Optical Science and Engineering,
Department of Physics & Astronomy, University of New Mexico
Albuquerque, NM 87131-0001, USA
The Fabry-Perot cavity is 1 mm long, of index n
0
=1.5, mirror reflectivity 99.9%.
The field transmission function of the Fabry-Perot cavity is given by
T

(1

)
 ikd
1

R e

2 ikd
The transmission linewidth (FWHM) is given by:
W

2(1

R ) c
R 2 n d
0
  8 rad / sec

0.2
/ , this value is in terms of

.
The free spectral range is given by
 
 c
  11 rad / sec

630 /
We have a pulse:
( )
  t e
 l

( )
 e
 t

( )
 e

 t
( )
The duration

is equal to 10 ns.
0
0
Then we take the Fourier Transform of this pulse, which is given by:
1

G
 

(

1
2
1
  2
)
We may see in the frequency domain the pulse is a Lorentzian with FWHM equal to

2
  8 rad / sec

0.2
/ range of the Fabry-Perot cavity.
, which is very small compared to the free spectral
Then we want to take the product of the Fourier transform of the pulse and the field transmission function of the Fabry-Perot cavity. Since the FWHM of Fourier transform of the pulse is very small compared to the free spectral range of the
Fabry-Perot cavity, we can make appropriate approximation, 2 kd 1 .
T

(1

)
 ikd
1

R e

2 ikd

(1

R )(1
 ikd )
1

R
 ikd )
In which k
 n
0
 c
.
Further simplify this equation,
T

(1
1


R
R )(1

 ikd ikd )
)

1
 
1
 
(1
2

) iRkd

(1

R )
2 
2 (1

)
2 2
R R k d
(1

R )
2 
(2 Rkd )
2
 i
( R
2 
1) kd
(1

R )
2 
(2 Rkd )
2
(1

R
(1
)
2


R )
(2
2
Rkd )
2
 i
(1

(
R
R
)
2
2


1)
(2 kd
Rkd )
2

(1

R )
2
(1


(2
R )
Rn
2
0
 d c
2
 i
(1

(
R
R
)
2
2


1)
(2 n
0

Rn
0 d c
 d c
2
Take the product of T and G, we will get
(1

R )
2
H

TG

[
(1

R )
2 
(2 Rn
0
 d c
2
(1

R ) /

2 

R
2
(1 ) /
[(2 Rn d c
0


) 2  

R i
( R
2 
1) n
0

/
(1

R )
2 
(2 Rn
0
 d c
2
) ]
 2
 i
(1

R ) /

2
]

(

1
2
1
  2
)
(

R
2
[(2

1) n
0
 d c


Rn d c
0
/ ) 2  
R ) ]
 2
Take the inverse Fourier Transform of H, using the derivative theorem of Fourier
Transform, we have
 out
( )

A exp(
 a t ) ( t

0)
In which a

[
(2

R
2
(1 ) /

2
/

)
2  
R )
2
]
1/ 2   16 1/ 2
(0.5 10 )
  8
2

And A is a constant which is not relevant to this problem.
So the duration of the pulse transmitted by this Fabry Perot is

'
  a
14 ns which is broader than the input pulse.
Now we have a Gaussian pulse

( )
 e
(1 ia t

2 with pulse chirp a. The frequency of the Gaussian pulse is 0.1 rad/ns below resonance.
We use the same method as above but solve it numerically.
The field transmission function of the Fabry-Perot cavity has the same form as the previous problem.
T

(1

R ) 2
(1

R )
2 
(2 Rn
0
 d c
2
 i
( R
2 
1) n
0
 d c
(1

R )
2 
(2 Rn
0
 d c
2
The Fourier Transform of the pulse envelope is
G

4(1
 ia )
 exp[
 2

2
/ 4(1
 ia )]
Since the frequency of the Gaussian pulse is 0.1 rad/ns below resonance, we write G as
G

4(1
 ia )

