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14/06/2015
1
412 PHYS
Lasers and their Applications
Department of Physics
Faculty of Science
Jazan University
KSA
Lecture-3
Laser levels and saturation
Two levels system
It is impossible to achieve population inversion and hence
get a laser light from a two level system by direct pumping.
In this case resonant absorption is the suitable way to
sustain the pumping but at its best we can only reach the
state of equal population in the two levels which is not also
possible if the spontaneous emission is taken into
considerations.
14/06/2015
2
The distribution of the state of atoms
under thermal equilibrium at any given
temperature is given by Boltzmann’s law
At relatively low temperatures, kBT << E practically no atoms
exist in the higher energy state.
At relatively high temperature, kBT >> E the populations of
the two energy levels are almost equal, that is, N1 ≈ N2.
As the temperature increases, N2 approaches N1 asymptotically, but
can never exceed it under thermal equilibrium conditions.
Population inversion is not possible in two level system:
W12N1
W21N2

Let’s assume a two-level system with radiative transition probabilities as
shown in the adjacent figure
W12 is the probability of absorption
W21 is the probability of stimulated
emission
A21=1/τ21 is he probability of spont. emission
The total number of atoms is
21 NNN 
12 21B B B 
  SEabs
12 21 ( )W W W B    
14/06/2015
3
At the steady state and neglecting the spontaneous emission because
of high spectral radiant energy, we can write the rate equation as:
2
2 2
1
( ) 0
2
N
W N N WN
N
    
That means the number of excited atoms to
higher level can not exceed 50%
The transition rate equation is now written as
2 2
1 2
21
dN N
W N W N
dt 
  
Laser gain saturation
To discuss the effect of gain saturation, let us consider the effect of
saturation of the laser beam as it grows exponentially within a gain
medium. For simplicity, let’s consider two level system by referring to the
last figure, we can write the rate equation as
As we have seen earlier the gain is
increasing by increasing the intensity
and length of the material
( )
( ) (0) ( ) (0)o z
I z I e G I 
   
0 ( ) ( ) N    
2 1 2 2/ ( - )- / .........(1)dN dt W N N N 
1 2 2 1 1 2; - ; / - / .....(2)tN N N N N N dN dt dN dt    
2 1 2( ) / ( ( - )) / 2 / - (2 1/ ) / ...(3)td N dt d N N dt dN dt N W N       
But in reality it is limited by saturation.
with
Substitute in (1)
14/06/2015
4
( / ) / (2 1/ ) / (1 2 )...(4)t tN N W N W      
At steady state 0
d N
dt


2W 
N
N
t
The variation of population inversion
with the ratio between W and (A=1/)
At saturation
/2N=N=N0,=N1;W t21
This condition
1 2{ 0, / 2}tN N N N   
is satisfied by letting
(W )or    
To maintain at a given population difference ΔN, the material absorb from
the incident radiation a power per unit volume given by
/ ( / (1 2 ))....(5)tdP dV h W N h N W W     
At saturation, the power per unit volume lost due to decay of /2N=N t2
( / ) / 2 ....(6)s tdP dV h N 
( )
tN/N =1/(1+2W )=1/(1+I / ) ......(7)
sat
I
( )
I
W= & / 2 ....(8)
h 2sat sat
h
I I I W 


 

 
  
where

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Laser lecture 03

  • 1. 14/06/2015 1 412 PHYS Lasers and their Applications Department of Physics Faculty of Science Jazan University KSA Lecture-3 Laser levels and saturation Two levels system It is impossible to achieve population inversion and hence get a laser light from a two level system by direct pumping. In this case resonant absorption is the suitable way to sustain the pumping but at its best we can only reach the state of equal population in the two levels which is not also possible if the spontaneous emission is taken into considerations.
  • 2. 14/06/2015 2 The distribution of the state of atoms under thermal equilibrium at any given temperature is given by Boltzmann’s law At relatively low temperatures, kBT << E practically no atoms exist in the higher energy state. At relatively high temperature, kBT >> E the populations of the two energy levels are almost equal, that is, N1 ≈ N2. As the temperature increases, N2 approaches N1 asymptotically, but can never exceed it under thermal equilibrium conditions. Population inversion is not possible in two level system: W12N1 W21N2  Let’s assume a two-level system with radiative transition probabilities as shown in the adjacent figure W12 is the probability of absorption W21 is the probability of stimulated emission A21=1/τ21 is he probability of spont. emission The total number of atoms is 21 NNN  12 21B B B    SEabs 12 21 ( )W W W B    
  • 3. 14/06/2015 3 At the steady state and neglecting the spontaneous emission because of high spectral radiant energy, we can write the rate equation as: 2 2 2 1 ( ) 0 2 N W N N WN N      That means the number of excited atoms to higher level can not exceed 50% The transition rate equation is now written as 2 2 1 2 21 dN N W N W N dt     Laser gain saturation To discuss the effect of gain saturation, let us consider the effect of saturation of the laser beam as it grows exponentially within a gain medium. For simplicity, let’s consider two level system by referring to the last figure, we can write the rate equation as As we have seen earlier the gain is increasing by increasing the intensity and length of the material ( ) ( ) (0) ( ) (0)o z I z I e G I      0 ( ) ( ) N     2 1 2 2/ ( - )- / .........(1)dN dt W N N N  1 2 2 1 1 2; - ; / - / .....(2)tN N N N N N dN dt dN dt     2 1 2( ) / ( ( - )) / 2 / - (2 1/ ) / ...(3)td N dt d N N dt dN dt N W N        But in reality it is limited by saturation. with Substitute in (1)
  • 4. 14/06/2015 4 ( / ) / (2 1/ ) / (1 2 )...(4)t tN N W N W       At steady state 0 d N dt   2W  N N t The variation of population inversion with the ratio between W and (A=1/) At saturation /2N=N=N0,=N1;W t21 This condition 1 2{ 0, / 2}tN N N N    is satisfied by letting (W )or     To maintain at a given population difference ΔN, the material absorb from the incident radiation a power per unit volume given by / ( / (1 2 ))....(5)tdP dV h W N h N W W      At saturation, the power per unit volume lost due to decay of /2N=N t2 ( / ) / 2 ....(6)s tdP dV h N  ( ) tN/N =1/(1+2W )=1/(1+I / ) ......(7) sat I ( ) I W= & / 2 ....(8) h 2sat sat h I I I W            where