Transcript PHYSICS 2310

Phys 2310Fri. Nov. 18, 2011
Today’s Topics
• Begin Chapter 10: Diffraction
• Reading for Next Time
1
Reading this Week
By Fri.:
Begin Ch. 10 (10.1 – 10.3) General Considerations,
Fraunhofer Diffraction, Fresnel Difraction
2
Homework this Week
Chapter 9 Homework
Chapter 9: #7, 8, 10, 14, 26 (due Monday Nov. 28)
Chapter 10: #8, 25, 30, 33 (due Friday,Dec. 1)
3
Chapter 10: Diffraction
• General Considerations
– There is no real physical distinction between diffraction and interference
– Huygens-Fresnel Principle
• Modification to Huygens Principle:
Every unobstructed point of a wave-front, at any given instant, serves as a source for
spherical, secondary wavelets (with the same frequency as that of the primary
wave). The amplitude of the optical field at any point beyond is the superposition
of all these wavelets (considering both their amplitudes and phases). This is really
a quantum mechanical effect since there is otherwise no physical explanation.
– Opaque Obstructions
• When an aperture is large compared to l, the effects of the boundary is minimal
since the fraction of waves affected is small and vice versa.
• Formally, diffraction occurs as a result of the boundary conditions for Maxwell’s
equations. The math is difficult but is solvable for a few special cases.
• Aperture: imagine a set of fictitious, non-interacting oscillators distributed over
the opening. Electrons at the edge interact with the E-field and dampen the EM
wave.
– Fraunhofer and Fresnel Diffraction
• Fresnel diffraction: when the distance between aperture and screen/detector is
small. Shape of wavefront is important.
• Fraunhofer diffraction: when the distance between aperture and screen/detector
is large (R > a2/l). This allows the assumption of plane waves.
4
Chapter 10: Diffraction
• General Considerations
– Light from Several Coherent Oscillators
• E-field will add according to amplitude and phase:
~
E  E 0 (r )e i ( kr1 t )  E 0 (r )e i ( kr2 t )  E 0 (r )e i ( kr3 t )    E 0 (r )e i ( krN t ) or :
~
E  E 0 (r )e ikr1 e it  [1  e ik ( r2  r1 )  e ik ( r3  r1 )  e ik ( r4  r1 )   e ik ( rN  r1 ) ]
but thephasedifferencearising fromadjacent oscillators is simply:
  k (r2  r1 ), 2  k (r3  r1 ), et c., where  kd sin  .
T hus theresultingfield is :
~
E  E 0 (r )e ikr1 e it  [1  (e i )  (e i ) 2  (e i ) 3  (e i ) 4   (e i ) N 1 ]
T hequantityin [] is a gemoetricseries :
 sinN/2 
[]  (e iN  1) /(e i  1)  e i ( N 1) / 2 
 thusif r is thedistancefromcent er toP :
 sin/2 
~
 sinN/2 
E  E 0 (r )e it e i[ kr1  ( N 1) / 2 ] 
 and so theintensityis thus:
 sin/2 
sin 2 ( N / 2)
I  I0
but for   2m we havemaximaat :
sin 2 ( N / 2)
d sin  m  ml
5
Chapter 10: Diffraction
• Fraunhofer Diffraction
– Small angle: r ~ R – ysin ~ R y
– Example of a single slit
• Position of maxima depend on l
Consider a slit element(ds) at theorigin.T hefield at some point P on a screen is :
E
sin(t  kr)dy where r is theactualdistanceof P fromeach element
R
(dy) and R is thedistanceof theorigin (midpoint)from thescreen.Expandingr in termsof
dE 
R and y :
r  R  y sin   ( y 2 / 2 R) cos2    (Fraunhofer : only first twoterms).
T husintegrating along theslit gives the totalfield at P :
E
EL
R
D/2
 sin[ωt-k(R-ysin θ)]dy or :
D / 2
E
E L sin[(kD / 2) sin 
sin(t  kR) but if   (kD / 2) sin  and k  2/l then:
R (kd / 2) sin 
E
E L D  sin  

 sin(t  kR) but intensityis thesquare of theamplitude:
R   
1E D
I ( )   L 
2 R 
2
2
 sin  

 since sin 2 (t  kR)  1 / 2. Simplifying :
  
