Theory study about quarter-wave-stack dielectric...

1

Theory study about quarter-wave-stack

dielectric mirrors

2

Theory study about quarter-wave-stack dielectric mirrors

Stratified Mediumy

zx

incident

reflectedtransmitted

EH

First, consider a wave propagation in a stratified medium. As we know, anyarbitrarily polarized plane wave could be resolved into two waves, one with itselectric vector perpendicular to the plane of incidence, we shall speak of atransverse electric wave (denoted by TE), another one with its magnetic vectorperpendicular to the plane of incidence we shall speak of a transversemagnetic wave (denoted by TM).

Here, we take the plane of incidence to be y,z-plane, z being the direction ofstratification.

Stratified Medium

x

incident

reflectedtransmitted

z

y

TE TM

3


So, now for a TE wave, Ey=Ez=0, Maxwell’s equations reduce to:

0

0

0

=!

!"

!

!

=!

!"

!

!

=#+!

!"

!

!

y

H

x

H

x

H

z

H

Ec

i

z

H

y

H

xy

zx

x

yz $%

0

0

0

=!+"

"

=!#"

"

=!

zx

yx

x

Hc

i

y

E

Hc

i

z

E

Hc

i

$µ

$µ

$µ

Hy Hz Ex are equations of y and z only. By eliminating Hy Hz we have:

0

0

2

2

0

2

2

2

2

2

2,

)(ln

!

"#$µ

µ

===

%

%&=+

%

%+

%

%

ckn

z

E

dz

dEkn

z

E

y

E xx

xx

4


Define: then, the above equation can be written as:

Now, the term on the left is a function of y only whilst the terms on the rightdepend only on z. Hence this equation only hold if each side is equal to aconstant (-K2 say):

It will be convenient to set

Then we get the solution of Y:

Consequently Ex is of the form

where U(z) is a function of z. From equations of last page, we see that Hy is givenby expression of the same form:

)()(),( zUyYzyEx =

dz

dU

Udz

dkn

dz

Ud

Udy

Yd

Y

1)(ln11 2

0

2

2

2

2

2 µ+!!=

UKUkndz

dU

dz

d

dz

UdK

dy

Yd

Y

22

0

2

2

22

2

2 )(ln,

1=+!""=

µ

22

0

2akK =

[ ] ayiketconsY 0tan !=

)( 0)(tayki

x ezUE!"

#=

)( 0)(tayki

y ezVH!"

#=

5


It had been derived in Born and Wolf’s «Principles of Optics» that:

Here, M is a transmission matrix and this matrix is unimodular, |M|=1. Thesignificance of M is to relate the x- and y-components of the electric (ormagnetic) vector in the plane z=0 to the components is an arbitrary planez=constant. (I skipped the complicated derivation here).

Now, we saw that the knowledge of U and V is sufficient for the completespecification of the field. Hence for the purposes of determining the propagationof a plane wave through a stratified medium, the medium only need bespecified by an appropriate 2×2 unimodular matrix M. So, we shall call M thecharacteristic matrix of the stratified medium.

!"

#$%

&'=!

"

#$%

&

)(

)(

0

0

z

z

V

UM

V

U

6


Stratified Medium

zx

z=0 z=z1

ε= ε(z)µ= μ(z)

incident

reflected transmitted

θ1

θ2

EH

ε1 μ1 ε2 μ2

y

A TE wave:

A

R T

Let A, R, T denote as before the amplitudes (possibly complex) of the electric vectorsof the incident, reflected, and transmitted wave. Further, let denoteas the above picture shows.

And the boundary conditions demand that the tangential components of E and H shallbe continuous across each of the two boundaries of the stratified medium.

212211,,,,, !!µ"µ"

7


If the relation

is also used, we get the following relations for a TE wave:

Assume the characteristic matrix to be:

so,

ESHrrr

!"=µ

#

!"

!#

$

%=

=

&=

+=

TV

TU

RAV

RAU

z

z

2

2

2)(

)(

1

1

10

0

cos)(cos1

1

'µ

('

µ

(

!"

#$%

&

2221

1211

mm

mm

!"

#$%

&'!"

#$%

&=!

"

#$%

&

)(

)(

2221

1211

0

0

1

1

z

z

V

U

mm

mm

V

U

!!

"

#

$$

%

&

'''!"

#$%

&=

!!

"

#

$$

%

&

(''

+

T

T

mm

mm

RA

RA

2

2

2

2221

1211

1

1

1 cos)(cos )µ

*)

µ

*

8


From the above equation, we can obtain the reflection and transmissioncoefficients of the film:

In terms of r and t, the reflectivity and transmissivity are

The phase δr of r may be called the phase change on reflection and the phaseδt of the phase change on transmission.

