Transcript Phys132 Lecture 5 - University of Connecticut

Physics 1502: Lecture 35
Today’s Agenda
• Announcements:
– Midterm 2: graded soon …
» solutions
– Homework 09: Wednesday December 9
• Optics
– Diffraction
» Introduction to diffraction
» Diffraction from narrow slits
» Intensity of single-slit and two-slits diffraction patterns
» The diffraction grating
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)
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.
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, ..