chapter 4. ray optics and gaussian...

Chapter 4.

Ray Optics and Gaussian Beam

Reading: Saleh Chapter 1 and Siegman Chapter 16 and 17

1. Ray Optics

We'll define "light rays" as directions in space, corresponding, roughly,

to k-vectors of light waves.

axisinput

output

Each optical system will have an axis, and all light rays will be assumed to propagate at small angles to the axis. This is called the Paraxial Approximation.

Ray optics describes light by rays that travel in different optical media in accordance with a set of geometrical rules (geometrical optics).

The Optic Axis

A mirror deflects the optic axis into a new direction.

This ring has an optic axis that is rectangular.

Optic axis A ray propagating

through this system

We define all rays relative to the relevant optic axis.

Rays 𝒙𝒊𝒏, 𝜽𝒊𝒏

𝒙𝒐𝒖𝒕, 𝜽𝒐𝒖𝒕

• its position, 𝑥• its slope, 𝜃

Optical axis

𝑥

𝜃

These parameters will change as the ray

propagates through an optical system.

A light ray can be defined by two coordinates:

These are often

written in vector

form:

a ‘ray vector’

𝑥𝜃

Ray Matrices

Ray matrices can describe both simple and complex systems.

These matrices are often called “ABCD Matrices.”

Optical system ↔ 2x2 Ray matrix

The effect on a ray is determined by multiplying its ray vector by the appropriate ray matrices.

For many optical components, we can define 2x2 “ray matrices.”

𝑥𝑖𝑛𝜃𝑖𝑛

𝐴 𝐵𝐶 𝐷

𝑥𝑜𝑢𝑡𝜃𝑜𝑢𝑡

Easiest example: rays in free space or a uniform medium

If 𝑥𝑖𝑛 and 𝜃𝑖𝑛 are the position and slope at 𝑧 = 0, and 𝑥𝑜𝑢𝑡and 𝜃𝑜𝑢𝑡 are the position and slope after propagating from 𝑧 = 0 to 𝑧 = 𝑧0, then:

𝑥𝑖𝑛, 𝜃𝑖𝑛

𝑧 = 0

𝑥𝑜𝑢𝑡, 𝜃𝑜𝑢𝑡

𝑧0

(notice the small angle approximation: tan 𝜃 ≅ 𝜃 )

𝑥𝑜𝑢𝑡 = 𝑥𝑖𝑛 + 𝑧0𝜃𝑖𝑛

𝜃𝑜𝑢𝑡 = 𝜃𝑖𝑛

Rewriting this expression in matrix notation:


=1 𝑧00 1


Ray Matrix for an Interface

At the interface, clearly:

𝑥𝑜𝑢𝑡 = 𝑥𝑖𝑛.

Now calculate 𝜃𝑜𝑢𝑡.

⇒ 𝜃𝑜𝑢𝑡 =𝑛1

𝑛2𝜃𝑖𝑛

𝜃𝑖𝑛

𝑛1

𝜃𝑜𝑢𝑡

𝑛2

𝑥𝑖𝑛 𝑥𝑜𝑢𝑡

Snell's Law says: 𝑛1 sin(𝜃𝑖𝑛) = 𝑛2 sin(𝜃𝑜𝑢𝑡)

which becomes for small angles: 𝑛1𝜃𝑖𝑛 = 𝑛2𝜃𝑜𝑢𝑡

𝑀𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒 =1 0

0𝑛1𝑛2

Ray matrix for a curved interface

Now the output angle depends on the input position, too.

𝑛1 𝑛2

𝜃1𝜃2If the interface has

spherical curvature:

𝑀 𝑐𝑢𝑟𝑣𝑒𝑑𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒

=

1 0𝑛1/𝑛2 − 1

𝑅

𝑛1𝑛2

For cascaded elements, multiply ray matrices

Notice that the order looks opposite to what it should be. Order matters!

