SV-nlt-c4-c5-Interference and diffraction and applications.pdf

Interference, Diffraction
& applications
Dr. Tran Thi Thu Hanh

HCMUT – Dr. HanhTTT
Outline
1. Wave Properties
2. Interference of Waves
3. Diffraction of Waves

Vocabulary
Wavelength, Amplitude, Period, Frequency
Optical path difference OPD
Optical path length OPL
Constructive Interference
Destructive Interference
Wavefront
Light rays

● Wavelength: The distance λ between identical points on the wave.
● Amplitude: The maximum displacement A of a point on the wave.
λ
Wavelength
Amplitude A
A
x
For a fixed time t :
Wave Properties

Wave Properties: continue
● Period: The time T for a point on the wave to undergo
one complete oscillation.
● Speed: The wave moves one wavelength λ in one period T
so its speed is v = λ / T.
T
v
λ
=
● Frequency: f = 1/T = cycles/second.
Angular frequency: ω = 2π f = radians/second
f
v λ
=
Period T
Amplitude A
t
For a fixed position x :
A

Example
displacement vs. time at x = 0.4 m
time in seconds
0 .02 .04 .06 .08 .1 .12 .14 .16
0
0.4
0.8
-0.4
-0.8
Displacement
in
mm
● What is the amplitude, A, of this wave?
● What is the period, T, of this wave?
And frequency f?
● If this wave moves with a velocity v = 50 m/s,
what is the wavelength, λ, of the wave?

Exercise 1: Wave Motion
The speed of sound in air is a bit over 300 m/s, and the speed
of light in air is about 300,000,000 m/s.
Suppose we make a sound wave and a light wave that both
have a wavelength of 3 meters.
1. What is the ratio of the frequency of the light wave to that of
the sound wave?
(a) Increases
(b) Decreases
(c) Stays the same
2. What happens to the wavelength if the light passes under water or
other medium (not in vacuum)?
(a) About 1,000,000
(b) About 0.000001
(c) About 1000
HCMUT – Dr. HanhTTT https://www.youtube.com/watch?v=yrRkJMHEcYE

The Wave Equation...
! We will assume waves of the form y(x,t) = f(x – vt) and
cos(kx - wt)
" What is the origin of these functional forms?
" These are solutions to a wave equation:
" Example: Sound waves = pressure waves:
2 2
2 2 2
1
h h
x v t
∂ ∂
∂ ∂
=
level)
sea
at
(air
m/s
30
v
th
wi
dt
p
d
v
dx
p
d
3
1
2
2
2
2
2
=
⋅
=

The Wave Equation...
2 2
0 0
2 2
y y
E E
x t
∂ ∂
µ ε
∂ ∂
=
8
0 0
1
3 10 m/s
= f =
k
c
µ ε
ω
λ
= = ×
0 sin ( )
y
E E kx t
ω
= −
0 sin ( )
z
B B kx t
ω
= −
For electromagnetic waves, the origin is Maxwell’s
equations, which lead to wave equations for the
electric and magnetic fields:

Harmonic Plane Waves
# Consider a wave that is
harmonic in x and has a
wavelength of λ.
#
( ) ⎟
⎠
⎞
⎜
⎝
⎛
λ
π
= x
2
cos
A
x
y
If the amplitude is maximum at
x = 0, this has the functional form:
y
x
λ
A
● And, if this is moving to the right with
speed v it will be described by:
( ) ( ) ( )
2
, cos cos
y x t A x vt A kx t
π
ω
λ
⎛ ⎞
= − = −
⎜ ⎟
⎝ ⎠
y
x
v
# Can you provide an example about a harmonic wave?
k ≡
2π
λ
● the “wave number”, K, is defined as:
HCMUT – Dr. HanhTTT https://www.youtube.com/watch?v=PVX4V5Adbzk

Summary
● The formula
describes a harmonic plane wave
of amplitude A moving in the
+x direction.
( ) ( )
t
kx
cos
A
t
,
x
y ω
−
= y
x
λ
A
● For a wave on a string, each point on the wave oscillates in
the y direction with simple harmonic motion of angular
frequency ω.
λ
π
=
2
k
● The wavelength is v
k
=
ω
● The speed is

● Sound waves or electro-magnetic (EM) waves that are created from a point
source are spherical waves, i.e., they move radially from the source in all
directions.
● These waves can be represented by circular arcs:
● These arcs are surfaces of constant phase (e.g., crests)
● Note: In general for spherical waves the intensity will
fall off as 1/r2.
λ
Spherical waves
# In next part of the course we will deal primarily with sound waves
and electromagnetic waves (radio frequency, microwaves, light).
# How bright is the light? How loud is the sound?

