20. gaussian beams · paraxial wave equation complex beam parameter gaussian beams. ray matrices...

20. Gaussian beams

|u(x,y)|2

x y

w = 3

Imaging and the lens law

When ray optics fails

Paraxial wave equation

Complex beam parameter

Gaussian beams

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

θ θ

=

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 θin. 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 do

2) A lens of focal length f3) 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 do > f, the object is more than one focal length away from the lens, then di > 0.

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.

1 1 1

o id d f

+ =

Object

f > 0

Virtual

image

Virtual images

This is a

magnifying

glass.

1 1 1

o id d f

+ =

When do < f, the object is less than one focal length away from the lens, then di < 0.

No real image occurs. But a virtual image is observed if you look back through the lens.

Failures of the ray optics approach

Thin lens,

focal length f

Input beam

(bundle of rays) Distance d = f

What happens in this

plane (focal plane)?

In particular, how big is

the illumination spot?

1 1 0

0 1 -1/ 1 0

=

out in

out

x f x

fθ

Each ray has a unique xin, but

all have θin = 0

(parallel rays)

0

- /

==

out

out in

x

x fθ

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?

Start with the wave equation:2 2 2 2

2 2 2 2 2

E E E 1 E

x y z c t

∂ ∂ ∂ ∂+ + =∂ ∂ ∂ ∂

Also, assume that the variation along the direction of

propagation (z axis) is given by:

ejkz × (a slowly varying function of z)

Assume a harmonic time dependence, of the form ejωt.

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

propagation direction (z axis)

E field amplitude changes along x,

slowly compared to the wavelength

E field envelope changes along z,

slowly compared to the wavelength

λ

optic axis

These assumptions imply: E(x,y,z,t) = u(x,y,z) · ejωt − jkz

and: ( )2 2 2

2 2 2, , and

∂ ∂ ∂ ∂<< =∂ ∂ ∂ ∂

u u u uk k c

z x y zω

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

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

propagation direction (z axis)

E field amplitude changes along x,

slowly compared to the wavelength

E field envelope changes along z,

slowly compared to the wavelength

λ

optic axis

What if ray optics is not good enough?

Paraxial wave equation

E(x,y,z,t) = u(x,y,z) · ejωt - jkz

( )

( )

( )

2 2j t kz

2 2

2 2j t kz

2 2

2j t kz2

2

E ue

x x

E ue

y y

Eu e

t

ω −

ω −

ω −

∂ ∂=∂ ∂∂ ∂=∂ ∂∂ = −ω∂

x, y, and t derivatives are easy: z derivative:

( ) ( )

( ) ( )

( )

j t kz j t kz

2j t kz j t kz2

2

2j t kz

2

E ujk ue e

z z

E uk ue 2jk e

z z

ue

z

ω − ω −

ω − ω −

ω −

∂ ∂= − × +∂ ∂

∂ ∂= − × − ×∂ ∂

∂+∂

wave equation becomes:

2 2 22

2 2 2

u u uk u 2jk u

x y z c

∂ ∂ ∂ ω+ − − = −∂ ∂ ∂

paraxial approximation

2 2

2 2

u u u2jk 0

x y z

∂ ∂ ∂+ − =∂ ∂ ∂

the “paraxial wave equation”

Paraxial and exact wave equations

2 2

2 2

u u u2jk 0

x y z

∂ ∂ ∂+ − =∂ ∂ ∂

2 2 2 2

2 2 2 2 2

E E E 1 E

x y z c t

∂ ∂ ∂ ∂+ + =∂ ∂ ∂ ∂

E(x,y,z,t) = u(x,y,z) · ejωt − jkz

exact wave equation Paraxial wave equation

( )( )

( )0

0

− −

=−

jk r re

E rr r

A well-known solution to the exact 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?

( ) ( )( )

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 ε ( ) ( )( )

2 2

0 0

2

0

− + −=

−

x x y y

z zεwhere

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:

11 1

2+ ≈ + +…ε ε

Therefore:

( )( )

( ) ( )( )

( ) ( ) ( )( )

2 20 0

0 02

2 2

0 0

0

0

, , ,

2

− + −−

− − −

=− + −

− +−

x x y yjk

jk z z z z

j te eE x y z t e

x x y yz z

z z

ω

Plug this approximate form for r − r0 into the expression

for the spherical wave:

We will ignore

this term

because it is

small.

Paraxial approximation to a spherical wave

A paraxial spherical wave:

( )

( ) ( )( )

( )

2 2

0 0

0

0

exp2

, ,

− + − − − =−

x x y yjk

z zu x y z

z z

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.

( ) ( )

( ) ( )( )

( )

2 2

0 0

0

0

exp2

, , , , , − + − +

− + − − − = =−

jkz j t jkz j t

x x y yjk

z zE x y z t u x y z e e

z z

ω ω

( )

( ) ( )( )

( )

2 2

0 0exp2

, ,

− + − − =

x x y yjk

R zu x y z

R z

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:

( ) ( ) ( )( )

2 2

0 0,

2

− + −=

x x y yx y k

R zφ

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.

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!

( )

( ) ( )( )

( )

2 2

0 0exp

2, , , −

− + − − = jkz j t

x x y yjk

R zE x y z t e

R z

ω

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”

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)

( ) ( ) ( )2 21

, , exp2

+= −

x yu x y z jk

q z q z

Complex source point

Also, without loss of generality, we can set x0 = y0 = 0.

complex real positions

(Note: q has

units of length)

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

Re Imjq z q z q z

= +

This part defines the phase, i.e., the wave fronts

This part is real! It tells us the x and y dependence of the amplitude!

Complex beam parameter q(z)

( ) ( ) ( )2 2

1 1 1exp Re Im

2

+ = − +

x yjk j

q z q z q z

This part plays the same role as R(z), the phase front radius of curvature.

This part is exp{real number}! It gives us the finite extent in the transverse dimension

( ) ( ) ( ) ( )2 2 2 21 1 1

, , exp Re exp Im2 2

x y x yu x y z jk k

q z q z q z

+ + = −

( ) ( ) ( )2 21 1

, , exp2

+= −

x yu x y z jk

q z 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!).

� the imaginary part of 1/q(z) MUST be negative! Otherwise,

u(x,y,z) diverges to infinity as x,y increase.

( ) ( ) ( )2

1 1= − jq z R z w z

λπ

Redefine R(z) and define w(z) such that

Constraints on q(z)

( ) ( ) ( ) ( ) ( )2 2

2 21 1 1, , exp Re exp Im

2

πλ

+ = − +

x yu x y z jk x y

q z q z q z

( ) ( ) ( ) ( )2 2 2 2

2

1, , exp exp

2

+ += − −

x y x yu x y z jk

q z R z w z

In that case, we have:

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

( ) ( ) ( ) ( )2 2 2 2

2

1, , exp exp

2

+ += − −

x y x yu x y z jk

q z R z w z

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 not 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)}

Gaussian beams

( ) ( ) ( ) ( )2 2 2 2

2

1, , exp exp

2

+ += − −

x y x yu x y z jk

q z R z w z

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.