    
2 / 4(1
 ia )]
Take the product of T and G, and do the inverse Fourier Transform of the product numerically, we can get the transmitted pulse for various chirp a.
The figures below show the pulse shape for various chirp a. (only t>0 shows). We may see that there is a value of a (around 100) for which the transmitted pulse has a minimum duration.
3

4

5

We consider now the same Fabry-Perot, but with a nonlinear index n
 n
0
 n I with n
2

10

15 cm
2
/ W .
We consider a signal that is triangular in time
  
0
(1
 t
100
)

100 t 100 ns in which
 corresponds to a peak intensity of
0
1 /
2
. The figure below shows the shape of the input pulse (intensity versus time).
6

The light radiation is centered 0.2
/ below resonance..
For the static approach, we assume the spectrum of the field is a

function, and
 
1
 m . Thus, for a given intensity, the intensity transmission T and the phase change

are inter-dependent.
 
4


( n
0
 n I in
) d

2 n
0
 off d c

4

 n TI d
2 0
(1

R )
 
0

TI
0 a in which

0

2 n
0
 off d
and c a


(1

R )
4
 n d
2
.
This equation above can be represented as T
 a
I
0
(
 
0
) .
Another relation is
T

1
1

F
2
 where F

4 R
(1

R )
2
.
7

By there two relations, we can solve the problem numerically for any given intensity.

One method to solve this problem is by plotting the two equations and find the intersections of these two curves. The three figures below shows this scenario in different scopes. We can see from these three figures that for a given input intensity, the output intensity could possess multiple states. However, when the input intensity is varied continuously, the output can only have one state depending on its former state. I have marked the process I expect it will undergo for our particular input pulse by ①-②-③-④. We can conclude from the figures that:
1. This process is not symmetric.
2. When the input is 0, the transmission is 0.2, and when the input is at its maximum, the transmission is roughly 0.002.
3. The transmission will jump at points ①-② and ③-④ where the input intensity is roughly I
0

0.002
a

0.00016
GW cm 2
8

9


The figure below gives the shape of the output pulse by computer program.
This shape corresponds very well with our graphical discussion. There are two jumps when the input intensity is very small and roughly equals to I
0

0.00016
/ 2
, the transmission will jump to almost unity at those points, so the output intensity is a little below 0.0002GW/cm
2
. When the intensity is at its maximum, the transmission is
0.002, so the output will be 0.002GW/cm
2
, which also agrees well. We may see that when the input is increased, the trend of the increase of the input intensity overwhelms the decrease of the transmission, so the output is also increased. The only point that cannot be shown in the figure is the asymmetrical character since this feature is not obvious for this particular pulse. We expect it will be obvious when the off-resonance is doubled. The figure below shows the transmitted pulse when the off-resonance is doubled. We can see this pulse is not symmetric. This point can be made even more clear when we plot the change of
δ in terms of time. We can see from the figure that when the input in increased, the phase change jumps from a
10

negative number directly to a positive number, while when the input in decreased, the phase changes falls to 0, and then jumps to a negative number, which justified our discussion above.
11

Now we solve the same problem by using the dynamic approach. We take into account the temporal dependence of the pulse and solve it in the frequency domain.
First, we take the Fourier transform of the pulse
  
0
(1
 t
100
)

100 t 100 ns
We have
G
  2
( ) 100 ( sin


)
2 where
 
100

2
.
The figure below shows the spectrum of the pulse.
12

Now, for a given field, the field transmission T and the phase change are inter-dependent.
T

(1

)
 i

  i

/ 2
 
0
 
T ( )
 in
( )
2 in which a

2 n n d
0 2 c (1

R )


0

.
By these two relations, we can solve the problem numerically for any given field.
After we solve it for every frequency component, we take the inverse Fourier
Transform to get the transmitted pulse shape.

