Review - Interference
(French pg 280-293 – but note French uses cos rather than sin)
→ The superposition of waves is called interference.
→ Consider a wave traveling right
y1 = A sin (kx − ωt )
general formula for a
transverse traveling wave
→ Consider another wave also traveling right with a
phase difference δ compared to the first
y 2 = A sin (kx − ωt + δ )
→ The two waves are shown below at some time t
→ The resultant wave is given by the sum:
y1 + y 2 = A sin (kx − ωt ) + A sin (kx − ωt + δ )
We showed that this was equal to a sum wave of the form:
δ
δ⎞
⎛
y1 + y2 = 2 A cos sin ⎜ kx − ωt + ⎟
2 ⎝
2⎠
1
R=
amplitude
for y1 + y2
angle for y1 + y2
Week 10 Lecture 2:
Problems 62 63 64 65 (courseware 91-95)
Two – Wave Interference (continued)
Summarizing the two wave interference stuff from
the last lecture:
you get the same result addition analytically or with
phasor addition.
•Analytical method is only really good for A1 = A2
and gets very messy for multiple numbers of waves
•Phasor addition can be used easily for more than 2
waves and also for unequal amplitudes and phases.
See Virtual
Physics website:
Double slit
experiment
We will use phasors
*************************************************************************************************
Intensity
As before – intensity is the square of the
resultant amplitude R [=2Acos(δ/2) for two
sources both with amplitude A]. (note that
Phase difference
this is time-independent)
between the two
∴
waves
2
2δ
Intensity = 4 A cos
Note that this is the result for two
sources, both amplitude A
2
2
Young’s 2-Slit Experiment
→ Consider 2 slits a distance ‘d’ apart - with a plane
wavefront hitting them from the left. The slits act as
spherical wave sources (Huygen’s stuff).
→ Both waves have the same amplitude and frequency.
→ They form an interference pattern on a screen – so
far away that the waves are effectively parallel. We
will examine what the pattern looks like as a function
of z away from the midpoint by considering the
interference (sum wave) at some point P.
Looking down
from the top
Plane
wave
P
barrier
x1
s1
d
s2
θ
θ
z
x2
dsinθ = path difference
screen
l
→ θ is the “observation angle” (angle that P is away
from the centreline).
→ The distance of wave 2 to point P is dsinθ further
than wave 1. So dsinθ is the “path difference”
between the two waves.
3
**See “Double slit interference” demo on Virtual Physics website
What is the intensity variation along the screen?
→ We could sum the waves, get the amplitude and
square it to get the intensity.
→ Or, looking at the intensity equation
for 2 wave sources
I = 4 A cos
2
2
δ
2
we could just determine δ (the phase difference at
any point) and put it in! And δ depends on the path
difference!!!!!
What is the phase difference between the 2 waves at P?
→ wave 1? y1 = A sin (kx1 − ωt )
→ wave 2?
y 2 = A sin (kx 2 − ωt )
x2 = x1 + d sin θ
but
path difference
∴ y 2 = A sin(k [x1 + d sin θ ] − ωt )
or
y 2 = A sin (kx1 − ωt + kd sin θ )
y1
So the phase difference is just
the path difference times k
dsinθ
this is the phase
difference δ
between 2 the two
waves at point P
4
So
δ = kdsinθ = phase difference between waves
Understand this! Very very important
→
you can use this expression ‘as is’ but when you
consider a screen far away you can simplify things a
bit...
sin θ ≈
z
δ = kd
l
but
k=
z
l
2π
λ
(provided you are at small θ’s)
so
2πdz
δ=
λl
This tells us how the phase difference between the two
waves changes as a function of z on a faraway screen
Now to find the intensity distribution along the screen
Intensity = I = 4 A2 cos 2
subbing in δ
δ
2
⎡ 2πdx ⎤
I = 4 A cos ⎢
⎣ 2λl ⎥⎦
2
2
⎡ πdz ⎤
I = 4 A2 cos2 ⎢
⎣ λl ⎥⎦
→
But don’t forget
that
d
z
= d sin θ
l
Intensity varies along the screen in a cos2 dependence
(see pictures on next slide)
5
6
→
So on a dark screen you will see light and dark fringes!
bright
When is the intensity highest and lowest?
using
πdz
I = 4 A cos
λl
2
2
Remember
this is δ/2
Highest?
when cos2 term = 1, I = 4A2
→ happens when
dz
Path
=
= d sin θ = 0, ± λ , ± 2λ , K, ± nλ
difference
l
= an integral multiple of the wavelength!
Lowest?
when cos2 term = 0, I = 0
→ happens when
Path = dz = d sin θ = ± λ , ± 3 λ , ± 5 λ , K
difference l
2
2
2
= an integral multiple of λ/2
See Virtual Physics Double slit interference!
