Optics and Microscopy II
Graph of Fourier Optics resolution (Goodman) removed due to copyright restrictions.
Two of tee mast hpmbmt and difbxlt to
aspects of'an
o l p i i d micmscoge are its Wtyta gmaak cwlmst d it albjlity
~ ~ 1 T R f i e R ~ s
What is rewllltim? RemMrnrkflues how hW E call !M!E
Two objects ane:dish-ble
if*
amtm are sep&
thm thpk fill1 nr%"M
.r) hslf rn~viminn
.- -
Iby
Huygens-Fresnel Principle
IXlTmdwmh ~ d a s a m n r e a ~ e d ~ o f
intakemeefEect- The basic physics oft
k two ~IBE
weYI-
M a l .W h i t e w e ~ ~ t e d i n ~ e a s ~ o r i ~
&mipint smmes (?ike the double slit q-t],
diEmchm
txmidm i x k c k r m of light hM
b
iesize. objects & as an
apertune-
Spherical Wave Solution
~ ~ ~ ~ b e ~ s i l y ~ w h ~ l . i g h t i s ~ ~ i n t o l
dimmiws that are ccmpmbk to its wavelength,- For coherent
light wmoe, like a laser, diflkxtim d k d s can be readily
observed A n ~ ~ l e i s ~ g l ~ l i g h t ~ a n a a r o l w s l i
-
The b-eatmmtof diflhctim efk& skated mthe 170Qs-1800swi
the intmdndkm dthe Huygens-Fmsnel Principle-
The Huygen's principle wl be derived d k d y from the wave
quation assuming the electric W d can k treated PS Scafaz
quantities. It is quite a bit o f w a and I will not go thm@ it
here. B(r,t ) = E,
sin(k,r -a)
Y
anti ck, = o
Diffraction I
Diffracted Light Intensity
Single slit diffraction is a result of the interference of light due to its wave nature
-3l
d
-2l
d
-l
d
Maximum Intensity
p
l
d
Detector Screen
Figure by MIT OCW.
2p
2l
d
3p
3l
d
Diffraction II La's considm the simplest diffraction situation that of a fmite size
slit:
The contribution of clement dy at position P can be calculated
from Hsvgen' s principle:
Diffraction Ill In the far field (Fm~mhoffea)limit, R3:3 D
We cm appr~~ximate
r by R and the distance r can be approximated
as a function of R y and E (usiny
of cosine):
The total field at P is:
Therefore, we have
E=-ED sin[(kD .!2 ) sin 81sia(c~- kR)
R
(kD1)rin8
Diffraction IV
DefutmgP =(kD/2)sinO duIcu]atingimemtemitytPwe
haw:
1 ED2 sinins 2
( 0 )= -(-1
(-1 p = I(0)sin c(P)
2 R
Uatethztinthis;deri~loe,wehavei~mear~~
@rmm!zl difbdm) as mtl as the vectar natwe of the electric
field-
Diffraction V
Courtesy of Paul Padley.
Source: Padley, Paul. Figure 3 in "Diffraction from a Circular Aperture."
Connexions, November 8, 2005, http://cnx.org/content/m13097/1.1/.
Microscopy imaging can be consider as the diffraction from a circular aperture
with a lens for focusing – diffraction results in “broadening” of the focal point.
Fourier Optics I
Recall
from
lecture
1, the interference
two plane
waves:
Recall
the
interference
of two of
plane
waves
r r
r r
r
E1 (r,t ) = E cos(k 1 .r + wt)
r r
r
r r
E 2 (r, t) = E cos(k 2 .r + wt )
r r r r
r
I(r,t ) = 2I + 2I cos(k1 .r - k 2 .r )
In the case where the waves incident symmetrically and looking at the intensity along the
y axis
r
k 1 = k sinqx̂ + k cosqŷ
r
k 2 = k sinqx̂ - k cosqŷ
r
r = yŷ
The intensity has a simple distribution depend on angle q:
r
I(r ,t ) = 2I (1 + cos(2k cosqy))
Note that when angle is zero degree (light wave counter propagating), the highest
2
frequency oscillation is observed at spatial frequency: 2k = 2p ( ) . When the waves are
l
parallel, angle is 90 degree, the spatial frequency is zero (constant intensity light).
Fourier Optics II
Consider two point source at the focal plane of a lens, the light rays
become collimated plane waves after the lens and interference is observed.
Fourier Optics III
What happen when the two sources coincide? Only parallel plane waves
are generated.
Fourier Optics IV
What happen if the point sources are made further apart?
Resolution viewed from Fourier Optics
Light emission from any object in the specimen plane can be
Decomposed into its Fourier components. Which Fourier component
will pass the finite aperture of the objective lens? Low frequencies!
Resolution viewed from Fourier Optics II
NA = n sin(q )
What is the maximum frequency that can be pass? Consider the case
of a very large lens (numerical aperture, NA approach one). The
waves will approach counter propagating and the maximum frequency
is:
k max
2
= 2p ( )
l
Note that maximum spatial frequency is a function of wavelength.
Shorter wavelength implies higher frequency (resolution) imaging.
At a given wavelength, we should expect a resolution of about
l
2
Resolution viewed from Fourier Optics III
Point Spread function
More quantitative analysis shows that:
2p r
2
p a
2p
J 1 (kasin
q
) =
J 1 (
a
) =
J 1 (
r
) =
J 1 (
NAr)
l
l
f
l
f
J 1 (x) = 0 at
x = 3.83 � rmin
0.6l
=
NA
Resolution viewed from Fourier Optics VI
Graph of Fourier Optics resolution (Goodman) removed due to copyright restrictions.