Page 85 - Modern Optical Engineering The Design of Optical Systems

68 Chapter Four

Exercises
1 A Gregorian telescope objective is composed of a concave primary mirror
with a radius of 200 and a concave secondary mirror with a radius of 50. The
separation of the mirrors is 130. Find the effective focal length and locate the
image. Figure 18.3 shows a Gregorian objective.
ANSWER: There are two ways to approach this: one is by raytracing; the other
using the two-component equations.
By raytracing:
(Note well the sign conventions for the radii, the index, and the spacing
between the mirrors.)
R 200 50
t 130
n 1.0 1.0 1.0
y 1.0 0.30 efl 1/( .002) 500.
nu 0 0.01 0.002 bfl 0.3/( .002) 150.
The focus is 150 130 20 behind (to the right of) the primary.
By separated component equations: The focal length of a mirror is R/2.
Concave mirrors act as positive focal length elements. So we use f 100
a
and f 25. In raytracing the 130 mirror spacing is regarded as a negative
b
distance; here we use the optical distance d n 130 ( 1.0) 130, and
the sign of the spacing is positive.
Eq. 4.5 f ab f f /(f f d)
a
a b
b
100 25/(100 25 130)
500
Eq. 4.6a B f ab (f d)/f a
a
500(100 130)/100
150

2 Find the effective, back, and front focal lengths of a system the front com-
ponent of which has a 10″ focal length and the rear component of which has
a 10″ focal length. The separation between them is 5″.
ANSWER:
Eq. 4.5 f ab f f /(f + f d)
b
a b
a
10( 10)/(10 10 5)
20
Eq. 4.6a B f ab (f d)/f a
a
20(10 5)/10
10.