Question

Considering the illustrations for Concave and Convex mirrors.

Prove using geometry that the reflected rays reach the focal point f=R/2 in the limit as the incoming rays approach the principal axis.

Hint: Consider the triangle formed by the radius of curvature, principal axis, and reflected ray, and use the law of sines.

Answer 1

There are two alternative methods of locating the image formed by a concave mirror. The first is purely graphical, and the second uses simple algebraic analysis.

The graphical method of locating the image produced by a concave mirror consists of drawing light-rays emanating from key points on the object, and finding where these rays are brought to a focus by the mirror. This task can be accomplished using just foursimple rules:

1. An incident ray which is parallel to the principal axis is reflected through the focus of the mirror.
2. An incident ray which passes through the focus of the mirror is reflected parallel to the principal axis.
3. An incident ray which passes through the centre of curvature of the mirror is reflected back along its own path (since it is normally incident on the mirror).
4. An incident ray which strikes the mirror at its vertex is reflected such that its angle of incidence with respect to the principal axis is equal to its angle of reflection.

The validity of these rules in the paraxial approximation is fairly self-evident.

Consider an object which is placed a distance from a concave spherical mirror, as shown in Fig. 71. For the sake of definiteness, let us suppose that the object distance is greater than the focal length of the mirror. Each point on the object is assumed to radiate light-rays in all directions. Consider four light-rays emanating from the tip of the object which strike the mirror, as shown in the figure. The reflected rays are constructed using rules 1-4 above, and the rays are labelled accordingly. It can be seen that the reflected rays all come together at some point . Thus, is the image of (i.e., if we were to place a small projection screen at then we would see an image of the tip on the screen). As is easily demonstrated, rays emanating from other parts of the object are brought into focus in the vicinity of such that a complete image of the object is produced between and (obviously, point is the image of point ). This image could be viewed by projecting it onto a screen placed between points and . Such an image is termed a real image. Note that the image would also be directly visible to an observer looking straight at the mirror from a distance greater than the image distance (since the observer's eyes could not tell that the light-rays diverging from the image were in anyway different from those which would emanate from a real object). According to the figure, the image is inverted with respect to the object, and is also magnified.

Figure 72 shows what happens when the object distance is less than the focal length . In this case, the image appears to an observer looking straight at the mirror to be located behind the mirror. For instance, rays emanating from the tip of the object appear, after reflection from the mirror, to come from a point which is behind the mirror. Note that only two rays are used to locate , for the sake of clarity. In fact, two is the minimum number of rays needed to locate a point image. Of course, the image behind the mirror cannot be viewed by projecting it onto a screen, because there are no real light-rays behind the mirror. This type of image is termed a virtual image. The characteristic difference between a real image and a virtual image is that, immediately after reflection from the mirror, light-rays emitted by the object converge on a real image, but diverge from a virtual image. According to Fig. 72, the image is upright with respect to the object, and is also magnified.

The graphical method described above is fine for developing an intuitive understanding of image formation by concave mirrors, or for checking a calculation, but is a bit too cumbersome for everyday use. The analytic method described below is far more flexible.

Consider an object placed a distance in front of a concave mirror of radius of curvature . In order to find the image produced by the mirror, we draw two rays from to the mirror--see Fig. 73. The first, labelled 1, travels from to the vertex and is reflected such that its angle of incidence equals its angle of reflection. The second ray, labelled 2, passes through the centre of curvature of the mirror, strikes the mirror at point , and is reflected back along its own path. The two rays meet at point . Thus, is the image of , since point must lie on the principal axis.

In the triangle , we have , and in the triangle we have , where is the object distance, and is the image distance. Here, is the height of the object, and is the height of the image. By convention, is a negative number, since the image is inverted (if the image were upright then would be a positive number). It follows that

(351)

Thus, the magnification of the image with respect to the object is given by

(352)

By convention, is negative if the image is inverted with respect to the object, and positive if the image is upright. It is clear that the magnification of the image is just determined by the ratio of the image and object distances from the vertex.

From triangles and , we have and , respectively. These expressions yield

(353)

Equations (352) and (353) can be combined to give

(354)

which easily reduces to

(355)

This expression relates the object distance, the image distance, and the radius of curvature of the mirror.

For an object which is very far away from the mirror (i.e., ), so that light-rays from the object are parallel to the principal axis, we expect the image to form at the focal point of the mirror. Thus, in this case, , where is the focal length of the mirror, and Eq. (355) reduces to

(356)

The above expression yields

(357)