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Wave and Beam optics
The fundamentals of optics at the introductory level are mostly treated using the “plane wave” model. The plane waves have an infinite extent in the transverse direction. Plane waves, however, never existed in the real world. This lecture is to establish the basics of waves of finite cross section.
Wave function Light propagates in the form of waves. In free space, the waves are governed by the wave equation,
(^1 ) 2 2 (^2) = ∂
∇ − ∂ t
U c
U (1)
where u is a function of r and t. In a one-dimensional system, the solutions are u(x-ct) and u(x+ct).
From the theory of electromagnetics, both the E and H fields of the electromagnetic waves in vacuum satisfies Eq. (1). The form of Eq.(1) for E and H can be derived from Maxwell’s equations.
The monochromatic harmonic waves can be expressed as
j r j t
U rt a r e
φ () 2 πν
−
where U is known as the complex wave function, and the real part, u(r,t), is given by
u ( r , t )= U + U^ * (3) Why using the harmonic waves?
The “ wavefronts” are the surfaces of equal phase φ ( r )= cons tan t. The optical intensity (power
per unit area) is given by 2 I ( r )= U ( r ).
The Helmhotz Equation: ( governing the spatial dependency of monochromatic waves)
0
0
0 '
0
∇⋅ =
∇⋅ =
∂
∂ ∇× =−
∂
∂ ∇× =
H
E
Maxqwellsequationinfreespace t
H E
t
E H
μ
ε
Try to prove that both the E and H satisfy equation (1).
For a given harmonic function U , the spatial-temporal differential equation in (1) can be reduced to a differential equation in space only. The function U in Eq.(2) satisfies the following equation:
c c
where k
k Ur
is the wave number.
Solution 1: plane waves
ik r j t U r t Ae
2 πν ( , )
− • −
(5)
where A represents the magnitudes of the electric and magnetic fields. The plane of constant phase is perpendicular to k. ( This can be easily proved by choosing one of the coordinate axes to be along the direction of k. Then the derivatives of A in the orthogonal directions are zero.)
From Maxwell’s equations, the vectors of E, H and k are orthogonal to each other.
Solution 2: Spherical waves in the outward direction from the origin.
jkr j t e r
A U r t 2 πν ( , ) − •− =
(6)
Can be proved by expressing the Laplacian in the spherical coordinates.
This is the wave originating from a point and propagating outward with diminishing intensity with the square of the distance.
Solution 3: Fresnel approximation of spherical waves (Paraboloidal waves)
At a point close to the z-axis but far from the origin so that ( x^2 + y^2 )^1 /^2 << z
z jkx y jkz e e z
A U r t^2
2 2
( , )
−^ + = (^) (7)
Paraxial rays The plane-wave solution propagating in the z-direction is given by
U ( r )= Ae^ ikz
If the constant amplitude is allowed to vary slowly,
U ( r )= A ( z ) e^ ikz (8)
where A(z) must vary slowly with position z within the distance of a wavelength. This is the expression for the paraxial wave in the z-direction. See picture Figure 2.2-
Now A(r) becomes the complex envelop of the Gaussian beam. The parameter z 0 is the Rayleigh range. ( What is the physical meaning of the Rayleigh range? )
To separate the real and imaginary parts of the parameter q(z), we express it using the following relation:
qz Rz j π W^2 z
(The meaning of R and W will be apparent later.)
From Eq.(8), (12), and (13), the wave function for Gaussian beam can be expressed as
( )] 2 ( )
] [ ()
[ 0 0
2 2
2
( )
( )
j z Rz
jkz jk W z e e W z
W U r A
− ρ^ − − ρ + ζ = (^) (14)
where
21 / 2 0
( ) 0 [ 1 ( z ) ]
z W z = W + Beam size at 1/e^2 (15)
( ) [ 1 (^0 )^2 ]
z
z R z = z + Radius of curvature (16)
0
0 π W
θ = Divergence angle (17)
0 1 /^2
λ z
W = Beam waist at minimum (18)
( ) tan ( ) 0
1 z
z
ζ z = − (19)
Properties of Gaussian beam Intensity at any position r : From (14),
] ( )
[^2 0 2 0
2
2
] ( )
( , ) [ W z e W z
W I z I
ρ
ρ
− = (^) (20)
where W (z) increases with z according to the following relation:
21 / 2 0
( ) 0 [ 1 ( z ) ] z W z = W +
Thus the bema expands with increasing z.
The axis is chosen so that the minimum waist occur at z =0. The beam expands and peak intensity decrease with increasing z , according to (20).
Rayleigh range : the distance over which the beam size expands by 2 times or the intensity reduces by a factor or two.
Depth of focus --- λ
π 02 0
W
z = (21)
Examples: For λ=1.0 μm W 0 = z0 = 1 cm 31400 cm 1 mm 314 cm 100 μm 3 cm Beam divergence Comparison with the case of plane waves restricted by an aperture:
D
θ = 1. 22 (half angle of divergence of plane waves of diameter D.
What is the difference in beam profile between Gaussian and truncated plane wave?
For an open aperture of 1 cm and λ=1.0 μm, the divergence, from the center to the first minimum , is 12.2 micro-rad.
For a Gaussian beam of 1-cm waist, the divergence, measured from the center to 1/e^2 is 29 micro-rad.
Definitions:
Far field z>>z 0 In the far field, the beam profile remains constant and expands linearly with z. The shape of the beam is simple and does not change with distance.