2
 sin  
 which is thesinc function:
I ( )  I (0)
  
I ( )  I (0) sin c 2  and it can be differentiated to find themaximaand minima(eq.10.19).
6
Chapter 10: Diffraction
• Fraunhofer Diffraction from2 Slits
– In this case the E-field is the sum of that from each slit:
a b / 2
b/2
E C

F ( z )dz  C
b / 2
 F ( z )dz
where F ( z )  sin[t  kR  2 ]
a b / 2
Integration yields :
 sin  
[sin(t  kR)  sin(t  kR  2 )] with   (ka / 2).
E  bC
  
Whensimplifiedand squared theintensitybecomes:
 sin 2  
 cos2 
I ( )  4 I 0 
  
Note themodulationof thecos (interference) termby thesinc (diffraction) term.
7
Chapter 10: Diffraction
• Diffraction by Many Slits
– Now generalize to N slits:
E C
b/2
a b / 2
2 a b / 2
3a b / 2
b / 2
a b / 2
2 a b / 2
3 a b / 2
 F ( z )dz  C
Na  b / 2
 C
 F ( z )dz
 F ( z )dz  C
 F ( z )dz  C
 F ( z )dz
with theapproximation r  R  z sin  the j - th termis :
Na b / 2
C
[sin(t  kR) sin(kz sin  )  cos(t  kR) cos(kz sin  )] jajabb 2/ 2
k sin 
which can be sinplified to :
Ej 
 sin  
 sin(t  kR  2j ) and upon evaluationof thegeometricseries (as before):
E j  bC
  
 sin   sin N 

E  bC
 sin[t  kR  ( N  1) ] and so theintensityis :
   sin  
2
 sin    sin N 
 
I ( )  I 0 
 with maximaat   0,   ,2 ,
    sin  
Notefrom thefigure thatas N increasestheindividual maximaget brighterand moredistinct.
2
8
Chapter 10: Diffraction
•
Diffraction Grating
– For a slit separation of a the location of each
order is:
– a sinm = ml (note l dependence)
Thus white-light produces a spectrum at each order
(m).
The angular dispersion can be computed via
differentiation.
A diamond is used to cut groves into glass and
aluminized to act as little mirrors.
It can be used in reflection without aluminizing
– Grating spectroscopy:
• A collimating lens can be used to produce an input
beam with i = constant. The grating equation:
a(sinm – sini) = ml
• The dispersed light from the grating can be
imaged using a lens acting as a camera. Thus the
camera “sees” light entering at different field
angles according to wavelength. Resulting image is
a series of “slit images” displaced according to
wavelength.
• See text for application examples.
9
Chapter 10: Diffraction
• Diffraction from a Square
Aperture
– Similar to a single slit but we
now integrate in 2-d
Using complexnotation he
t disturbance at P is given by expressingthe
contribution fromeach differential element(as before)but now integrating
over twodimensions. Fortunately, we can just split theintegralinto two
parts,adding theeffectof of the waveletsin theverticaland hroizontal
directions. Specifically thedifferential disturbance producedby distribution of
waveletsover a surface (S), with  A as thesource strengthper unit area,is :
 
dE   A e i (t  kr ) dS (thegeneralcase)
 r 
We next approximate r as :
r  R[1  2(Yy  Zz ) / R 2 ] (see textpg. 464 for a justification).We now
consider the specificcase of a retangularaperture(see figure).T hus,
substituting and factoringout theR - termin common:
E
 A e i (t  kR)
R
b/2
a/2
b / 2
a / 2
ikYy / R
 e dy
e
ikZz / R
dz
If we let  '  kbY / 2 R and  '  kaZ / 2 R we have:
E
 A e i (t  kR)  sin  '  sin  ' 
 and thus theintensitybecomes:


R
  '   ' 
2
 sin  '   sin  ' 

