How do the evanescent waves contribute/cause a diffraction limit?

Question

In the following article, under the section Physics of the Superlens, it is stated:

The light emitted or scattered from an object includes not only propagating waves but also evanescent waves, which carry the subwavelength detail of the object. The evanescent waves decay exponentially in any medium with a positive refractive index so that they cannot be collected at the image plane by a conventional lens, and this results in a diffraction-limited image.

Why do the evanescent waves need to be collected? How do they cause the diffraction limit? Non-technical as well as technical answers are appreciated.

Answer 1

Suppose our object is described by the function $g(x,y,z=0)$. We can write this function as a superposition of plane waves (Fourier decomposition), $$ g_{object}(x,y,z=0) = \iint df_x df_y \;G(f_x, f_y,z=0)e^{-2\pi i (f_x x + f_y y)} $$ where $f_x$ and $f_y$ are the spatial frequencies. Hence, the object consists of many different spatial frequencies.

Now, if we use the fact that any imaging system (=lenses) possesses a finite resolution, we conclude that after propagating the distance $z$ some spatial frequencies, which were present in the object and thus in $G(f_x, f_y,0)$, have vanished from the image. Hence, the propagation of the light changes the function $G(f_x, f_y,z)$ -- which is why we included the variable $z$. In fact, the free space propagator is given by $$G(f_x, f_y,z) =G(f_x, f_y,0) \cdot \exp{\left( ikz\sqrt{1 - \lambda^2(f_x^2 + f_y^2)} \right)} $$ where you see that for large enough spatial frequencies the term under the square root becomes negative. These are the evanescent waves.

A full discussion can be found in the textbook from Goodman "Fourier Optics".