Transcript Fresnel Diffraction Geometrical optics… …light can`t turn a corner. I

Chapter 13: Fresnel Diffraction
Diffraction
Geometrical optics…
I
…light can’t turn a corner.
Diffraction
Physical optics…
Francesco Maria Grimaldi
(1618 - 1663)
I
…actually, it can.
Huygens-Fresnel principle
every point on a wavefront may be regarded as a secondary
source of spherical wavelets
The propagated
wave follows the
periphery of the
wavelets.
Huygens, just add the
wavelets considering
interference!
Huygens-Fresnel principle
If one perturbs a plane wavefront, the Huygens wavelets
will no longer constructively interfere at all points in
space. Adding the wavelets by physical optics explains
why light can turn corners and create fringes around
images of objects.
Obliquity factor
(oblique  obscure?)
- wavelets propagate isotropically—in forward and reverse directions
- to use the Huygens approach, modify amplitude of wavefront as a
function of q:
Calculating the diffracted wave
spherical waves:
EO 
Es
r
e
i ( kr  t )
Fresnel-Kirchhoff diffraction integral:
EP 
 ikE S
2
e
 i t
 F q 
phase shift
i  e
 i / 2
e
ik ( r  r  )
rr 
dA
obliquity factor
F (q ) 
1  cos q
2
In general, not an easy task. Lucky for you, Fresnel made it simpler.
Diffraction dudes: Fresnel and Fraunhofer
Contemporaries, but not collaborators (nor competitors).
Fresnel vs. Fraunhofer diffraction
S
P
Fraunhofer:
both incident and diffracted
waves may be considered to
be planar (i.e. both S and P
are far from the aperture)
Fresnel:
occurs when either S or P are
close enough to the aperture
that wavefront curvature is not
negligible
Fresnel diffraction from an edge
Fresnel diffraction from a slit
Irradiance
irradiance vs. position, just after a slit illuminated by a laser
x1
From Fresnel to Fraunhofer diffraction
Incident
plane wave
Fresnel vs. Fraunhofer diffraction
view from source:
 
h
2
2r

h
view from point of interest:
2
2p
 
near field 
h
2

2r
h
2
2q
 
11 1 2
   h  
2 p q
A
d 

where d represents p or q (=distance
from source or point to aperture)
A is aperture area
Fresnel number, F
source:
Fresnel diffraction occurs when:
F 
h
2
d
 1
point:
Fraunhofer diffraction occurs when:
F 
h
2
d
1
where h = aperture or slit size
 = wavelength
d = distance from the aperture (p or q)
Fresnel diffraction from an array of slits:
the Talbot effect
Screen with
array of slits
Diffraction
patterns
ZT = 2d2/
• one of the few Fresnel diffraction problems that can be solved analytically
• beam pattern alternates between two different fringe patterns
http://www.flickr.com/photos/gaeloso/4256791024/
Rolling out the optical carpet: Talbot effect
http://physicsworld.com/
cws/article/print/133
Getting into the zone
Fresnel’s approach to diffraction from circular aperatures
zone spacing = /2:
r1 = r0 + /2
r2 = r0 + 
r3 = r0 + 3/2
…
rn = r0 + n/2
these are called the
Fresnel zones
Stay in the zone
as we draw a phasor diagram where each zone is subdivided into 15 subzones
5½
4
3
2
1
half-period zones
• obliquity factor shortens successive phasors
• circles do not close, but spiral inwards
• amplitude a1 = A1 : resultant of subzones in 1st half-period zone
• composite amplitude at P from n half-period zones:
An  a1  a 2 e
i
 a3e
i 2
An  a1  a 2  a 3  a 4 e
 a4e
i 3
i 3
 ... a n
 ... a n e
i ( n 1 ) 
Adding up the zones
individual phasors
composite phasors
for large N,
resultant amplitude=
half that of zone 1
Implications of Fresnel zones
For N contributing Fresnel zones,
If N is small,
a1 ~ aN
for odd N,
for even N,
AN ~ a1
AN ~ 0
If N is large (i.e. huge aperture)
aN  0
for any N,
AN ~ ½ a1
strange Implications of Fresnel zones. Part 1
A circular aperture is matched in size
with the first Fresnel zone:
What is amplitude of the wavefront at P?
AP = a1
Now open the aperture wider to also
admit zone 2:
AP ~ 0 !
Now remove aperture, allowing all
zones to contribute:
AP = ½ a1 !!!
(Irradiance only ¼ !)
strange Implications of Fresnel zones. Part 2
A circular disk is matched in size with
the first Fresnel zone:
What is amplitude of the wavefront at P?
• all zones except the first contribute
• first contributing zone is the second
AP = ½ a2
• irradiance at center of shadow
nearly the same as without the disk
present!
How absurd!
Siméon Denis Poisson (1781-1840)
Poisson/Arago spot
The Fresnel zone plate
16 zones
An  a1  a 2  a 3  a 4 e
i 3
 ... a n
If the 2nd, 4th, 6th, etc. zones are blocked, then:
A16  a1  a 3  a 5  a 7  a 9  a11  a13  a15
Amplitude at P is 16 times the amplitude of a1 /2
Irradiance at P is (16)2 times!
An alternative to blocking zones
Fresnel vs. plano-convex
lens
lens
phases of adjacent Fresnel zones changed by 
Kewaunee, Wisconsin
Fresnel lighthouse lens
other applications: overhead projectors
automobile headlights
solar collectors
traffic lights
Fresnel’s treatment of straight edges
cylindrical wavefronts diffracted by rectangular aperture:
consider source
to be a slit
zones are now rectangular strips
edge view:
Fresnel’s treatment of straight edges
• again, zone spacing = ½ 
• adding the phasors gives the
endpoints of a Cornu spiral
Phasors trace a Cornu spiral
2 zones:
• areas of Fresnel strip zones
decrease rapidly with n
• successive phasor
amplitudes of zones are
much shorter—half circle
never reached
• phasors continue to spiral to
limit point E (eye)
• zones of lower half produce
twin spiral in 3rd quadrant
Applications of the Cornu spiral
straight edge:
wire:
Other Cornu spirals
Cornucopia
Cornu spiral pendant
$34.99
Roman Cornu horn
http://www.mboot.net
Cornu aspersum, garden snail
Exercises
You are encouraged to
solve all problems in the
textbook (Pedrotti3).
The following may be
covered in the werkcollege
on 13 October 2010:
Chapter 13:
1, 2, 3, 5, 6, 7, 10, 18