21. Propagation of Gaussian beams

How to propagate a Gaussian beam

Rayleigh range and confocal parameter

Transmission through a circular aperture

Focusing a Gaussian beam

Depth of field

Gaussian beams and ABCD matrices

Higher order modes

2 21 1, , exp

2

x yu x y z jkq z q z

Gaussian beams

, , , , , j kz tE x y z t u x y z e

A laser beam can be described by this:

where u(x,y,z) is a Gaussian transverse profile that varies slowly along the propagation direction (the z axis), and remains Gaussian as it propagates:

2

1 1 j

q z R z w z

The parameter q is called the “complex beam parameter.”It is defined in terms of two quantities:

w, the beam waist, andR, the radius of curvature of the

(spherical) wave fronts

Gaussian beams

2 2

2

1, , , , , exp x yE x y z t u x y zq z w

2 2 2

2

1, , ~ exp 2 x yI x y zq z w

The magnitude of the electric field is therefore given by:

x axis

on this circle, I = constant

The spot is Gaussian:And its size is determined bythe value of w.

with the corresponding intensity profile:

Propagation of Gaussian beamsAt a given value of z, the properties of the Gaussian beam are described by the values of q(z) and the wave vector.

So, if we know how q(z) varies with z, then we can determine everything about how the Gaussian beam evolves as it propagates.

Suppose we know the value of q(z) at a particular value of z.

e.g. q(z) = q0 at z = z0

This determines the evolution of both R(z) and w(z).

Then we can determine the value of q(z) at any subsequent point z1 using:

q(z1) = q0 + (z1 - z0)

Propagation of Gaussian beams - exampleSuppose a Gaussian beam (propagating in empty space, wavelength ) has an infinite radius of curvature (i.e., phase fronts with no curvature at all) at a particular location (say, z = 0).

Suppose, at that location (z = 0), the beam waist is given by w0.

Describe the subsequent evolution of the Gaussian beam, for z > 0.

We are given that R(0) = infinity and w(0) = w0. So, we can determine q(0):

2

20

1 10 0 0

jq R w

jw

NOTE: If R = at a given location, this implies that q is a pure imaginary number at that location: this location is a focal plane.

2

00 wq j

Thus

The Rayleigh range

where zR is the “Rayleigh range” - a key parameter in describing the propagation of beams near focal points.

If w0 is the beam waist at a focal point, then we can write q at that value of z as:

20

Rwz

focus Rq j z

It is something like the focal spot area divided by .

Now, how does q change as z increases?

Propagation of Gaussian beams - exampleA distance z later, the new complex beam parameter is:

0 Rq z q z jz z

22

1 1

R

R R

z jzq z z jz z z

22

22

1 1Re

R

R

zq z R zz z

z zR zz

2 22

2

0 2

1Im

1

R

R

R

zq z w zz z

zw z wz

• At z = 0, R is infinite, as we assumed.• As z increases, R first decreases from infinity, then increases.• Minimum value of R occurs at z = zR.

• At z = 0, w = w0, as we assumed.• As z increases, w increases.• At z = zR, w(z) = sqrt(2)×w0.

2

1 1 j

q z R z w z

Reminder:

beam waist at z = 0: R =

w0w(z)

R(z)

Rayleigh range and confocal parameter

Confocal parameter b = 2 zR

zR

b = 2zR

Rayleigh range zR = the distance from the focal point where the beam waist has increased by a factor of (i.e., the beam area has doubled).2

w002w 02w

Propagation of Gaussian beams - example

R(z)/zR

0 2 4 6 8 100

2

4

6

8

10

z/zR

w(z)/w0

22

Rz zR zz

2

0 21 R

zw z wz

Note #2: positive values of Rcorrespond to a diverging beam, whereas R < 0 would indicate a converging beam.

Note #1: for distances larger than a few times zR, both the radius and waist increase linearly with increasing distance.

When propagating away from a focal point at z = 0:

A focusing Gaussian beam

zRzR

At a distance of one Rayleigh range from the focal plane, the wave front radius of curvature is equal to 2zR, which is its minimum value.

What about the on-axis intensity?We have seen how the beam radius and beam waist evolve as a function of z, moving away from a focal point. But how about the intensity of the beam at its center (that is, at x = y = 0)?

10,0, u z

q z

210,0, I z

q z

-10

0

10

01

23

0

1

inte

nsity

propagation distance away

from focal point (in units of zR)

transverse beam

profile

At a focal point, w = w0

Here, w = 3.15w0

0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

propagation distance

peak

inte

nsity

The peak drops as the width broadens, such that the area under the Gaussian (total energy) is constant.

The peak intensity drops by 50% after one Rayleigh range.

Suppose, at z = 0, a Gaussian beam has a waist of 50 m and a radius of curvature of 1 cm. The wavelength is 786 nm. Where is the focal point, and what is the spot size at the focal point?