_____ light rays
- - - - wave fronts
O
Fundamental Notions: Rays – Wave front
Wave front is the locus of all adjacent points at which the phase
of the wave is the same
A ray is an imaginary line along the direction of travel of the wave.
When waves travel in a homogeneous isotropic material, the rays
are always straight lines perpendicular to the wave fronts

Fundamental Notions: Amplitude and Intensity
SOUND WAVE: peak differential pressure, po power transmitted/area (loudness)
EM WAVE: peak electric field, Eo power transmitted/area (brightness)
Amplitude, A Intensity, I
● We will rarely (if ever) calculate the magnitudes of p or E, and we
will generally calculate ratios of intensities, so we can simplify our
analysis and write:
I ≡ A2 or A = √ I
Intensity Amplitude
Transmitted power per
unit area (W/m2)
Power transmitted is proportional to the square of the amplitude.
2
tim e-averag ed p o w er = I area (am p litu d e)
P
< > = × ∝
2
o
E
I
c
µ
〈〉
=

INTERFERENCE
● Application of the phenomenon: Interferometers
● Precise distance and rotation measurements
● Noninvasive microscopy
#Interference: what happens when you add waves?
# e.g., light + light = interference ??
Interference!

INTERFERENCE
!Interference is a phenomenon that occurs when
there is the superposition of two or more
coherent waves, resulting in bright and dark
fringes.
!Two waves are coherent if they have
- The same frequency,
- The same oscillation direction
- And the constant phase difference

Superposition of Waves
# Question: What happens when two waves “collide?”
# Answer: They ADD together!
# We say the waves are in a “superposition”
(super_pulse1)
(super_pulse2)
HCMUT – Dr. HanhTTT https://www.youtube.com/watch?v=eW5VGGJuWtQ

Pulse 2 has four times the peak intensity of pulse 1, i.e., I2 = 4 I1.
(a) 4 I1
(b) 5 I1
(c) 9 I1
Exercise
1. What is the maximum intensity, Imax?
2. What is the minimum intensity, Imin?
(a) 0
(b) I1
(c) 3 I1
(super_pulse1)
(super_pulse2)

Adding Waves with Different Phases
# Suppose we have two waves with the same amplitude A1 and angular frequency
ω $ their wave numbers k are also the same. Suppose that they differ only in
phase φ :
A1 cosα +cosβ
( )= 2A1 cos
β +α
2
⎛
⎝
⎜
⎞
⎠
⎟cos
β −α
2
⎛
⎝
⎜
⎞
⎠
⎟
y1 + y2
φ / 2
( )
kx −ωt +φ / 2
( )
Trig identity:
y1 = A1 cos(k x - ω t) and y2 = A1 cos(k x - ω t + φ)
Spatial dependence of 2
waves at t = 0:
Resultant wave:
Amplitude Oscillation
y = y1 +y2
φ

Interference of Waves
# What happens when two waves are present at the same point in
space and time? (single w)
# Always add amplitudes (pressures or electric fields).
# What we observe however is Intensity (absorbed power).
Stereo speakers Listener
For 2 equal waves, A = 2A1
cos(φ / 2)
Example:
● “Constructive interference”:
waves are “in phase” (φ = 0,
2π, 4π, ..)
● “Destructive interference”:
waves are “out of phase”
(φ = π, 3π, 5π, …)
Of course, f can take on an infinite number of values. We won’t
use terms like “mostly constructive” or “slightly destructive”!!!
Terminology
r1
r2
I = A2

)
t
cos(
A
E
)
t
cos(
A
E
2
2
2
1
1
1
ϕ
+
ω
=
ϕ
+
ω
=
)
t
cos(
A
)
t
cos(
A
E
E
E 2
2
1
1
2
1 ϕ
+
ω
+
ϕ
+
ω
=
+
=
)
t
cos(
A
E Φ
+
ω
=
)
cos(
A
A
2
A
A
A 2
1
2
1
2
2
2
1 ϕ
−
ϕ
+
+
=
)
cos(
I
I
2
I
I
A
I 2
1
2
1
2
1
2
ϕ
−
ϕ
+
+
=
=
2
2
1
1
2
2
1
1
cos
cos
sin
sin
tan
ϕ
ϕ
ϕ
ϕ
A
A
A
A
+
+
=
Φ
The resultant wave function is :
A1
A2
A
ϕ1
ϕ2
Φ
Consider 2 coherent waves arriving at a point M:
Interference of two coherent waves
Intensity