The figure below shows the result by using the dynamic approach ( only when t>0 is shown, the shape when t<0 is symmetric with t>0).
13

The shape of the pulse looks good and it has a jump somewhere. But the amplitude of the transmitted intensity seems too small compared to the static approach. Since I don’t have other method to tell whether it is right or wrong, it has to be left for further study.
When the triangular pulse is 10 times shorter, the figure below shows the transmitted pulse.
14

15

clear; tao=1e-8;
R=0.999; a=2000; shift=1e8; n=1.5; d=0.001; c=3*1e8;
N=128; omiga=linspace(0,1e9,N); f=((1-R)^2./((1-R)^2+(2*R*n.*omiga*d/c).^2)+i*(R^2-1)*n.*omiga*d/c./((1-R)^2+(2*R*n.*omi ga*d/c).^2))*tao/sqrt(4*pi*(1+i*a)).*exp(-(omiga+shift).^2*tao^2/(4*(1+i*a)));
Ws=omiga(2)-omiga(1);
Ts=2*pi/Ws;
H=ifft(f); hp=H(1:N/2+1)*Ws;
T=Ts*(0:N/2)/N;
%hold; h1=figure; plot(T,abs(hp));
%plot(T,hp); xlabel('time,sec'), ylabel('sqrt(intensity)')
16

clear; lamda=1e-6;
R=0.999;
F=4*R/(1-R)^2; d=0.001; n0=1.5; n2=1e-6; shift=-2e8; c=3*1e8; a=lamda*(1-R)/(4*pi*n2*d); delta0=2*n0*shift*d/c;
N=10001; t=linspace(-10,10,N); f=1-0.2*abs(t)+0.01*(t).^2; step=0.0000001; delta(1)=delta0-step; for n=1:(N+1)/2
deltaR(n)=100;
while abs(delta(n)-deltaR(n))>0.00000025
delta(n)=delta(n)+step;
T(n)=1/(1+F*sin(delta(n)/2)^2);
deltaR(n)=delta0+T(n)*f(n)/a;
end
delta(n+1)=delta(n);
G(n)=T(n)*f(n); end for n=(N+3)/2:N
deltaR(n)=100;
while abs(delta(n)-deltaR(n))>0.00000025
delta(n)=delta(n)-step;
T(n)=1/(1+F*sin(delta(n)/2)^2);
deltaR(n)=delta0+T(n)*f(n)/a;
end
delta(n+1)=delta(n);
G(n)=T(n)*f(n); end
%plot(t,f); h1=figure; plot(t,G) xlabel('time,ns'), ylabel('intensity, GW/cm2')
17

clear; lamda=1e-6;
R=0.999;
F=4*R/(1-R)^2; d=0.001; n0=1.5; n2=1e-6; shift=-2e8; c=3*1e8;
%a0=2*n2/(1-R)/pi/376.7*n0; a0=n0*n2*d/c/(1-R)/pi; delta0=2*n0*shift*d/c; omiga0=2*pi*c/n0/lamda;
N=128; omiga=linspace(-0.4,0.4,N); f=sin(50*omiga).^2./omiga.^2/1.2;%parameter 1.2 is strange plot(omiga,f) xlabel('omiga,rad/ns') for n=1:N
a=a0*omiga0*abs(f(n))^2;
b=-delta0;
c=-(1-R)*delta0/2;
T(n)=1/6/a*((-108*c+12*3^(1/2)*((4*b^3+27*c^2*a)/a)^(1/2))*a^2)^(1/3)-2*b/((-108*c+12*3^(1
/2)*((4*b^3+27*c^2*a)/a)^(1/2))*a^2)^(1/3);
G(n)=T(n)*f(n); end
%inverse Fourier Transform
Ws=omiga(2)-omiga(1);
Ts=2*pi/Ws;
H=ifft(G); hp=H(1:N/2+1)*Ws;
TA=Ts*(0:N/2)/N; h1=figure; plot(TA,abs(hp).^2); xlabel('time,ns'), ylabel('intensity, GW/cm2')
%h2=figure;
%plot(T)
18