Week 10 Lecture 3:
Multiple source interference
Problems F2003Q4c, 66, 67 (courseware 96-99)
→ If we can add 2 waves we can add more!
→ Fortunately this is only of practical interest for waves
with equal ω, A and successive δs.
Phasor Addition for N Waves – see next slide (just a head
to tail construction of phasors…)
→ Convince yourself this is the same as the two vector
case:
 (kx-ωt) is the angle to the 1st phasor (although axis is
not shown).
 δ is the phase difference between successive waves.
 R is the resultant amplitude.
 We must determine α - as before with 2 waves.
In the Head to Tail Construction
→ It maps out a circle radius r (or an arc of a circle)
→ Each phasor subtends an angle δ, for a total
subtended angle of Nδ (similar triangles).
→ Previously: for 2 phasor case need to find R and β
For N phasor case – need to find R and also α to1
get an equation representing the sum wave.
R2 is the intensity
along the line in
the observation
angle θ direction.
R
es
ul
ta
nt
A
m
pl
it
ud
e
=
R
For waves having the
same
A, ω and successive δ’s
2
ASIDE…
Try the phasor addition program on the website (links).
You can get the diagram on the previous page using
N=10 (sources) – then try δ= 2ο, 11o, 30o, 36o (this is
the phase difference between adjacent sources).
So what is happening physically when you do this?
You are simply increasing the observation angle!!
36 deg
30 deg
N
sources
δ not θ
11 deg
2 deg
From phasor addition
Program… For:
δ = 0o then I = 100%,
δ= 2o then I = 99%
δ= 11o then I = 73%
δ= 30o then I = 3.8%
δ = 36o then I = 0%
Hang on – I thought you got a minimum
when δ=π, 3π, 5π, etc…. How can you get a
minimum at δ=36o? BECAUSE this is 10
sources! The earlier δ minima were for 2
sources.
Remember that
δ=kdsinθ so this
angle isn’t θ but it
is related to θ.
Note that we
don’t know θ at
this point
because we
haven’t set d or λ
With N=5 and δ=144o you get a cool star!!
This is a minima, because R=0! This means
that at the θ angle corresponding to δ=144o
you have a minima (no intensity)
3
Now back to our story…
In order to determine resultant phasor- need α and R
What is α?
→ compare with the 2 phasor case:
→ 2 phasors with δ phase difference – this angle is δ/2
→ N phasors with same δ diff – this is α = ( N − 1) δ
2
why N-1? Because N = 1
corresponds to only one phasor
∴ no phase difference.
What is R?
→ from triangle ODP we can use the cosine rule:
R 2 = r 2 + r 2 − 2r 2 cos Nδ
= 2r 2 (1 − cos Nδ )
⎛ 1 − cos Nδ ⎞
R = 4r ⎜
⎟
2
⎝
⎠
2
take square root
2
1 − cos Nδ
R = 2r
2
Nδ
R = 2r sin
2
=sinNδ/2
Now, we don’t know what “r” is, so let’s get rid of4
it…
To do this Æ get an expression for A
→ from triangle ODB – do same cosine rule analysis as
we did for R.
A = 2 r sin
dividing
Nδ ⎞
2 r sin⎛⎜
⎟
R
2
⎝
⎠
=
δ
A
2r sin
2
δ
2
Nδ
sin
2
∴ R=A
δ
sin
2
This is it!!
An
expression
for R!
So for the resultant wave the amplitude is
given by R above and the angle with respect to
the x-axis is (kx-ωt + α)
where α is:
α = ( N − 1)
Note I deleted some
words here
δ
2
5
Putting everything together:
∴So on a faraway screen at the sum of all
waves at a particular point P (or angle θ) is
Nδ
2 sin ⎛⎜ kx − ωt + ( N − 1) δ ⎞⎟
y1 + y2 + K yn = A
δ
2
⎝
⎠
sin
2
sin
Amplitude R of
sum wave
Angle of sum wave
wrt the x axis
Intensity (the square of the amplitude)
Nδ
sin
2
I = A2
δ
sin 2
2
2
where
δ = kd sin θ =
2π
λ
d sin θ
But A2 = source intensity Is (for each individual wave source) so
⎞
2 ⎛ Nπ
Nδ
sin
d
sin
θ
⎜
⎟
sin
λ
⎝
⎠
2 =I
I = Is
s
⎛π
⎞
2 δ
sin
sin 2 ⎜ d sin θ ⎟
2
⎝λ
⎠
2
6
How does the pattern change with increasing Ns?
