Diffraction of a extensive source

[viko]

Homework Statement

There is an extensive source (since it isn't important it can be monochromatic). The waves goes through a condenser lens that makes them converge. So, what we have is the "image" reproduced at the other side of the lens, at the focal plane.
So now: I understand that each dot of that image works again as a point source. Is that correct? (I assume that's correct.)

Now supose we have a single slit at the focal plane.
---->How can I solve the diffraction?<---- (I'll use, for example, the kirchhoff integral in the Fraunhofer aproximation).

Homework Equations

Kirchhoff integral in fraunhoffer aproximation:

[tex] E= \frac{-1}{2 \pi \lambda} [/tex] [tex] \frac{e^{i(wt-kr)}}{r}[/tex] [tex] \frac{1+ \cos( \Theta )}{2} [/tex] [tex] \int_{ }^{ } \int_{S}^{ } E_{incident}(r') [/tex][tex]e^{ik\frac{x}{r}x'}[/tex] [tex]e^{ik\frac{y}{r}y'} dx' dy' [/tex]

[tex]E_{incident}(r') = [/tex] [tex]\frac{A}{z'} [/tex] [tex]e^{i(wt-kz') + ik [/tex] [tex]\frac{(x'^2+y'^2)}{z'} [/tex]

The Attempt at a Solution

Considering that I have a single source at each point of the slit,
trying to solve the integral, I found that I cannot evaluate the waves in the slit, because the waves cannot be evaluated at the convergence point.

So, what to do?

I don't know why, but I think in consider the waves at the focal plane as a plane wave. Does that make any sense?

 

[AvroArrow]

.If it does, then:E_{incident}(r') = \frac{A}{z'} e^{i(wt-kz')}Then I can solve the integral.Am I right?