Laser lecture 04

14/06/2015
1
6/14/20151
412 PHYS
Lasers and their Applications
Department of Physics
Faculty of Science
Jazan University
KSA
Propagation of Gaussian Beams
& Optical Resonators
Lecure-4
Propagation of laser beams cab be described by the Helmholtz equation
one of the possible solutions to this equation is given by
2 2
0
( )
( , , ,) exp { ( ) }
2 ( )
k x y
x y z i p z
q z
 

  
0 Constant depends on the beam amplitude and can be determined by
boundary conditions
zki
ezyxzyxU 
 ),,(),,( 0),,()( 22
 zyxUk

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2
( ) ln(1 / )op z i z q  )(zp
( )q z complex beam parameter, given as
2
0
0 0( )
i w
q z z iz z q z


     
Complex function given as
Rayleigh range at which he beam waist is given by0z 0( ) 2w z w
0w Is the minimum beam waist
For a spherical wave
0
2 2 2 2 2
0 0 0
1 1 1
( ) ( )
zz
i i
q z iz z z z z R z n w z


    
  
The real part represents the wavefront curvature, with a radius given
by
2
0( ) [1 ( / ) ]R z z z z 
)(z Beam waist radius, given by
2 1/2
0 0( ) [1 ( / ) ]w z w z z 
0w Is the minimum beam waist radius, given in terms of the beam
divergence as
1/2
0
0
log(2)
2 tan( / 2)
z
w
 
  
  
    
   

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Laser resonators
 The feedback in lasers is achieved by placing the amplifier (active medium)
between mirrors, a construction we call an optical cavity or resonator.
The resonator is the space of optical amplifier that contains the feedback elements
The resonator
When the population inversion occurs in the active medium, the spontaneous
emission produces a photon that propagates along the optical axis of the
active medium and the resonator
The photons interacts with the excited atoms and the stimulated emission will
occur and hence a wave with amplified amplitude will propagate through the
medium towards one of the mirrors
Upon reflection from the mirror, the wave will be further amplified by passing
through the medium due to the resonance with the excited atoms (because
they both have same energy )
Eventually the wave will be oscillated between mirrors and get amplified in
every pass and loss some photons in the output mirror as the output beam

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Laser resonator stability
1 2
0 1 1 1L L
r r
       
  
12 r،r Radii of mirror’s
curvatures
L Length of the
resonator
The Condition for resonator stability
• The cavity is an essential part of a laser. It provides the
positive feedback that turns an amplifier into an oscillator.
• The design of the cavity is therefore very important for
the optimal operation of the laser.
Types of resonators
Plane parallel resonator
Confocal resonator
Hemispherical resonator
Large radius resonator
Concentric resonator

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Example:
Sol.
Determine whether or not the following mirror arrangements lead to stability:
a. Two mirrors with radii of curvature of 1.8 m, separated by a distance of 2 m
b. One mirror with radius of curvature of 2m and the other with radius of 3m,
separated by a distance of 2.3m
c. One mirror with radius of curvature of 5m and the other with radius of 3m,
separated by a distance of 4m
d. Two mirrors with radius of curvature of 0.5 m, separated by a distance of 0.5m
1 20 (1 / )(1 / ) 1L r L r   
a. r1=r2=1.8m, L=2m
2 2
1 2(1 / )(1 / ) (1 2 /1.8) (1 1.11) 0.121L r L r        Cavity is stable
b. r1=2 m, r2=3m, L=2.3m
1 20 (1 / )(1 / ) 1L r L r   
(1 2.3/ 2)(1 2.3/3) ( 0.15)(0.25) 0.0345     
Cavity is unstable
c. r1=5 m, r2=3m, L=4m
(1 4/5)(1 4/3) (0.2)( 0.33) 0.067      Cavity is unstable
d. r1=0.5 m, r2=0.5m, L=0.5m
(1 0.5/ 0.5)(1 0.5/ 0.5) 0   Cavity is on edge of stability
---confocal cavity

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Laser modes
•The cavity determines the properties of the beam of light that is emitted
by the laser.
• This beam is characterized by its transverse and longitudinal mode
structure.
Transverse modes are created in cross section of the beam,
perpendicular to the optical axis of the laser.
Longitudinal modes only specific frequencies are possible inside
the optical cavity of a laser, according to standing wave condition.
Transverse mode structure
A transverse mode is a field configuration on the surface of one reflector that propagates
to the other reflector and back, returning in the same pattern,
apart from a complex amplitude factor (that gives the total phase shift and loss of the
round trip.