|
A
|A
A
PHASE
OF
OUT
are
waves
two
:
1)π
(2m
Δ
1
)
cos(Δ
:
ce
Interferen
e
Destructiv
A
A
A
PHASE
IN
are
waves
two
:
2mπ
Δ
1
)
cos(Δ
:
ce
Interferen
ve
Constructi
2
1
min
2
1
max
−
=
=>
+
=
ϕ
=>
−
=
ϕ
+
=
=>
=
ϕ
=>
+
=
ϕ
ϕ
Δ
+
+
= cos
A
A
2
A
A
A 2
1
2
2
2
1
ϕ
ϕ
Δ
+
+
=
Δ
+
+
=
cos
I
I
2
I
I
I
cos
A
A
2
A
A
I
2
1
2
1
2
1
2
2
2
1
2
1 ϕ
−
ϕ
=
ϕ
Δ
2
2
1
2
2
1
max
min
2
1
2
1
max
min
)
A
(A
I
)
A
(A
I
I
I
A
A
A
|
A
|A
A
A
A
+
≤
≤
−
≤
≤
+
≤
≤
−
≤
≤
Constructive Interference & Destructive Interference

Summary
● The resultant intensity of two equal-intensity
waves at the same point in space is:
I = 4 I1cos2(φ/2)
● For nonequal intensities, the maximum and
minimum intensities are
Imax = |A1 + A2|2 Imin = |A1 - A2|2

ConstructiveInterference: Δϕ = 2mπ ⇒ δ = mλ
Destructive Interference: Δϕ = (2m +1)π ⇒ δ = (m +
1
2
)λ
φ = Δϕ =
2π
λ
δ
Relation between Phase difference and
Optical Path Difference (OPD)
The phase difference between the two waves may be due to
a difference in their source phases or a difference in the
path lengths to the observer:
With δ = r2 – r1 : path difference

Huygen’s principle

What happens when a plane wave meets a small aperture?
Answer: The result depends on the ratio of the wavelength λ
to the size of the aperture a :
Huygen’s principle
All points on wavefront are point
sources for ‘spherical secondary
wavelets’ with speed, frequency
equal to initial wave.
Wavefront
at t=0
Wavefront
at time t
λ >> a
- Similar to a wave from a point source.
“Diffraction”: Interference of
waves from objects or apertures.
λ << a
- The transmitted wave is still
concentrated in the forward
direction, and at near distances
wavefronts have the shape of the
aperture.
https://www.youtube.com/watch?v=wSFfTM4jAwk

Double-slit interference
# Light (wavelength λ) is incident on a two-slit
(two narrow, rectangular openings) apparatus:
● If either one of the slits is closed,
a diffuse image of the other slit will
appear on the screen. (The image will
be “diffuse” due to diffraction. We
will discuss this effect in more detail
later.)
Monochromatic light
(wavelength λ)
S1
S2
screen
Diffraction
profile
I1
● If both slits are now open, we see
interference “fringes” (light and
dark bands), corresponding to
constructive and destructive
interference of the electric-field
amplitudes from both slits.
I
S1
S2
HCMUT – Dr. HanhTTT https://www.youtube.com/watch?v=9D8cPrEAGyc

Two-Slit Interference
m = 0, ±1, ±2,...
δ=dsinθ=mλ Constructive
Interference
δ=dsinθ=(m + 1/2)λ Destructive
Interference
Basic result:
m=0
m=1
m=2
m=-1
m=-2
θ
θ = sin-1(mλ/d)
“lines” of
constructive
interference:
d
θ
θ
I
2λ/d
λ/d
0
-λ/d
r
Usually we care about the linear
(as opposed to angular)
displacement y of the pattern:
L
y
y = L tanθ

Two-Slit Interference, small angles
y ≈ m(λ/d)L
y ≈ (m + 1/2)(λ/d)L
d
θ
Y ≈ Lθ
I
2λL/d
λL/d
0
-λL/d
L
The slit-spacing d is often large compared to λ, so that θ is small.
Then we can use the small angle approximations to simplify our results:
y = L tan θ ≈ L θ (in radians)
m = 0, ±1, ±2,...
θ ≈ m(λ/d)
Constructive
Interference:
(bright)
θ ≈ (m + 1/2)(λ/d)
Destructive
Interference:
(dark)
For small angles: (θ << 1 radian)
sin θ ≈ θ ≈ tan θ (only in radians!)

