When Equation (\ref{9.31}) is introduced into the expression for the electric field, Equation (\ref{9.27}), the imaginary part of 1/\(\tilde{\text{q}}\) gives rise to a Gaussian spatial variation

\[\text {Exponent}=-\left(\frac{\text{w}_{0}^{2}}{4}+\frac{i \text{z}}{2 \text{k}_{0}}\right)\left[\text{p}-\frac{2 i \text{x}}{\left(\text{w}_{0}^{2}+\frac{2 \text{i} x}{\text{k}_{0}}\right)}\right]^{2} \label{9.41}\]

The radius of curvature is infinite at z=0 corresponding to a plane wave-front. For z ≫ zR the radius of curvature approaches the distance z.

\[\frac{1}{w_{0}^{2}}\left[x^{2}+i p w_{0}^{2} x\right]=\frac{1}{w_{0}^{2}}\left[x+\frac{i p w_{0}^{2}}{2}\right]^{2}+\frac{p^{2} w_{0}^{2}}{4}. \label{9.37}\]

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Gaussianbeamradius

Notice that the amplitude function (\ref{9.24}) becomes very small if p2 or q2 is greater than \(4 / \text{w}_{0}^{2}\): : this means that the waves in the bundle describing the radiation beam that have transverse components p,q much larger than ±2/w0 can be neglected. In a typical case the laser beam radius is ∼1 mm so that the amplitude A(p,q) becomes small for | p |, | q | larger than 2 × 103 m−1 . But at optical frequencies λ ∼ (1/2)×10−6 m so that k0 ∼ 2\(\pi\)/λ ∼ 4\(\pi\) ×106 m−1 . Thus the important values of the transverse components p,q of the plane waves that make up the beam are very small compared with the total wavevector k0. The longitudinal component of the wave-vector, the z-component k, is given by

It is impossible to generate an unbounded plane wave, of course. Nevertheless, the concept of unbounded plane waves is a very useful one because a finite beam of radiation can be described as the superposition of plane waves having different amplitudes and phases and propagating in slightly different directions, see Figure (9.3.8). To simplify matters let us assume that the amplitude function, A(p,q), is symmetric in p,q: ie. A(-p,-q)= A(p,q). This simplification allows one to construct a beam in which the electric field is polarized along a particular direction in the plane- along the x-direction, say. Eqn.(\ref{9.21}) illustrates how such a beam could be constructed:

Gaussianbeamdivergence

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\[\text{A}(\text{p}, \text{q})=\frac{\text{E}_{0} \text{w}_{0}^{2}}{4 \pi} \exp \left(-\frac{\text{w}_{0}^{2}}{4}\left(\text{p}^{2}+\text{q}^{2}\right)\right). \label{9.24}\]

\[\begin{aligned} \frac{\text{E}_{0} \text{w}_{0}^{2}}{4\left(\frac{\text{w}_{0}^{2}}{4}+\frac{\text{i} \text{z}}{2 \text{k}_{0}}\right)} &=\frac{\text{E}_{0}}{\left(1+\frac{i \text{z}}{\text{z}_{\text{R}}}\right)} \\ &=\frac{\text{E}_{0}\left[1-i \text{z} / \text{z}_{\text{R}}\right]}{\left(1+\left[\text{z} / \text{z}_{\text{R}}\right]^{2}\right)} \\ &=\frac{\text{E}_{0}}{\sqrt{1+\left(\text{z} / \text{z}_{\text{R}}\right)^{2}}} \exp (-i \psi) \end{aligned}\]

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This means that as one moves along the beam the radius of the beam slowly increases and becomes greater by \(\sqrt{2}\) at z = zR: ie. at one Rayleigh range removed from the minimum beam radius, or beam waist.

\[\text{I}=\exp \left(\frac{-\text{x}^{2}}{\left(\text{w}_{0}^{2}+\frac{2 \text{i} \text{z}}{\text{k}_{0}}\right)}\right) \int_{-\infty}^{\infty} \text{du} \exp \left(-\left[\frac{\text{w}_{0}^{2}}{4}+\frac{i \text{z}}{2 \text{k}_{0}}\right] \text{u}^{2}\right), \nonumber \]

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\[\text{E}_{\text{x}}(\text{x}, \text{y}, \text{z})=\frac{\text{E}_{0}}{\sqrt{1+\left(\text{z} / \text{z}_{\text{R}}\right)^{2}}} \exp \left(\frac{i \text{k}_{0}}{2 \tilde{\text{q}}}\left(\text{x}^{2}+\text{y}^{2}\right)\right) \exp \left(i\left[\text{k}_{0} \text{z}-\psi\right]\right), \label{9.27}\]

