Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Engineering Optics and Optical Techniques
Lecture Note No. 7.
Interferometer (Chapter 9)
Fraunhofer Diffraction (Chapter 10)
SUPERPOSITION OF PLANE WAVES (Sections 9.1, 9.2)


E1 r ,t   E o1 cosk1  r  t   1 


E 2 r ,t   E o 2 cosk 2  r  t   2 
E  E1  E2 , whose intensity is equal to the ensemble average of E 2 :







I  E 2  E1  E2  E1  E2



 
 
 
E2 E2
E12  E 22  2 E1  E2  o1  o 2  2 E1  E2  I1  I 2  2 E1  E 2
2
2
and the trigonometric identity (p.387) shows that
 
1  
1  
E1  E 2  E o1  E o 2 cosk1  r   1  k 2  r   2   Eo1  E o 2 cos 
2
2
where  represents the phase differential.
If E o1  E o 2    / 2, I  I1  I 2 , i.e., the flux density remains constant.
Furthermore, if I 1  I 2  I o ,
I  I1  I 2  2I o .
 
If Eo1  Eo 2   0 , I  I1  I 2  Eo1  Eo 2 cos   I1  I 2  2 I1 I 2 cos 
Furthermore, if I 1  I 2  I o ,
I  2 I o 1  cos    4 I o cos 2

2
I max  4I o at   0,  2 ,  4 , …
I min  0 at     ,  3 , …
*Condition of Interference:
Optical Path Differential (OPD) < coherence length l c
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Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Young’s Fringes … Wave Front Splitting Interferometry (Sec. 9.3)
I  4 I o cos 2

with   k1 r1   1  k 2 r2   2  k r1  r2   ka sin  
2
2 ay
 s
(  1   2 for a single coherent source)
( r1  r2   lc for the interferometry condition)
I  4 I o cos 2
ay
s
Locations of the brightest fringes are for  
y max 
ay
 0,   ,  2 ,...,  m :
s
ms
, m  0,  1,  2, ...
a
*If OPD r1  r2  > l c , the contrast of the fringes will degrade or vanish.
2
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Fizeau Fringes … Amplitude Splitting Interferometry (Sec. 9.4)
Eir = Internal Reflection
E2r = External Reflection
Eir and E2r constitute a relative phase shift of  (Fig. 4.44)
1

OPD = 2 nfd =  m   o for maximum or constructive interference.
2

1

Thus, d m   m   , m  0, 1, 2, 3,...
2 2

And d m 

4
for m = 0 [Criterion for optical flat:  
*Example of a soap bubble:
3

… no random fringes]
4
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
OPD of Dielectric Coating or Thin Liquid Film
OPD = n1 (AB + BC) – no AD, Snell’s law: no sin  i  n1 sin  t
AB = BC =
d
cos  t
AD = AC sin  i  AC
Therefore, OPD =

n1
n
sin  t  2d tan  t 1 sin  t
no
no

2n1d
1  sin 2  t  2n1d cos t
cos t
Phase differential,  
OPD
o
 2  OPD  k o 
4n1d
o
cos  t 
4d

cos  t ~
With    , 3  , 5 ... for destructive interference, i.e., d 
Non-reflection coating thickness: o / 4ncoating
4
m aterial
 3
,
4 4
4d

5
4
;
 , ...
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
THIN FILM COATING (Sec. 9.7) More rigorous analysis
For E Plane-of-Incidence, the boundary conditions constituting a conservation of tangential
E
component of E-field and H-field give:
[Goal: rI  rI  0 for non-reflection]
EiI
'
E I  EiI  E rI  EtI  E rII
At boundary I:


'
H I  H iI  H rI  cos  iI  H tI  H rII
cos  iII



*
o

EiI  E rI no cos  iI  o EtI  E rII' n1 cos  iII
o
o

1


,   o ,  H 
*  B  E   E  H , n 
c
o


At boundary II:
E II  EiII  ErII  EtII
H II 
o

EiII  ErII n1 cos iII  o EtII ns cos tII
o
o
The phase differential for the traverse through the film once is given as:
  OPDI II  k o =
so that
 o 
En
 o 
k o 2n1d cos  iII 
 k o h [1/2 of the OPD on p.4]
2
EiII  EtI e i  EtI e iko h
and
5
'
ErII
 ErII e iko h
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Combining ALL the above equations shown on p. 5 gives:
E I  E II cos k o h 
H I  E II
H II i sin k o h
o
n1 cos  iII
o
o
n1 cos  iII i sin k o h  H II cos k o h
o
In matrix form (for E-field perpendicular to the plane-of-incidence)

cos k o h

 EI  
H   
 I 
i sin k h   o n cos 
o
1
iII

o


o
n1 cos  iII   E II 
E 

o
  I  II 

  H II 
 H II 

cos k o h

i sin k o h
*  I : Characteristic Matrix that relates the fields at the two adjacent boundaries
Similarly, for E-field parallel to the plane-of-incidence,

cos k o h

 EI  
H   
 I 
i sin k h   o n / cos 
o
1
iII

o


o
n1 / cos  iII   E II 
E 

o
  I  II 

  H II 
 H II 

cos k o h

i sin k o h
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Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Antireflection Coatings
Note that the characteristic matrix for antireflection coating condition is expressed for both E