𝑴𝟏 𝑴𝟑𝑴𝟐


= 𝑀3𝑀2𝑀1𝑥𝑖𝑛𝜃𝑖𝑛



A thin lens is just two curved interfaces.We’ll neglect the glass in between (it’s a really thin lens!), and we’ll take n1 = 1.

n=1

R1 R2

n≠1

n=1𝑀𝑅 =

1 0

−𝑛2 − 𝑛1𝑛2𝑅

𝑛1𝑛2

𝑀 = 𝑀𝑅2𝑀𝑅1 =

1 0

−1 − 𝑛

𝑅2𝑛

1 0

−𝑛 − 1

𝑛𝑅1

1

𝑛

=

1 0

− 𝑛 − 11

𝑅1−

1

𝑅21

This can be written:

“Lens-Maker’s Formula”

where:1 0

−1

𝑓1

1

𝑓= 𝑛 − 1

1

𝑅1−

1

𝑅2

Ray matrix for a lens

The quantity, 𝑓, is the focal length of the lens. It’s the most important parameter of a lens. It can be positive or negative.

If 𝑓 > 0, the lens deflects rays toward the axis.

𝑓 > 0𝑅1 > 0𝑅2 < 0

If 𝑓 < 0, the lens deflects rays away from the axis.

𝑓 < 0𝑅1 < 0𝑅2 > 0

different types of lenses:

1

𝑓= 𝑛 − 1

1

𝑅1−

1

𝑅2

𝑀𝑙𝑒𝑛𝑠 =

1 0

−1

𝑓1

Real lens systems can be very complicated

The ability to describe a complicated lens system using a single 2-by-2 matrix is invaluable!

Example: a 13-element telephoto lens (Olympus)

A lens focuses parallel rays to a point

one focal length away.

1 1 0

0 1 1/ 1 0

out in

out

x f x

f

f

f

A lens followed by propagation by one focal length:

Assume all input

rays have in = 0

The opposite occurs if the

arrows are all reversed. Rays

diverging from a point are made

parallel by the lens.

For all rays

xout = 0!

00

/1/ 1 0

in

in

f x

x ff

At the focal plane, all rays

converge to the z axis (xout = 0)

independent of input position.

Lenses focus light!

f

f

d1

f / # = 1

f

f

d2

f / # = 2

Small f-number lenses collect more light but are harder to engineer.

The f-number of a lens

The f-number, “f / #”, of a lens is the ratio of its focal length and its diameter.

It is a dimensionless number that is a quantitative measure of lens speed,

and an important concept in photography.

f / # = f / d

Abberations: imperfections in lenses

There are numerous different kinds of lens abberations. Most can

be corrected with clever lens design.

Rays which are far from the optic axis focus

to a different point than rays close to the axis.

Off-axis rays focus to different points.

The focal length of a lens can depend

on the wavelength of the light.

Abberations: imperfections in lenses

Astigmatism

Astigmatism is when the horizontal and vertical planes of a

lens have different focal lengths.

This can happen when a lens is used in an off-axis

configuration, as shown here:

It can also happen if the lens is not axially symmetric, so that the

curvature is ellipsoidal rather than spherical. Roughly 1/3 of the

population suffers from astigmatism of the cornea.

Near-sightedness (myopia)

In nearsightedness, a person can

see nearby objects well, but has

difficulty seeing distant objects.

Objects focus before the retina.

This is usually caused by an eye

that is too long or a lens system

that has too much focusing power.

Myopia is corrected with a

negative-focal-length lens. This

lens causes the light to diverge

slightly before it enters the eye.

Near-sightedness

Cylindrical lenses

A "spherical lens" focuses in both transverse directions.

A "cylindrical lens" focuses in only one transverse direction.

This is astigmatism on purpose!

Examples of cylindrical lenses:

When using cylindrical lenses, we must perform two separate Ray

Matrix analyses, one for each transverse direction.

Ray tracing

Geometric ray tracing - the

process of following a

collection of rays through

an optical system to

characterize its

performance.

This concept has been adapted

by the computer graphics world

for rendering of three-dimensional

scenes. One must accurately

account for reflection, refraction,

and absorption.

Imaging and Ray Matrices


We describe the propagation of rays through a paraxial

optical system using ray matrices.

out in

out inD

B x

C

x A

Something interesting

occurs if this

coefficient happens

to be equal to zero.

Let’s compute the

ray matrix for this

configuration, Rimage.

A system images an object when B = 0.

All rays from a point xin arrive at a point xout, independent

of the input angle in. This is what we mean by “image”.

xout = A xinWhen B = 0, we have:

B = 0 is a necessary

condition for imaging.

And when B = 0, then

A is the magnification.