So, the point now is to get the characteristic matrix of the stratified medium,then we can obtain reflection and transmission coefficients. Consequently, weobtain the reflectivity and transmissivity and phase shift introduced byreflection and transmission.

i

i

iip

pmmppmm

p

A

Tt

pmmppmm

pmmppmm

A

Rr

!µ

"cos

)()(

2

)()(

)()(

22221121211

1

22221121211

22221121211

#

$$%

$$&

'

=

+++==

+++

+(+==

2

1

22, t

p

pr == TR

9


Now, consider a homogeneous dielectric film. In this case areconstants. Assume the media to be nonmagnetic (µ=1). θ denotes the anglewhich the normal to the wave makes with the z-axis in the film. As shown in«Principles of Optics» , a homogeneous has the following characteristic matrix:

This characteristic matrix describes one dielectric film. For a multilayer dielectricmirror, we consider a stack of N pairs alternate layers with refractive indices n1

and n2 and with respective thickness d1 and d2 , deposited upon a substrate withrefractive index ns.

The same, we assume all the medias to be nonmagnetic (µ=1).

!µµ! =n,,

!!

"

#

$$

%

&

'

'=

((

((

cossin

sincos)(

ip

p

i

M z

!"

#!$ cos

2cos

0

0nznzk %==

!!µ

"coscos #=#= np

10


Substrate······

······

0n

1n

1n

2n

2n

sn

1d

1d

2d

2d

pair 1 pair 2 pair N

z

y

0!

S!

1!

1!

2!

2!

For this structure, the characteristic matrixes corresponding to n1 and n2 are:

!!

"

#

$$

%

&

'

'=

!!

"

#

$$

%

&

'

'=

222

2

2

2

2

111

1

1

1

1

cossin

sincos

cossin

sincos

((

((

((

((

ip

p

i

M

ip

p

i

M

111

0

1cos

2!

"

#$ dn%=

222

0

2cos

2!

"

#$ dn%= 222111

coscos !! npnp ==

11


Let Mp be the characteristic matrix corresponding to one period:

N periods:

Once we get the matrix MNp, we can obtain the reflection and transmissioncoefficients by:

In general condition, it is very complicated to calculate a Nth power of a matrix.

!!!!

"

#

$$$$

%

&

'''

'''

=(=

21

2

1

21212211

21

1

21

2

21

1

2

21

21

sinsincoscossincoscossin

cossinsincossinsincoscos

))))))))

))))))))

p

pipip

p

i

p

i

p

p

MMM p

!"

#$%

&'''' ('=

2221

1211)(

MM

MMMM betoassumeN

pNp

!!"

!!#

$

+++=

+++

+%+=

)()(

2

)()(

)()(

222101211

0

222101211

222101211

ss

ss

ss

pMMppMM

pt

pMMppMM

pMMppMMr 000

cos!np =

sssnp !cos=

12


A quarter-wave-stack multilayer dielectric mirrors

In particular case, the alternate layers could be fabricated with a appropriatethickness corresponding to specified wavelength and incident angle to let theoptical path length equal to quarter-wavelength.

Then, it has:

So, the characteristic matrixes corresponding to n1 and n2 are reduced to:

4coscos

0

222111

!"" == dndn

221

!"" ==

!!

"

#

$$

%

&

'

'=

!!

"

#

$$

%

&

'

'=

0

0

0

0

2

22

1

11

ip

p

i

M

ip

p

i

M

222111

2

1

1

2

21coscos

0

0

!! npnp

p

p

p

p

MMM p ==

""""

#

$

%%%%

&

'

(

(

==

13


!!!!

"

#

$$$$

%

&

'

'

==N

N

N

pNp

p

p

p

p

MM

)(0

0)(

)(

2

1

1

2

The reflection and transmission coefficients could be shown as:

Ns

N

s

NN

Ns

Ns

s

NN

s

NN

p

p

p

p

p

p

pp

pp

p

p

pt

p

p

p

p

p

p

p

p

pp

pp

p

p

pp

pp

p

p

r

2

2

1

0

2

1

2

10

1

2

0

2

2

1

0

2

2

1

0

2

10

1

2

2

10

1

2

)(1

)(2

)()(

2

)(1

)(1

)()(

)()(

!+

"!

=

!"+!"

=

!+

!"

=

!"+!"

!""!"

=

14


Assume the light to be normal incidence:

The reflection and transmission coefficients could be reduced as:

The reflectivity and transmissivity are:

With the increase of N, and if n1<n2 then R→1, T→0. So, stack quarter-wavemultilayer dielectric is a familiar method to fabricate a very high reflectivitymirror.

22221111210cos,cos0 nnpnnp

s======== !!!!!! o

Ns

N

Ns

Ns

n

n

n

n

n

n

t

n

n

n

n

n

n

n

n

r2

2

1

0

2

1

2

2

1

0

2

2

1

0

)(1

)(2

)(1

)(1

!+

"!

