Diffraction: rectangular aperture and gaussian beam

[Enialis]

Is the Fresnel-Kirchhoff formula (FKF) valid also for gaussian beams? I a book starting from the gaussian intensity:
$$U_0(x,y)=\sqrt{\frac{2}{\pi\omega_0^2}}\exp\left(-\frac{t^2+s^2}{\omega_0^2}\right)$$
it said that using the FKF in free space the gaussian beam spreads (in far-field assumption). It concludes that the field can be expressed with a standard Gaussian distribution but with complex exponential rather than real. How can be it proved?
Now my question is: what happen if there is a diffraction through rectangular aperture to this gaussian beam? What is the correct theory to use in far-field assumption?

 

[Valerie Park]

The Fresnel-Kirchhoff formula (FKF) is indeed valid for gaussian beams. In the far field assumption, the FKF can be used to describe the diffraction of a gaussian beam through a rectangular aperture. The field can be expressed with a standard Gaussian distribution but with complex exponential rather than real. This can be shown by using the convolution theorem, which states that the product of two Fourier transforms is equal to the Fourier transform of their convolution. Thus, the diffracted field can be expressed as the convolution of the incident beam and the impulse response of the aperture, which can be calculated using the FKF.

 

[astrokreng]

Yes, the Fresnel-Kirchhoff formula (FKF) is valid for gaussian beams as well. The FKF is a general formula that describes the diffraction of any type of wave, including gaussian beams.

To prove this, we can start with the FKF equation:

U(x,y) = \frac{1}{i\lambda}\int\int_{S} U_0(x',y')\frac{e^{ikr}}{r}dxdy

Where U(x,y) is the diffracted field, U_0(x',y') is the incident field, r is the distance from the aperture to the observation point, and S is the aperture.

We can substitute the gaussian beam expression for U_0(x',y'):

U(x,y) = \frac{1}{i\lambda}\int\int_{S} \sqrt{\frac{2}{\pi\omega_0^2}}\exp\left(-\frac{(x'-x)^2+(y'-y)^2}{\omega_0^2}\right)\frac{e^{ikr}}{r}dxdy

Then, we can use the identity:

\int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}

To simplify the integral:

U(x,y) = \frac{1}{i\lambda}\sqrt{\frac{2}{\pi\omega_0^2}}e^{ikr}\int\int_{S} \frac{e^{-\frac{(x'-x)^2+(y'-y)^2}{\omega_0^2}}}{r}dxdy

Next, we can use the Gaussian integral:

\int_{-\infty}^{\infty} e^{-ax^2+bx}dx = \sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}}

To further simplify the integral:

U(x,y) = \frac{1}{i\lambda}\sqrt{\frac{2}{\pi\omega_0^2}}e^{ikr}\int\int_{S} \frac{e^{-\frac{(x'-x)^2}{\omega_0^2}}e^{-\frac{(y'-y)^2}{\omega_0^2}}}{r}dxdy

We can then