Path:  physics insights > basics > simple optics > Ideal Lenses II:  How Any Ray is Bent

We already discussed the basic properties of ideal lenses in Ideal Lenses and Mirrors.  However, all we learned there was what happens to rays which are entering the lens parallel to the axis, or are traveling on a path which passes through the front or back focus of the lens.  On this page we'll determine the path followed by any ray passing through the lens, and along the way we'll show that an ideal lens has a flat focal plane.  (In our earlier work we assumed there was a flat focal plane; we'll be using much weaker assumptions on this page.)

We will proceed by assuming that an ideal positive lens produces an image (of some sort) of a scene at infinity.  In particular, this implies that parallel rays must be brought to a focus at a single point, no matter what angle they are traveling at relative to the lens axis.  We don't need to assume anything beyond that about the details of the image produced.

Formally, we'll assume the following:

1. The lens has a "front focus", which is a particular point located somewhere on the axis of the lens.  Rays which enter the lens after passing through the front focus exit the lens on paths parallel to the axis.   (For a symmetric lens, due to the reversibility of light paths, this is equivalent to assuming the lens brings rays entering parallel to the axis to a common focus.)
   
   
2. Rays passing through the center of the lens are not deflected.
   
   
3. Parallel rays entering at any angle are brought to a common focus (somewhere).  In other words, any collection of parallel rays which enter the lens will be bent such that their paths will all intersect at a single point, somewhere behind the lens.  We won't assume anything about where the "focus" of a particular bundle of parallel rays must lie; merely that it must have one.

Ideal Positive Lenses

In our initial discussion we'll be considering a cross section of the lens, and we'll assume that all rays lie in the plane of the cross section.

Figure 1 -- A ray passing through an ideal lens:

In figure 1 we show an ideal lens which lies on the x axis.  The center of the lens is on the y axis, and the "front" focus is shown with a blue dot, above the lens on the y axis.

We show a red ray, "ray 1", passing through the lens.  It hits the lens with angle θ₀ from a line perpendicular to the surface of the lens, and after passing through it emerges with angle θ₁ from the line perpendicular to the lens.  The point at which the ray enters the lens is at distance x₀ from the center of the lens.  We want to find θ₁ as a function of θ₀ and x₀.

We know how rays which pass through the center of the lens are deflected (they just go straight), and we know how rays which pass through the front focus are deflected (they emerge from the lens traveling parallel to the axis of the lens).  So, we've drawn rays 2 and 3 parallel to ray 1, with ray 2 passing through the center of the lens, and ray 3 passing through the front focus.  We also know that rays which enter the lens traveling parallel to each other will all converge to a single point, so we know that ray 1 must be bent such that it arrives at the intersection of rays 2 and 3, and that is how we've shown it in figure 1.

The  distance from the lens to the point at which the rays focus is f₂.  Comparing Triangle T1 with Triangle T2, we can see that

1)   

Consequently the focal point of the parallel rays is at distance f from the lens, regardless of the angle at which the rays enter the lens.  So, the focal plane of an ideal lens is, indeed, a plane.

From Triangle T1, we can also see that

2)   

And from equation (1) and Triangle T3 we can see that

3)   

from which we obtain the result we wanted:

4)

This fully characterizes an ideal positive lens (for rays which are coplanar with the axis).

The Z Axis

We have said nothing about the z axis.  We showed a cross section of an ideal lens, and assumed that all the rays we were discussing lie in a single plane, which contains the axis of the lens.  What about rays which are skew to the axis?

We won't prove this rigorously (at least, not at this time) but skew rays can be understood by drawing a line from the center of the lens to the point where the ray enters the lens.  That line, along with the lens's axis, define a plane; call that the axis plane for the ray.  That plane contains the lens axis, along with a cross section of the lens.

Project the ray into the axis plane.  Now the picture in that plane looks exactly like what we've been discussing, and in fact the behavior of the ray, projected into that plane, goes according to equation (4).

Now, what about the leftover component of the ray?  The ray also had a component of motion which was perpendicular to the axis plane.  What happens to it?  Answer:  Nothing.  Imagine another plane, perpendicular to the axis plane, which contains the point where the ray enters the lens.  Call it the "tangential plane" (it's tangential to a circle drawn on the lens which is centered on the origin).  If we project the ray into the tangential plane, we find that its projection in that plane does not bend -- it is not deflected in that plane by passing through the lens.

With this final argument, we really have fully characterized an ideal positive lens.

While this last section really is straightforward enough, it would benefit a great deal from some pictures and some clearer reasoning.  Unfortunately the 3-d pictures required would take substantial time to draw, and the reasoning, while not arcane, is hard to state clearly.  I may eventually add some images and a clearer argument.

Ideal Negative Lenses

Not done yet.

Negative lenses also follow formula (4).  However, I have not shown that on this page, and for the time being I'm not going to -- if I find the time I may come back and add another image and some more trig to complete the proof.  (In fact the formulas are essentially identical with the positive case; only the diagram needs to be redrawn.)



Page created on 01/14/2009.  Updated, with clarified assumptions, 20 Jan 2009