and E  for  iI   iII   tII  0 and k o h  :
2

0


I 

i   o n1

o

 o 
n1
o 

0 

i
MASAGING ALL above equations, together with  iI   iII   tII  0 ,




E rI n1 no  ns  cos k o h  i no ns  n12 sin k o h
r1 

EiI n1 no  ns  cos k o h  i no ns  n12 sin k o h
With k o h 

, which is equivalent to d 
 film
2
4
the reflectance at the boundary I is given as:
R1  r r
*
1 1
n n

n n
o
o
s
s
 n12


2
and
2 2
1
n
(called a “quarter-wave antireflection coating”),
R1  0 when n12  no n s
*For example, no  1, ns  1.80 (flint glass in Table 4.1), the coating material must have
n1  1.342 and MgF2 ( n1  1.38 ) may be selected to use. The coating thickness is calculated as:
d
o for d  587.5618 nm from Table 6.1
4n1
7
= 106.44235 nm
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
For a double-layer of n1 and n 2 ,
   I M II

0




i   o n1

o

0
 o  
n1
o 


0  i  o n2
 
o
i

 o 
n2
 n / n
o    2 1
  0
0 

i

 n1 / n2 
0
Also, the reflectance is given as,
 n 2 n  ns n12 
R2   22 o
2 
 n2 no  ns n1 
2
and
R2  0
n
when  2
 n1
2

n
  s
no

Note that n2  n1 .
The layers are referred as s (substrate: g-l-s)-H-L-a (air).
 For underwater lens, the coating thickness (d) and materials (n1, n2) must be different
because of different no.
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Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
OPTICAL DIFFRACTION (Sections 10.1 and 10.2)
R: smaller of S and P
If R 
a2

… Fraunhofer diffraction (Far-field); similar diffraction patterns
with increasing distance (Today)
If R 
a2

… Fresnel diffraction (Near-field); varying shape of diffraction
patterns with varying distance – (Lecture Note 9)
9
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Coherent Line Source [the same  , and the same polarization]
Ei 
L
ri
sin t  kri yi
 L : Source strength per unit length

sin t  kr
dy  L
D / 2
r
R
E  L 
D/2
D/2
D / 2
sin t  krdy
10
for R >> y
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
SINGLE SLIT (slit width b) … R >> b and b << 1
E
L
R 
b/2
b / 2
sin t  krdy
and substituting r  R  y sin  (This is also called a Fraunhofer condition)
Carrying out the integration gives
E
I   
 oc
2
 L b  sin  

 sin t  kR ,
R   
E
2
 E
2
1 b
  L 
2 R 
2

2
kb
b
sin  
sin 
2

2
 sin  
 sin  

  I 0
  I o sin c 2  (Table A-1)






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Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
DOUBLE SLIT (slit width of b and separation of a)
E
L
R
b/2
b / 2
sin t  k R  z sin  dz 
L
R
sin t  k R  z sin  dz
a b / 2
a b / 2
Carrying out the integration gives,
 sin 2  
 cos 2 
I    4 I o 
2
  
with  
kb
ka
sin  and  
sin 
2
2
ya
… Young’s fringes
s
If b  0,
I    4 I o cos 2   4 I o cos 2
If a  0 ,
 sin 2  
 sin 2  


 … single-slit of width 2b


I    4 I o 

I
0
2
2

  
  
FOR N-EQUALLY SPACED SLITS:
 sin 2 
I    I o 
2
 
 sin N  2

 sin  

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Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Arbitrary Shape Aperture
R >>, and r >> aperture dimension
dE 

A
r
e i t kr  dS 
A
R
e i t kr  dS … (1)
  

r  X 2  Y  y   Z  z   R 1  y 2  z 2 / R 2  2Yy  Zz  / R 2
2
2

since R  X 2  Y 2  Z 2

 R1   

 
 f ' ' 0  ...  R1   ]
2!
2


Combining (1) and (2), and integrating w.r.t. the total aperture area will give,
 e
ik Yy  Zz  / R
Aperture
13
dS
 f  
1/ 2
r  R 1  Yy  Zz  / R 2 … (2)
~  A e i t kR 
E
R
1/ 2

Taylor series expansion of r gives, [ f    f 0    f ' 0 

1/ 2
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
CIRCULAR APERTURE (radius a)
z   cos  , y   sin 
Z  q cos , Y  q sin , dS  dd
Then,
i t  kR 
~  Ae
E
R
2
a
 
0
0
~  e i t kR 
E A
R
+
 e
ik Yy  Zz  / R
Aperture
e i kq / R  cos   dd
~  A e i t kR 
Integrate it for   0 (axi-symmetry), E 
2a 2 R / kaqJ 1 kaq / R 
R
1 ~ ~ * 2 A2 A 2  J 1 kaq / R 
Since I  EE 


2
R 2  kaq / R 
 2 J ka sin  
I    I 0 1

 ka sin  
2
2
… Airy function,
J1(x) in Table 10.1
I 0 
 A2 A 2
2R
2
,
and
14
q
 sin 
R
dS
Engineering Optics and Optical Techniques- 2007 Spring
Lecture note No. 7 by Professor Kenneth D. Kihm
Homework Assignment #8
E-o-C Problems:
Ch.9:
Ch. 10:
3, 5, 27, 32, 47
7, 8, 22
*Special Assignment: Plot the three-dimensional Airy pattern
(Fig. 10.23) using any computer plotting software
Due: March 29 (Thursday), 2006
15