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I found out information that these add-on fish eye lenses for smartphones have a focal length of about 2mm. I bought a lens and tried it with several different phones and it worked. I found out that these phones have different focal lengths (3.85mm and 5mm). My question is what is the mathematic behind this optical combination (fish eye lens + phone lens). These are my calculations: $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$; I assume that distance between both lenses is 1mm. The image of the fish eye lens is the object for the second lens so: $\frac{1}{4}=\frac{1}{-1}+\frac{1}{d_i}$, or $d_i=0.8\rm{mm}$. So the focal point of the total system has changed from 4mm (without fish eye lens) to 0.8mm. Am I wrong?

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It is not quite this simple. The assumption you made is that the image of the fish eye lens is always on the focal point of the lens. This is only true for an object that is infinitely far away. In reality, the location of the image of the fish eye lens will depend on the distance the object is away from the fish eye lens.

In practice, however, the object you are taking a picture of is likely to be at a distance $d\gg 2\rm{mm}$ away, and your math is a good approximation. You should keep in mind however that in general two lens systems cannot be simplified as a single lens with one focal point.

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  • $\begingroup$ Thank you very much for your comment! Indeed, the objects I have in mind are at least 10 meters away, so I thought that would be enough to approximate with the focal length. I am quite green with the lens system calculations and this is why I try to understand things deeper. $\endgroup$ – D. K. Feb 9 '16 at 0:39

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