chp 27 Gassian laser beam.pdf

ME 587 Engineering Optics, Prof. X. Xu, School of Mechanical Engineering, Purdue University

Propagation of a Gaussian Laser Beam

2 22

2 2

n EEc t

Gaussian beam propagation is a complicated 3-D problem

1-D wave equation:2 2 2

2 2 2

E n Ez c t

3-D wave equation:

A plane wave: ( )0( , )i k r tE E r t E e

A spherical wave: 0 ( ),( , , , ) i k r tEE r t er


Paraxial approximations of laser beam propagation: EM wave propagation is close to the optical axis

Examine wave propagation close to the z-direction

x

y

z

rR

2 2 2 2( )r R x y 2 2

2 1/ 22

( )[ (1 )]x yr RR

If the laser beam is close to the z axis,

2 2 2x y R 2 2 2 2

2 1/ 22 2

( ) ( )[ (1 )] ~ (1 )2

x y x yr R RR R

2 2

2x yikikr ikR Re e e

Will be discussed later.


( ), , , ( , , ) i kz tE x y z t U x y z e

Let 2

2 22tEE E

z

2t : Laplacian operator in the transverse plane (x,y), or (r,).

Gaussian Laser Beam Solution

2 22

2 2( , , , )n EE x y z tc t

Need to solve:

Look for a solution in the form of:

Spherical wave like Plane wave like

2 2 ( ) 2 ( )[ ]i kz t i kz tt t tE Ue U e

2( ) ( )

2i kz t i kz tE U e ikUe

z zz

22 ( )

2( )i kz tU U Uik k U ik e

z z z

22 ( )

22i kz tU Uk U i k e

z z


22

2

E Et

2 2 2 2 22 ( )

2 2 2 22 0i kz tU U U U ni k k U U e

zx y z c

Combining all the terms:

=0

2

2 2U Ui k

zz

Assume:

2 2

2 2 2 0U U Ui k

zx y


Now write the equation in the cylindrical coordinate system:

2

2 2

1 1 2 0U U Ur ikr r r r z

Due to cylindrical symmetry, 0

1 2 0U Ur ikr r r z

Assume a trial solution: (Look for simplest solution, the so-called TEMoo mode)

2

0 exp exp 2ikrU E iP zq z

Longitudinal radialphase amplitude distribution


20 exp exp 2U ikr ikrE iPr q q

20 2exp exp 2U ikr ikrr E iP

r r q q

2 2

0 exp exp 2ikr ikr ikrE iP

q q q

2 22 20 01 22. . . exp exp exp exp2 2ikE k r Eikr ikrL H S iP iPq q q q

2. . . 2UL H S ikz

2 2 222 'exp exp exp exp2 2 2

ikr ikr ikrik iP iP iP qq q q

Insert into

is derivative with respect to z

2

0 exp exp 2ikrU E iP zq z

1 2 0U Ur ikr r r z


Cancel all exp (iP) and exp(ikr2/2q) we obtain:

2 2 2 2

2 2

2 2 0ik k r k rkP qq q q

1 oq q = q z

iPq

Guess a simplest solution:

2 22

2 2

2( 2 ) ( ) 0ik k kkP q rq q q

and


Choose z = 0 at the focal point, or waist of a beam

z=0

2

0 exp( )exp 2ikrU E iP

q

The second term needs to be real at z = 0, meaning all rays are in plane at z = 0, therefore qo is imaginary.

R z a numbero Rq iz is real

Since


Define: R, as: 22 2 1R Rz z zR z z

z z

which is Real

22 2 2 22 1RRR R

z zz z zkz k z

which is also Real

2 2

0 2exp exp exp 2r ikrE iP

R zz

The second term on the RHS indicates a Gaussian function,

2 2 2 2 2

1 1 1RR R R

zz i iq z iz Rz z z z

From

2

0 exp( )exp 2ikrU E iP

q

q = z -i zR


at z = 0, is minimum (at the focal point), called o . 2

2 2 1Rk

z zk z

2 2 2

2R

22 z2 2o o oR

okz or

k

Physical meaning of ZR, more to be discussed later

From

(z) - the width of a laser beam , which changes with z


Now solve for P:

1 , - R R

i iP iPq z iz z iz

ln RiP z z iz C

1tan

exp Rzi

zoiP ez

2 2

0 2exp exp exp 2r ikrU E iP

z R z


Complete solution for TEMoo mode Gaussian laser beam:

( )i kz tE Ue

2 21

2exp exp tan exp 2i to

oR

r z krE i kz i ez z z R z

amplitude longitudinal radial phase

phase

21 RzR z zz

2 2

2o o

Rkz

22 222( ) 1 1

2

o

R

R R

z z zzk z z

1/ 2222

21 1o oR R

z zz z

(z) varies with z as:

Important relations:


For a wave propagating close to the z - axis

x

y

zrz

2 2r z2 2

2 1/ 22 2[ (1 )] ~ (1 )2

r rq z zz z

2

2rikikq ikz ze e e

Therefore for a wave near the z-direction, 2

2krkz

z

q

2 2 2q z r

is a constant

for

which can be considered as radius of curvature


2 2

12exp exp tan exp 2

i too

R

r z krE E i kz i ez z z R z

The phase term is 2

1tan2R

z krkzz R z

For z >> zR

The phase term ~ 2

2krkzR z

Compared to the previous discussion on radius of curvature

R(z) is the radius of curvature

z=0

R


If z >> zR,

oR o

z zz

full angle = 2/z = 2 0.636o o

far field divergence angle. z=0

Divergence is proportional to and 1/o.

2

( )oRz

Divergence of a Gaussian beam

1/ 22

1oR

zz


Focusing of a Gaussian beam using a lens

Far field divergence angle:

2 0.636o o

Far field divergence angle can also be approximated as 2/f2 2

o f 0

f

043

f

Practically, some corrections are needed:

Equation to find the radius at the focal point, 0 ~ f, ~ , ~ 1/


1/#2 . .

ffd N A

1/#2 2 . .ff

N A

Relation with f number (f/#) and numerical aperture (N.A.):

04

3 . .N A

Note: in order to obtain the smallest spot, one must expand the beam to fill the diameter of the lens (d) to achieve a smallest spot

043

f

When 2 = d,

Again, the beam should fill the lens (2 = d).


Depth of focus

z=0

1/ 22

0/ 1 1.05R

zzz

0.32

R

zz

200.32z

2 2

2o o

Rkz

Total depth of focus: 200.64 0.64 Rz Z

Let

1/ 22

1oR

zz


Often, ZR is used for characterizing the depth of focus, also called the Rayleigh range:

Z2o

R

At z = zR, = 1.4o

1/ 22

1oR

zz

Since


Power carried by a Gaussian beam:

212 oavg avg

S ExH c E

Total power P:2 *1

2 oavg o oP S c EE rdrd

2 2 22

2 2

1 2exp2

o oo o o

E rc rdrdz z

221

2 2o

o ocE f z Energy is conserved.

[W/m2]

2 2

12exp exp tan exp 2

i too

R

r z krE E i kz i ez z z R z


Beam intensity (per unit area [W/m2]) distribution (variation of power intensity with r):

2 2 2*

2 2

o

1 1 2( , ) exp2 2

z

o o oo

cE rI r z c EEz z

z

2 4 2 2

* o2 2 2

E1 1 2( , ) exp2 2

o oo

c rI r z c EEz z

Elliptically shaped beam:

Circular beam:2 2 2

2 2

2exp exp 2r x yE

Elliptical beam:2 2

2 2exp 2x y

x yE

- Gaussian distribution

(when z is large)


Knife-edge method for measuring the Gaussian beam radius or diameter

2 4 2 2

* o2 2 2

E1 1 2( , ) exp2 2

o oo

c rI r z c EEz z

2

0 2

2( ) exp rI r I

2 2

0 2

2( )( , ) exp x yI x y I x

2 2

0 2

2( )( ) expx

x yI x dx dyI

Scan direction

At a given z location:


Propagation of a Gaussian beam in optical systems (lens, free space, etc.)2 2

12exp exp tan exp 2

oo

R

r z krE E j kz jz R

2 2 2 2 2 22 o RR RR

z zz R z zz kz

q(z) = z - izR 21 1 i

q z R

With the knowledge of q, the Gaussian beam properties are completely determined, such as z, zR, o, and .


12

1

,Aq BqCq D

A, B, C, D are the same as in the transfer matrix in geometry optics

This is called the ABCD Law.

Let q1 and q2 are the values of q before and after an optical element, then

Use q to describe propagation of a Gaussian beam


Example:

For free space: B 1 dC D 0 1A

2 11

1q dq q d

o

compare with q(z) = z -izR.

Thin Lens: 1 01- 1f

12

2 11

0 1 1 1 1 q1

qqf qq

f

chp 27 Gassian laser beam.pdf

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