Interference pattern This is just my pure imagination.
 Let's consider two slits having width $w$ for each separated by a distance $d$. If the width of one slit is reduced to $\frac{w}{2}$.
What will happen to the interference pattern of the light from the two slits?
 A: If the center to center distance of the slits is unchanged, the spacing of the interference pattern will be unchanged. However, the effective intensity of the two slits will now be different (twice as much light through one slit than the other), so there will not be perfect extinction in the troughs of the pattern.
So the first thing is - the contrast of the fringe pattern will be reduced (peak to valley ratio of 3:1 - namely (1+0.5)/(1-0.5)). 
Depending on the ratio of $w$ and $d$, there will be a secondary effect: for Fraunhofer (far field) diffraction, the fringe pattern is actually the Fourier transform of the aperture function. When the aperture function is the convolution of a wide slit (of width $w$) with two narrow slits of spacing $d$, the convolution theorem tells us that the fringe pattern is the product of the Fourier transforms of the individual components: that is, a $\rm{sinc}$ function (with spacing depending on $w$) multiplied by a $\cos$ function (with spacing determined by $d$). That nice analysis is harder to do when the two slits have a different shape (width), so you would have to calculate the Fourier transform explicitly.
Here is an example of plots showing the effect: top plot has "same width" slits, with a 5:1 d:w ratio, while bottom plot is "half width", with one slit half as wide as in the top example. Horizontal and vertical axis are arbitrary. I plotted amplitude: if you plot intensity (amplitude squared), the contrast ratio is 9:1 instead of 3:1, obviously.

