In this section, we will discuss lenses, after which we should have enough theoretical knowledge to speak of some applications of optical systems, and difficulties that arise from their use.

The lenses that we will discuss will be thin, in the sense that their widths are small enough that we can basically ignore the behaviour of the rays inside the lenses, except for refraction at the surfaces. Also, assuming slimness allows us to make measurements of radii, foci, and source and image points independent of the side of the lens that we are interested in.

Lenses are classified as one of two types, converging and diverging.

A converging lens focusses rays from infinity so that they intersect at a common point, more or less. A diverging lens bends rays from infinity so that it seems that they originate from the same point. In either case, we refer to the point of convergence or divergence as the focus of the lens.

In our diagrams, we will generally use two convex surfaces to denote a converging lens,
and two concave surfaces to denote a diverging
lens^{1}.

In order to develop theory for thin lenses, we need to adopt a convention about how to refer to the two different surfaces of a lens, because the final image formed by the lens depends on the interaction of the source object with each surface of the lens.

Given a source point *S* (whose image we want to find), we define Surface 1 as the
surface of the lens on the side that the source lies. Surface 2 is the surface opposite to the
source. Distances, though measured from the nearest surface of the lens, will be independent
of the sides because of our assumption of thinness.

Consider a lens of index *n*_{2} surrounded by some uniform material with
index *n*_{1}. Although in the diagrams that follow, the lens is of the
convergent type, the formulas are equally valid for diverging lenses when we utilise
our sign conventions.

Now, from our development of refraction at spherical surfaces, we can find the image of the source point with respect to interaction with Surface 1. We know that the positions of the source and image satisfy

where we have used subscripts for *R*, *s* and *i* to denote interaction with
the first surface of the lens.

The key observation at this point is that the image point *i*_{1} acts as
the source point *s*_{2} for the interaction with the second surface of the lens.
Applying our knowledge of spherical refraction yet again, we get

where we have been careful to switch the indices, as we are now travelling from the lens
to the surrounding region. We have also substituted *i*_{1} for
*s*_{2}, but with a minus sign. This can be explained via our sign
conventions. A priori, we do not know which side of the lens the first image will land on.
If *i*_{1} is positive, then in our diagram, it is to the right of Surface 1.
Hence, by our thin lens assumption, the first image point is also to the right of Surface 2.
The first image point is defined to be the second source point, which
we now know is on the side to which light passes. By our sign conventions,
*s*_{2} must be negative, so we must tack on a negative sign to
*i*_{1} when we make our substitution. If *i*_{1} is negative,
then it is to the left of Surface 1 (Surface 2), and so is our new source point. The source
is positive by our sign conventions, so we again need to tack on a negative sign. In either
case, we must take the negative in making our substitution, so the move is valid.

Substitution of our 1/*i*_{1} result for interaction with Surface 1 into
the equation for interaction with Surface 2 yields

where we have simplified using *s* to denote the object position and *i*
to denote the final image position.

If we let *n*_{1} = 1 and *n*_{2} = *n* (i.e. we have a glass
lens of some index in air), then we get a fairly accurate way of predicting the location of
images.

This result is called the *lens-makers' equation*. The sharp reader will notice the
striking similarity between this equation and the position equation for mirrors. In fact,
as we let *s* get larger and larger, its reciprocal tends to zero, and we can find
the focus of the lens in air.

For the rest of our discussion, we will be working with lenses in air; this is the case
that applies for the applications we will look at, and it is conceptually easier to carry
around a single index *n* anyway.

We close our discussion of image positions with an observation. The equation

is symmetric, in the sense that if we move the source to the other side of the lens and shine light from the source's side, the focus is exactly the same. (This shows that a lens has two foci, on on each side of the lens, with equal distance to the lens.) This symmetry is most easily seen from the lens-makers' equation.

Reversing the direction of travel is
equivalent to switching the roles of the radii *R*_{1} and
*R*_{2} in the lens-makers' equation. However, the signs of
*R*_{1}, *R*_{2}
are now the reverse of what they were when the light travelled "forward." This is because
of our sign conventions; Sides A and B have swapped places. Thus, when we go forward,
the lens-makers' equation gives the quantity 1/*R*_{1} a positive sign,
and when we go backward, the equation gives the quantity 1/*R*_{1} a negative
sign, but *R*_{1} is now the negative of what it was going forward.
In essence, we are applying a double
negative, and they cancel. A similar result holds for *R*_{2}.

Finding the magnification of a thin lens is simple. We need only to trace two rays from a source point, and find their intersection, then use similar triangles as we did with mirrors. We will use a ray from infinity, and a ray that goes through the physical centre of the lens.

The ray from infinity crosses the focus; assuming that the thin lens
acts as if incoming rays are bent at the middle of the lens, we can easily find the locus
any incoming ray parallel to the optical axis. After making a simplification, the ray through
the centre of the lens is also easy. Since the lens is thin, a ray that goes through the
centre of the lens will act very much like it is hitting a flat plane of glass. In our
treatment of Snell's Law, we saw that a ray hitting a flat plane of glass was displaced, but
did not change direction. Since our lens is thin, the displacement will be very small, and
we can assume that a ray going through the centre continues on as if not refracted at all.
Now, the magnification is of course the ratio of the heights of the image to the source; from
our development of the positions of sources and images we see that the familiar
result *M* = -*i*/*s* holds.

The facts presented in this subsection are readily verified using the source-image and
magnification equations developed above. The distance from the lens to the foci *F*
and *F'* are equal.

For a diverging lens, the image is always virtual, diminished, and upright, as illustrated in the following diagram.

Converging lenses are divided into three cases, which depend on the source location.

If the source is more than two focal lengths away from the lens,

we see that the image is inverted, diminished, and real.

If the source is between one and two focal lengths away from the lens,

the image is magnified, and is again real and inverted.

For a source that is within one focal length of a converging lens,

the image is virtual, right side up, and magnified.

<< Refraction at Spherical Surfaces

Summary of Mirrors and Lenses >>

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