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Lecture 4
Physics 1502: Lecture 34
Today’s Agenda
• Announcements:
– Midterm 2: graded soon …
– Homework 09: Friday December 4
• Optics
– Interference
– Diffraction
» Introduction to diffraction
» Diffraction from narrow slits
» Intensity of single-slit and two-slits diffraction patterns
» The diffraction grating
In
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Lecture 4
A wave through two slits
In Phase, i.e. Maxima when ΔP = d sinθ = nλ
Out of Phase, i.e. Minima when ΔP = d sinθ = (n+1/2)λ
d
θ
ΔP=d sinθ
Screen
A wave through two slits
In Phase, i.e. Maxima when ΔP = d sinθ = nλ
+
Out of Phase, i.e. Minima when ΔP = d sinθ = (n+1/2)λ
+
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Lecture 4
The Intensity
What is the intensity at P?
The only term with a t dependence is
sin2( ).That term averages to ½ .
If we had only had one slit, the
intensity would have been,
So we can rewrite the total intensity as,
with
The Intensity
We can rewrite intensity at point P
in terms of distance y
Using this relation, we can rewrite
expression for the intensity at point
P as function of y
Constructive interference occurs at
where m=+/-1, +/-2 …
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Lecture 4
Phasor Addition of Waves
Consider a sinusoidal wave whose
electric field component is
E0
E2(t)
E1(t)
ωt+φ
E0
ωt
Consider second sinusoidal wave
The projection of sum of two
phasors EP is equal to
E0
φ/2
E2(t)
φ
ER
EP(t)
E1(t)
E0
ωt
Phasor Diagrams for Two
Coherent Sources
ER=2E0
E0
ER
450
E0
E0
ER=0
E0
E0
E0
ER
E0
E0
ER
900
E0
2700
E0
ER=2E0
E0
E0
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Lecture 4
SUMMARY
2 slits interference pattern (Young’s experiment)
How would pattern be changed if we add one or more slits ?
(assuming the same slit separation )
3 slits, 4 slits, 5 slits, etc.
Phasor: 1 vector represents 1 traveling wave
single traveling wave
2 wave interference
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Lecture 4
N-slits Interference Patterns
Φ=0
Φ=90
Φ=180
Φ=270
Φ=360
N=2
N=3
N=4
Change of Phase Due to Reflection
Lloyd’s mirror
P2
S
P1
L
I
Mirror
The reflected ray (red) can be considered as
an original from the image source at point I.
Thus we can think of an arrangement S and I
as a double-slit source separated by the
distance between points S and I.
An interference pattern for this experimental
setting is really observed …..
but dark and bright fringes
are reversed in order
This mean that the sources S and I are different in phase by 1800
An electromagnetic wave undergoes a phase change by 1800 upon
reflecting from the medium that has a higher index of refraction than
that one in which the wave is traveling.
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Lecture 4
Change of Phase Due to Reflection
n1
n2
n1
1800 phase change
n1<n2
n2
no phase change
n1>n2
Interference in Thin Films
1800 phase
change
1
Air
Film
Air
no phase
change
2
A wave traveling from air toward film
undergoes 1800 phase change upon
reflection.
The wavelength of light λn in the medium
with refraction index n is
t
The ray 1 is 1800 out of phase with ray 2 which is equivalent
to a path difference λn/2.
The ray 2 also travels extra distance 2t.
Constructive interference
Destructive interference
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Lecture 4
Chapter 34 – Act 1
Estimate minimum thickness of a soap-bubble film (n=1.33) that
results in constructive interference in the reflected light if the film is
Illuminated by light with λ=600nm.
A) 113nm
B) 250nm
C) 339nm
Problem
Consider the double-slit arrangement shown in Figure
below, where the slit separation is d and the slit to
screen distance is L. A sheet of transparent plastic
having an index of refraction n and thickness t is
placed over the upper slit. As a result, the central
maximum of the interference pattern moves upward a
distance y’. Find y’
where will the
central
maximum
be now ?
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Lecture 4
Solution
Phase difference for going
though plastic sheet:
Corresponding path length
difference:
Angle of central max is approx:
Thus the distance y’ is:
gives
Phase Change upon Reflection from a Surface/Interface
Reflection from
Optically Denser Medium (larger n)
180o Phase Change
Reflection from
Optically Lighter Medium (smaller n)
No Phase Change
by analogy to reflection of traveling wave in mechanics
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Lecture 4
constructive: 2t = (m +1/2) λn
destructive: 2t = m λn
Examples :
constructive: 2t = m λn
destructive: 2t = (m +1/2) λn
Application
Reducing Reflection in Optical Instruments
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Lecture 4
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t
c
D
Experimental Observations:
(pattern produced by a single slit ?)
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Lecture 4
How do we understand this pattern ?
