Single Slit Diffraction Experiment How is possible for an interference pattern to be created from a single wave?
Why is the central maximum twice as wide as the others?
 A: In interference, we have only two significant overlapping sources. So we just add the electric field vectors to get the resultant field, whose square is directly proportional to the Intensity of light on the screen.
However in case of diffraction, we have to consider electric fields from continuous points. So, the phases will change continuously from till the phase is the same as the other point of the slit.
Consider the setup. The distance between the screen and slit is $D$ and the width of slit is $a$

Electric field at C will be the sum of all electric field vectors due to all the points between A and B. We can place all the infinitesimal electric field vectors as shown in the figure:

Note that the total arc length represents the magnitude of total electric field through the slit $=E_0$.
As you consider finer pieces, this becomes a circular arc. The angle $\alpha$ and $\beta$ are equal (Proof left to the reader) and it is the same as the phase difference between $dE_A$ and $dE_B$. The resultant of all the vectors can be obtained by the polygon rule, i.e., the closing segment of the polygon as shown:

We have the arc length $=E_0$ and the angle $=\alpha=\beta=\Delta\phi$. By little geometry, We can easily solve the value of
$$E_{Resultant}=\frac{2E_0}{\alpha}\sin{\frac\alpha2}=E_0\left(\frac{\sin\left(\frac{\alpha}{2}\right)}{\left(\frac{\alpha}{2}\right)}\right)=E_{0}\left(\frac{\sin\left(m\right)}{\left(m\right)}\right)$$
Where $m=\frac{\alpha}{2}=\frac{\Delta\phi}{2}$
Now, intensity is proportional to the square of the net electric field.
$$I=E_{0}^{2}\left(\frac{\sin\left(m\right)}{m}\right)^{2}$$
The variation of intensity is similar to that of $\left(\frac{\sin\left(m\right)}{m}\right)^{2}$. It is zero when $\sin m=\sin\frac{\Delta\phi}{2}=0$ except for the case where $\Delta\phi=0$, i.e. the central point. At that point, we need to take the limit of the function which is $1$. However the other solutions are equally spaced at $n\pi$. This results in the central maxima being double the width of the other maxima. The graph of the function $\left(\frac{\sin\left(m\right)}{m}\right)^{2}$ is as follows.

A: there is no interference if the slit is infinitesimal. 
But in this regard infinitesimal is what's smaller than a wavelength (maybe half wavelength as I don't remember the exact solution). 
But if it is larger than that you can say that it is composed of several infinitesimal slits next to each other with 0 distance between them. You do get interference.
I think you can understand the second part, if you look at the distance between the dark parts. They have the same width and their distances from the middle are the same
A: This happens because now the slit behaves as multiple small slits, and you easily figure it out, see this (let's say the slit as AB and where ever you form fringe be X):
For central maxima: all slits will be in nearly same phase, as all need to travel same nearly distance, so the add-up giving brightest maxima. (AX = BX) 
For first dark fringe: there will some X on screen such that half part of light (say AO for some O in AB) is in positive phase, and corresponding to each phase there is equally negative phased wave (from OB).
The important part, for next bright fringe: let's divide AB into AM + MN + NB for some M, N in AB. Now the case is that AM and MN cancels each other, but NB is left, giving a bright fringe but lesser bright.
Further, you can divide it into 4 sections for next dark fringe, and 5 sections for bright one (obviously less bright).
i hope it helped :)
