As discussed earlier, light diverges as it leaves any real object. We shall use this as the definition of a real object – one for which the light diverges from the object’s position, called the object point, O, as it approaches the thin lens.  If the light approaching a thin lens converges toward some point (the object point,  O) on the other side of the lens, then the light cannot possibly be coming directly from a real object.  In this case, we say that the object for the lens is a virtual object.  (In such a case, the  light must be coming from another lens or a mirror which has taken the diverging light from a real object and is making it converge toward the thin lens under consideration. We are not concerned at this point with how the light approaching the lens came to be converging. Rather, we are simply acknowledging the fact that it is converging as it approaches the lens by saying that the object distance–the distance from the center of the lens to the point to which the light is converging–is that for a virtual object.)

Now consider the light as it leaves the other side of the lens.  This light will either converge or diverge as it leaves the lens.

If the light is converging to some point as it leaves the lens, then that point, called the image point, I, is real and the corresponding image position is positive. 

Thus, if the distance from the lens to the image point is D_i, then it follows that, for a real image, the image position d_i is given by: d_i = +D_i. 

Light that is focused to a real  image point can be projected onto a screen or a piece of paper placed at the image point. (Hence the term “real” image.  The light will, of course, diverge once it moves beyond this real image point, just as if it were leaving a real object....) 

On the other hand, if the light as it leaves the lens tends to diverge, then it must be diverging from some point, called the image point, I, on the other side of the lens. Light after leaving the lens does not form a real image–that is, it does not focus to a point.  The image point in this case is therefore said to be a virtual point, and the corresponding image position is negative. 

If the light is diverging from some point as it leaves the lens, then that point, called the image point, I, is virtual and the corresponding image position is negative. 

Thus, if the distance from the lens to the image point is D_i, then it follows that, for a virtual image, the image position d_i is given by: d_i = –D_i. 

Light from a virtual image point will not form a clear picture on a screen or piece of paper, but it can be focussed by our eye if we look directly into the light as it leaves the virtual image point.

It sometimes confuses people to hear that our eyes cannot focus light that is converging toward a real image point, but we can readily focus on light that is diverging from a virtual image point.  If you think about it for a minute the reason should be clear:  Our eyes are configured to view real objects.  Light  diverges as it leaves real objects, which means that the optics of our eyes is made to converge the diverging light entering our pupils onto the retinas in the back of our eyes.  Light coming from a virtual image point is also diverging. This means that our eyes can readily focus it onto the retina so we can see the image. 

However, light heading toward a real image point is converging; our eyes cannot focus light onto the retina that is already converging before it enters the optical system of  our eye.  For example, your image in a flat (plane) mirror is a virtual image – you have no trouble viewing it.  But have you ever stood directly in front of a screen onto which the image of a slide was being projected? If you stand back and look at the image once it has been projected onto the screen, you can readily view it. However, if you stand in front of the screen and look at the light coming from the projector before it reaches the screen, you cannot make out the picture that is being projected. You cannot make out a real image from the light as it approaches the image point (the image point on the screen in the case of the projector).

Now that we have defined the object and image positions, d_o and d_i, we can define the magnification, m of the image.  The magnitude of the magnification, |m|, gives us exactly what we would expect it to give us: it tells us the ratio of the image height to the object height.  (How much bigger is the image than the object? ...by how much is it magnified?)

The sign of the magnification tells us whether the image is right-side up or upside down relative to the object – that is, it tells us whether the image is upright or inverted relative to the object: 

If m > 0, then the image is upright.

If m < 0, then the image is inverted.

The magnification for a given thin lens can be computed from the object and image positions as follows:

Note that you must be very careful to use the appropriate signs for the image and object positions in Eq. (2.3) above!