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Separating the fundamental Gaussian mode
Lasers operating with the fundamental Gaussian mode TEM00 are preferred due to
the following reasons:
1. TEM00 has a symmetrical, uniform circular configuration with the greatest
intensity at its center: this suits many applications that requires high
accuracy
2. Contains about 85% of the total output intensity
3. Can be easily separated from higher order modes by using a pin hole
aperture with a diameter that allow only photons propagated along the
optical axis to be incident on mirrors
It is preferred to get the laser operated in the fundamental mode TEM00
The intensity of the Gaussian mode is given by
  2 2
0 0exp( 2 / )I r I w w 
The total power 2
/ 2P w I
02 /div w  Divergence
0w Radius of Gaussian beam (the radius at which
the intensity reduced by 2
/1 e

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Example 1:
A He-Ne Laser with a Gaussian fundamental mode TEM00 operates with a
wavelength of 632.8 nm . A lens is used to collimate the beam to pass through a
glass plate of a thickness of 12 mm and a refractive index of 1.46. suppose the
plate is place at the focal point of the lens where the beam diameter is 2m
Calculate the beam diameter at the other end of the plate?
0 0
2
12 0.012 , 632.8 , 2 1
2 2
d m
z mm m nm d w w m

        
 
1
2 2
0
0
1
z
w z w
z
  
    
   
 
 
2
62
60
0 9
1
2 2
6 3
6
3
1.46 10
7.25 10
632.8 10
0.012
10 1 1.66 10
7.25 10
beam dimeter 2 ( ) 3.32 10
n w
z m
w z m
w z m





 


 
   

  
      
   
  
Sol.
We have
A fundamental Gaussian beam from a Ti:Sapphire laser with a wavelength of 759nm
and a power of 1mW incident on a target far from the minimum waist point by
100m. If the radius of the minimum waist is 2mm, find the beam waist at the
target?
Calculate the radius of wavefront curvature and peak intensity of the beam
Sol.
Waist radius at the target
1/22
2 3 2
0
0 0 9
0
(2 10 )
( ) 1 , ( ) 3.14 16.55
759 10
wz
w z w z Rayleigh range
z




   
       
   
1/22
3 100
( ) 2 10 1 12
16.55
w z mm
  
      
   
Example 2:

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Usually the spot size is represented by the beam waist and the area of the laser
spot (if it is circular) is give by 2 3 2
( ) 3.14 12 10 0.0314A w z m 
    
The radius of wavefront curvature is
2
2
0
16.55
( ) [1 ( / ) ] 100 1 102.74
100
R z z z z m
  
      
   
peak intensity
3
3 2
2 3 2
1 10
4.4 10 /
( ) / 2 3.14 (12 10 ) / 2
p
P P
I W m
A w z



    
  
Note: is the effective area
2
{ ( ) / 2}A w z 
The longitudinal modes determine the emission spectrum of the laser.
The light bouncing repeatedly off the end mirrors sets up standing waves inside the
cavity.
LONGITUDINAL MODES
More general
where n is the average refractive index of the cavity.
The last Equation implies that only certain frequencies which satisfy will oscillate

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Plane-mirror resonators
• This type of cavity consists of two flat, parallel mirrors separated by a distance L . It is also called
a Fabry-Perot resonator (F-P).
• In this type of cavity, the beam fills the space between the mirrors nearly uniformly, unlike in the
spherical mirror cavities where the beam is focused somewhere inside (or outside of) the cavity.
The finesse coefficient of the Fabry-Perot cavity Is given as
1 2
1/4 1/4
1/2 1/2
1 21
c
R R
f
R R