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\[\text{A}(\text{p}, \text{q})=\frac{1}{4 \pi^{2}} \int \int_{-\infty}^{\infty} \operatorname{dxdy} \text{E}_{\text{x}}(\text{x}, \text{y}, 0) \exp (-i[\text{px}+\text{qy}]). \label{9.23}\]

A time dependence exp (−iωt) is assumed, but this factor will be suppressed in the following. The output beam from a typical gas laser, a He-Ne laser for example, exhibits the spatial variation (\ref{9.22}) at the output mirror with w0 approximately equal to 1 mm. Such a beam profile is called a Gaussian beam profile. The spatial Fourier integral in (\ref{9.21}) can be inverted for z=0 to obtain

\[\text {Exponent}=-\frac{\text{w}_{0}^{2}}{4}\left(1+\frac{2 i \text{z}}{\text{w}_{0}^{2} \text{k}_{0}}\right)\left[\text{p}^{2}-\frac{4 i \text{px}}{\text{w}_{0}^{2}\left(1+\frac{2 i \text{z}}{\text{w}_{0}^{2} \text{k}_{0}}\right)}\right]. \label{9.40}\]

\[\text{E}_{\text{x}}(\text{x}, \text{y}, \text{z})=\frac{\text{E}_{0} \text{w}_{0}^{2}}{4\left(\frac{\text{w}_{0}^{2}}{4}+\frac{i \text{z}}{2 \text{k}_{0}}\right)} \exp \left(\frac{i \text{k}_{0}\left[\text{x}^{2}+\text{y}^{2}\right]}{2 \tilde{\text{q}}}\right) \exp \left(i \text{k}_{0} \text{z}\right), \label{9.43}\]

\[\text{E}_{\text{x}}(\text{x}, \text{y}, \text{z}, \text{t})=\int \int_{-\infty}^{\infty} \operatorname{dpdq} \text{A}(\text{p}, \text{q}) \exp (i[\text{px}+\text{qy}+\text{kz}-\omega \text{t}]), \label{9.21}\]

\[\text{A}(\text{p}, \text{q})=\frac{\text{E}_{0}}{4 \pi} \text{w}_{0}^{2} \exp \left(-\frac{\text{w}_{0}^{2}}{4}\left[\text{p}^{2}+\text{q}^{2}\right]\right). \nonumber\]

\[\text{A}(\text{p}, \text{q})=\frac{\text{E}_{0}}{4 \pi^{2}} \int \int_{-\infty}^{\infty} \text{d} \text{xdy} \exp \left(-\left[\frac{\text{x}^{2}}{\text{w}_{0}^{2}}+i \text{px}\right]\right) \exp \left(-\left[\frac{\text{y}^{2}}{\text{w}_{0}^{2}}+i \text{qy}\right]\right). \label{9.35}\]

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Gaussianbeamprofile

\[\text{I}=\int_{-\infty}^{\infty} \text{dp} \exp \left(-\frac{\text{w}_{0}^{2} \text{p}^{2}}{4}+i \text{px}-i \frac{\text{p}^{2} \text{z}}{2 \text{k}_{0}}\right). \label{9.39}\]

See Section(9.4.2) for the details of the calculation. The variable \(\tilde{\text{q}}\) is called the complex radius of curvature of the beam. This nomenclature stems from the description of a spherical wave-front, Figure (9.4.9) as will be explained in the next paragraph. The length zR is called the Rayleigh range.

Gaussianbeamformula

\[\text{I}=\int_{-\infty}^{\infty} \text{d} \text{x} \exp \left(-\left[\frac{\text{x}^{2}}{\text{w}_{0}^{2}}+i \text{px}\right]\right). \label{9.36}\]

Gaussianbeampdf

Interested readers can learn more about Gaussian beams and Gaussian beam optics in the book ”An Introduction to Lasers and Masers” by A.E. Siegman, McGraw-Hill, New York, 1971; chapter 8.

\[\frac{1}{\tilde{\text{q}}}=\frac{1}{\text{z}-\text{z}_{\text{R}}}=\frac{\text{z}+\text{i} \text{z}_{\text{R}}}{\text{z}^{2}+\text{z}_{\text{R}}^{2}}. \label{9.31}\]

The beam radius at the output mirror, the position of the minimum beam radius, is usually \(\text{w}_{0} \cong 1 \text{mm}\) for a typical gas laser operating in the visible. For a wavelength of λ = 5 × 10−7 meters the Rayleigh range for such a laser is zR= 6.28 meters. Therefore the beam diameter will have expanded by only \(\sqrt{2}\) = 1.41 at a distance of 6.28 meters from the laser output mirror.

\[\text {Exponent}=-\frac{\text{w}_{0}^{2}}{4}\left(\text{p}^{2}-\frac{4 i \text{px}}{\text{w}_{0}^{2}}+\frac{2 i \text{p}^{2} \text{z}}{\text{w}_{0}^{2} \text{k}_{0}}\right), \nonumber \]

