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Fundamental of Optical Engineering
Lecture 9

The amount of light reflected when a beam
moves from one media to another can be
reduced by placing a thin coating layer
between them.
2
R  A122  A23
 2 A12 A23 cos 

A12A23 > 0 and we want Rmin.  cos = -1.
n2  n1n3
  (2 N  1) 
t2 
 2 N  1 
4 n2 t2

4n2
N  0: Very thin film
t2 

4n2

n1 = 1.5, n3 = 1.7. What should be n2 for
antireflection film?

Find the thinnest film to be coated to prevent
the reflected light give n1 = 1 and n3 = 3.6 if
λ=0.83μm.
D
qi
qi
qi
n1
t
A
C
qt
E
F
qt
n2
t
qt
B


Consider the case of non normal incidence as
shown in the previous figure.
The emerging beam travels with an opticalpath difference between them as
  n2  AB  BC   n1 ( AD)
  n2  AE  FC   n1 ( AD)   n2  EB  BF 

By Snell’s law, AE  AG sin qt  

and AD  AC sin qi , this yields
 n0
2 AE  AD 
 nf


AC 
 sin qt
2 



Then we have
n0 AD  2n f AE  n f  AE  FC 

So that, an optical-path difference is
  n f  EB  BF   2n f EB

As EB = tcosqt , finally, we have
  2n f t cos qt

Therefore, a round trip phase shift in this
case equals to
  2 

4 n2t cos qt

Therefore,
 4 n2t cos qt 
R  A  A  2 A12 A23 cos 




2
12
2
23

Consider a film of thickness t and refractive index 1.6
sandwiched between two media of refractive index 1.5.
◦ (a) determine all values of t for which the reflectance will be
a maximum at normal incidence for λ = 1 μm and calculate
the reflectance.
◦ (b) For an angle of incidence of 20 relative to the normal,
calculate the wavelength at which the reflectance will
maximum. Use the smallest value of t determined in (a).
◦ (c) Calculate the reflectance for both s- and p-polarization
for the case considered in (b).


These are instruments which utilize coherent
summation of wave amplitudes.
Two beam interferometer:
  A1ei  A2 ei
1
2
PA  A12  A22  2 A1 A2 cos 1  2 
Ax2  A12  A22  2 A1 A2 cos 1  2  BSx 
Az2  A12  A22  2 A1 A2 cos 1  2  BSz 
BSx ( z )  phase shift due to second baeam splitter
for light leaving from splitting in x- or z-direction.


In general, BSx =  + BSz
Assume they are lossless beam splitters.
Ax2  A12  A22  2 A1 A2 cos 1  2  BSx 

Ax2  Az2  2 A12  A22


For 50:50 beam splitter.
Pin
PAx 
2
Pin
PAz 
2
1  cos 1  2  BSx  
1  cos 1  2  BSz  
1 
2 n1 L1

1  2 
, 2 
2 n2 L2

2  n1 L1  n2 L2 


Suppose in a MZ interferometer for λ = 0.6328 μm, PAx = 0
and PAz = Pin. Then, a microscope slide 2 mm thick with a
reflective index of 1.55 is placed in one arm of the
interferometer. What are the new values of Pax and Paz.
Pin
1  cos 1  2 
Pout 
2
1 
4 n1 L1

1  2 
, 2 
4 n2 L2

4  n1 L1  n2 L2 

 1  2   2 for L1 

2n1

For a Michelson interferometer in air with λ = 1.06 μm, Pout =
0.5 Pin. One of the mirrors is displaced by increasing L1
continuously and Pout increases continuously to a final value
of 0.65 Pin. How large is the displacement?

After one round trip
A  Ain 1  R R Re  ei  Ain 1  R
  optical loss coefficient
  round trip phase shift 

A  Ain 1  R 1  Re  ei


4 nL

After 2 round trips

A  Ain 1  R 1  Re  ei  R 2 e 2 e 2i


After n round trips
N
A  Ain 1  R  a n
n0
a  Re  ei

Steady state (N )
N
1
a 

1 a
n 0
n
A
Ain 1  R
1 a
2
P A 
Ain
2
1  R 
1 a
2


1  a  1  Re cos 
2
   Re
2

sin 

2
 1  R 2 e 2 cos 2   2 Re  cos   R 2 e 2 sin 2 
1  a  1  R 2 e 2  2 R cos  e 
2

Therefore,
Pout 
Pin 1  R 
2
1  R 2 e 2  2 Re  cos 2 
max Pout for   2 N 
min Pout for    2 N  1 

If  = 0 (lossless resonator), e- = 1
 Pout max 
 Pout min 
Pin 1  R 
2
1  R  2R
2
Pin 1  R 
2
1  R  2R
2
 Pin

Pin 1  R 
1  R 
2
2
Light from a laser of wavelength λ is transmitted through a
lossless Fabry-Perot interferometer in air. The mirror
reflectances are equal to R. As the mirror separation is
increased from an initial value, the transmitted power
increases to a maximum of 21 mW for a mirror separation D.
As the mirror separation is further increased D+0.25 μm, the
transmitted power decreases to a minimum of 0.3 mW.
(a) What is λ in μm?
(b) What is R?
(c) What is the transmitted power when the mirror separation
is D + 0.99 μm?

Soln