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.

Virtual Images

We saw in the previous post how by a ray diagram we can find the position and size of the image of an object placed behind a converging lens at a distance greater than the focal length of  the lens.  If the object is placed at a distance from the lens less than the focal length then an image is still seen, but it can no longer be projected onto a screen.   We must therefore now distinguish between types of image. If the  light, passes through the image it will be said to be real.  In contrast, one for which the light by which it is viewed does not pass through, will be said to be virtual.    The questions now arise as to the nature of this image and how its properties may be found.
We may answer these questions by applying what we have already learnt to the lens system in the  figure below.

We show two identical lenses placed parallel to one another, their focal length $f$ apart.  An object (a miniature tree) is shown at a distance $u$ from the first lens, where $u$ is less than $f$.  A real inverted image of the tree may be seen or projected upon a screen and its position and size may readily be found  by drawing just two rays.  We draw one ray  from P at the top of tree parallel to the optical axis to meet the lens at Q, where it is deviated to pass through the focus F.  Since this point is the optical centre of the second lens it passes through undeviated.  The second ray is drawn through the optical centre of the first lens, where it is undeviated, to meet the second lens at R.  Here since C is at the focus of the second lens,  the ray is deviated to be  parallel to the optical axis.   The image of P is at J, where the two rays we have constructed meet, as shown.  To find its position we note that the triangles FCQ  and JRF are similar and FR is equal to $h_1$.  Hence $h/h_1=f/w$.  We can also see that triangles COP and CFR are similar, so that $h/h_1=u/f$.  Equating the two expressions for $h/h_1$ we find
\begin{equation}  f^2=uw \end{equation}.
 We now observe that in the absence of the first lens the same image at I could be obtained by placing a suitable object at M,  its position being found by producing the rays PR and QF back to meet at K, as shown.   Through the second lens this object would look precisely similar to the image, of the original tree at O, produced by the first lens.  We therefore locate this latter image at M.  It is a virtual image,  since no rays of light actually pass through or originate from it. We may find its location either using equation (1), or more simply from the figure below.

We wish to find the value of $v$, giving the location of the image. We shall see the reason for the minus sign presently.  Triangles KLM and QCF are similar , so that $-(h_2-h)/h=-v/f$.  Triangles KMC and POC are similar and so $-v/u=h_2/h$.  Comparing the two equations we arrive at the result
\begin{equation}\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \end{equation}
and we now see the reason for introducing the minus sign as this equation is exactly the form of the lens equation obtained for real images in the last post. We accordingly adopt the convention that the distance of a virtual image from the lens is negative. 

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.