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Gaussianbeamsoftware

\[\exp \left(\frac{-\text{k}_{0} \text{z}_{\text{R}}\left(\text{x}^{2}+\text{y}^{2}\right)}{2\left(\text{z}^{2}+\text{z}_{\text{R}}^{2}\right)}\right)=\exp \left(-\frac{\left(\text{x}^{2}+\text{y}^{2}\right)}{\text{w}^{2}}\right), \label{9.33}\]

This page titled 9.4: Gaussian Light Beams is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by John F. Cochran and Bretislav Heinrich.

\[\text{E}_{\text{x}}(\text{x}, \text{y}, \text{z})=\frac{\text{E}_{0}}{\sqrt{1+\left(\text{z} / \text{z}_{\text{R}}\right)^{2}}} \exp \left(\frac{i \text{k}_{0}}{2 \tilde{\text{q}}}\left[\text{x}^{2}+\text{y}^{2}\right]\right) \exp \left(i\left[\text{k}_{0} \text{z}-\psi\right]\right), \label{9.46}\]

where R is the radius of curvature. A comparison of this expression with Equation (\ref{9.27}) shows why \(\tilde{\text{q}}\) is called the complex radius of curvature. One can separate the reciprocal of the complex radius of curvature into its real and imaginary parts:

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Gaussianbeamintensity formula

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Now using the approximation Equation (\ref{9.25}) in Equation (\ref{9.21}) investigate the beam profile at some arbitrary value of z:

\[\text{I}=\frac{\sqrt{\pi}}{\sqrt{\frac{\text{w}_{0}^{2}}{4}+\frac{i z}{2 \text{k}_{0}}}} \exp \left(\frac{i \text{k}_{0} \text{x}^{2}}{2\left[\text{z}-i \text{k}_{0} \text{w}_{0}^{2} / 2\right]}\right). \label{9.42}\]

\[\text{E}_{\text{x}}(\text{x}, \text{y}, 0)=\text{E}_{0} \exp \left(\frac{-\left(\text{x}^{2}+\text{y}^{2}\right)}{\text{w}_{0}^{2}}\right). \label{9.22}\]

\[\begin{aligned} \text { Exponent } &=-\frac{\text{w}_{0}^{2}}{4}\left(1+\frac{2 i \text{z}}{\text{w}_{0}^{2} \text{k}_{0}}\right) \\ & \cdot\left(\left[\text{p}-\frac{2 i \text{x}}{\left(\text{w}_{0}^{2}+\frac{2 i \text{z}}{\text{k}_{0}}\right)}\right]^{2}+\frac{4 \text{x}^{2}}{\left(\text{w}_{0}^{2}+\frac{2 \text{i} z}{\text{k}_{0}}\right)^{2}}\right), \end{aligned}\]

\[\text{I}=\exp \left(-\frac{\text{p}^{2} \text{w}_{0}^{2}}{4}\right) \int_{-\infty}^{\infty} \text{du} \exp \left(-\frac{\text{u}^{2}}{\text{w}_{0}^{2}}\right)=\text{w}_{0} \sqrt{\pi} \exp \left(-\frac{\text{p}^{2} \text{w}_{0}^{2}}{4}\right). \label{9.38}\]

\[\text{k}^{2}=\left(\frac{\omega}{\text{c}}\right)^{2}-\text{p}^{2}-\text{q}^{2}=\text{k}_{0}^{2}-\text{p}^{2}-\text{q}^{2}. \nonumber \]

Eqn.(\ref{9.21}) is an example of a Fourier Integral. The amplitude function A(p,q) can be chosen to give the required beam profile in the x-y plane at some plane z=constant; it is convenient to choose this plane to be at z=0. The beam profile at any other position z can be obtained using the integral (\ref{9.21}). As an example of how this works let us treat a specific case for which the mathematics can be easily worked out. Suppose that at z=0 the beam cross-section can be described as a plane wave whose amplitude falls off exponentially along x and y:

Gaussianbeamcalculator

\[\begin{align} \mathrm{E}_{\mathrm{x}}(\mathrm{x}, \mathrm{y}, \mathrm{z})=& \frac{\mathrm{E}_{0} \mathrm{w}_{0}^{2}}{4 \pi} \int \int_{-\infty}^{\infty} \operatorname{dpdq} \exp \left(-\frac{\mathrm{w}_{0}^{2}}{4}\left[\mathrm{p}^{2}+\mathrm{q}^{2}\right]\right). \label{9.26} \\ & \cdot \exp (i[\mathrm{px}+\mathrm{qy}]) \exp \left(-i \frac{\left[\mathrm{p}^{2}+\mathrm{q}^{2}\right] \mathrm{z}}{2 \mathrm{k}_{0}}\right) \exp \left(i \mathrm{k}_{0} \mathrm{z}\right) \nonumber \end{align}\]

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