/

1/ 1/ 1/

o i o i

o i o i

B d d d d f

d d d d f

11

1/ 1 /0 1

1 / /

1/ 1 /

oi

o

i o i o i

o

dd

f d f

d f d d d d f

f d f

From the object to

the image, we have:

1) A distance do2) A lens of focal length f

3) A distance di

This is called the “Lens Law.”(only valid if the imaging system consists of a single lens)

1 1 1

o id d f

So B = 0 if:

The Lens Law

1 1 0 1

0 1 1/ 1 0 1

i o

image

d dR

f

1 11 / 1

i i

o i

A d f dd d

i oM d d

(writing A = M, magnification)

1 11 / 1

o o

o i

D d f dd d

1/ o id d M

1 1 1

o id d f

If the imaging condition,

is satisfied, then:

1 / 0

1 / 1 /

i

image

o

d fR

f d f

0

1/ 1/image

MR

f M

So:

Imaging

magnification

When the object is more than one focal length away from a lens,

a positive lens projects a real image on the opposite side of the

lens from the object.

Object

f > 0

Real images

Image

Projectors and camera lenses work this way.

So does the lens in your retina. The inverted

image projected onto your cornea is inverted

again by your brain, so that things look right-

side-up.

𝑑𝑖 > 𝑓 → 𝑑𝑜 =𝑑𝑖 − 𝑓

𝑑𝑖𝑑𝑓> 0

When the object is less than one focal length away from a lens,

no image occurs, but a virtual image is said to occur if you look

back through the lens.

Object

f > 0

Virtual

image

Virtual images

This is how a

magnifying

glass works.

𝑑𝑖 < 𝑓 → 𝑑𝑜 =𝑑𝑖 − 𝑓

𝑑𝑖𝑑𝑓< 0

2. Gaussian Beams

|u(x,y)|2

x y

w = 3

When ray optics fails

Paraxial wave equation

Complex beam parameter

Gaussian beams

Failures of the ray optics approach

Thin lens,

focal length f

Input beam

(bundle of rays) Distance d = f

What happens

in this plane?

(the focal plane)

1 1 0

0 1 -1/ 1 0

out in

out

x f x

f

Each ray has a

unique xin, but

all have in = 0

(parallel rays)

0

- /

out

out in

x

x f

According to this

analysis, the spot size

in the focal plane is

identically zero!

This is obviously wrong: if we want to compute the spot size, then

we need something better than ray optics!

0

-1/ 1 0

inf x

f

What if ray optics is not good enough?

These assumptions imply: 𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡

and: 2 2 2

2 2 2, , and

u u u uk k c

z x y z

Start with the wave equation:

Assume: the variation along the direction of propagation (𝑧axis) is given by: 𝑒𝑖𝑘𝑧 × 𝑓(𝑧) (a slowly varying function of 𝑧)

Also, assume a harmonic time dependence, of the form 𝑒−𝑖𝜔𝑡.

Now, plug this form for 𝐸(𝑥, 𝑦, 𝑧, 𝑡) into the wave equation.

(slowly varying with respect to both 𝑒𝑖𝑘𝑧 and also the transverse variation)

(Note: this is, essentially, the paraxial approximation)

𝜕2𝐸

𝜕𝑥2+𝜕2𝐸

𝜕𝑦2+𝜕2𝐸

𝜕𝑧2=

1

𝑐2𝜕2𝐸

𝜕𝑡2

Paraxial wave equation

𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡

𝑥, 𝑦, and 𝑡 derivatives:

𝜕2𝐸

𝜕𝑧2= −𝑘2𝑢𝑒𝑖(𝑘𝑧−𝜔𝑡) + 2𝑖𝑘

𝜕𝑢

𝜕𝑧𝑒𝑖 𝑘𝑧−𝜔𝑡

+𝜕2𝑢

𝜕𝑧2𝑒𝑖 𝑘𝑧−𝜔𝑡

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2− 𝑘2𝑢 + 2𝑖𝑘

𝜕𝑢

𝜕𝑧= −

𝜔2

𝑐2𝑢

Wave equation

becomes:

paraxial approximation

𝜕2𝐸

𝜕𝑥2=𝜕2𝑢

𝜕𝑥2𝑒𝑖 𝑘𝑧−𝜔𝑡

𝜕2𝐸

𝜕𝑦2=𝜕2𝑢

𝜕𝑦2𝑒𝑖 𝑘𝑧−𝜔𝑡

𝜕2𝐸

𝜕𝑡2= −𝜔2𝑢𝑒𝑖 𝑘𝑧−𝜔𝑡

the “paraxial

wave equation”