1. What is the spacing Δy between fringe maxima on a screen 2m away?
a. 1 µm b. 1 mm c. 1 cm
2. If we increase the spacing between the slits, what will happen to Δy?
a. decrease b. stay the same c. increase
3. If we instead use a green laser (smaller λ), Δy will?
a. decrease b. stay the same c. increase
Exercise: 2-slit interference
A laser of wavelength 633 nm is
incident on two slits separated
by 0.125 mm.
I
S1
S2
Δy

Multi-Slit Interference
What changes if we increase the
number of slits (e.g., N = 3, 4,
1000?)
# First look at the “principal maxima”:
# If slit 1 and 2 are in phase with
each other, than slit 3 will also be in
phase.
Conclusion: Position of
“principal interference
maxima” are the same!
(i.e., d sinθ = m λ)
S3
S2
P
Incident wave
(wavelength λ) y
L
d
S1
Atot = 3 A1
Itot = 9 I1
For N slits
Itot = N2 I1

General properties of N-Slit Interference
• The positions of the principal maxima of the intensity patterns always
occur at φ = 0, ±2π, ±4π, ... [φ is the phase between adjacent slits]
(i.e., dsinθ = ±mλ, m = 0, 1, 2,…).
• The principal maxima become taller and narrower as N increases.
• The intensity of a principal maximum is equal to N2 times the maximum
intensity from one slit. The width of a principal maximum goes as 1/N.
• The # of zeroes between adjacent principal maxima is equal to N-1. The
# of secondary maxima between adjacent principal maxima is N-2.
0 2π
-2π
I
0
16I1
N=4
0 2π
-2π
I
0
25I1
N=5
0 2π
-2π
I
0
9I1
N=3
10 0 10
0
10
20
25
0
h5( )
x
10
10 x
-λ/d 0 λ/d
φ
θ
10 0 10
0
5
9
0
g( )
x
10
10 x
φ
θ
-λ/d 0 λ/d
10 0 10
0
10
16
0
h( )
x
10
10 x
-λ/d 0 λ/d
φ
θ
For N slits the first zero is at 2π/N.

1. What is the total number of principal maxima can be observed on a
screen 2m away?
2. What is the width of the principal maxima on a screen 2m away if
N=25?
3. Number of slits per unity of the length is equal to 1/d. What is the
number of slits per 1 cm?
Exercise: N-slit interference
A laser of wavelength 760 nm is
incident on N slits separated
by 0.125 mm.

Exercise
Light interfering from 10 equally spaced slits initially illuminates
a screen. Now we double the number of slits, keeping the
spacing constant.
What happens to the net power on the screen?
a. stays the same b. doubles c. increases by 4

N-Slit Interference – Summary
# The Intensity for N equally spaced slits is given by:
L
y
and
d
sin
d
≈
⋅
≈
=
= θ
λ
θ
λ
θ
λ
δ
π
φ
2
* Note: we can not be able to use the small angle approximations if d ~ λ.
y
L
d
θ
φ is the phase difference between adjacent slits.
● As usual, to determine the pattern at the screen
(detector plane), we need to relate φ to θ or y = Ltanθ:
*

The Young interference of light is considered: wavelength of light is
of 600 nm, the distance between two slits is of 1 mm and the
distance from the slits to screen is of 1m. Determine:
● The width of the light band of interference pattern, i?
● The position of the 3th light band (or maximum) and the 4th dark
band (or minimum)?
● Could you propose the experiment for determination of the
wavelength of the light in general? Please, explain in details.
Exercise

Thin – film interference
Self-Reading: Interference of thin film !

DIFFRACTION
# Diffraction is the deviation of light from a straight-line
path when the light passes through an aperture or around
an obstacle.
# Diffraction is due to the wave nature of light.

# So far in the N-slit problem we have assumed that each slit is
a point source.
# Point sources radiate equally in all directions.
# Real slits have a non-zero extent – a “slit width” a. The transmission
pattern depends on the ratio of a to λ.
# In general, the smaller the slit width, the more the wave will diffract!
Laser Light
(wavelength λ)
screen
Diffraction
profile
I1
Small slit:
Laser Light
(wavelength λ)
screen
Diffraction
profile
I1
Large slit:
Let’s examine this effect quantitatively!
Single-slit “Diffraction”
https://www.youtube.com/
watch?v=uohd0TtqOaw

