Diffraction through a circular hole
 
I   , z   Io 

4
l
z


2
2
 J1  kD  2 z  
2

  kD  2 z  
 J1  kD 2  
I    I max  2

  kD 2  
Diffraction of single l
2
2
J1 1.22   0
kD
 1.22
2
Imaging diffraction with a lens/mirror
A lens images the far-field I  x , y  on a screen at f from
the lens, but  x , y must be measured from ______
So far-field can be observed easily even if the far-field d is
too far away to be convenient
Imaging diffraction with a lens/mirror
The lens/mirror may be placed anywhere at/after the
aperture.
In fact the lens/mirror diameter itself could be the aperture!
Diffraction limitations on imaging
D
Every star gives the same shape on the screen
 J1  kD 2  
I    I star  2

  kD 2  
2
Diffraction limitations on imaging
Rayleigh limit: Peaks are over the other star’s first zero
J1 1.22   0
kD
 1.22
2
 min
1.22l

D
Diffraction orders
Spectrometer
Put slits at f from mirrors.
Grating in center
Diffraction orders may overlap
Diffraction orders may overlap
Have to use short-pass or long-pass filters to “clean out”
overlapping parts.
Gratings cut by an “engine”
Typical grating
for visible to UV
light: 1000-2000
grooves/mm, 10
cm wide105
lines.
Modern methods
Blazed gratings put
more light into a
nonzero order
Resolving power of a grating
How much must l change so peak moves by half its width?
 h 
sin  N

l 


2 a
I    I peak sinc 

 l  N 2 sin 2   h  


 l 
2
l  peak
RP 

 mN
l

Understand from phases arriving from aperture
We consider now only the center of the screen I(0,0,d)
How does the phase of the light
from general x’ differ from that
coming from x’=0?
For small enough d (large
enough a), it can differ by  to
many  (Fresnel regime).
E  x, y, z   
ikz
ie e
i

k 2 2
x y
2z
lz
E  0, 0, d 


E  x, y, 0  e
i

k
x 2  y  2
2z

e
i
k
 xx yy
z
aperture

E  x, y, 0  e
i

k
x 2  y 2
2z

dxdy
aperture
The phasor diagram
approximates this integral
dxdy 
Fresnel zones
Light from the first “zone” in the aperture all tends to
make the screen brighter at (0,0,d)
Fresnel zones on a wavefront
Phasor addition from Fresnel zones on aperture
In our minds, we color aperture
areas white and black when we
cross these  boundaries.
Fresnel zones for a slit
Resultant diffraction
Intensity on screen
Aperture zones
d,a are such that the above
aperture is 9 zones wide:
Bright at center
Fresnel zones for circular apertures
circular aperture zones have
almost identical areas
d,a are such that the above
aperture is filled with 9 zones:
Bright at center
Resultant diffraction
Intensity on screen
If we move the screen back, there
are ________ circular zones in the
same aperture
a) more b) fewer
Eventually, if d grows or a shrinks,
we get less than one zone in the
aperture. This is the Fraunhofer
regime
After that, nothing much changes
for bigger d or smaller a.
Show this occurs when d > a2/ l
Draw the phasor on the spiral that shows the E-field
with no obstacle there ( zones open..represents the
incoming light).
A. I got it mostly right
B. I got it mostly wrong but I tried
Poisson’s spot in shadow of ball bearing
Suppose a ball blocks zones 1-3.
Show the beginning and ending points
on the phasor diagram of the light that
reaches center spot on the screen
Estimate I at center of screen in terms
of the incident Io.
A. I got it mostly right
B. I got it mostly wrong but I tried
Center is always bright, and
is about as bright as the
incoming light!
Zone plates used to focus
ultrasonic waves
Suppose you make a zone
plate that lets only zones
2, 5 and 8 through.
Find the strength of I at center of
screen in terms of Io that hits the
plates.
Fresnel zones are widely used in antenna-receiver tech
L+l
L+l/2
L
No aperture: The zones here have to do with the phase of
reflections from other objects
How high to raise an antenna from the ground?
If the ground reflects zone 2, you may do better lowering
the antenna