Lecture 28
Physics 1202: Lecture 28
Today s Agenda
•  Announcements:
–  Midterm 2: solutions
•  HW 8 this Friday
•  Diffraction
–  Review
•  Polarization
–  Reflection by surface
D
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Lecture 28
Experimental Observations:"
(pattern produced by a single slit ?)"
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 28
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)
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
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Lecture 28
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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Lecture 28
Polarization of light
•  Recall E&M wave
y
x
22_all_imgs_in_ppt
z
•  This is an example of linearly polarized light
–  Electric field along a fixed axis ( here y )
E
•  Most light source are nonpolarized
– Electric field along random axis
Polarizers
•  Made of long molecules (organic polymers)
–  Block electric field along their length
–  Electric field perpendicular passes through
E
E
•  So Eafter=E cosθ
•  Recall that I ~ E2
I = I0 cos2θ
Malus’s law
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Lecture 28
Polarization of Electromagnetic Waves (light)
as view along direction
of propagation
unpolarized
Ε
linearly polarized
Ε
Polarization by Absorption
E = Emax cos Θ
I = Imax cos2 Θ
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Lecture 28
EXAMPLE:
Two polarizers have polarization angles set at Θ1 =15o and Θ3 = 90o
with respect to vertical axis, as show on the Figure below.
(a)  By what factor is the intensity of unpolarized light attenuated
going through both polarizers (If/Ii = ? ).
(b)  Does the attenuation factor increase or decrease if a third
polarizer is inserted in-between the two polarizers, with
polarization angle Θ2 =45o ?
Unpolarized light through linear polarizer: If/Ii = <cos 2(Θ)> = 0.50
I = Imax cos2 Θ
If/Ii = cos2(90-15) = 0.07
If/Ii = 0.50 x 0.07 = 0.04 !
If/Ii = 0.5
If/Ii = cos2(45-15) = 0.75
If/Ii = cos2(90-45) = 0.55
unpolarized
polarized
polarized If/Ii = 0.5x0.75 x 0.55 = 0.21 !
Polarization by Reflection
Brewster s angle:
n1 sin Θp = n2 sin Θ2
note : Θ2 = 90 - Θp
using :
n1 = 1 and n2 = n
n = sin Θp / cos Θp
Or: n = tan Θp
When unpolarized light is
incident on a reflecting surface,
the reflected and refracted
beams are partially polarized.
The reflected beam is completely
polarized when the angle of
incidence equals the polarizing
angle (Brewster s angle)
for n =1.55
Θp = 57o
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Lecture 28
EXAMPLE
How far above the horizon is the moon when its image,
reflected in a calm lake, is completely polarized ?
(nwater = 1.33)
a.  22.2°
b. 7.7°
c. 16.6°
d. 36.9°
e. 53.1°
n = tan Θp
Θp = tan-1(1.33) = 53.1o
Polarization by Double Refraction
• There are materials where the speed of light is not the same in all directions.
• Such (birefingent) materials thus have two indexes of refraction
• And light beam splits into two beam, ordinary (O) and extraordinary (E) ray
• E and O rays are also polarized in mutually perpendicular directions.
A calcite crystal produces a
double image because it is a
birefringent (no=1.658,
nE=1.486) material.
Unpolarized light incident on a calcite crystal splits into an
ordinary (O) ray and an extraordinary (E) ray which are
polarized in mutually perpendicular directions.
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Lecture 28
Application
•  Some materials become birefringent when stressed (glass, plastic,..)
•  Since the birefringence effectively rotates the polarization of the light
when such an object is placed between the polarizer and analyzer oriented
at 90o, only the light passing thorough stressed portion of the material will be
observed. This provides a way to image stress in model plastic structures.
A plastic model of an arch
structure under load
conditions observed between
perpendicular polarizers. Such
patterns are useful in the
optimal design of
architectural components.
Bright area ==> max stress
Dark area => minimum stress
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