Near field : z on the same order of z 0 The amplitude and phase changes with distance rapidly.
Phase of of Gaussina beam From Eq.(14), the imaginary part of the exponents is
2
R z
k z kz z
ρ φ ρ = −ζ + (22)
Along the axis,when ρ=0, the phase has two components. The first component is the phase of plane waves and the second is the correction factor given by (19). The correction is negative. The total accumulated phase deviation from the plane wave by propagating from z=-∞ to z=+∞ is π. For a Gaussian beam, the wave is a superposition of plane waves in various directions, resulting a delay in the z-direction of the composite waves.
A Review of the ABCD Law of the ray optics ( Chapter 1 )
In geometric optics, an arbitrary optical system can be represented by a 2x2 matrix, which transforms the position and angles of the incident beam into the position and angle of the outgoing beam.
^ =
' ' in
in
out
out r
r
C D
A B
r
r (23)
To s how a diagram
Common optical systems:
Homogeneous medium of length d (^)
1 d (24)
Thin lens of focal length f
f
Dielectric interface
2
n
n (^) (26)
Spehrical mirror of radius (concave)
R
A medium with quadratic index profile
− sin( ) cos( )
cos( ) sin( )
2 2 2
2 2
2
l k
k l k
k k
k
l k
k k
k l k
k
(28)
where the index of refraction varies with the lateral position according to
= 0 −^22
(ρ ) 1 ρ k
k n n and l is the length. (29)
Examples of application of the ABCD law to ray optics.
- A ray of a given position and slope propagating in free space.
- Snell’s law
- Parallel rays incident on converging lens.
- Special case for a ray parallel to the axis entering a quarter-pitch, half-pitch graded-index lens.
The ABCD law for the beam optics The q parameter of the transmitted Gaussian beam through an optics is related to that of the incident beam by the following relations:
Cq D
Aq B q
1
1 2
Where q 1 is the incident and q 2 is the transmitted and A,B, C and D are the elements of the ABCD matrix.
Example:
- A Gaussian beam has a waist of W 0. Find the q -parameter, beam size and radius of curvature after propagating through a uniform medium of distance z.
The waist at z=0 (^2) 0
( 0 )^2 ( 0 )
W
j W
j
q R π
From the ABCD law 0 1
( )^0
q z q z (30)
q(z) is also defined through Eq. (13) By equating the real and imaginary parts of the Eqs (13) and (30), the result should be the same as Eqs.(15) and (16)
- A lens of focal length f is placed at the waist of a Gaussian beam of waist W 0. Find the location and size of the minimum spot of the transmitted beam.
f
q(0) q 1 q^2
W 0 l
W 2
Approach: the beam is transformed by the lens followed by the propagation in free-space of distance l after the lens. The q-value may be calculated for each step.
( 1 ) ] ( )^2 ( )
[
0 , 1
2 2
2
) ( )
) ( 2 ( )
( 2 ( )
( , )
− ρ^ − ρ − + + + ζ
= ×
jkz jl m R z
jk W z
lm l m
e
W z
y H W z
x H W z
W U x y A
Where Hl and Hm are the Hermite polynomial of order l and m. The rest of the parameters are defined the same as for the fundamental Gaussian mode.
The phase shift along the axis is the larger. The transverse variation follows the Gaussian envelop function with a modulation defined by Hl and Hm.
HOMEWORK
- Starting from Maxwell equations, prove that the electric and magnetic fields in free space follows the wave equation Eq.(1). List all the assumptions and how the speed of the light is related to vacuum susceptibility and permeability. The derivation can be found in any book in the Chapter of electromagnetic waves. You need to do the derivation once.
- A lens of focal length f is placed at the waist of a Gaussian beam of waist W0. Find the location and size of the minimum spot of the transmitted beam. Determine the minimum spot size of the focused beam.
- Problem 1.4.2: Ray-Transfer Matrix of a GRIN Plate. Determine the ray-transfer matrix of a SELFOC
plate [i.e., a graded-index material with parabolic refractive index ) 2
n ( y )≈ n 0 ( 1 − α^2 y^2 ]
of width d. Problem 1.4.3: The GRIN Plate as a Periodic System. Consider the trajectories of paraxial rays inside a SELFOC plate normal to the z axis. This system may be regarded as a periodic system made of a sequence of identical continuous plates of thickness d each. Using the result of Problem 1.4.2, determine the stability condition of the ray trajectory. Is this condition dependent on the choice of d?
- Problem 3.1.1: Beam Parameters. The light from a Nd:YAG laser at wavelength 1.06 μ m is at a Gaussian beam of 1-W optical power and beam divergence 2 θ 0 = 1 mrad. Determine the beam waist radius, the depth of focus, the maximum intensity, and the intensity on the beam axis at a distance z = 100 cm from the beam waist.
Problem 3.1.2: Beam Identification by Two Widths. A Gaussian beam of wavelength λ 0 = 10. 6 μ m (emitted by a CO (^) 2 laser)has widths W 1 (^) = 1. 699 mm and W (^) 2 = 3. 38 mm at two points separated by a distance d = 10 cm. Determine the location of the waist and the waist radius.
Problem 3.2.1: Beam Focusing. An argon-ion laser produces a Gaussian beam of wavelength λ = 488 nm and waist radius W (^) 0 = 0. 5 mm. Design a single-lens optical system for focusing the light to a spot of diameter 100 μ m .What is the shortest focal-length lens that may be used?
