Concave and Convex Mirrors

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Spherical
mirrors can be divided into two classes: concave and convex. A spherical mirror is said to be concave if
the reflecting surface and the center of curvature (the center of the sphere of
which the mirror forms part) are on the same side of the mirror; convex if the
reflecting surface and the center of curvature are on opposite sides of the
mirror. The center of the mirror *P* is called the pole and the radius
through the pole is called the axis of the mirror.

Let us
first consider the image formed by a concave mirror of a point *O* on the axis. It was assumed earlier that an optical device forms a sharp point
image of a point object. In the case of
a concave mirror, this is true only if the rays of light make small angles with
the axis, i.e. strike the mirror close to the pole. Such rays are said to be *paraxial*. Rays outside the paraxial region do not
converge to a sharp point, with the result that the image is blurred
considerably, a defect known as *spherical
abberation*. If the image is sharply
defined, its location can be found by tracing out the paths of *any* two rays emanating from *O*.
One of these may be conveniently chosen to be normal to the mirror,
since it is just reflected back along the original path. Let the other ray strike the mirror at *Q*, making an angle *q* with the radius *CQ ,
C *being the centre of curvature of the mirror. The image *I* is located at the intersection of the
two reflected rays. We shall assume, for the moment, that *OP* is greater than the radius of curvature . Denote by *U* and *V*, respectively,
the *distances* of the object *O *and the image *I* from the pole *P*. It follows from the geometry that

(1.1) |

whence |

Now because of the paraxial condition ( ), we have

(1.3) |

and Equation (1.2) becomes

This formula gives the image distance *V* in terms of the object distance *U* and the radius of curvature *R,
*assuming that . We note the important special case in which ,
when Equation (1.4)
reduces to . This describes a situation in which a ray of
light moving parallel to the axis strikes the mirror and is reflected so as to
pass through a point *G* midway between
the centre of curvature and the pole.
Indeed, all parallel rays (within the paraxial region) are brought to a
focus at the point *G, *called the
principal focus of the mirror. A
concave mirror is therefore a converging mirror.

We are now
in a position to use the ray-tracing method to determine the location and
nature of the image of an *extended*
object produced by a concave mirror.
Because of the paraxial

condition, only a very small portion of the mirror is
illuminated, and this can be represented, to a good approximation, by a flat
surface, the concavity or convexity of the mirror being indicated by means of
the shaded triangles at the ends, as shown.
In the interest of clarity it is, of course, necessary to exaggerate
considerably the dimensions of the mirror perpendicular to the axis. Let be an erect object of height located a distance *U* from the pole. The image *I* of the point *O* clearly lies somewhere along the axis of the mirror. The image of may be found by tracing out the paths of two
rays, one passing through the center of curvature *C* (i.e. normal to the
mirror) which is reflected back along itself, the other making an angle *q*
with the axis, as shown. This image lies at the intersection of the two
reflected rays. We note for future
reference that the image *II'* is *inverted*. The magnification *m* of
the image may be defined as the ratio of the height of the image to the height of the object:

(1.5) |

From the triangles *POO'
*and *PII'*, we have

(1.6) |

whence |

It is instructive to repeat the above calculations in the case where the distance of the object from the pole of the mirror is less than the radius of curvature. Using the same notation as before, we get

(1.8) |

(1.9) |

or |

Note the difference
between Equation (1.10)
which locates the image on the *right*
of the mirror when the object is closer to the pole than the center of
curvature, and Equation (1.4), which locates the image
on the *left* of the mirror when the
object is further from the pole than the center of curvature. The formula for obtaining the position of
the image would appear to depend on the position of the object. To see how this undesirable situation might
be remedied, let us use the ray-tracing method to find the magnification of an
extended object . The image is found to be *upright* and to be located on the *right* of the mirror. From the triangles *POO'* and *PII'*, we note
that

(1.11) |

and the magnification is given by

This is identical to Equation (1.7) for
the magnification produced when the object is further from the pole than the
center of curvature, even though in that case the image was *inverted*.

In order to
resolve this confusing situation, a suitable sign convention must be
introduced. In each of the two cases considered above, both
the object (which takes the form of an arrow oriented at right angles to the
axis) and the center of curvature are on the *left* of the mirror, so the difference between the mirror equations (1.4)
and (1.10)
must be a reflection of the fact that the image is on the *left *in the former case and on the *right* in the latter.
Further, Equation (1.12) must be amended so as to
indicate whether the image has been inverted, with respect to the object, or
not. These considerations suggest that
the sign convention might be based on a Cartesian coordinate system in which
the pole of the mirror is chosen as the origin and the axis of the mirror as
the *x*-axis. Let us denote by *u*, *v*, and *r* the *x*-*coordinates *of the the object, image and
center of curvature, respectively, and by the *y*-coordinates
of the tips of the object and image arrows, respectively, the tails being
located on the axis. The (lateral)
magnification may then more usefully be
defined as the ratio .
In the first example (on p. 77), the center of curvature is on the left of the
mirror ( ) and an erect object on the left ( ) produces an inverted image on the left ( ).
Equation (1.4) may then be written as

(1.13) |

where is the *x*-coordinate
of the principal focus. The quantity is called the focal length of the
mirror. It is clearly a distance, hence
inherently positive, while the coordinate *f*
may be either positive or negative. The
lateral magnification becomes

(1.14) |

In the second example, just discussed, an upright object on the left ( ) produces an upright image on the right of mirror ( ). Thus Equation (1.10) again becomes

and the (lateral) magnification is

(1.16) |

as before.

We have
thus obtained a single mirror equation (1.15)
and a single formula (1.17)
for the (lateral) magnification which hold in both of the situations we have
examined. These results do, in fact,
hold in all cases and will be used exclusively in what follows. The sign convention used is called the *Cartesian
convention*. It has the virtue of
extreme simplicity, though of course care must be taken to observe the correct
signs of the coordinates. It does not
require *a priori* knowledge of any
aspect of the image (e.g. whether it is real or virtual), since the information
gleaned from an application of Equations (1.15)
and (1.17) is
sufficient to describe the image completely.

If *v* turns out to be positive (negative),
the image is located to the right (left) of the mirror. If *m*
is positive, the object and the image are on the same side of the mirror axis.
Thus an erect object produces an erect image, and an inverted object produces
an inverted image. On the other hand,
if *m* is negative, the object and image
are on opposite sides of the axis, i.e. an erect object produces an inverted
image, and vice versa. To simplify
matters, however, it may be convenient to assume that the object is always
erect. If *m* is numerically greater (less) than unity, the image is magnified
(diminished). A real (virtual) image is
produced if the reflected rays converge (diverge). Thus a real (virtual) image produced by a concave mirror must be
on the same (opposite) side of the mirror as the object.