the position of image plane in optical imaging
I agree with user lxg that it is a notational difference, as for the pinhole camera we always have: 

$$S_2 = f$$

That is not necessarily true for a camera with a lens. A pinhole camera is not the same as a camera with a lens, because a pinhole camera is sharp at all $S_1$. What makes focal length a descriptive quality of both camera types is that the magnification $M$ of a distant object imaged onto the image plane when the object is in focus can be estimated just from the focal length and the distance to the object. For a single-lens camera:

$$M = -\frac{S_2}{S_1}$$

When an object is in focus:

$$\frac{1}{S_1}+\frac{1}{S_2} = \frac{1}{f}\\
\Rightarrow M = \frac{f}{f-S_1}$$

If $f$ is very small compared to $S_1$, as it is for distant objects, it follows that:

$$M \approx -\frac{f}{S_1}$$

For the pinhole camera we have similarly:

$$M = -\frac{S_2}{S_1} = -\frac{f}{S_1}$$

The magnification is negative because the image is upside-down compared to the object.

In mobile phones the focal length is in the order of 1.5 mm to 7 mm. If you are not doing close-up, then the pinhole model approximates the magnification well, with an error under 0.15 % to 0.7 % for objects at a distance of 1 m or greater.