Download ABCD law for Gaussian laser beams

Physics 571 Lecture #6
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ABCD law for Gaussian beams
The ABCD law for Gaussian beams enables one to predict the parameters of a Gaussian beam
that exits from an optical system, given the parameters of an input Gaussian beam as well as the
ABCD matrix representing the optical system. The system may be arbitrarily complex with many
optical components.
Keep in mind that ABCD matrices were introduced to describe the propagation of rays. On
the other hand, Gaussian beams are governed by the laws of diffraction. So it may seem surprising
that the ABCD matrices can be helpful here. For example, consider a collimated Gaussian beam
that traverses a converging lens. By ray theory, one expects the Gaussian beam to focus at the
focal point of the lens. However, a collimated beam by definition is already in the act of going
through focus. In the absence of the lens, there is a tendency for the beam to grow via diffraction,
especially if the beam waist is small. This tendency competes with the focusing effect of the lens,
and a new beam waist can occur at a wide range of locations, depending on the exact outcome of
this competition. The ABCD law for Gaussian beams takes care of all of this.
A Gaussian beam is characterized by its Rayleigh range z0 = kw02 /2. From this, the beam waist
radius w0 may be extracted if the wavelength is known. Suppose that a Gaussian beam encounters
an optical system at position z, referenced to the position of the beams waist as shown in Fig. 1.
The beam exiting from the system, in general, has a new Rayleigh range z00 . The waist of the new
beam also occurs at a different location. Let z 0 denote the location of the exit of the optical system,
referenced to the location of the waist of the new beam. If the exiting beam diverges as in Fig. 1,
then it emerges from a virtual beam waist located before the exit point of the system. In this case,
z 0 is taken to be positive. On the other hand, if the emerging beam converges to an actual waist,
then z 0 is taken to be negative since the exit point of the system occurs before the focus.
z
z′
Figure 1: Gaussian laser beam traversing an optical system described by an ABCD matrix. The
dark lines represent the incoming and exiting beams. The gray line represents where the exiting
beam appears to have been.
The ABCD law is embodied in the following relationship:
z 0 − iz00 =
A(z − iz0 ) + B
C(z − iz0 ) + D
(1)
where
√ A, B, C, and D are the matrix elements for the optical system. The imaginary number
i ≡ −1 imbues the law with complex arithmetic. It makes two equations from one because the
real and imaginary parts of Eq. 1 must separately be equal.
We now prove the ABCD law. We begin by showing that the law holds for two specific ABCD
matrices. First, consider the matrix for propagation through a distance d:
A B
1 d
=
.
(2)
C D
0 1
Of course, we know that simple propagation has minimal effect on a beam. For instance, the
Rayleigh range is unchanged, so we expect that the ABCD law should give z00 = z0 . The propagation
through a distance d modifies the beam position by z 0 = z + d . We now check that the ABCD law
agrees with these results by inserting Eq. 2 Eq. 1:
z 0 − iz00 =
1(z − iz0 ) + d
= z + d − iz0 .
0(z − iz0 ) + 1
(propagation through a distance d)
Obviously, the law holds in this case.
Next we consider the ABCD matrix of a thin lens (or
A B
1
=
C D
−1/f
a curved mirror):
0
.
1
(3)
(4)
A beam that traverses a thin lens undergoes the phase shift −kρ2 /2f , as shown in the appendix.
This modifies the original phase of the wave front −kρ2 /2R(z) of the solution to the paraxial wave
equation. The phase of the exiting beam is therefore
kρ2
kρ2
kρ2
=
−
,
2R(z 0 )
2R(z)
2f
(5)
1
1
1
=
−
R(z 0 )
R(z) f
(6)
which we may simplify to
In addition to this relationship, the local radius of the beam w(z) cannot change while traversing
the “thin” lens. Therefore the following relationship is also true:
z 02
z2
0
0
w(z ) = w(z) ⇒ z0 1 + 02 = z0 1 + 2 .
(7)
z0
z0
If the ABCD law for a thin lens is correct for Gaussian beams, we have the relation
z 0 − iz00 =
z − iz0
. (traversing a thin lens with focal length f )
− iz0 ) + 1
− f1 (z
(8)
Let’s assume for the moment that Eq. 8 is correct. By setting the real and imaginary parts of the
left-hand and right-hand sides of this equation equal, we get two useful relationships:
2 − zf + z 2
f
z
0
z0 = − 2
(9)
z − 2zf + f 2 + z02
and
z00 =
f 2 z0
z 2 − 2zf + f 2 + z02
(10)
We can use these to find a relatively simple expression for R(z 0 ) and then show that Eq. 6 is
correct. This completes the proof of the ABCD law for a Gaussian beam traversing a thin lens.
