Ray diagrams for the thin converging lens

In the previous post we saw that the telescope was invented by what must have been a happy accident, as there was very little theoretical foundation to guide design. Once the telescope did appear, Kepler began the laying of such a foundation but before discussing his and other work it will be helpful to give a short account of the foundations of Geometrical Optics as it stands today.

The figure above the passage of light from a small part of an object (an apple on a miniature tree) through a converging lens, the lens being at a distance from the object greater than its focal length.   Kepler would have understood that light from the object would form an image (Kepler called it a pictura) on the back of an eye (shown schematically on the right) and would also have been aware that an inverted image of the tree would be formed upon a screen suitably placed between the lens and the eye.  However, he did not consider the significance of this intermediate image for determining what is directly seen or, for example, the effect of placing a second lens between the image and the eye.  Since all rays  of light going through both lenses from a particular apple go through the image then if the position of the image has been found we can take it to be the object of the second lens in determining the paths of the rays through it.  Let us consider, then, how we may find a relation between the distances of image and object.  We shall proceed semi-empirically, not seeking to prove that the lens does produce an image, or how it does so, but simply assume that it does. We define the optical axis as a line perpendicular to the (circular) lens and passing through its centre, termed in this context the optical centre. We shall  assume

•  A beam of light parallel to the optical axis and incident on the face of a lens is brought to a focus (the principal focus). There are two principal foci, since the beam may be incident on either face.
• The distance of the focus from the optical centre, termed the focal length, is the same on either side of the lens.
• The principle of the reversibility of light is true.  Thus if  a ray of light travels from A to B, light from B to A can travel on the same path.
• All light passing through the lens from a particular point P of an object are brought to a focus J.  That is, all rays passing through the lens from P meet at a point J, the image of P. These two points are said to be conjugate.
• The  lens is thin, by which we mean that the value of its thickness is small compared with the other dimensions in the problem.

In the figure above we represent a lens perpendicular to the plane of the figure.   The line OI is the optical axis and C the optical centre. The object is represented by the arrow from O to P at a distance x from the focus F; The ray PQ is parallel to the optical axis and hence passes through F₂, as shown. Since we are assuming the lens is thin, we neglect the change in the lateral position of the ray as it passes through the lens.   A second ray PR passes through the focus F₁ and, invoking the principle of the reversibility of light, travels parallel to OI after deviation by the lens. The point J where the two rays meet is, under our assumptions, the image of P.  Now triangles POF₁ and RCF₁ are similar so that $h/h'=x/f$, where $h$=OP, $h'$=CR, and $x$=OF₁.   Furthermore, triangles QCF₂ and JIF₂are similar, where I is such that IJ is perpendicular to the optical axis. Then  since CQ=$h$ and IJ=$h'$, we find $h/h'=f/x'$.  Equating the two expressions we have found for $h/h'$, we see that $x/f=f'/x'$ or
\begin{equation}  f^2=xx' \end{equation}.
This is Newton's equation relating the distance of object and image from the foci. 
It will be noted that for a  given value of $x$ the position of I (given by the value of  $x'$) is independent of  of $h$, so that if the object is perpendicular to the optical axis, then so will be the image.  Furthermore, $h/h'$ is independent of $h$, so that there is no lateral distortion of the image.
If $u=f+x$, the perpendicular distance of the object from the lens, and $v=f+x'$, the perpendicular distance of the image, then is readily shown from Newton's equation that
\begin{equation}\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \end{equation},
which is known as the lens equation.  It is also readily shown that
\begin{equation} \frac{h}{h'}=\frac{u}{v} \end{equation}.
It will be useful to note that it can now be seen that a ray (as shown dotted in the figure) from any point on the object to its conjugate passes through the optical centre.