# Does a camera measure intensity? (picture of diffraction pattern on a wall)

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.

• I'm guessing that the red line is a fit to the data? If you know that the intensity model is sinc${}^2$, why did you use a sinc as model? Also, your fit doesn't hit the peaks and zeros, so it looks like it's not the best fit. You should check to see what ImageJ does to get the profile. Did you simply take a photo of the screen? Did you start with a jpg? Something in the chain of processing might be compressing the data. You need to establish what happens in every step of your processing. – garyp Oct 24 '16 at 0:40
• What @garyp said. Plus, how are you getting the horizontal scale for the data? I ask because in the single figure you show us there is no apparent scale to use for a reference. Now, assuming you are using a tripod or other fixed mount for the camera you can establish the scale with a different image (say one with the lights one so you can image a ruler of some kind), but sometimes this step is neglected. – dmckee --- ex-moderator kitten Oct 24 '16 at 0:43
• @garyp nothing amazingly helpful imagejdocu.tudor.lu/doku.php?id=gui:analyze:plot_profile – Vladimir Vargas Oct 24 '16 at 3:42
• How big is the beam that illuminates the slit? – flippiefanus Oct 24 '16 at 4:28
• @flippiefanus I think this is a very important question that was not pointed out by my instructor at the laboratory (and that didn't occurred to me). We didn't use a beam spreader to make sure that the slit was illuminated uniformly. – Vladimir Vargas Oct 24 '16 at 16:09

So you might get a better fit by using $2^{Dx}$ instead of $x$, where $x$ is the pixel value and $D$ is the dynamic range of your camera in f-stops. I think this will be between about 10 and 20 depending on how good your camera is. (The pixel value $x$ should range from 0 to 1 in this formula, so for example if it ranges from 0 to 255 you should divide it by 256 first.)