Page 69 - Modern Optical Engineering The Design of Optical Systems

52 Chapter Three

n′u′ u′ nu y(n′ n)/R 0.214286 1.0 ( 0.5)/( 10) 0.164286,
and the image location can be found from l′ y/u′ 1/0.164286 6.086961.
The magnification can be found from an oblique raytrace from the top of a
1 mm object, or from Eq. 3.15b, m h′/h nl′/n′l. Thus m 1.5 ( 6.086961)/
1.0 7 1.304349 and the image size is 1.304349 mm.
Tabulating the results for the three thicknesses, we have:
(a) th 7.0 u′ 0.164286 l′ 6.08961 h′ 1.304349 mm
(b) th 10.0 0.100000 10.00000 1.500000 mm
(c) th 16.66 . . . 0.090000 11.11111 1.000000 mm
Note that the image size can also be found from Eq. 4.16, m nu/n′u′.

2 Given an equiconvex lens, radii 100, thickness 10, and index 1.50,
trace a ray (parallel to the axis) through the lens, beginning at a ray height of
(a) 1.0, and (b) 10.0, and determine the axial intersection of the ray.
ANSWER:
R 100. 100.
t 10.
n 1.0 1.5 1.0
(a) y 1.0 0.9666 . . . l′ y/u′ 98.3051
nu 0.0 0.005 0.0098333 . . .
(b) y 10.0 9.666 . . . l′ y/u′ 98.3051
nu 0.0 0.050 0.098333 . . .

3 Determine the effective and back focal lengths of the lens in Exercise 2.,
(a) from the raytrace data, and (b) using the thick lens equations.
ANSWER:
(a) Eq. 3.19 efl y /u′ 1.0/( 0.0098333) 101.6949
1 k
Eq. 3.20 bfl y /u′ .9666/( 0.0098333) 98.3051
K k
(b) Eq. 3.21 (1/f) (n 1) [c c tc c (n 1)/n]
1 2 1 2
0.5[0.01 ( 0.01) 10 0.01 ( 0.01) 0.5/1.5]
0.5[0.02 0.000333] 0.0098333 . . .
efl 101.6949
Eq. 3.22 bfl f ft (n 1)c /n]
1
101.6949 101.6949 10 0.5 0.01/1.5
101.6949 3.3898
98.3051
4 What is the focal length of the lens in Exercise 3 if it is treated as a thin
lens?
ANSWER: Eq. 3.25 (1/f ) (n 1)[c c ]
1 2
0.5[0.01 ( 0.01)] 0.5 0.02 0.01
efl 1/ 100.0