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2+ 2𝑖𝑘

𝜕𝑢

𝜕𝑧= 0

𝑧 derivative:

𝜕𝐸

𝜕𝑧= 𝑖𝑘𝑢𝑒𝑖(𝑘𝑧−𝜔𝑡) +

𝜕𝑢

𝜕𝑧𝑒𝑖 𝑘𝑧−𝜔𝑡

Paraxial and exact wave equations

exact wave equation Paraxial wave equation

a spherical wave emerging

from the point r = r0.

Can we use this as a guide

to find a solution to the

(approximate) paraxial

equation?

𝜕2𝑢

𝜕𝑥2+𝜕2𝑢

𝜕𝑦2+ 2𝑖𝑘

𝜕𝑢

𝜕𝑧= 0

𝜕2𝐸

𝜕𝑥2+𝜕2𝐸

𝜕𝑦2+𝜕2𝐸

𝜕𝑧2=

1

𝑐2𝜕2𝐸

𝜕𝑡2


A well-known solution

to the exact equation:

𝐸 𝑟 =𝑒𝑖𝑘(𝑟−𝑟0)

𝑟 − 𝑟0

2 2

0 0

2

0

1

x x y y

z zIn that case:

Note that: 2 2 2

0 0 0 0r r x x y y z z

2 2

0 0

0 2

0

1x x y y

z zz z

In the spirit of the paraxial approximation, we assume that the relevant

values of (x x0) and (y y0) are always small compared to (z z0).

Paraxial approximation to a spherical wave

0 0 1 r r z z

2 2

0 0

2

0

x x y y

z zwhere

𝒖 𝒙, 𝒚, 𝒛

2 2

0 0

0 0

02

x x y yr r z z

z z

Taylor expansion of a square root:1

1 12

Therefore:

Plug this approximate form for r r0 into the expression for the spherical wave:

Paraxial approximation to a spherical wave

𝐸 𝑥, 𝑦, 𝑧, 𝑡 =𝑒𝑖𝑘 𝑧−𝑧0 𝑒

𝑖𝑘𝑥−𝑥0

2+ 𝑦−𝑦02

2 𝑧−𝑧0

𝑧 − 𝑧0 +𝑥 − 𝑥0

2 + 𝑦 − 𝑦02

2 𝑧 − 𝑧0

𝑒−𝑖𝜔𝑡We will ignore

this term

because it is

small.

A paraxial spherical wave:

Paraxial spherical waves

This is an approximation of the solution to the exact equation. Is it

also an exact solution to the approximate equation?

Well, it’s a bit tedious to prove, but YES it is.

𝑢 𝑥, 𝑦, 𝑧 =𝑒𝑖𝑘

𝑥−𝑥02+ 𝑦−𝑦0

2

2 𝑧−𝑧0

𝑧 − 𝑧0

𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡 =𝑒𝑖𝑘

𝑥−𝑥02+ 𝑦−𝑦0

2

2 𝑧−𝑧0

𝑧 − 𝑧0𝑒𝑖𝑘𝑧−𝑖𝜔𝑡

R(z) = z z0To make things a bit more compact, we define:

Gaussian spherical waves

We can then see that the phase of the wave, in a

fixed transverse plane (i.e., at fixed z), varies as:

Just as with the spherical wave we started with, the surfaces of

constant phase are spherical. The radius of these spheres is 𝑅 𝑧 .

The radius of curvature increases linearly

with increasing propagation distance.

𝑢 𝑥, 𝑦, 𝑧 =𝑒𝑖𝑘

𝑥−𝑥02+ 𝑦−𝑦0

2

2𝑅 𝑧

𝑅(𝑧)=

1

𝑅 𝑧𝑒𝑖𝜙 𝑥,𝑦

𝜙 𝑥, 𝑦 =𝑥 − 𝑥0

2 + 𝑦 − 𝑦02

2𝑅 𝑧

If the radius of curvature at 𝑧 = 𝑧0 is 𝑅0, then we should have defined 𝑅 𝑧 = 𝑅0 + (𝑧 − 𝑧0).

