Single slit diffraction - wave vs particle view If monochromatic light is shot through a single slit onto a screen, we can analyze the pattern on the screen using wave properties. This analysis is done assuming the wavelength is constant.
But with a particle view... there's uncertainty in the momentum after going through the slit... hence with DeBroglie's relation $p=\dfrac{h}{\lambda}$, uncertainty in the wavelength.
So is there any contradiction here... the first analysis has no uncertainty in wavelength (allowing us to calculate minimums and maximums using the wavelength of light and the width of the slit), the second view does have uncertainty in wavelength. 
How does one reconcile these views?
EDIT: The impression I'm getting is that the magnitude of the momentum of the photon doesn't change... only the direction does. But with heisenberg's uncertainty principle:
$\Delta x\Delta p_{x}\ge \dfrac{\hbar}{2}$, can't I arbitrarily make the slit narrower and narrower, imparting greater and greater uncertainty to the momentum along that direction. Wouldn't the magnitude of the momentum of the photon change at some point?
 A: The following quote is relevant to whether in quantum mechanical terms there exists a monochromaticity possible , i.e. exact knowledge of momentum for the photon:

instead of a slit, there is an electron.
So the problem "photon impinging on slit" is a quantum mechanical problem, and there exists an uncertainty on the momentum of the impinging photon from the Heisenberg Uncertainty,  and an uncertainty in the size of the slit due to the fact that it is defined by quantum mechanical objects like the outer electrons of the sides of the slit.
It is a scattering problem and could in principle be solved , but the HUP allows for estimates rather than going through the complicated mathematics of solving for the scattering.
Each individual photon may be elastically scattered, but its momentum is uncertain within the HUP of the wavelength times momentum and the scatter will be uncertain to that extent. 
The classical view with the quantum mechanical are reconciled because h is such a small number that the HUP is easily satisfied.
A: There's not necessarily uncertainty in the wavelength of the particle. The magnitude of the momentum vector could be the same for every particle, but its direction could be different. The particle's speed is certain, but the direction it heads in is not.
In real light and particle sources, however, there is always uncertainty in the wavelength. Even lasers emit a small range of wavelengths instead of being monochromatic.
A: Assume that the photon or electron moves in the xy plane along the y direction. A screen is paced in the xz plane with a slit of width dx. The particle is constrained to the width of the slit, hence after the screen there is an uncertainty in $p_x$. This uncertainty can be calculated from the wave function and is actually the probability distribution that produces the pattern.
You can say that the single slit diffraction pattern visualizes the uncertainty. Note that if the transmission of the slit is not a rect function but a gaussian, then the interference pattern is also a Gaussian. This corresponds more closely to the popular image of uncertainty.
A: There is definitely change in p>h/dx. But if the slit is 10 nm (not possible practically) and h=7.10^-34 you get dp=10^-25. This so low that is absolutely monochromatic.
