable us to define a property of a
given optical medium by measuring θ1 and θ2
when medium 1 is a vacuum
The constant is then the property of medium 2
alone and it is called the refractive index (n).
We usually write
n = (Sin i) / (Sin r)
n is also a ratio of the speeds in the 2
mediums i.e. n = cvacuum / vmedium

48. Combining them!
Rearranging n2/n1 = sin θ1 / sin θ2
But n1 = 1
n2 = sin θ1 / sin θ2
Therefore anb = nb / na

49. Using Refractive Index
Refractive index is written for materials in the
form of light entering from a vacuum or air into
the material.
The refractive index of a vacuum or air is 1
It can also be shown that, for two mediums (1
and 2)
n1 sin θ1 = n2 sin θ2
Care needs to be taken when dealing with light
leaving a material

50. Eg 5
Light strikes a glass block at an angle of 60o
to
the surface. If air
nglass
= 1.5, calculate:
(a) the angle of refraction (R) and
(b) the angle of deviation (D).

51. Part (a)
i = 300
air
nglass
= 1.5
air
nglass
sin R =
sin R =
R = 19.5o
R = 20o
(2 sig digits)
Part (b)
i = 300
R = 19.5o
D = i - R
D = 30 – 19.5
D = 10.5o
D = 11o
(2 sig digits)

52. Eg 6
Light of wavelength 500 nm is incident on a
block of glass (air
nglass
) at 60o
to the glass surface.
Calculate (a) the velocity, and (b) the
wavelength of the light waves in the glass.

53. Part a
vair
= 3.0 x 108
m s-1
air
nglass
= 1.5
vair/
vglass = air
nglass
vglass =
vair/ air
nglass
vglass =
3.0 x 108
/1.5
vglass
= 2.0 x 108
m s-1

54. Part b
air
nglass
= 1.5
i = 30o
λair
= 500 nm = 5.0 x 10-7
m
vair
= 3.0 x 108
m s-1
λair/
λglass
= air
nglass
λglass
= 3.3 x 10-7
m

55. Critical angle
The critical angle is the
angle of incidence that
results in the refracted
ray travelling along the
boundary between the
two media.
nt/ni = sinθc
Image from Physics classroom

56. Total internal reflection
Total internal reflection occurs when
the angle of incidence is greater than
the critical angle.
Image from Hyper Physics

57. Questions
1.The refractive index of water is 1.5. A swimming pool is filled to a depth of
1.8m with water. How deep would it appear to someone standing on the
side of the pool?
2. A microscope is focussed on a scratch at eh bottom of a beaker.
Turpentine is poured into the beaker to a depth of 4cm, it is found that
the lens must be raised by 12.8mm in order to bring back into focus.
What is the refractive index of turpentine?
3. A ray of light enters a pond at an angle of 30 deg to the horizontal. What
is its direction as it travels through the water? The refractive index of
water is 1.33
1. Ang = real/apparent
1.5 = 1.8/x
Therefore,
x = 1.2 m
2. Ang = real/apparent
= 4/2.72
Therefore,
= 1.47
3. Sin i / sin r = sin 60 / sin r = 1.33
sin r = sin 60/ 1.33 = 0.651
Angle is 41 degrees

58. In a ripple tank this is achieved by using a flat
piece of plastic, giving two regions of different
depth
As the wave passes over the plastic it enters
shallow water and slows down.
As v = f λ,
if v decreases
And f is constant (the source hasn’t changed)
λ must also decrease
So the waves get closer together

59. If the waves enter the shallow area at an angle
then a change in direction occurs.
Shallow water

60. This is because the bottom of the wavefront as
drawn, hits the shallow water first so it slows,
and hence travels less distance in the same
time as the rest of the wavefront at the faster
speed travel a larger distance!

61. Deep water
If the waves enter the deep area at an angle
then a change in direction occurs

62. This is because the top of the wavefront hits
the deep water first so it speeds up, and hence
travels more distance in the same time as the
rest of the wavefront at the slower speed travel
a smaller distance!

63. Refraction of Sound
A sound wave is also able to be refracted.
This is due to the fact that the speed of
sound is affected by temperature and the
medium through which it travels.

64. Diffraction
Diffraction is the spreading out of a wave as it
passes by an obstacle or through an aperture
When the wavelength is small compared to the
aperture the amount of diffraction is minimal
Most of the energy associated with the waves
is propagated in the same direction as the
incident waves.

65. When the wavelength is comparable to the
opening then diffraction takes place.
There is considerable sideways spreading, i.e.
considerable diffraction

69. examples
Light through net curtains
Fog Horns – two frequencies to ‘fill in’ gaps
AM radio – can bend round small buildings
Dolphins use two different frequency waves to
‘see’ and ‘fine tune’ their surroundings

71. Using Huygens’ Principle
Remember that Huygens' idea was to consider
every single point on the wavefront of the wave
as itself a source of waves.
In other words a point on the wavefront would
emit a spherical wavelet or secondary wave,of
same velocity and wavelength as the original
wave.

72. Therefore as a wave goes through a gap or
passed an obstacle the wavelets at the edges
spread out.
Huygens’ construction can be used to predict
the shapes of the wave fronts.

74. The new wavefront would then be the surface
that is tangential to all the forward wavelets
from each point on the old wavefront.
We can easily see that a plane wavefront
moving undisturbed forward easily obeys this
construction.

75. The Principle of Linear Superposition
Pulses and waves (unlike particles) pass
through each other unaffected and when they
cross, the total displacement is the vector
sum of the individual displacements due to
each pulse at that point.
Try this graphically with two different waves

76. Interference
Most of the time in Physics we are dealing with
pulses or waves with the same amplitude.
If these cross in a certain way we will get full
constructive interference, here the resultant
wave is twice the amplitude of each of the
other 2
+ =

77. If the pulses are 180o
(π) out of phase then the
net resultant of the string will be zero. This is
called complete destructive interference.
+ =

78. Path difference
Constructive interference occurs when the
path difference between the rays is equal to a
whole number of wavelengths.
C.I. = nλ
Destructive interference occurs when the path
difference between the rays is equal to a
multiple of half a wavelength.
D.I. = (n + ½) λ

79. Double slit interference
s
D
λ
d
monochromatic
light
Young’s double-slit experiments = λD / d

80. Coherent light having a wavelength of 675 nm
is incident on an opaque card having two
vertical slits separated
by 1.25 mm. A screen is located
4.50 m away from the card.
What is the distance between the central maximum
and the first maximum?
SOLUTION: Use s = λD / d.
λ = 675×10-9
m, D = 4.50 m, and d = 1.25×10-3
m.
Thus
s = λD / d
= 675×10-9
×4.50 / 1.25×10-3
= 0.00243 m.
Example