Ray diagrams for the thin diverging lens.

The telescope that Hans Lippershey attempted to patent in 1608 (see post for 18th September)) had a converging objective and diverging (concave) eyelens.  We term such an instrument a Galilean telescope because Galileo made a number of them and used them to such good effect in observing the Moon, Jupiter and Venus.  The objective takes a parallel beam of light from an object (effectively) at infinity to the left and brings it to a focus, as shown in the first figure below. When correctly positioned between the objective and its

focus(we say,when the telescope is focussed) the eyepiece will then give back a parallel beam of light, as shown in the second figure. The focus of the converging lens is shown at F. We say this is

also the (principal) focus of the diverging lens. Any ray which would pass through this focus if it were not deviated by the lens is rendered parallel to the optical axis by its passage through the lens.  Conversely, because of the reversibility of light, a ray parallel to the optical axis is deviated so that appears to pass through the focus.  We accordingly analyse the optical properties of the diverging lens making the following assumptions.

•  A beam of light parallel to the optical axis diverges when incident on the face of a lens and all rays of the beam appear to originate from a single point, the 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, with optical centre at C.  The object is represented by the arrow OP.  In accordance with the discussion above a ray of light from P directed to the far focus F₂ is rendered parallel to the optical axis when it strikes the lens at R.  A second ray from P, parallel to the optical axis, is deviated at Q so that it appears to have passed through the the focus F₁. The rays QT and RS meet at J, when produced back, so that it is the image point of P. Triangles F₁IJ and F₁CQ are similar, from which we find $h'/h=x'/f$. Furthermore, triangles F₂CR and F₂ are similar, so that $h'/h=f/x$. Comparing the two equations we see that \begin{equation} f^2=xx' \end{equation} This is Newton's equation for the thin concave lens.
Taking the distances of object and image form the lens as $u$ and $v$ respectively, we readily find from Newton's equation that \[1/u-1/v=-1/f\]. With the convention that the distances are measured from left to right, and the focal length of a divergent lens is negative we change the signs of $v$ and $f$ to bring the equation in the standard form
 \begin{equation} \frac{1}{u}+ \frac{1}{v}=\frac{1}{f} \end{equation}
 (c.f. our previous equation for the converging lens). Finally we should note that, as with the converging lens, rays through the optical centre are undeviated.