Chapter 7

Spherical Mirrors

Convex and concave spherical reflecting surfaces have image-forming qualities which can be discovered from the laws of reflection stated in chapter 1. In what follows we shall apply the matrix method to the case of reflection, again limiting consideration to paraxial rays.

The difficulty that often arises in discussing image formation by mirrors is that of applying the sign convention used for refraction. In the case of refraction the rays continue in the same (positive x) direction after refraction but in reflection the rays are of course reversed. To avoid the difficulties, one may define a different convention for reflection or specify the positive image direction in terms of the on-going emergent rays. Or, in terms of the mathematical formulation, one can regard reflection as a special case of refraction in which Snell's law, n theta = n ' theta ', (for small angles), becomes the law of reflection, theta = - theta ' for n = -n '.

In the matrix formulation it is perhaps easier to maintain all our previous sign conventions for refraction and treat the reflected rays as having a slope angle equal to - alpha '. This in effect allows us to treat the rays after reflection as if they were proceeding on in the same direction as the incoming rays. Of course the final calculated quantities such as a positive image distance or a positive focal distance, for example, will actually lie to the left of the reference point in constructing a ray diagram. The actual path of a ray after reflection from a convex mirror is shown in Fig. 30 and the corresponding diagram for deriving the reflection matrix is shown in Fig. 31.

Figure 30.

Figure 31.

The transformation equations for reflection are easily derived from Fig. 31. From the figure,

$$\theta=\delta+\alpha$$

and
$$\alpha'= \theta+\delta.$$

Therefore
$$\alpha'= 2\delta+\alpha=2y/R+\alpha$$

since
$$\delta=y/R=y'/R$$

The transformation equations are then,
$$ y = y'+ 0 \alpha'$$


$$\alpha=-2/Ry'+\alpha'$$

or
$$\bmatrix{y\cr\alpha}_o=\bmatrix{1&0\cr -2/R&1} \bmatrix{y'\cr\alpha}$$

where
$$\bmatrix{1&0\cr -2/R&1}$$

is the reflection matrix.

The Single Reflecting Surface

For a single reflecting surface the [A]OO matrix refers simply to the vertex of the reflecting surface and from (28) with [A]OO equal to the reflection matrix,
$$ f = -R / 2 \eqno(71)$$ (71)

and from (30) f ' = f as the matrix determinant is unity. Of course Newton's equations for image formation apply as always and may be written here as
$$ x x' = R^2/ 4$$

and
$$ m = R / 2x = 2x' / R. $$

Alternatively,
$$f'/s'+f/s=1$$

becomes
$$1/s'+1/s=-2/R$$

and
$$m=-s'/s$$

The distances x and x ', s and s ', and R follow the same convention as before for refraction. The final answer however must then be "reflected" back to its true position. Let us illustrate this with the following example:

Example 1

A concave mirror has a radius of 12 cm and an object 2 cm high is placed 24 cm to the left of the vertex O of the mirror. Where is the image formed and how large is it?

Solution: (1) f = -R/2 = -(-12)/2 = 6 cm.

OF is 6 cm to the left of O.

OF' is 6 cm to the right of O.

Since x = 18 cm, x'(18) = 36, x ' = 2 cm to the right of F ' or 8 cm to the right of O (I '(calc) in Fig. 32).

Figure 32.

Now the actual rays pass through the point 8 cm to the left of O, the point I ' in Fig. 32. The magnification m = -f/x = -6/18 and therefore the image is inverted and 2/3 cm in size.

(2) We may also use

$$1/s'+1/s=-2/R$$

which gives
$$1/s'=(-2/-12)-1/24=1/6-1/24=1/8$$

Then s ' = 8 cm to the right of the matrix plane at O (I '(calc) in Fig. 32) or the real image is 8 cm to the left of O, (I' in Fig. 32).