7
Summary Young’s 2-Slit Experiment –
Superposition of 2 Waves from Identical Sources
(same A, ω, phase)
P
x1
s1
θ
θ
d
z
x2
l
s2
path difference = dsinθ = extra distance wave 2 travels
to point P compared with wave 1.
∴
screen far
away
Even though these 2 sources are in phase initially, at point
P they are out of phase because wave 2 travels further!
At point P:
y1 = A sin (kx1 − ωt )
y 2 = A sin (kx 2 − ωt )
x 2 = x1 + d sin θ
∴ y 2 = A sin (kx1 − ωt + kd sin θ )
y1
∴ δ = kd sin θ = kd
Intensity
phase difference between
the 2 waves at point P = δ!
z
l
if θ is small
= R 2 = 4 A2 cos 2
at screen
δ
2
minima:
maxima:
cos2δ/2 = 0
cos2δ/2 = 1
8
Phasor Description for 2-Slit
Interference?
Max I when cos2(δ/2) = 1
Path diff = dsinθ =
k=
2π
λ
dz
= 0, λ , 2λ , K
l
This is when
so
kdsinθ = δ = 0, 2π , 4π , K
∴ phasor
addition
since
δ
2
=
πdz
λl
maximum R = 2A so I = 4A2
For any given
observation angle
θ we can draw
phasor diagrams
like these. These
two happen to be
for the special
cases of maxima
and minima
δ = 0, 2π, 4π
Minimum I = 0 when
cos2(δ/2) = 0
Path diff = dsinθ =
dz λ 3λ 5λ
,
= ,
K
2 2 2
l
This is when
δ = π , 3π , 5π , K
δ = π , 3π , 5π , K
For any other phase
difference besides these
simple ones - do a head to 9
tail and determine the
resultant phasor amplitude
Distance between centres of 2 bright fringes on
a screen? (Intensity Peaks)
Brightest
when
distance to nth fringe
dsinθ =
distance
between 2
fringes
(adjacent
peaks)
dzn
= nλ
l
nλl
zn =
d
so
= zn +1 − zn = ((n + 1) − n )
Fringe
spacing
= zbetween fringes
=
λl
λl
d
=
λl
d
d
You can also express this more generally in terms
dzn
of sinθ.
= d sin θ = nλ gives a bright fringe
l
So in terms of θ, fringes are
Note:
sin θ =
λ
d
apart.
n=2
n=1
centre, n=0
•n is the order of interference of the fringes
•Spacing is independent of n close to θ=0
•Spacing ↑ as λ increases
•Spacing ↓ as d increases
•Intensity same for all fringes, since n does not10
change I.
problem
11
Playing with laser/slit demo:
Laser wavelength: λ = 630 nm
Distance to screen: L = 5.0 m
Slit distance: d = 0.2 mm
Distance between intensity peaks:
Calculate distance between peaks:
630 x10−9 x5
=
= 16mm
−3
d
0.2 x10
λL
What is the path difference between waves for n=1
and n=2?
= λ !!!
What is the phase difference between adjacent
waves for the n=1 and n=2 peaks?
Peaks are 1 λ apart.
Phase difference = kdsinθ = kλ = δ
=(2π/λ) x λ = 2π = δ
What is the observation angle for n=1 and n=2?
For n=1
dsinθ = λ so θ = sin-1(λ/d) = 0.003 rad = 0.18o
For n=2
dsinθ = 2λ so θ = sin-1(2λ/d) = 0.006 rad = 0.36o
Why are there so few peaks? (bonus question)
12
Stay tuned for future lectures…..
More on phase difference δ:
Let’s start with a summary: When you create
interference using slits, the light from the slits
always starts out in phase (because both
‘sources’ are formed from the same wave). Then
the interference pattern occurs because at any
given observation angle θ the light from one
‘source’ follows a longer path than the other one.
Therefore the phase difference δ at any given θ is
due to the path difference:
Phase
difference
= δ = k dsinθ
2π/λ
Path difference
The intensity is related to δ in the following way:
I = R = 4 A cos
2
2
2
δ
2
thus when δ = kdsinθ = 0, 2π, 4π, .....You have a max
intensity
dsinθ = 0, λ, 2λ ...
and when δ = kdsinθ = π, 3π, 5π, … You have a minimum
intensity 13
dsinθ = λ/2, , 3λ/2, 5λ/2 ...
So now, what would happen if, instead of
using slits, you used some sort of (single
wavelength) source – like a radio beacon.
You could make these beacons out of phase
with one another. So the overall phase
difference δ at any observation angle θ would
have two components:
Source
Phase difference = δ = kdsinθ + Phase
difference
phase diff
due to path
difference
The intensity formula is the same as on the previous page, but
now we just have to calculate δ using the formula above, so
where are the max and min positions?:
when δ = kdsinθ + s.p.d. = 0, 2π, 4π, .....You have a max
intensity
and
when δ = kdsinθ + s.p.d. = π, 3π, 5π, … You have a min
intensity 14