Fresnel Approximation

FRESNEL APPROXIMATION

Revisiting our transfer function of free space (given in last sub-sub-section):

$$H(k_x,k_y;z) = e^{-i\beta{z}}$$

where $\beta = \sqrt{k^2 – k_x^2 – k_y^2}$. If we consider the spatial frequency along the x and y directions to be low i.e. the ray through the wavefronts is almost parallel to z, then we can approximate $sin\theta \approx \theta$. In this case, we can Taylor expand the propagation constant $\beta$:

$$\beta = k\sqrt{1-\frac{k_x^2}{k^2}-\frac{k_y^2}{k^2}} \approx k[1-\frac{1}{2}(\frac{k_x^2+k_y^2}{k^2})]$$

The propagation description for paraxial waves may be:

(a) Frequency domain: each spatial frequency acquires a quadratic phase $(k_x^2+k_y^2)z/2k$ on propagation over a distance z

(b) Spatial domain: the initial field is convolved with a quadratic phase k(x^2+y^2)/2z

In both pictures, what this means is that paraxial waves acquire a quadratic phase on propagation.

GAUSSIAN BEAMS

Suppose we have a flat wavefront and Gaussian amplitude profile for both the x- and y-directions in the input plane A, such that the field can be given as:

$$\varepsilon(x,y;0) = E_0e^{-(x^2+y^2)/\omega_o^2}$$

where $\varepsilon$ was used to denote that this would be the envelope of the spatial plane wave, and $\omega_0$ is known as the beam waist and is the radius of which the field drops by 1/e.

The angular spectrum for this Gaussian propagator would be given by the usual Gaussian transform from the spatial to angular domain:

\begin{align} A(k_x,k_y;0) & = -E_0\int\int_{-\infty}^{\infty}e^{-(x^2+y^2)/\omega_0^2}e^{i(k_xx+k_yy)}dxdy \\ & = E_0\pi\omega_0^2e^{-\omega_0^2(k_x^2+k_y^2)/4} \end{align}

Then, if we want to propagate the plane at A in the z-direction, we have:

$$A(k_x,k_y;z) = E_0\pi\omega_0^2e^{-\omega_0^2(k_x^2+k_y^2)/4}e^{-ikz}e^{i(k_x^2+k_y^2)z/2k}$$

The field is then given as:

$$E(x,y;z) = \frac{E_0}{4\pi}\omega_0^2e^{-ikz}\iint^{\infty}_{-\infty}e^{-(k_x^2+k_y^2)(\frac{\omega_0^2}{4}-\frac{iz}{2k})}\times e^{-i(k_xx+k_yy)}dk_xdk_y$$

Since this is the equation for the Fourier Transform of a Gaussian, the integral becomes:

$$= \frac{E_0}{1-i\frac{2z}{k\omega_0^2}}e^{-(x^2+y^2)/\omega_0^2(1-i\frac{2z}{k\omega_0^2})}$$

Rayleigh range can then be defined as:

$$z_R = \frac{\pi\omega_0^2}{\lambda}$$

which gives the length scale over which a beam spreads due to diffraction. We can use this to rewrite the phase factor $1-i2z/k\omega_0^2$ as:

\begin{align} 1-i\frac{z}{z_R} & = \sqrt{1+(\frac{z}{z_R})^2}e^{-itan^{-1}(z/z_R)} \\ \sqrt{1+(\frac{z}{z_R})^2}e^{-i\phi(z)} \\ & = \frac{\omega(z)}{\omega_0}e^{-i\phi(z)} \end{align}

It may also alternatively be expressed as:

$$\frac{1}{1-i(z/z_R)} = \frac{z_R}{z_R-iz} = \frac{iz_R}{z+iz_R} = \frac{iz_R}{q(z)}$$

Here we define $q(z) = z+iz_R$ as the “q-parameter” which is essentially the complex radius. We then define the wavefront radius as:

$$R(z) = z[1+(\frac{z_R}{z})^2]$$

$$\omega^2(z) = \omega_0[1+(\frac{z}{z_R})^2]$$
Thus, the Gaussian beam envelope in terms of the parameters R(z) and $\omega(z)$:
$$\varepsilon(x,y,z) = E_0\frac{\omega_0}{\omega(z)}e^{-i[\phi(z)+k(x^2+y^2)/2R(z)]}e^{-(x^2+y^2)/\omega^2(z)}$$
$$\varepsilon(x,y,z) = E_0[\frac{iz_R}{q(z)}]e^{-ik(x^2+y^2)/2q(z)}$$