Download Lecture 34 - UConn Physics

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
A wave through two slits
In Phase, i.e. Maxima when DP = d sinq = nl
Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l
d
q
DP=d sinq
Screen
A wave through two slits
In Phase, i.e. Maxima when DP = d sinq = nl
+
Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l
+
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 …
Phasor Addition of Waves
Consider a sinusoidal wave whose
electric field component is
E2(t)
E1(t)
E0
wt+f
E0
wt
Consider second sinusoidal wave
E2(t)
The projection of sum of two
phasors EP is equal to
EP(t)
E1(t)
f/2
f
ER
wt
E0
E0
Phasor Diagrams for Two
Coherent Sources
ER=2E0
ER
450
E0
E0
E0
ER=0
E0
E0
E0
ER
E0
E0
ER
900
E0
2700
E0
ER=2E0
E0
E0
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
N-slits Interference Patterns
F=0
F=90
F=180
F=270
F=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.
Change of Phase Due to Reflection
n1
n2
1800 phase change
n1<n2
n1
no phase change
n1>n2
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 ln 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 ln/2.
The ray 2 also travels extra distance 2t.
Constructive interference
Destructive interference
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 l=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 ?
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
constructive: 2t = (m +1/2) ln
destructive: 2t = m ln
constructive: 2t = m ln
destructive: 2t = (m +1/2) ln
Examples :
Application
Reducing Reflection in Optical Instruments
Experimental Observations:
(pattern produced by a single slit ?)
How do we understand this pattern ?
First Destructive Interference:
(a/2) sin Q = ± l/2
sin Q = ± l/a
Second Destructive Interference:
(a/4) sin Q = ± l/2
sin Q = ± 2 l/a
mth Destructive Interference:
sin Q = ± m l/a
m=±1, ±2, …
See Huygen’s Principle
So we can calculate where the minima will be !
sin Q = ± m l/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 ?
Phasor Description of Diffraction
Let’s define phase difference (b) between first and last ray (phasor)
b = S (Db) = N Db
central
max.
1st
min.
2nd
max.
Can we calculate the intensity
anywhere on diffraction pattern ?
(a/l) sin Q = 1: 1st min.
Db / 2p = Dy sin (Q) / l
b = N Db
= N 2p Dy sin (Q) / l
= 2p a sin (Q) / l
Yes, using Phasors !
Let take some arbitrary point on the diffraction pattern
This point can be defined by angle Q or
by phase difference between first and last ray (phasor) b
The resultant electric field magnitude
ER is given (from the figure) by :
sin (b/2) = ER / 2R
The arc length Eo is given by : Eo = R b
ER = 2R sin (b/2)
= 2 (Eo/ b) sin (b/2)
= Eo [ sin (b/2) / (b/2) ]
So, the intensity anywhere on the pattern :
I = Imax [ sin (b/2) / (b/2) ]2
b = 2p a sin (Q) / l
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)
q
Lens
Incoming
wave
Screen
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 Q = l / a
Qmin ~ l / a
Resolution
(circular aperture)
Diffraction patterns of two point sources for various angular
separation of the sources
Rayleigh’s criterion
for
circular aperture:
Qmin = 1.22 ( l / a)
EXAMPLE
A ruby laser beam (l = 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
Qmin = 1.22 ( l / a)
R / 3.84 108 = 1.22 [ 6.943 10-7 / 2.7 ]
R = 120 m !
Two-Slit Interference Pattern with a Finite Slit Size
Interference (interference fringes):
Iinter = Imax [cos (pd sin Q / l)]2
Diffraction (“envelope” function):
Idiff = Imax [ sin (b/2) / (b/2) ]2
b = 2p a sin (Q) / l
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 l ?
d
No!
a
a
1st minimum interference:
d sin Q = l /2
1st minimum diffraction:
a sin Q = l
The same place (same Q) :
l /2d = l /a
a /d = 2
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 Q = m l m = 1, 2, ..