I was performing a diffraction experiment with a slit of width 0.08mm (reported by the manufacturer). The conditions of my experiment allowed the Fraunhofer approximations. Thus, doing the Fourier transform of my slit in 1D (since the other dimension is so large that diffraction cannot occur) produces the known intensity:
$$I(x) \propto \text{sinc}^2\left(\frac{a\pi}{\lambda z}x\right)$$
where $a$ is the width, $\lambda$ the wavelength of the source, and $z$ is the distance between the slit and the screen (the wall).
This is my picture:
I used ImageJ to obtain the profile of the pattern (I suppose it is an intensity profile since it's the same as averaging the RGB channels of the camera on a line). What I obtain is
where the $x$ axis is in cms and the $y$ axis in arbitrary units. Clearly the second peaks of the sinc (shown in red) are not as big as the first one (even more if we consider sinc$^2$!). Moreover I got a 0.00123 cm value for $a$ with the fit. The only thing I can think about is that the camera does not measure light intensity, but other thing. Do you know what other thing it's measuring? Or how do I relate what the camera measures with the intensity? Thanks a lot.