8.5 Lens Diagrams

Lens Diagrams, like other ray diagrams, help us make calculations or determine precisely what a lens will do in a given situation. They can be made even more precise if they are drawn to scale (as they usually are). In order to draw these diagrams successfully, a few pieces of information about the lens are needed: the type of lens, its focal length, and also the distance of the object from the lens (for scale diagrams this becomes essential).

Just as in previous ray diagrams, we consider an object to be a series of points, and by obeying a set of rules, the images of the points, created by constructed rays passing through a lens, can be found. Thus the image of the object can be constructed. From each point, at least two different rays of light are drawn, and where they intersect is the location of the image of the point. There are basically three rules to be followed by (light) rays in converging lens diagrams:

1. A ray parallel to the principal axis that passes through the lens will be bent so it goes through the focal point on the other side of the lens.

2. Similarly, a ray that goes through a focal point before passing through the lens will emerge from the lens parallel to the principal axis

3. A ray that passes through the optical center (the point intersected by the principal axis) of the lens will continue in the same direction.

We will now look at a few examples to see how these rules work in practice. Here we have our same pencil, which we want to produce an image of by using a lens.

Now, we will create three rays going through point A. Because we already know what happens to three particular kinds of rays (mentioned above), why not make the rays in our diagram of the same kind?

Where the three rays meet, we have our projection of A: A¢ . Actually, only two of these rays were really needed to find A¢ , but a third was used to double check. Point B¢ must be found differently because it is on the principal axis. Since all three rays through this point (parallel to axis, through lens center, and through first focal point) follow the path of the axis, point B¢ is found simply by drawing a line perpendicular to the principal axis, that intersects point A¢ .

Let’s look at what happens when an object is placed in front of the focal point, as in this second diagram, where a pencil is placed between it and the lens.

We will do this diagram to scale, in order to see the relationship between image and object. The focal length, object distance, and object height are written in green, in centimeters.

This time, when two rays are drawn, they do not meet at a point, instead they diverge. How can an image be found for A? By following these two lines to their apparent origin, where they intersect.

Notice that the image is on the same side of the lens as the object. Finally, we add all relevant numbers to the diagram.

As you may have noticed, the ratio of the image distance to the object distance is equal to the ratio of the image height to the object height, or

This equation holds true for all lenses, including diverging ones. The ratio of image height to object height is significant on its own as well. It indicates how much an object will appear to have grown or shrunk. This ratio is also known as the (linear) magnification of the lens. Remember that this number is not a constant, as one lens can produce images of many sizes.

A second relationship exists, though in this case it is less evident. The focal length, f, the distance of the object from the lens, u, and the distance of the image from the lens, v, are also related for all lens diagrams. This is shown in the following equation, known by many physicists as the lens equation:

Note that any units of length may be used, but must be consistent for all three lengths.

Employing the diagram above, we get u = 2, v = 3, and f = 6. However, substituting this into the equation, we find that 1/2 + 1/3 does not equal 1/6. Is there something wrong with the lens? No! We must take into account established convention that says focal length and image distance must be negative, if they are on the same side of the lens as the object, or, if the image formed is imaginary (see section 8.6). Since the image distance, v, for our diagram is indeed located on the same side of the lens as the object distance, we know that it is considered negative. Putting this value back into the equation, we find it to be true: 1/2 - 1/3 = 1/6.

One final special case of the converging lens diagram occurs when the object is placed directly on the focal point, as shown below.

The ray that is originally parallel to the principal axis bends towards the far focal point at the same angle as the ray that goes through the center of the lens. The result is that two parallel rays are produced. This means that no image is formed, as the rays from the top point of the object will never intersect. Let us check this using the lens equation, to try to find the image distance, v. So far we have f = a (the focal length), and u = a. Putting this in the equation, we have 1/a + 1/v = 1/a, which means 1/v = 0. This situation can only happen when v = ¥ , which once again translates to there being no image.

A diverging lens of course does not work in the same way as a converging lens, and therefore does not follow the same set of rules. For instance, rules 1 and 2 above are no longer applicable. In a diverging lens, rays that are parallel to the principal axis will diverge in such a way (upon leaving the lens) that they appear to be originating from the focal point on the side of the lens they entered from.

In using the lens equation for the diverging lens, the focal length, f, is always negative, because the focal point on the same side of the lens as the object is the only one that is used. This is also why the power of a diverging lens is negative. If the focal length is always negative, this means the 1/f term in the lens equation must also be always negative, and 1/u + 1/v likewise. Since u, or 1/u can never be negative, as it is the starting point, 1/v is therefore negative, and must also be greater (in absolute value) than 1/u. Or, 1/u < 1/v, which means v < u. The implications of this are that the image is always between the object and the lens, and that it is therefore always smaller than the object.

In this diagram, the green ray has been constructed in order to determine the amount of divergence of the ray that is parallel to the principal axis.

What happens when the object is placed on a focal point of a diverging lens? We will use the lens equation to find out. If the focal length is given the variable a, the equation looks like this: 1/a + 1/v = -1/ a. This means that 1/v = -2/a, or v = -a/2; the object is located exactly halfway between the lens and the focal point once again on the first side of the lens.

Because ratio of the image distance to object distance is 1/2, the image size compared to the object size, or its magnification, is also 1/2.