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 ' ',
(for small angles), becomes
the law of reflection,
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
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,
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
The transformation equations for reflection are
easily derived from Fig. 31. From the figure,
The transformation equations are then,
is the reflection
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
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
The distances x and x ', s and
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:
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?
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).
Now the actual rays pass through the point 8 cm
to the left of O, the point
I ' in Fig. 32. The magnification m =
and therefore the image
is inverted and 2/3 cm in size.
(2) We may also use
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).