Magnification
So far we have discussed the focusing action only for points on the axis. Now let us discuss also the imaging of objects not exactly on the axis, but a little bit off, so that we can understand the properties of magnification. When we set up a lens so as to focus light from a small filament onto a “point” on a screen, we notice that on the screen we get a “picture” of the same filament, except of a larger or smaller size than the true filament. This must mean that the light comes to a focus from each point of the filament. In order to understand this a little better, let us analyze the thin lens system shown schematically in Fig. 27–7. We know the following facts: This is all we need to establish formula (27.12) by geometry, as follows: Suppose we have an object at some distance $x$ from the focus; let the height of the object be $y$. Then we know that one of the rays, namely $PQ$, will be bent so as to pass through the focus $R$ on the other side. Now if the lens will focus point $P$ at all, we can find out where if we find out where just one other ray goes, because the new focus will be where the two intersect again. We need only use our ingenuity to find the exact direction of one other ray. But we remember that a parallel ray goes through the focus and vice versa: a ray which goes through the focus will come out parallel!  So we draw ray $PT$ through $U$. (It is true that the actual rays which are doing the focusing may be much more limited than the two we have drawn, but they are harder to figure, so we make believe that we can make this ray.) Since it would come out parallel, we draw $TS$ parallel to $XW$. The intersection $S$ is the point we need. This will determine the correct place and the correct height. Let us call the height $y'$ and the distance from the focus, $x'$. Now we may derive a lens formula. Using the similar triangles $PVU$ and $TXU$, we find \begin{equation} \label{Eq:I:27:13} \frac{y'}{f}=\frac{y}{x}. \end{equation} Similarly, from triangles $SWR$ and $QXR$, we get \begin{equation} \label{Eq:I:27:14} \frac{y'}{x'}=\frac{y}{f}. \end{equation} Solving each for $y'/y$, we find that \begin{equation} \label{Eq:I:27:15} \frac{y'}{y}=\frac{x'}{f}=\frac{f}{x}. \end{equation} Equation (27.15) is the famous lens formula; in it is everything we need to know about lenses: It tells us the magnification, $y'/y$, in terms of the distances and the focal lengths. It also connects the two distances $x$ and $x'$ with $f$: \begin{equation} \label{Eq:I:27:16} xx'=f^2, \end{equation} which is a much neater form to work with than Eq. (27.12). We leave it to the student to demonstrate that if we call $s = x + f$ and $s' = x' + f$, Eq. (27.12) is the same as Eq. (27.16).