Page 190 - Modern Optical Engineering The Design of Optical Systems

Characteristics of the Human Eye 173

1
3 Assuming a depth of focus of
diopter, over what range of distance is
4
vision clear when the eye is focused at 10 in?
1
ANSWER: A distance of 10 in is about
m, and may be expressed as 4 diopters,
4
so the depth of focus, expressed in diopters is 3.75 to 4.25 diopters. The corre-
sponding distances are 1/3.75 = 0.2667 m and 1/4.25 = 0.2353 m. Thus the
depth of focus is 0.2667 0.2353 = 0.0314 m = 31.4 mm = 1.235 in.
4 It is desired to set an optical vernier to a precision of 0.0001 in. Assuming
that the vernier projects an image of a ruled scale on to a screen which is
viewed from a distance of 10 in, and that the setting is made by aligning a
scale line with a cross hair on the screen, what magnification must the pro-
jection lens of the optical vernier have? Use 10 seconds of arc for the vernier
acuity of the eye. (Note: 1 second equals 0.000005 radians.)
ANSWER: At a distance of 10 in, 10 seconds of arc equals (10 0.000005)
10 in = 0.0005 in. For this to be equivalent to 0.0001 in at the scale, the scale
must be magnified by 0.0005/0.0001 5.0 .

5 A convex reflector with a radius of curvature 10 in is mounted on a spin-
dle and rotated. (a) What is the largest amount that its center of curvature
can be displaced from the axis of rotation without the motion of the reflected
image being detected by the naked eye? (Assume that the reflected image is
viewed from a distance of 10 in.) (b) What are the fastest and slowest speeds
of rotation at which the motion caused by a decentration of 0.02 in can be
detected?
ANSWER: (a) If the displacement is e, then the total up and down is 2e, and
viewed from a distance of 10 in the motion subtends an angle of 0.2e radians.
If the eye can detect a motion of 10 seconds, or 0.000050 radians, then 0.2e
equals 0.000050, and e equals 0.000050/0.2 0.000250 in. (b) If the displace-
ment is 0.02 in and the image rotates in a circle, the image travels 2 0.02
0.12 in per revolution of 360 . At a viewing distance of 10 in this is 0.012
radians per revolution. If the slowest detectable motion is one (or 2) minutes
of arc per second (0.0003 rad/sec), then the rate of rotation is (0.0003
rad/sec)/(0.012 rad/rev), or 0.024 rev/sec = 1.4 rev/min (or 2.9 rpm if we use 2
min). If the fastest visible is 200 /sec ( 3.5 rad/sec), then the rate of rotation
is 3.5/0.012 280 rps.

6 (a) If a plane parallel plate is specified to have zero 10 millidiopters pow-
er, what is the shortest tolerable focal length it may have? (b) Assuming that
one surface is truly flat, what is the strongest (shortest) acceptable radius for
the other surface, if the index is 1.6? (c) If the piece has a diameter of 20 mm,
how many Newton’s rings will be visible when this surface is tested against a
true flat? (Assume a wavelength of 555 nm. One fringe occurs for each half
wavelength change in the thickness of the airspace.)