Look at the case when θ=0 degrees
N
sources
When θ
then
approaches
0
Nπ
λ
(i.e. dsinθ
d sin θ
approaches
0)
is small so the small angle
approx holds: sin2() = ( )2
⎛ Nπ
⎞
d sin θ ⎟
⎜
λ
⎠
I ≈ Is ⎝
2
π
⎛
⎞
⎜ d sin θ ⎟
⎝λ
⎠
2
= IsN2
This tells us that the intensity increases
with N2. ----More slits or sources give a
higher intensity of the principle maxima 7
(see next slide)
8
Important Points to Note
1. Principal maxima shape: gets narrower and
sharper with increasing N according to I α N2
2. Principal maxima position for N sources is same
as for N = 2.

ie. maxima when dz = nλ
l
this is when δ =
λ
d sin θ = nλ
(since z/l was for small angles)
d sin θ = 0, 2π , 4π , K
Nδ
sin
2
I = Is
δ
sin 2
2
3. Minima occur?
Try program
N=30, δ=2π/30
= 360/30=12o
2π
or
2
Try phasor
program
N=30, δ=0
2π 4π 6π
,
,
,K
N N N
top term = 0 so minimum
when
δ=
(only works for n≠N) -can also
do with phasors – when they
sum to zero
1. Secondary maxima? - roughly halfway between
the adjacent minima though not exactly. Try program
δ=18o since next
min at 24o
2. Spacing between principle maxima? (like N=2)
λl
z
λ
z=
=
sin
θ
=
spacing
d or
l
d
→ ∴ principle max spacing given by
sin θ =
λ
d
→ As d/λ gets larger you get many principle maxima.
9
Summary slide: MSI
Summary: Multiple Source Interference
“N” sources a distance “d” apart all with amplitude
“A” and the same “ω”
for any given observation
angle θ we calculate δ then
overall path
difference from
top to bottom
overall phase difference
= Nkdsinθ = Nδ
path difference between
each source so δ = phase
difference between each
source = δ = kdsinθ
get resultant
amplitude R for this
angle θ and R2 = I 10
→ problem
11
Polar plot of intensity (radial) vs θ (angle):
I
Is
θ
Note that you could NOT
get this pattern for light
waves (λ~1µm) but could
for radio waves with λ
from 1cm to 100 m.
plotting:
Nπ
sin 2 ⎛⎜
d sin θ ⎞⎟
I
⎠
⎝ λ
=
π
Is
sin 2 ⎛⎜ d sin θ ⎞⎟
⎝λ
⎠
as a function of θ
Or d/λ = ½
(source spacing
is half the
wavelength)
Here λ/d=sinθ = 2. How does this work? This means sin
12 θ
changes by 2. Why? well, sin θ changes by 1 going from 0
to 90o and then again by 1 from 90 to 180o. 1+1 = 2!
Spacing
increases
relative to the
wavelength
Plots of θ vs I/Is for N=2
Primary peak
spacing is
sinθ = λ/d = 1
Or θ = 0ο , 90o
Peak spacing is
sinθ = λ/d = 0.5 or
θ = 0o, 30o and 90o
Peak spacing is
sinθ = λ/d = 0.1 or
θ = 0o, 5.7o,11.5o,17.5o…
All of these peaks
should have the
same intensity (but
they didn’t plot that
way on my printer)
13
Note the Youngs 2-slit demo gave
a d/λ of ~200 so lots of peaks!!
Plots of θ vs I/Is for d/λ = ½ (spacing half the wavelength)
(Note change
in scale of plot
as N
increases)
14
Nπ
sin ⎜
d sin θ ⎞⎟
⎠
⎝ λ
Again recall: I =
π
Is
sin 2 ⎛⎜ d sin θ ⎞⎟
⎠
⎝λ
2⎛
Further points to note: (following on from earlier points)
→ Critical parameters are N and d/λ.
 Large N means more narrow and higher peak
intensity at principle maxima.
AND
If d/λ is small – get
very few principle
maxima.
small d/λ and high N gives
a very intense, directional
signal. Can do this for radio
waves where λ is large (1cm
– 100m) to get intense
directional beam. (see next
slide on broadside array).
If d/λ is large – get
very many principle
maxima.
for visible light the
wavelength λ is ~ 1µm or
10-6m. ∴ so even the
smallest spacing (say
d=0.2mm like in the laser
demo) will give d/λ~200.
This will give a huge number
of primary peaks which is
why we see so many peaks
15
when we do the
demonstration with light!