Lets assume that a signal $x(n)$ is up-sampled by adding 1 zero between two adjacent samples to form a signal $y(n)$. How does a digital LPF give an oversampled version $z(n)$, with more data points? (As show in the image below) I know that the question might seem stupid but the books that I am using don't really explain that part.
@user29568: Your question is very far from stupid. It's a darned good question. To see why time-domain "zero stuffing" generates the additional spectral images, that must be filtered out as described by Fat32, you're welcome to see my explanation of this effect at: http://www.dsprelated.com/showarticle/761.php
A digital low pass FIR filter might change each added (with value zero) sample to become a weighted average of nearby original points. A suitably weighted average is a close approximation to interpolating along a reconstructed "smooth" (or bandlimited) curve that would pass thru all the original points. So you end up with more data points along that hypothetical curve.
Effect of an ideal lowpass filter (LPF) on the output of an expander (interpolator) can be described in either of the time or frequency domains.
For the frequency-domain approach, the key observation is that the DTFT of the expanded sequence $y[n]$ of the middle figure, includes frequency-scaled and shifted copies (a.k.a images) of the original DTFT of the input $x[n]$. The images are centered at $\omega = 2\pi k/M$, where $M$ is the oversampling factor and $k=0,1,...,M-1$.
The LPF, whose passband includes only the base copy of the expanded spectrum, removes all the remaining surplus images, and retains only the baseband. This is accomplished in the time-domain by convolving filter's impulse reponse and signal $y[n]$, to produce $z[n]$. The cutoff frequency of the ideal LPF is given by $\omega_c = \pi/M$. In your case $w_c = \pi/2$ radians per sample, and the filter also must have a passpand gain of $M=2$ if it's to be used as an interpolation filter.