At z = 0, we have: 240

1 1 786 10 50

nmjq m m

= 10-4 j10-4

So:4

010 1

mq

j

The focal point is the point at which the radius of curvature is infinite, which (as we have seen) implies that q is imaginary at that point.

q(z) = q0 + z Choose the value of z such that Re{q} = 0

410 1

2

m j

410 2

mz At

410 2

mq z j

4 2

1 2 0.786 10

mj j

q z m w

This gives w = 42 m.

Propagation of Gaussian beams - example #2

Aperture transmissionThe irradiance of a Gaussian beam drops dramatically as one moves away from the optic axis. How large must a circular aperture be so that it does not significantly truncate a Gaussian beam?

Before aperture After aperture

Before the aperture, the radial variation of the irradiance of a beam with waist w is:

2

22

2

2

r

wPI r ew

where P is the total power in the beam: 2, P u x y dxdy

Aperture transmissionIf this beam (waist w) passes through a circular aperture with radius A (and is centered on the aperture), then:

fractional power transmitted 2 2

2 22 2

20

2 2 1

A r A

w wr e dr ew

A = w

~86%

~99%

A = (/2)w

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

A/w

fract

iona

l pow

er tr

ansm

itted

Focusing a Gaussian beamThe focusing of a Gaussian beam can be regarded as the reverse of the propagation problem we did before.

d0

DA Gaussian beam focused by a thin lens of diameter D to a spot of diameter d0.

How big is the focal spot?

Well, of course this depends on how we define the size of the focal spot. If we define it as the circle which contains 86% of the energy, then d0 = 2w0.

Then, if we assume that the input beam completely fills the lens (so that its diameter is D), we find:

02 2 #fd f

D

where f# = f/D is the f-number of the lens.

It is very difficult to construct an optical system with f# < 0.5, so d0 > .

Depth of field

a weakly focused beam

a tightly focused beam

small w0, small zRlarger w0, larger zR

Depth of field = range over which the beam remains approximately collimated

= confocal parameter (2zR)

Depth of field 22 2 # Rz f

If a beam is focused to a spot N wavelengths in diameter, then the depth of field is approximately N2 wavelengths in length.

What about my laser pointer? = 0.532 m, and f# ~ 1000

So depth of field is about 3.3 meters.

f/# = 32 (large depth of field)

f/# = 5 (smaller depth of field)

Depth of field: example

The q parameter for a Gaussian beam evolves according to the same parameters used for the ABCD matrices!

Gaussian beams and ABCD matrices

Optical system ↔ 2x2 Ray matrix

system

BDC

MA

inout

in

Aq BqCq D

If we know qin, the q parameter at the input of the optical system, then we can determine the q parameter at the output:

Gaussian beams and ABCD matrices

Example: propagation through a distance d of empty space

10 1

dM

The q parameter after this propagation is given by:

in

out inin

Aq Bq q dCq D

which is the same as what we’ve seen earlier: propagation through empty space merely adds to q by an amount equal to the distance propagated.

But this works for more complicated optical systems too.

Gaussian beams and lenses

1 1

in inout

inin

Aq B qqCq D q

f

Example: propagation through a thin lens:1 01 1

Mf

1 11 1 1

in

out in in

qf

q q q f

So, the lens does not modify the imaginary part of 1/q.The beam waist w is unchanged by the lens.

The real part of 1/q decreases by 1/f. The lens changes the radius of curvature of the wave front: 1 1 1

out inR R f

Assume that the Gaussian beam has a focal spot located a distance do before the lens (i.e., at the position of the “object”).

Gaussian beams and imaging

Example: an imaging system

2in

in Rwq j z j

Then win = beam waist at the input of the system

Question: what is wout? (the beam waist at the image plane)

0 0ray matrix

1/ / 1/ 1/

i o

o i

d d Mf d d f M

(M = magnification)

(matrix connecting the object plane to the image plane)

1

R

R

jz Mjz f M

If we want to find the beam waist at the output plane, we must compute the imaginary part of 1/qout:

2 2 2

1 1 1Imout R inq M z M w

This quantity is related to the beam waist at the output, because:

So the beam’s spot size is magnified by the lens, just as we would have expected from our ray matrix analysis of the imaging situation.

1in

outin

Mqqq f M

2

1Imout outq w

wout = M win

Gaussian beams and imaging

Interestingly, the beam does not have at the output plane. R

in Rq j z

Higher order Gaussian modes

2

2

2, exp

n nx xu x z H

w z w z

The Gaussian beam is only the lowest order (i.e., simplest) solution. The more general solution involves Hermite polynomials:

where the first few Hermite polynomials are:

0

1

22

1

2

4 2

H x

H x x

H x xAnd, in general, the x and y directions can be different!

, , , , nm n mu x y z u x z u y z

The Gaussian beam we have discussed corresponds to the values n = 0, m = 0 - the so-called “zero-zero” mode.

Higher order Gaussian modes - examples

x y

|u(x,y)|2

10 mode

x y

|u(x,y)|2

01 mode

x y

|u(x,y)|2

12 mode

x y

|u(x,y)|2

A superposition of the 10 and 01 modes: a “donut mode”