Chapter 3. Ray Optics and Gaussian...
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Chapter 3. Ray Optics and Gaussian Beam Reading: Saleh Chapter 1 and Siegman Chapter 16 and 17
Transcript of Chapter 3. Ray Optics and Gaussian...
Chapter 3.
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.
axis input
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, qin
xout, qout
• its position, x
• its slope, q
Optical axis
x
q
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.”
A B
C D
Optical system ↔ 2x2 Ray matrix
in
in
x
q
out
out
x
q
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.”
Easiest example: rays in free space or a uniform
medium
If xin and qin are the position and slope at z = 0, and xout
and qout are the position and slope after propagating
from z = 0 to z = z0, then:
0out in in
out in
x x z
q
q q
xin, qin
z = 0
xout qout
z0
01
0 1
out in
out in
x xz
q q
Rewriting this expression
in matrix notation:
(notice the small angle approximation: tan q q )
Ray Matrix for an Interface
At the interface, clearly:
xout = xin.
Now calculate qout.
qout = [n1 / n2] qin
qin
n1
qout
n2
xin xout
1 2
1 0
0 /interfaceM
n n
Snell's Law says: n1 sin(qin) = n2 sin(qout)
which becomes for small angles: n1 qin = n2 qout
Ray matrix for a curved interface
1 2 1 2
1 0
( / 1) / /
curvedinterface
Mn n R n n
Now the output angle depends on the input position, too.
n1 n2
q1 q2 If the interface has
spherical curvature:
For cascaded elements, multiply ray
matrices
3 2 1
out in
out in
x xM M M
q q
Notice that the order looks opposite to what it should be.
Order matters!
M1 M3 M2 in
in
x
q
out
out
x
q
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=1
1 0
1/ 1f
This can be written:
“Lens-Maker’s Formula”
where: 1 2
1 1 11
n
f R R
𝑀𝑅 =
1 0
−𝑛2 − 𝑛1𝑛2𝑅
𝑛1𝑛2
𝑀 = 𝑀𝑅2𝑀𝑅1 =
1 0
−1 − 𝑛
𝑅2𝑛
1 0
−𝑛 − 1
𝑛𝑅1
1
𝑛
=
1 0
− 𝑛 − 1 1
𝑅1−1
𝑅21
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.
1 0=
1/ 1
lensMf
If f > 0, the lens deflects
rays toward the axis.
f > 0 R1 > 0
R2 < 0
If f < 0, the lens deflects
rays away from the axis.
f < 0 R1 < 0
R2 > 0
different types of lenses:
1 2
1 1 11
n
f R R
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
fq
f
f
A lens followed by propagation by one focal length:
Assume all input
rays have qin = 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
Optical system ↔ 2x2 Ray matrix
We describe the propagation of rays through a paraxial
optical system using ray matrices.
out in
out inD
B x
C
x A
q q
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 qin. This is what we mean by “image”.
xout = A xin When 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 do
2) 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 / 1i i
o i
A d f dd d
i oM d d
(writing A = M, magnification)
1 11 / 1o 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
fq
Each ray has a
unique xin, but
all have qin = 0
(parallel rays)
0
- /
out
out in
x
x fq
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 (𝑧 axis) 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
𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡
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−𝑖𝜔𝑡 We 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 =𝑒𝑖𝑘𝑥−𝑥0
2+ 𝑦−𝑦02
2 𝑧−𝑧0
𝑧 − 𝑧0
𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡 =𝑒𝑖𝑘𝑥−𝑥0
2+ 𝑦−𝑦02
2 𝑧−𝑧0
𝑧 − 𝑧0𝑒𝑖𝑘𝑧−𝑖𝜔𝑡
R(z) = z z0 To 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 =𝑒𝑖𝑘𝑥−𝑥0
2+ 𝑦−𝑦02
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
, ,0,0
E x y tR
It has a constant transverse profile, even for x,y
Not a particularly realistic description of a “beam”
𝐸 𝑥, 𝑦, 𝑧, 𝑡 =𝑒𝑖𝑘𝑥−𝑥0
2+ 𝑦−𝑦02
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
𝑞 𝑧exp 𝑖𝑘
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
𝑞 𝑧exp 𝑖𝑘
𝑥2 + 𝑦2
2Re
1
𝑞 𝑧exp −𝑘
𝑥2 + 𝑦2
2Im
1
𝑞 𝑧
Define: R(z) and w(z) such that 1
𝑞 𝑧=1
𝑅 𝑧+ 𝑖
𝜆
𝜋𝑤2 𝑧
In that case, we have:
𝑢 𝑥, 𝑦, 𝑧 =1
𝑞 𝑧exp 𝑖𝑘
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
𝑞 𝑧exp 𝑖𝑘
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
𝑞 𝑧exp 𝑖𝑘
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
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:
𝐸 𝑥, 𝑦, 𝑧, 𝑡 = 𝑢 𝑥, 𝑦, 𝑧 𝑒𝑖𝑘𝑧−𝑖𝜔𝑡
𝑢 𝑥, 𝑦, 𝑧 =1
𝑞 𝑧exp 𝑖𝑘
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−2𝑥2+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 0 Thus 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 ≡𝜋𝑤02
𝜆
𝑞 0 = −𝑖𝑧𝑅
Propagation of Gaussian beams - example
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 = ∞
𝑤0 𝑤(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.
𝑤0 2𝑤0 2𝑤0
Radius and waist of Gaussian beams
𝑅 𝑧
𝑧𝑅
Note #2: positive values of 𝑅 correspond
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: 𝑅 𝑧 =𝑧2 + 𝑧𝑅
2
𝑧 𝑤 𝑧 = 𝑤0 1 +
𝑧2
𝑧𝑅2
0 2 4 6 8 10 0
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
𝑧𝑅 𝑧𝑅
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 6 0
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.
Propagation of Gaussian beams - example
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
2 μm 𝑞 𝑧 = 𝑖
104
2 μm
1
𝑞 𝑧= −𝑖
2
104μm= −𝑖0.786μm
𝜋𝑤2
Aperture transmission The 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 = 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 2 0
0.2
0.4
0.6
0.8
1
𝐴/𝑤
fractional pow
er
transm
itte
d =2
𝜋𝑤2 2𝜋𝑟𝑒
−2𝑟2
𝑤2 𝑑𝑟𝐴
0
= 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 𝑧𝑅 larger 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
The q parameter for a Gaussian beam evolves according
to the same parameters used for the Ray matrices!
Gaussian beams and Ray matrices
Optical system ↔ 2x2 Ray matrix
system
B
DCM
A
inout
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)
Then 𝑤𝑖𝑛 = beam waist at
the input of the system 𝑞𝑖𝑛 = −𝑖𝑧𝑅 = −𝑖
𝜋𝑤𝑖𝑛2
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+ 𝑖𝑧𝑅𝑓
Im1
𝑞𝑜𝑢𝑡=1
𝑀2𝑧𝑅=1
𝑀2×𝜆
𝜋𝑤𝑖𝑛2
Im1
𝑞𝑜𝑢𝑡=𝜆
𝜋𝑤𝑜𝑢𝑡2 → 𝑤𝑜𝑢𝑡 = 𝑀𝑤𝑖𝑛