I ( )  I 0 
 
 '   ' 
Noticethat theformof theintensitypatternis theproduct of two sinc functions,
one in each dimension.
2
10
Chapter 10: Diffraction
• Diffraction from a Circular Aperture
– Similar but we integrate over the aperture in azimuth angle and r.
Similar tothesingle slit or retangularaperturewe integratethedifferential,
complexformfor thedisturbance at P but here we use sphericalcoordinates :
E
 A e i (t  kR)
R
a
2
 e
i ( kq / R ) cos(   )
dd
 0  0
T heintegralover is known as theBessel Function( J n ) and can be
evaluatednumerically viaseries expansion.See any textbo
ok on differential
equationsas theyare a well - known solution whereaxialsymmetryis involved.
Specifically :
J m (u ) 
E
i m
2
2
e
d and so thesolution:
0
 A e i (t  kR)
R
iu cos
2πa 2(R/kaq)J1 (kaq / R) and thus:
2
I ( ) 
2 2 A A 2  J 1 (kaq / R) 

 or :
R 2  kaq / R 
 2 J (ka sin  ) 
I ( )  I 0  1

 ka sin  
2
11
Chapter 10: Diffraction
• Implications of Diffraction in
Optical Systems
– Diffraction Limits the
Resolution of Optical Systems
• The larger the aperture the
smaller the core and the “Airy
Rings”
• The larger the wavelength the
larger the core and the “Airy
Rings”
• Regardless of magnification the
resolution of given aperture is
limited.
– Difficult for Hubble to see
planets around nearby stars.
min = 1.22 l/D
12
Chapter 10: Diffraction
• Fresnel Diffraction
– When either the screen or the source is at a finite distance plane waves are
insufficient and Huygens wavelets are located along a curved surface. The
math is much more complicated since we must specify the directionality of the
secondary wavelets from the source itself [K() = ½(1+cos)]. The secondary
wavelets around an annulus are in phase relative to the source and hence the
phase at P is t-k(+r).
T hus thedisturbance at point P for a source strength A will be :
A
cost  k (   r )dS
r
Consider the skinny triangle formedby theannulus and rays to point P .
T helaw of cosinesgives :
dE  K
r 2   2  (   r0 ) 2  2  (   r0 ) cos which we differentiate to give :
2rdr  2  (   r0 ) sin d but since thearea of annulus is :
dS  d 2 (  sin  ) we can substitutefor d :
dS  2

(   r0 )
El  K l 2
rdr and so thedisturbance from thel - th annulus is :
 A
cos[t  k (   r0 )]dr which is :
(   r0 ) r
rl
l 1
 Kl A  l
sin(t  k  kr)rr rrll 1
El 
(   r0 )
Since rl 1  r0  (l - 1)l/2 and rl  r0  ll/2 thisreduces to :
El  (1) l 1
2K l  A  l
sin(t  k (   r0 )
(   r0 )
A geometricseries or phasoraddition can be used to computetheresult (pg. 488- 490)
13
Fresnel’s Half-Period Zones
• Fresnel’s Approach
– Consider a series of zones s1,
s2, s3 … around point O, each
l/2 further from point P
– Consider phase difference D
For small angles we have:
s2 s2
ab


s
and
2a 2b
2ab
l
ab
m  s m2
so area and intensit yare :
2
2ab
S m   ( s m2  s m2 1 ) and so
l 2ab
a
bl
2 ab ab
But everyhalf - period theamplitudeinverts:
S m  Am  

A  A1  A2  A3  A4    (1) m 1 Am
and amplitudedecreases wit h distance(d m ) :
Am  C
Sm
(1  cos )
dm
Expandingseries and grouping yields :
A1 Am