Gaussian spherical waves

Laser

beam

Reference plane

at z = z0

radius of

curvature R0

𝑅0 = radius of curvature at 𝑧 = 𝑧0

Radius of curvature increases with

increasing propagation distance z

The size of the beam is still infinite

What have we gained from all of this manipulation?

Seemingly not much! This is still not a very useful expression!

How does this E-field vary at a particular value of z, say z = 0?

In other words, what is the transverse profile of the field?

1, ,0,

0E x y t

R

It has a constant transverse profile, even for x,y

Not a particularly realistic description of a “beam”

𝐸 𝑥, 𝑦, 𝑧, 𝑡 =𝑒𝑖𝑘

𝑥−𝑥02+ 𝑦−𝑦0

2

2𝑅 𝑧

𝑅 𝑧𝑒𝑖 𝑘𝑧−𝜔𝑡

A simple modification to fix this: let the source location be complex!

(Is this legal? YES - this parameter cancels out when this form is

inserted into the wave equation - it has no physical meaning!)

Replace the real R(z) by the complex quantity q(z) = q0 + (z - z0)

Complex source point

Also, without loss of generality,

we can set x0 = y0 = 0.complex real positions

(Note: q has

units of length)𝑢 𝑥, 𝑦, 𝑧 =1

𝑞 𝑧𝑒𝑖𝑘𝑥2+𝑦2

2𝑞 𝑧

If q is complex, then so is 1/q:1

𝑞= Re

1

𝑞 𝑧+ 𝑖Im

1

𝑞 𝑧

This part contributes to

the phase of the wave

This part is real! It tells us

the x and y dependence of

the amplitude!

Complex beam parameter q(z)

This part plays the same role as R(z),

the phase front radius of curvature.

This part gives us the finite extent in

the transverse dimension!

𝑢 𝑥, 𝑦, 𝑧 =1

𝑞 𝑧exp 𝑖𝑘

𝑥2 + 𝑦2

2Re

1

𝑞 𝑧+ 𝑖Im

1

𝑞 𝑧

𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2Re

1

𝑞 𝑧exp −𝑘

𝑥2 + 𝑦2

2Im

1

𝑞 𝑧

Re{1/q(z)} can take on any value, except infinity

(since that would correspond to a zero radius of

curvature, which makes no sense!).

BUT: the imaginary part of 1/𝑞 𝑧 MUST be positive! Otherwise, u(x,y,z) diverges to infinity as x,y increase.

Constraints on 𝒒(𝒛)

𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2Re

1

𝑞 𝑧exp −𝑘

𝑥2 + 𝑦2

2Im

1

𝑞 𝑧

Define: R(z) and w(z) such that1

𝑞 𝑧=

1

𝑅 𝑧+ 𝑖

𝜆

𝜋𝑤2 𝑧

In that case, we have:

𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2𝑅 𝑧exp −

𝑥2 + 𝑦2

𝑤2 𝑧

Properties of the complex beam parameter

By replacing the (real valued) radius of curvature R(z) for a spherical

wave emerging from a real source point z0 with a complex radius of

curvature q(z), we convert the paraxial spherical wave into a Gaussian

beam. This is still an exact solution to the paraxial wave equation, but

with the desirable property of finite extent in the x-y plane, and

therefore finite total energy.

radius of curvature (beam waist, or spot size)2

Note: this is not the same R(z) as we had before, although it still

plays the role of the radius of curvature of the phase fronts. But

it is no longer true that R(z) evolves with propagation distance

according to R(z) = R0 + (z - z0). Instead, q(z) evolves:

q(z) = q(z0) + (z - z0)

…and 1/R(z) = Re{1/q(z)}

𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2𝑅 𝑧exp −

𝑥2 + 𝑦2

𝑤2 𝑧

Gaussian Beams

Plots of |u|2 (which is proportional to the irradiance) versus x and y:

|u(x,y)|2

x y

w = 3 |u(x,y)|2

x y

w = 1

The value of w determines the width of the beam.

𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2𝑅 𝑧exp −

𝑥2 + 𝑦2

𝑤2 𝑧

3. 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

Ray matrices

Higher order modes

http://upload.wikimedia.org/wikipedia/commons/0/01/Jonquil_flowers_at_f5.jpghttp://upload.wikimedia.org/wikipedia/commons/5/51/Jonquil_flowers_at_f32.jpg

Gaussian beams

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:

The parameter q is called the “complex beam

parameter.” It is defined in terms of w, the beam

waist, and R, the radius of curvature of the (spherical)

wave fronts:


𝑢 𝑥, 𝑦, 𝑧 =1


𝑥2 + 𝑦2

2

1

𝑞(𝑧)

1

𝑞 𝑧=

1

𝑅 𝑧+ 𝑖

𝜆

𝜋𝑤2 𝑧

Gaussian beams

𝑢 𝑥, 𝑦, 𝑧 =1

𝑞 𝑧𝑒𝑖𝑘𝑥2+𝑦2

2𝑅 𝑧 𝑒−𝑥2+𝑦2

𝑤2 𝑧

The magnitude of the electric

field is therefore given by:

𝑢 𝑥, 𝑦, 𝑧 =1

𝑞 𝑧𝑒−𝑥2+𝑦2

𝑤2 𝑧

The corresponding intensity profile:

𝐼 𝑥, 𝑦, 𝑧 ∝1

𝑞 𝑧 2𝑒−2

𝑥2+𝑦2

𝑤2 𝑧

= 𝝓 𝒙, 𝒚

x axis

on this circle,

I = constant

The spot is Gaussian:

𝒆−𝒙𝟐

𝒘𝟐 𝒆−𝟐𝒙𝟐

𝒘𝟐

Propagation of Gaussian beams

At 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 - example

Suppose a Gaussian beam (propagating in empty space, wavelength l)

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.

NOTE: If 𝑅 0 = ∞ at a given location, this implies that q is a pure imaginary number at that location: this is a focal point.

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

1

𝑞 0=

1

𝑅 0+ 𝑖

𝜆

𝜋𝑤2 0

= 𝑖𝜆

𝜋𝑤2 0Thus 𝑞 0 = −𝑖

𝜋𝑤02

𝜆

The Rayleigh range

The “Rayleigh range” is 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 define a new constant:

At a focal point:

𝑧𝑅 ≡𝜋𝑤0

2

𝜆

𝑞 0 = −𝑖𝑧𝑅


A distance z later, the new complex beam parameter is:

• 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 = 2𝑤0.

Reminder:𝑞 𝑧 = 𝑞 0 + 𝑧 = −𝑖𝑧𝑅 + 𝑧

1

𝑞 𝑧=

1

𝑧 − 𝑖𝑧𝑅=𝑧 + 𝑖𝑧𝑅

𝑧2 + 𝑧𝑅2

1

𝑞 𝑧=

1

𝑅 𝑧+ 𝑖

𝜆

𝜋𝑤2 𝑧

Re1

𝑞 𝑧=

𝑧

𝑧2 + 𝑧𝑅2 =

1

𝑅(𝑧)

𝑅 𝑧 =𝑧2 + 𝑧𝑅

2

𝑧

Im1

𝑞 𝑧=

𝑧𝑅

𝑧2 + 𝑧𝑅2 =

𝜆

𝜋𝑤2 𝑧

𝑤 𝑧 = 𝑤0 1 +𝑧2

𝑧𝑅2

Rayleigh range and confocal parameter

Confocal parameter: 𝑏 = 2𝑧𝑅

𝑧𝑅

𝑏 = 2𝑧𝑅

beam waist at 𝑧 = 0: 𝑅 = ∞

𝑤0𝑤(𝑧)

𝑅 𝑧

Rayleigh range 𝑧𝑅: the distance from the focal point where the beam

waist has increased by a factor of 2 - the beam area has doubled.

𝑤02𝑤0 2𝑤0

Radius and waist of Gaussian beams

𝑅 𝑧

𝑧𝑅

Note #2: positive values of 𝑅 correspond to a diverging beam, whereas 𝑅 < 0would 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: 𝑅 𝑧 =𝑧2 + 𝑧𝑅

2

𝑧𝑤 𝑧 = 𝑤0 1 +

𝑧2

𝑧𝑅2

0 2 4 6 8 100

2

4

6

8

10

𝑤 𝑧

𝑤0

𝑧/𝑧𝑅

𝑅 𝑧

𝑧𝑅=

𝑧

𝑧𝑅+𝑧𝑅𝑧

𝑤 𝑧

𝑤0= 1 +

𝑧2

𝑧𝑅2

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

10

0

1

2

3

0

1

inte

nsity

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,

𝐼𝑧

The peak drops as the

width broadens, such

that the area (the total

energy) is constant.

