Page 218 - Modern Optical Engineering The Design of Optical Systems

Stops, Apertures, Pupils and Diffraction 201

Exercises
1 Find the position and diameter of the entrance and exit pupil of a 100-mm
focal length lens with a diaphragm located 20 mm to the right of the lens. The
lens diameter is 15 mm and the diaphragm diameter is 10 mm.
ANSWER: Obviously the exit pupil is actually the diaphragm, so it is 10 mm in
diameter and is located 20 mm to the right of the lens.
The entrance pupil is the image of the diaphragm formed by the lens. Using
Eq. 2.5, and turning the system around, we have s 20, and

s′ sf/(s f) 20 100/( 20 100) 2000/80 25
and if we turn the system back to its original orientation, the pupil is 25 mm
to the right of the lens. The magnification is m s′/s 25/ 20 1.25, so
the entrance pupil diameter is 1.25 10 12.5.

2 What is the relative aperture of the lens of exercise #1 with light incident
(a) from the left, and (b) from the right?

ANSWER: (a) f-number 100/12.5 f/8
(b) f-number 100/10 f/10

3 A telescope is composed of an objective lens, f 100 in, diameter 1 in
and an eyelens f 1 in, diameter 0.5 in, which are 11 in apart. (a) Locate
the entrance and exit pupils and find their diameters. (b) Determine the object
and image fields of view in radians. Assume both object and image are at
infinity.
ANSWER: (a) The entrance pupil is at the objective (which is the aperture
stop), and has the same diameter as the objective (since a ray from the
object passing through the rim of the objective will pass through the eyelens
at y 0.1 in, well within the eyelens diameter.) The exit pupil is the image of
the aperture stop (objective) formed by the eyelens, and using Eq. 2.5 we find
the image at
s′ sf/(s f) 11 1/( 11 1) 1.1 in (to the right of the eyelens—
this is the eye relief )
The magnification m s′/s 1.1/( 11) 0.1 and the exit pupil diameter
is h′ mh 0.1 1 0.1 in, where the minus sign indicates an inverted
image.
(b) For zero vignetting the ray from the bottom of the objective to the top of the
eyelens determines the field. The slope of this ray is from 0.5 in at the objective
to 0.25 at the eyelens in a distance of 11 in. Thus u 0.75/11 0.0681818 . . .
and in 10 in the ray rises 0.681818 in to a height of 0.681818 0.5 0.181818 in
above the axis. With a 10 in objective the “real” field is thus 0.0181818 radian,
and with a 1 in eyepiece, the apparent field is 0.181818 radians.
For total vignetting, the ray from the top of the objective to the top of the
eyelens will determine the field. Its slope is from 0.5 in at the objective to 0.25 in
at the eyepiece in a distance of 11 in. Thus u 0.25/11 0.022727 . . . and