# Slit of width a. Where are the minima?
# The first minimum is at an angle such
that the light from the top and the
middle of the slit destructively
interfere:
# The second minimum is at an angle such
that the light from the top and a point
at a/4 destructively interfere:
Location of nth-minimum:
P
Incident Wave
(wavelength λ) y
L
a
θ
a/2
δ
min
sin
2 2
a λ
δ θ
= =
min
sin
a
λ
θ
⇒ =
min,2
sin
4 2
a λ
δ θ
= =
min,2
2
sin
a
λ
θ
⇒ =
min,
sin n
n
a
λ
θ =
θ
a/4
δ
(n = ±1, ±2, …)
Single-slit “Diffraction”

Diffraction: Exercise
Suppose that when we pass red light (λ = 600 nm)
through a slit of width a, the width of the spot (the
distance between the first zeros on each side of the bright
peak) is W = 1 cm on a screen that is L = 2m behind the
slit. How wide is the slit?
2 m
a 1 cm = W

Exercise
Which of the following would broaden the
diffraction peak?
a. reduce the laser wavelength
b. reduce the slit width
c. move the screen further away
2 m
a 1 cm = W

Optical spectroscopy – how we know about the world
• Quantum mechanics $ definite energy levels, e.g.,
of electrons in atoms or molecules.
• When an atom transitions between energy levels $
emits light of a very particular frequency.
• Every substance has it’s own “signature” of what
colors it can emit.
• By measuring the colors, we can determine the
substance, as well as things about it’s surroundings
(e.g., temperature, magnetic fields), whether it’s
moving (via the Doppler effect), etc.
Optical spectroscopy is invaluable in materials
research, engineering, chemistry, biology, medicine…
But how do we precisely measure wavelengths???

Diffraction Gratings
Diffraction gratings (Interference gratings) allow us to resolve
sharp spectral signals.
0 λ2/d sin θ
N-slit Interference:
IN = N2I1
0 λ1/d sin θ
λ1
λ2
Shift of
the peak:
(the basis for optical spectroscopy)
https://www.youtube.com/watch?v=oae5fa-f0S0

# Rely on N-slit interference.
# Consist of a large number of evenly spaced parallel slits.
# Recall that the intensity pattern produced by light of wavelength λ
passing through N slits with spacing d is given by:
2
1
2
2
⎭
⎬
⎫
⎩
⎨
⎧
=
)
/
sin(
)
/
N
sin(
I
IN
φ
φ
where:
10 0 10
0
10
20
25
0
h5( )
x
10
10 x
φ
0 2π
-2π
I
0
25I1
N=5
d sin
2
θ
φ π
λ
=
Consider very narrow slits (a << d), so I1 is roughly constant.
y
L
d
θ
● The position of the first principal maximum is given by sin θ = λ/d
(can’t assume small θ!) $ Different colors $ different angles.
● Width of the principal maximum varies as 1/N – improves ability to
resolve closely spaced lines.

# How effective are diffraction gratings at resolving light
of different wavelengths (i.e. separating closely-spaced
‘spectral lines’)?
# Concrete example: Na lamp has a spectrum with two yellow “lines”
very close together: λ1 = 589.0 nm, λ2 = 589.6 nm (Δλ = 0.6 nm)
# Are these two lines distinguishable using a particular grating?
IN = N2I1
0 λ1/d sin θ
0 λ2/d sin θ
λ1
λ2

min
min
sin
d d
d Nd
λ λ
θ θ
λ λ
θ
Δ
= → Δ ≈
Δ
→ Δ ≈ =
● We assume “Rayleigh’s criterion”: the minimum wavelength separation we can
resolve Δλmin ≡ λ2-λ1 occurs when the maximum of λ2 overlaps with the first
diffraction minimum of λ1. (Δθmin=λ/Nd)
IN = N2I1
N
1
m in
=
λ
λ
Δ Larger N $ Smaller Δλmin
(Higher spectral resolution)
0 λ1/Nd sin θ
θ
λ1/d
“Rayleigh Criterion”
N = number of illuminated
lines in grating.
Δθmin
λ2/d

● We can squeeze more resolution out of a given grating by working in “higher
order”. Remember, the principal maxima occur at sinθ = mλ/d, where m =
1,2,3… designates the “order”. (Δθmin≈ λ/Nd still*) You can easily show:
Nm
1
m in
=
λ
λ
Δ
0 λ/d 2λ/d 3λ/d sin θ
First order Second order Third order
m = 1 m = 2 m = 3
* To be precise: Δθmin= λ/(Nd cosθ), (but Δλ/λ = 1/Nm is correct.)
Larger Nm Smaller Δλmin
(Higher spectral resolution)

National University of Hochiminh city
Hochiminh University of Technology

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