Thin lens equation

Remarks

A proof uses ray transfer matrices. The matrix from the object plane to the image plane is $$\begin{pmatrix} 1 & s_i \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1/f & 1 \end{pmatrix}\begin{pmatrix} 1 & s_o \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 - s_i/f & s_o + s_i - s_o s_i / f \\ -1/f & 1 - s_o/f \end{pmatrix}.$$ The image plane is where all rays from one object point meet, that is, where the entry $B$ vanishes, which is the equation above. The entry $A = 1 - s_i/f$ is then the magnification, and it equals $-s_i/s_o$.

Example: an object $30$ cm in front of a lens with $f = 10$ cm has $1/s_i = 1/10 - 1/30 = 1/15$, so $s_i = 15$ cm and $m = -1/2$: a real, inverted image half the size. For $s_o = f$ the equation gives $1/s_i = 0$: the rays leave parallel and the image is at infinity. For $0 < s_o < f$ with $f > 0$, $s_i < 0$: the image is virtual, as in a magnifying glass.

Other sign conventions write $\frac{1}{s_i} - \frac{1}{s_o} = \frac{1}{f}$ with signed coordinates along the axis.

Going deeper

For a student

The thin lens equation is the simplest case of first-order imaging. For a thick lens or a system of lenses it keeps its form when the distances are measured from the principal planes, which are among the cardinal points. The Newtonian form $x_o x_i = f^2$, with distances measured from the focal points, is equivalent.