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.