Interference
• Principle of Superposition- Constructive Interference
• If the crest of one wave meets the crest of another wave then the
resulting amplitude is the sum of the amplitude of these two waves.
• Example of “Constructive Interference”
Wave 1
Wave 2
Resultant
The
amplitude of
the resultant
wave is the
sum of the
amplitudes
of waves
one and two
above.
Interference - 2
• Principle of Superposition
• If the crest of one wave meets the trough of another wave then the
resulting amplitude is the sum of the amplitude of these two
waves.
• Example of “Destructive Interference”
Wave 1
Wave 2
Resultant
The
amplitude of
the resultant
wave is the
sum of the
amplitudes
of waves
one and two
above. In
this case
they cancel
each other.
Interference of Sound Waves
• Compression Waves in the air emitted by a speaker.
High pressure - crest
Low pressure (rarefaction) - trough

Interference of Sound Waves
• Sound is emitted coherently from the two speakers.
Where is there constructive and destructive
interference?
Constructive
Destructive

Interference of Sound
Q:
A:
What did you hear?
Loud and quiet spots.
Loud spots are areas of
constructive interference.
Quiet spots are areas of
destructive interference
Light also shows interference!
Thomas Young (1801)
What do you think you would see if
you were to arrange for this
experiment to be done with light?
Bright
S
c
r
e
e
n
Laser Light
Dark
Small
Slits
Interference of Light
Light areas = Constructive Intereference
Dark areas = Destructive Intereference
But why are there lots of bright
and dark fringes?
Interference of light
Count the crests to
this point from
each slit!
Crest + Crest
The wave from the
bottom slit travels
exactly one
wavelength further
than the waves
from the top slit, so
a crest and crest
still meet! –Hence
Constructive
Interference occurs
Crest + Crest
Crest + Crest
Interference of light – Conditions for maxima
• Assume P is the
first Maximum
(bright fringe)
• The path difference
(Δ) from S1 to P and
S2 to P is d2-d1.
• But since we know
that P is a maximum
we know the path
difference must be
one wavelength
• i.e Δ = d2-d1= λ
• If P was the second
fringe then Δ would
be 2 λ and so on.
• Thus for a maximum
Δ=n λ
where n=0,1,2,3 etc
n.b. n is often referred to as the
‘order’ of the maxima
Interference of light – Conditions for minima
• The case for the areas of
destructive interference (or minima)
is similar to that for maxima except
the minima are half way between
the maxima.
• This happens when the waves from
the bottom slit travels ½ λ more
than the wave from the top slit and
so the waves meet completely out
of phase producing destructive
interference. (a crest meets a
trough)
• Thus for minima: Δ=(n+
½)λ
where n=0,1,2,3 etc
• The first order minimum occurs
between the central maxima and 1st
order maxima i.e. when n=0
Minima
Double Slit Interference
1.
In the microwave experiment shown below, C is the zero order
maximum and D is the first order maximum. AD = 52 cm and BD
= 55 cm.
(a) What is the path difference at point D?
D
Path difference  55 - 52  3 cm
A
(b) What is the wavelength of the microwaves?
  3 cm
C
B
(c) What is the path difference to the second order maximum?
Path difference to 2nd order max  2  3  6 cm
Screen
(d) What is the path difference to the minimum next to C?
Path difference to 1st min  λ
2
 1.5 cm
(e) What is the path difference to the next minimum?
Path difference to 2nd min  (1  1 )λ  4.5 cm
2
(f) What is the path difference at point C?
Zero
Click the mouse to continue
Double Slit Interference
2.
In a microwave interference experiment, H is the second minimum, that is
there is one other minimum between H and G. Measurement of distances
EH and FH gives: EH = 42.1 cm and FH = 46.6 cm. Calculate the wavelength
and frequency of the microwaves used.
H
Path difference  FH  EH
 46.6 - 42.1  4.5 cm
E
G
F
Path difference to min  (1  1 )λ  4.5 cm
2
3
2  4.5
λ  4.5 cm
λ

 3 cm
2
3
f 
v

3 108
10
f 

1
.
0

10
Hz
2
3 10
Screen
Path Difference and Geometry
• Can we relate the fringe pattern to the geometry of
Note because the
the set up?
screen is relatively very
far away the rays can
be considered to be
parallel
S
θ
S1
D
d
θ
1)
S2
Δ
sinθ=Δ/d
Δ=dsinθ
2) tanθ = S/D
or
But for such small
angles tanθ=sinθ
n λ=dsinθ
So sinθ= nλ/d =S/D
By measuring θ and d
you can calculate the
wavelength of light.
Thus if n=1 then
S=λD/d
Double Slit Interference
Light of wavelength 4.2x10-7m falls on a double slit and creates a
interference pattern on a screen 3m away. The fringe separation
is 4.1mm. Calculate the slit separation?
S = λD/d
d = λD/S
d = (4.2x10-7 x 3)/4.1x10-3
d = 0.0003m
Interference of Sound Waves
• Conditions for Constructive and destructive
interference
Constructive: Path lengths from
each speaker differ by an integral
number of wavelengths - where the
blue circles intersect or the black
dotted circles intersect.

Destructive: Path lengths
from each speaker differ by
/2, 3 /2, 5 /2 etc. - where
the blue and black circles
intersect