First Destructive Interference:
(a/2) sin Θ = ± λ/2
sin Θ = ± λ/a
Second Destructive Interference:
(a/4) sin Θ = ± λ/2
sin Θ = ± 2 λ/a
mth Destructive Interference:
sin Θ = ± m λ/a
m=±1, ±2, …
See Huygen’s Principle
So we can calculate where the minima will be !
sin Θ = ± m λ/a
m=±1, ±2, …
So, when the slit becomes smaller the central maximum becomes ?
Why is the central maximum so much stronger than the others ?
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Lecture 4
Phasor Description of Diffraction
Let’s define phase difference (β) between first and last ray (phasor)
β = Σ (Δβ) = N Δβ
central
max.
1st
min.
(a/λ) sin Θ = 1: 1st min.
Δβ / 2π = Δy sin (Θ) / λ
β = N Δβ
= N 2π Δy sin (Θ) / λ
= 2π a sin (Θ) / λ
2nd
max.
Can we calculate the intensity
anywhere on diffraction pattern ?
Yes, using Phasors !
Let take some arbitrary point on the diffraction pattern
This point can be defined by angle Θ or
by phase difference between first and last ray (phasor) β
The resultant electric field magnitude
ER is given (from the figure) by :
sin (β/2) = ER / 2R
The arc length Eo is given by : Eo = R β
ER = 2R sin (β/2)
= 2 (Eo/ β) sin (β/2)
= Eo [ sin (β/2) / (β/2) ]
So, the intensity anywhere on the pattern :
I = Imax [ sin (β/2) / (β/2) ]2
β = 2π a sin (Θ) / λ
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Lecture 4
Other Examples
Light from a small source passes by
the edge of an opaque object and
continues on to a screen. A
diffraction pattern consisting of
bright and dark fringes appears on
the screen in the region above the
edge of the object.
What type of an object would create a
diffraction pattern shown on the left,
when positioned midway between screen
and light source ?
• A penny, …
• Note the bright spot at the center.
Fraunhofer Diffraction
(or far-field)
θ
Lens
Incoming
wave
Screen
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Lecture 4
Fresnel Diffraction
(or near-field)
Lens
Incoming
wave
P
Screen
(more complicated: not covered in this course)
Resolution
(single-slit aperture)
Rayleigh’s criterion:
• two images are just resolved WHEN:
When central maximum of one image falls on
the first minimum of another image
sin Θ = λ / a
Θmin ~ λ / a
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Lecture 4
Resolution
(circular aperture)
Diffraction patterns of two point sources for various angular
separation of the sources
Rayleigh’s criterion
for
circular aperture:
Θmin = 1.22 ( λ / a)
EXAMPLE
A ruby laser beam (λ = 694.3 nm) is sent outwards from a 2.7m diameter telescope to the moon, 384 000 km away. What is
the radius of the big red spot on the moon?
a.
b.
c.
d.
e.
500 m
250 m
120 m
1.0 km
2.7 km
Moon
Earth
Θmin = 1.22 ( λ / a)
R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ]
R = 120 m !
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Lecture 4
Two-Slit Interference Pattern with a Finite Slit Size
Interference (interference fringes):
Iinter = Imax [cos (πd sin Θ / λ)]2
Diffraction (“envelope” function):
Idiff = Imax [ sin (β/2) / (β/2) ]2
β = 2π a sin (Θ) / λ
Itot = Iinter . Idiff
smaller separation
between slits
=> ?
The combined effects of two-slit and single-slit
interference. This is the pattern produced when
650-nm light waves pass through two 3.0- mm
slits that are 18 mm apart.
smaller slit size
=> ?
Animation
Example
The centers of two slits of width a are a distance d apart. Is it
possible that the first minimum of the interference pattern
occurs at the location of the first minimum of the diffraction
pattern for light of wavelength λ ?
d
No!
a
a
1st minimum interference:
d sin Θ = λ /2
1st minimum diffraction:
a sin Θ = λ
The same place (same Θ) :
λ /2d = λ /a
a /d = 2
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Lecture 4
Application
X-ray Diffraction by crystals
Can we determine the atomic
structure of the crystals, like
proteins, by analyzing X-ray
diffraction patters like one shown ?
Yes in principle: this is like the problem
of determining the slit separation (d)
and slit size (a) from the observed
pattern, but much much more
complicated !
A Laue pattern of the enzyme
Rubisco, produced with a wide-band
x-ray spectrum. This enzyme is
present in plants and takes part in
the process of photosynthesis.
Determining the atomic structure of crystals
With X-ray Diffraction (basic principle)
Crystals are made of regular
arrays of atoms that
effectively scatter X-ray
Scattering (or interference)
of two X-rays from the crystal
planes made-up of atoms
Bragg’s Law
Crystalline structure of sodium
chloride (NaCl). length of the cube
edge is a = 0.562 nm.
2 d sin Θ = m λ m = 1, 2, ..
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