=

!+

!"

=

2

0

2, t

n

nr

s== TR

15


Now, go back to the equations of reflection and transmission coefficients.Obviously, the reflection and transmission coefficients are real numbers. Itmeans that under ideal condition, no fabrication error, no absorption, nowavelength deviation, there will not be phase shift introduced by reflection andtransmission from a quarter-wave-stack dielectric mirrors for a TE wave. And itcan be derived easily for a TM wave that the reflection and transmissioncoefficients are real numbers, too.

So, no phase shift for both components of a polarized light means no changefor polarization status.

Anyway, we should consider some error factors as we can.

16


We got some basic information for the fabrication and error of the mirror fromthe company. But, they can not offer much more detailed description about therefraction index and thickness of the dielectric layer due to the protection oftechnology.

In general, a scheme of 37 layers Ta2O5/SiO2 was applied. Use the refractionindices like:

5098.1

455.1

0411.2

2

52

2

1

=

==

==

s

SiO

OTa

n

nn

nn

Error of n1 and n2 : 0.1%---0.01%

Error of thickness: 0.3% (totally for 37 layers)

Laser wavelength deviation: Δλmax<0.5nm (from laser test report)

17


Now, for a normal incidence, consider the worst condition (each error reachesits biggest value) for a TE wave. Then,

Assume δ=0.01rad, the characteristic matrix of layers are:

)(007.0:

2

2cos

2

max

max

radget

ndnd

!

+!"="=

#

#$

%

$&

%

$'

!!!!

"

#

$$$$

%

&

++'

+'+

=

!!!!

"

#

$$$$

%

&

++'

+'+

=

)01.02

cos()01.02

sin(

)01.02

sin()01.02

cos(

)01.02

cos()01.02

sin(

)01.02

sin()01.02

cos(

2

22

1

11 ((

((

((

((

in

n

i

M

in

n

i

M

[ ] !"

#$%

&

''

'=(=

2716.15266.894

2597.02716.15

1

18

21

i

iMMMM

18


So, we can get the reflection coefficient as:

It is shown above that the TE wave will have a common 180 degree reflectionshift and an additional about 1.96 degree phase shift introduced by small errorof the mirror fabrication and laser wavelength deviation for one time reflectionfrom the quarter-wave-stack dielectric mirror.

999992.0

96.1180arg

874.8933287.38

658.8947855.7

)()(

)()(

2

1

222101211

222101211 0

==

+==

!!

+""#"

+++

+!+=

=

rR

r

i

i

nMMnnMM

nMMnnMMr

r

n

ss

ss

oo$

19


For a TM wave, the same matrix hold, with p replaced by q:

!!

"

#

$$

%

&

'

'=

!!

"

#

$$

%

&

'

'=

222

2

2

2

2

111

1

1

1

1

cossin

sincos

cossin

sincos

((

((

((

((

iq

q

i

M

iq

q

i

M

nqq

n 1cos

,0,1=!!!!!! "!=

==== #µ#$µ$

#

µ o

!!!!

"

#

$$$$

%

&

++'

+'+

=

!!!!

"

#

$$$$

%

&

++'

+'+

=)01.0

2cos()01.0

2sin(

)01.02

sin()01.02

cos(

)01.02

cos()01.02

sin(

)01.02

sin()01.02

cos(

2

2

1

1

1

2

((

((

((

((

n

i

in

M

n

i

in

M

[ ] !"

#$%

&

'

''=(=

2716.152597.0

266.8942716.15

1

18

21

i

iMMMM

20


999991.0

96.1arg

048.5923866.25

567.5921566.5

)(1)(

)(1)(

2

1

2221

0

1211

2221

0

1211

0

==

==

!!

!!""#"

+++

+!+

==

rR

r

i

i

n

MM

nn

MM

n

MM

nn

MM

r

r

n

ss

ss

o$

Compare the results of TE an TM we could find out that the phase shift introducedby small error of the mirror fabrication and laser wavelength deviation for one timereflection will be same. It means the polarization status will be maintained (samephase shift for TE and TM wave) and only the polarization direction will change (TEwave has a 180 degree phase shift but TM has not).

Notice: Only under the condition of normal incidence, the TE and TM waves willhave the same phase shift introduced by small error of the mirror fabrication andlaser wavelength deviation

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Three points had been assumed for this calculation:

1. No absorption when laser pass through mirrors. (Could we ignore it?)

2. The multilayer dielectric layer’s material is isotropic. (Is it? Need certified)

3. Approximately describe the gaussian beam as a plane wave. (How to

deal with a real gaussian beam?)

Theory study about quarter-wave-stack dielectric...

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