(if m is odd) and
2
2
A A
A  1  m (if m is even)
2
2
A
14
Fresnel Diffraction from a Circular
Aperture
• Imagine zones within a circular
aperture
– If radius corresponds to the outer
edge of first half-period zone: A =
2Ainf and intensity is 4x higher.
– If radius corresponds to the outer
edge of second half-period zone:
A = A1 – A2 = 0!
– Intensity drops even though hole is
bigger!
– Increasing hole size further results
in periodic maxima and minima.
15
Fresnel Zone Plates
• Alternately blocking either
the even or odd zones using
a mask (right).
– Configured for a specific
source.
• Result is a lens that will
image a distant source!
– “Focal length” will be:
s m2
s12
f 

ml l
16
Fresnel Diffraction from a Circular
Aperture
• Now let’s add the amplitudes
from small circular zones
within an aperture
• Divide each half period zone
into 8 subzones:
1l
2l
l
, b
,  b
82
82
2
First amplitudeis a1 at right, thenadd a 2 , etc.
d b
After8 subzones we get vectorAB
Repeat for the8 subzones in next half - period
and theresult is vectorCD and when added
produces vectorAD (almostzero amplitude).
Adding successive zonesproducesleft hand
figure. T he" vibrationcurve"for a circular aperture.
Withinfinitesimal subzones we get theright hand
figure.
17
Fresnel Diffraction from a Single Slit
• Now consider a slit.
– Half-period zones on
cylindrical wavefront:
2l
,
2
2
Strip dividion of the wavefront.
Similar tocircular case but now area
b, b 
l
, b
is proportion
al to width so larger
obliquity factor.Divissioninto9 parts
now yields half - periodand A1  OB.
Second strip yields A 2  BC with A 2  A1 .
Continuingresultsin thespiralconverging
to OZ. Infinitesimal stripsproduce the
right hand figure. Vectoris a phasor.
T headdition of theamplitudesfrom the
lower half of theslit also producesa
phaselag and a similar but invertedspiral.
Quantitatively :
2
 ( a  b)



s 2
l
abl
2
Wherewe introducea new variable:
 s
2(a  b)
abl
18
Chapter 10: Fresnel Integrals
• Fresnel Integrals
– Derivation of Fresnel
integrals:
For theX and y coords.on Cornu's sprial :
dx  d cos  cos
dy  d sin   sin

x   cos
0
 2
2
 2
2
 2
2
d and
d and so :

d and y   sin
0
 2
2
d
T heycannotbe integratedin closed form
but can be numerically evaluated.
19
Chapter 10: Fresnel Diffraction from Edge
• Fresnel Diffraction from Edge
– Start at point P where the
amplitude is OZ
– Along the screen the verctor
head remains fixed an the tail
moves along the spiral to a
maximum at b’.
– Continuing along the screen
the amplitude goes through B’
and a minimum at c’.
– Secondary maxima (fringes)
occur at d’.
– Ampiltude decreases and
converges to OZ’.
– Scale of the pattern:
Let distancesa  b  100 cm, and  500 nm..
So thedistancealong the wavefrontis :
s 
l
abl
 0.0354 and along thescreen :
2( a  b )
ab
bl (a  b)
s 
 0.0708 cm
a
2a
20
Chapter 10: Fresnel Diffraction from Slit
• Fresnel Diffraction from Slit
– Procedure is similar but each
side acts as an opaque edge.
– For a = 100cm, b = 400cm, l =
400 nm, and s = 0.02 cm, 
= 0.5
– Intensity at P’ found by
drawing vector  = 0.5 at
different positions along the
spiral and measuring the
corresponding amplitude (A
and A’)
21
Chapter 10: Fresnel Diffraction
• Fresnel Diffraction from Rectangular (slit) Aperture
– See Hecht sec. 10.3.8 for alternative explanation of how to
use the spiral to get the complex amplitude given the silt
width (positions on the spiral).
22
Chapter 10: Fresnel Diffraction
• Fresnel Diffraction from Semi-opaque Aperture
23
Reading this Week
By Fri.:
Begin Ch. 10 (10.1 – 10.3) General Considerations,
Fraunhofer Diffraction, Fresnel Difraction
24
Homework this Week
Chapter 9 Homework
Chapter 9: #7, 8, 10, 14, 26 (due Monday Nov. 28)
Chapter 10: #8, 25, 30, 33 (due Friday,Dec. 1)
25