If the two mirrors have similar reflectivities R1=R2 then
1
c
R
f
R



The finesse coefficient has great importance for determining the fluorescence
line shape 
/cf   
 Frequency spacing between modes
 Laser fluorescence line width
2 (2 )kL m 
2 4
2 (2 )
n L L
kL m
C
 


  
Note, that the condition of standing wave is
2
m
L


2
m
c
m
n L
 
Frequency spacing between modes
or Free Spectral Range
1
2
m m
c
n L
   
Phase displacement for one round trip oscillation
Frequency of the mode of order m
Derivation of the number of Longitudinal modes

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2
22
c
or
n Ln L
  
The number of operating
longitudinal modes is given by
( / 2 )
m
c n L


The quality factor of Fabry-Perot
resonator is
0
Q




Calculate the width of the frequency mode for a Fabry-Perot resonator
consists of two similar plane mirrors separated by a distance of 1 cm
Assume the reflectivities 70% 99.9% 95%
Calculate the finesse factor for the wavelength of 800 nm from a GaAs laser
And find also the value of the resonator quality factor for each case?
8
103 10
1.5 10
2 2 1 0.01
c
Hz
n L


   
 
8
14
9
3 10
3.75 10
800 10
c
Hz
 

    

1 0.9999R For 10
4 5
4
0.9999 1.5 10
3.141 10 4.78 10
1 0.9999 3.14 10
c
c
f Hz
f
 


        
 
Example 3:
Sol.

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Resonator quality
14
80
5
3.75 10
7.85 10
3.78 10
Q



   
 
2 0.95R For
10
80.95 1.5 10
61.2 2.45 10
1 0.95 61.2
c
c
f Hz
f
 


       

14
60
8
3.75 10
1.53 10
2.45 10
Q



   
 
For 3 0.75R 
10
9
14
5
9
0.75 1.5 10
10.87 1.38 10
1 0.75 10.87
3.75 10
2.72 10
1.38 10
c
c
f Hz
f
Q
 




       


    
 
Example 4:
For a He-Ne laser with a wavelength of 632.8 nm, if the length of the
resonator is 30 cm, find:
a. Frequency difference between longitudinal modes (mode spacing
b. Number of modes
c. Frequency of the laser light
Sol. 8
3 10
0.5
2 2 0.3
c
GHz
L


  

a. Mode spacing
b. Number of modes 6
6
2 2 2 0.3
0.948 10
0.6328 10
m
L L
m
m

 

     

c. Frequency of the laser
6 9 14
0.948 10 0.5 10 4.74 10m Hz        
or
8
14
6
3 10
4.74 10
0.6328 10
c
Hz
 

   


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Note: 1) and 2) allow only low powers to be obtained (of no practical use)
1. Reducing laser cavity length to make the mode
spacing large and hence allow only one mode to
operate 2
c
n L
 
2. Reduce pumping power and hence allow the
amplification only for the central mode
Laser operation in single longitudinal mode
There are many methods that can be used to force the laser oscillation in
single longitudinal mode
4. Using Prism to select
one longitudinal mode
5. Using grating to select
one longitudinal mode
3. Using Etalon
Generate additional losses for the extra modes by
placing frequency selective optical elements in the
laser resonator
The lasing mode gets some of the gain of the killed
modes higher power/mode

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Example 5: An ion Argon laser with a wavelength of 514.5 nm with a spectral
bandwidth of 2GHz and a cavity length of 50 cm. find the number of possible
longitudinal modes?
Sol.
9
8 2
2 10
6mod
/ 2 3 10 / (2 50 10 )
m es
c L


 
  
  
Refractive index of
argon gas =1
Example 6: If a spectral filter with a bandwidth of 0.1 nm is used to obtain a single
longitudinal mode from a He-Ne laser, what should be the length of the laser cavity?
8
9 10
2 9 2
3 10
(0.1 10 ) 7.5 10
(632.8 10 )
c
Hz 




      

Sol. The fluorescence frequency bandwidth is given by
For a single mode oscillation,
we should have
8
10
3 10
0.002
2 2 7.5 10
c
L L m
L


     
 

Laser lecture 04

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Laser lecture 04