Now envision the situation in which we have a point source in the vicinity of some arrangement of reflecting and refracting surfaces representing an optical system. Of the infinity of rays emanating from O, in general only one will pass through an arbitrary point in space. Even so, it is possible to arrange for an infinite number of rays to arrive at a certain point I.

Thus, if for a cone of rays coming from O there is a corresponding cone of rays passing through I, the system is said to be stigmatic for these two points. The energy in the cone reaches I, which is then referred to as a perfect image of O. The wave could conceivable arrive to form a finite patch of light, or blur spot, about I; it would still be an image of O but no longer a perfect one.
Nonideal images are formed in practice because of

The requirement that every ray from O, like OPI, refracts and passes through the image I. Another such ray is OVI, where V is normal to the surface at its vertex point V. By Fermat's principle, these are isochronous rays. Since the media on either side of the refracting surface are characterized by different refractive indices, however, the isochronous rays are not equal in length. The transit time of a ray through a medium of thickness x with refractive index n is
.
(8.1)
Therefore, equal times imply equal values of the product nx, called the optical path length. For this problem, Fermat's principle requires that
.
(8.2)
In terms of the (x,y) coordinates of P, the first term in (8.2) becomes
.
(8.3)
The constant in the equation is determined by the middle member of equation (8.2), which can be calculated once the specific problem is defined. Equation (8.3) describes the Cartesian ovoid of revolution.
In most cases, however, the image is desired
in the same optical medium as the object. This goal is achieved by a lens
that refracts light rays twice, once at each surface, producing a real
image outside the lens. Thus it is of particular interest to determine
the Cartesian surfaces that render every object ray parallel after the
first refraction. Such rays incident on the second surface can then be
refracted again to form an image. Depending on the relative magnitudes
of the refractive indices, the appropriate refracting surface is either
a hyperboloid (ni > n0) or an ellipsoid
(ni > n0). It should be noted that
the aberration free imaging so achieved applies only to the object point
I at the correct distance from the lens and on axis. For nearby
points, imaging is not perfect. The larger the actual object, the less
precise is its image. Because images of actual objects are not free from
aberrations and because hyperboloid surfaces are difficult to grind exactly,
most optical surfaces are spherical. The spherical aberrations so introduced
are accepted as a compromise when weighed against the relative ease of
fabricating spherical surfaces. We will concentrate on the case of spherical
reflecting and refracting surfaces with a radius of curvature R.
In the limit
,
we deal with the special case of a plane surface.

In the figure the mirror shown is convex, and two rays of light originating at O are drawn, one normal to the spherical surface at its vertex V and the other an arbitrary ray incident at P. The first ray reflects back along itself; the second reflects at P as if from a tangent plane at P, satisfying the law of reflection. The two reflected rays diverge as they leave the mirror. The intersection of the two rays (extended backward) determines the image point I conjugate to O. The image is virtual, located behind the mirror surface. Object and image distances from the vertex are shown as s and s' respectively.
which combine to give
.
(8.4)
We can write these angles in terms of the distances by approximating them as small angles. In this case we have that

and

so that
.Returning to (8.4), we get
.
(8.5)
Because we approximated the effect of the angles, this approximation leads to first order, or Gaussian, optics.
If the spherical surface is chosen to be concave instead, the center of curvature would be to the left. For certain positions of the object point O, it is then possible to find a real image point also to the left of the mirror. In these cases, the resulting geometric relationship consists of terms that are all positive.
It is possible, employing a sign convention, to represent all cases by the single equation
The sign convention is as follows. Assume that
the light propagates from left to right. Then
For an object at infinity, incident rays are
parallel and
.
The image distance is defined as the focal length of the mirrors.
Thus,
(8.6)
and the mirror equation can be written, more compactly, as
.
(8.7)

The object is an extended object of transverse dimension h0. The image of the top of the arrow is located by two rays whose behavior on reflection is known. The ray incident at the vertex must reflect to make equal angles with the axis. The other ray is directed toward the center of curvature along the normal and so must reflect back along itself. The intersection of the two reflected rays occurs behind the mirror and locates a virtual image there. Because of the equality of the three angles shown, it follows that
.
(8.8)
The lateral magnification is determined by the ratio of lateral image size to corresponding lateral object size, giving
.
(8.9)
Extending the sign convention to include magnification, we assign a plus magnification to the case where the image has the same orientation as the object and a negative magnification when the image is inverted relative to the object. Applying this to the figure, in order to produce a plus magnification since s' must be negative, we must modify equation (8.9) to give a general form of
.
(8.10)



In all cases, three rays are drawn from the object to the mirror. The first ray moves horizontally from the object to the mirror. This ray will pass through the focal point, since parallel rays focus at the focal point. By a similar argument, the second ray can be drawn. This ray runs from the object to the focal point, and then reflects from the mirror along a path parallel to the central axis. Finally, the third ray is drawn from the object to the center of curvature. This is the ray which is unaffected by the mirror and continues to move in a straight line. The point at which the three rays again cross show where the image will form. If the object and mirror are drawn to scale, then the position of the object can be determined directly.
Probably the most common mirror in our everyday
living is the planar mirror. The spherical mirror described by equation
(8.5) yields a plane mirror with
,
s' = -s. The negative sign implies a virtual image for a
real object. Similarly, the magnification is found from equation (8.10)
to be equal to unity. This tells us that the resulting image is a full
sized, virtual erect image. Each point of an extended object, at a perpendicular
distance s from the mirror, is imaged at the same distance behind
the mirror. The only obvious change is the 180
rotation about the optical axis, an effect known as reversion. Notice
that if we image a coordinate axis, it also undergoes a revision.

The coordinate system, which was originally described by a right hand rule, now requires a left hand rule to properly orient the axes. The process which converts a right handed coordinate system into a left handed one is known as inversion, and it has important uses in other areas of physics as well.
We saw earlier that a spherical mirror will reflect a point into a perfect image only when the object and image both lie at the center of curvature of the mirror. This rarely happens in reality. With other distances, a perfect image will result only if an aspherical mirror is used. In order to understand this, consider the following drawing:

Here, the surface
represents plane waves traveling though a region with index of refraction
n2, and then impinging on a Cartesian ovoid with an index
of refraction n1. By Fermat's principle, in order for
an arbitrary point D on the wavefront to focus at the point F1
in minimum time, the optical path lengths must all equal a constant, as
stated in equation (8.2). Comparing the equation to the figure, we see
that
or
.
(8.11)
From analytic geometry, the surface
can be associated with the directrix of the ellipse, so that
(8.12)
where e is the eccentricity. Solving (8.12)
for
and substituting into (8.11) shows that we must have
.
(8.13)
It should be noted that even though we used an ellipse to construct this result, it applies equally well to any conic section.
Returning to mirrors, we can apply this result to see that any conic shape can be used to make a mirror. Indeed, telescopic mirrors are actually parabolic in shape, and hyperbolic and ellipsoidal elements are used in the construction of Cassegrainian and Gregorian secondaries.
Last updated: July 22, 1997
Comments to: D-Suson@tamuk.edu