So far we have shown that the ABCD law works for two specific examples, namely propagation
through a distance d and transmission through a thin lens with focal length f (or a concave mirror
with focal length f ). From these elements we can derive more complicated systems. The ABCD
matrix for a thick lens cannot be constructed from just these two elements. However, we can
construct the matrix for a thick lens if we sandwich a thick window (as opposed to empty space)
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between two thin lenses. With these relatively few elements, essentially any optical system can be
constructed as long as the beam propagation begins and ends up in the same index of refraction.
To complete our proof of the general ABCD law, we need only show that when it is applied to
the compound element
A B
A2 B2
A1 B1
A2 A1 + B2 C1 A2 B1 + B2 D1
=
=
,
(11)
C D
C2 D 2
C1 D 1
C 2 A1 + D 2 C 1 C 2 B 1 + D 2 D 1
it gives the same answer as when the law is applied sequentially. In other words, we need to prove
that the ABCD law for Gaussian beams obeys the rules of matrix multiplication. Explicitly, we
have
h
i
A1 (z−iz0 )+B1
A
0
0
2
C1 (z−iz0 )+D1 + B2
A2 (z − iz0 ) + B2
h
i
z 00 − iz000 =
=
C2 (z 0 − iz00 ) + D2
C A1 (z−iz0 )+B1 + D
2
C1 (z−iz0 )+D1
2
=
A2 [A1 (z − iz0 ) + B1 ] + B2 [C1 (z − iz0 ) + D1 ]
C2 [A1 (z − iz0 ) + B1 ] + D2 [C1 (z − iz0 ) + D1 ]
=
(A2 A1 + B2 C1 )(z − iz0 ) + (A2 B1 + B2 D1 )
(C2 A1 + D2 C1 )(z − iz0 ) + (C2 B1 + D2 D1 )
=
A(z − iz0 ) + B
,
C(z − iz0 ) + D
(12)
completing the proof. Thus, we can construct any ABCD matrix that we wish from matrices that
are known to obey the ABCD law. The resulting matrix also obeys the ABCD law.
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Appendix: Phase Shift from a Thin Lens
Consider a monochromatic light field that goes through a thin lens with focal length f . In traversing
the lens, the field undergoes a phase shift. Let us reference this phase shift to that experienced
by the light that goes through the center of the lens. In the Fig. 2, R1 is a positive radius of
curvature, and R2 is a negative radius of curvature, according to our previous convention. We take
the distances `1 and `2 to be positive.
Light passing through the off-axis portion of the lens experiences less material than the light
passing through the center. From the geometry shown in Fig. 3, and using the paraxial ray
approximation, we see that `1 + `2 + d = t, where t is the thickness of the (“thin”) lens. The
difference in optical path length is (1 − n)(`1 + `2 ). This means that the phase of the field passing
through the off-axis portion of the lens relative to the phase of field passing through the center is
∆φ = −k(n − 1)(`1 + `2 ).
(13)
The negative sign indicates a phase advance (i.e. same sign as −ωt) in contrast to a phase delay,
which takes a positive sign. Off axis, the phase advances because the light travels through less
material and gets ahead of the light traveling through the center of the lens.
We can find expressions for `1 and `2 from the equations describing the spherical surfaces of
the lens. As shown in Fig. 4, the Pythagorean theorem gives us:
(R1 − `1 )2 + x2 + y 2 = R12
and
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(R2 − `2 )2 + x2 + y 2 = R22 .
(14)
ℓ1
(x2 + y2)1/2
ℓ2
t
ℓ1 ℓ2
R2
R1
d
Figure 2: A thin lens, which modifies the phase of a field
passing through it.
Figure 3: A zoomed-in
view of Fig. 2.
(x2 + y2)1/2
ℓ1
R1
Figure 4: Another view of Fig. 2 to emphasize the geometry used to find Eqs. 14
In the Fresnel approximation, the light propagation takes place in the paraxial limit. It is therefore
appropriate to neglect the terms `21 and `22 in comparison with the other terms present. Within this
approximation, the Eqs. 14 are easily inverted, and they render
`1 ≈
x2 + y 2
,
2R1
`2 ≈
x2 + y 2
.
2R2
(15)
We are now able to evaluate the phase advance from Eq. 13 in terms of x and y. Substitution
of Eq. 15 into 13 yields
1
1 x2 + y 2
∆φ = −k(n − 1)
−
.
(16)
R1 R2
2
We note that Eq. 16 contains the focal length f of a thin lens according to lens-maker’s formula.
With this identification, the phase introduced by the lens becomes
∆φ =
k
x2 + y 2 .
2f
In summary, the light traversing a lens experiences a relative phase shift given by
k
2
2
E x, y, zafter lens = E x, y, zbefore lens exp −i
x +y
.
2f
4
(17)
(18)
Equation 18 introduces a wave-front curvature to the field. For example, if a plane wave (i.e.
a uniform field E0 ) passes through the lens, the field emerges with a spherical-like wave front
converging towards the focus of the lens.
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