Page 189 - Modern Optical Engineering The Design of Optical Systems

172 Chapter Eight

very small. (Actually, for the very best vision, it’s about a half a
degree.) You might also try it with the numbers below. Stare at the x.

9123456789123456789123456789123456789
9123456789123456789123456789123456789
9123456789123456789123456789123456789
123456789123456789X123456789123456789
9123456789123456789123456789123456789
9123456789123456789123456789123456789
9123456789123456789123456789123456789

5. In part, vision is a learned skill. You probably feel that you see
your entire field of vision quite distinctly. Try this:
(a) Stare straight ahead and extend your arm out to the side with
your fingers spread out. Q: How many fingers do you see?
A: The proper answer is “What fingers?” If you wiggle them or
wave you will see motion, buy you can’t resolve the fingers.
(b) Now extend your arm at about 45 to your line of sight. Now
you can detect fingers but you can’t tell how many.
(c) Extend your arm straight ahead and, with one eye, stare at
your thumb. You are now quite sure there are several fingers,
but you still can’t tell how many.

Exercises
1 What power telescope is necessary to enable a person with “normal” vision
to read letters 1-mm high at a distance of 300 ft? (Note: 1 minute of arc is
0.0003 radians.)
ANSWER: One millimeter at 300 ft is 0.04 in at 3600 in, which subtends
0.0000111 radians. For “normal” vision we need the letter to subtend 5 minutes,
or 0.0015 radians. Thus the magnification power must be 0.0015/0.0000111,
or 135 .

2 What power corrective lens would be prescribed for a nearsighted person
who could not focus clearly on an object more than 5 in away from his eye?
ANSWER: A lens with a focal length of 5 in will produce a virtual image
of a distant object at 5 in from the lens. Five inches is about 1/8th of a
meter, and such a lens has a power of 1/f 1/( 1/8) 8 diopters.
However, since the lens will not be in contact with the eye, the lens focal
length plus the lens-to-eye distance must equal 5 in. A spacing of 1.5 in
would require a 3.5 in focal length, or 88.9 mm, or 0.0889 m, and its
power is 11.2 diopters.