First of all, light waves and matter waves may be treated together, using the same maths, because the waves associated with light and the waves associated with matter are fundamentally the same thing.
Second, all the waves before they interfere and after they interfere may be written in terms of the probability current $j^\mu (x,y,z,t)$, and its transformation from one frame to another is obtained by a simple Lorentz transformation of the coordinates $(t,x,y,z)$. In the non-relativistic limit, the Galilean transformation will work as a good approximation of the Lorentz transformation.
Third, whenever a wave – locally or globally – has a well-defined energy-momentum vector of the corresponding particle (whether it's a photon, electron, or something else), then the wave has a simple plane-wave form
$$ \psi = \exp (ip^\mu x_\mu) $$
At least the frequencies and wavelengths are determined by this simple complex exponential. Note that $x_\mu$ and $p^\mu$ transform according to the Lorentz transformation, in the same way, and their inner product – the argument of the exponential – is therefore invariant under Lorentz transformations which is why the phase above is invariant, too. One doesn't need to make "explicit" checks; it is clear that the predictions for the locations of the interference minima and maxima will be the same in all reference frames.
Now, if the momentum in the direction between "slit and the photographic plate" which I will call the $z$-direction is $p_z$ and the total energy is $E$ for the interfering particle, then $(E,p_z)$ is transformed by the standard Lorentz transformation when boosted by the velocity $v$. This transformation of $(E,p_z)$ to $(E',p'_z)$ is equivalent to the calculation of the velocity $V$ of the (wave-represented) particle
$$ V = \frac{p_z c^2}{E} $$
(the denominator is the total energy so in the non-relativistic limit, it has to be replaced by $m_0 c^2$ and not by $mv^2/2$) and composing this $V$ with the $v$ by the relativistic composition formula,
$$V'=(V+v)/(1+Vv/c^2)$$
For photons and other particles moving at the speed of light, the speed is $V=c$ and we get the "new" speed as $V'=c$ again. That's not too informative because for photons, the frequency or energy is not encoded in the speed (the latter is always $c$).
The transformation of $E,p_z$ – that obey $E=|p_z c|$ for $z$-directed photons – under the Lorentz transformation given by the motion at speed $v$ is given by the Doppler shift – both the momentum and the frequency get rescaled by
$$ \sqrt{ \frac{c+v}{c-v} }$$
The wavelength in the $z$-direction gets contracted by this factor. However, the distance between the slits and the plate gets contracted as well, and the time gets shortened because the plate is moving "against" the photons. These three modifications are consistent with each other because the number of wavelengths between the slit and the plate gets multiplied by
$$ \frac{ {\sqrt{1-v^2/c^2}} }{ \sqrt{ \frac{c+v}{c-v} } } = 1-\frac vc$$
which is exactly the factor by which the time spent between the slit and the plate will shorten because the plate is moving against the photons.
Let me emphasize once again that if we only change the speed in the $z$ direction, the transverse momentum components $p_x,p_y$ and the corresponding components of the wave vector $k_x,k_y$ will remain unchanged. That's true for relativity and non-relativistic physics, massless as well as massive particles.
Best Answer
The experiment looks like:
You can calculate the path lengths $A$ and $B$ using Pythagoras' theorem. The 100th bright fringe is positioned where the different in path length is 100$\lambda$ i.e.
$$ B - A = 100\lambda$$
The resulting equation is messy but easily solved numerically. I got $s \approx 1.51$m at the 100th fringe, so the spacing between the 100th and -100th fringes would be about $3.02$m.