The peak intensity drops by

50% after one Rayleigh range.

1

𝑞 𝑧=

1

𝑧 − 𝑖𝑧𝑅

𝑢 0,0, 𝑧 =1

𝑞 𝑧→ 𝐼 𝑧 ∝

1

𝑞 𝑧

2

=1

𝑧2 + 𝑧𝑅2

Suppose, at 𝑧 = 0, a Gaussian beam has a waist of 50 mm 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:

So:

The focal point is the point at which the radius of curvature is infinite.

This implies that 𝑞 is purely imaginary at that point.

𝑞 𝑧 = 𝑧0 + 𝑧 Choose the value of 𝑧 such that Re 𝑞 = 0

This gives

𝑤 = 35 μm.


1

𝑞0=

1

−104 μm+ i

0.786 μm

𝜋 50 μm 2= −1 + 𝑖 10−4μm−1

𝑞0 = −1

21 + 𝑖 104μm

At 𝑧 =104

2μm 𝑞 𝑧 = 𝑖

104

2μm

1

𝑞 𝑧= −𝑖

2

104μm= −𝑖

0.786μm

𝜋𝑤2

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:

𝐼 𝑟 =2𝑃

𝜋𝑤2𝑒−2𝑟2

𝑤2

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

Aperture transmission

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

fractional power transmitted

𝐴 = 𝑤

~86%

~99%

𝐴 =𝜋

2𝑤

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

fra

ctio

na

l p

ow

er

tra

nsm

itte

d=

2

𝜋𝑤2න0

𝐴

2𝜋𝑟𝑒−2𝑟2

𝑤2 𝑑𝑟 = 1 − 𝑒−2𝐴2

𝑤2

Focusing a Gaussian beam

The focusing of a Gaussian beam can be regarded as the reverse of

the propagation problem we did before.

𝑑0

𝐷

A 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: 0

22 #

fd f

D

ll

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 > l.

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

Depth of field 2

2 2 # Rz f l

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? l = 0.532 mm, 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

http://upload.wikimedia.org/wikipedia/commons/5/51/Jonquil_flowers_at_f32.jpghttp://upload.wikimedia.org/wikipedia/commons/0/01/Jonquil_flowers_at_f5.jpg

The q parameter for a Gaussian beam evolves according

to the same parameters used for the Ray matrices!

Gaussian beams and Ray matrices


system

B

DCM

A

in

out

in

Aq Bq

Cq 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

1

0 1

dM

The q parameter after this propagation is given by:

inout in

in

Aq Bq q d

Cq 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

11

in inout

inin

Aq B qq

Cq Dq

f

Example: propagation through a thin lens:1 0

1 1

M

f

11

1 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

𝑅𝑜𝑢𝑡=

1

𝑅𝑖𝑛−1

𝑓

Assume that the Gaussian beam has a focal spot located a distance

𝑑0 before the lens (i.e., at the position of the “object”).

Gaussian beams and imaging

Example: an imaging system

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

0 0ray matrix

1/ / 1/ 1/

i o

o i

d d M

f d d f M

(M = magnification)

(matrix connecting

the object plane to

the image plane)

Then𝑤𝑖𝑛 = beam waist at the input of the system

𝑞𝑖𝑛 = −𝑖𝑧𝑅 = −𝑖𝜋𝑤𝑖𝑛

2

𝜆

If we want to find the beam waist at the output plane, we must

compute the imaginary part of 1/𝑞𝑜𝑢𝑡:

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.

Gaussian beams and imaging

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

𝑞𝑖𝑛 = −𝑖𝑧𝑅 𝑞𝑜𝑢𝑡 =𝑀𝑞𝑖𝑛

−𝑞𝑖𝑛𝑓+1𝑀

=−𝑖𝑧𝑅𝑀

1𝑀+ 𝑖

𝑧𝑅𝑓

Im1

𝑞𝑜𝑢𝑡=

1

𝑀2𝑧𝑅=

1

𝑀2×

𝜆

𝜋𝑤𝑖𝑛2

Im1

𝑞𝑜𝑢𝑡=

𝜆

𝜋𝑤𝑜𝑢𝑡2 → 𝑤𝑜𝑢𝑡 = 𝑀𝑤𝑖𝑛

chapter 4. ray optics and gaussian...

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