:
1. Estimate the smallest dot that can be seen from 20 cm. Assume
that the dot cannot be seen if its image stimulates only a single cone in the
fovea, and that the fovea is 1.5 mm in diameter and that the effective lens
center of the eye is 14 mm from the retina.
Textbook (p. 17) says that a cone is 0.5 micron in diameter, with
cone-to-cone spacing of 3.0 microns (I assume center-to-center spacing here). In
worst case, image of dot will be centered on one cone and could have a radius of
(3.0 - 0.25) microns without stimulating a second cone. So max diameter of image
without being seen could be 5.5 microns. Magnification M = di/do
where image and object distances are measured from the effective lens center. M
= 14/200, so ho = hi/M = (200)(5.5)/14 = 78.6 microns is
minimum diameter of dot to guarantee it will be seen.
2. About how long will it take to transmit a 256 level (8 bits)
grayscale image of size 512 x 512 at a baud rate (bits/s) of 9600?
[(512 x 512 x 8) bits]/[9600 bits/s] = 218 s = 3.6 minutes
3. A 4 x 4 image has values
{{3,1,2,1(q)},{2,2,0,2},{1,2,1,1},{1(p),0,1,2}}. Compute the D4,
D8, and Dm distances between p and q subject to V{0,1}
[i.e., considering only pixels with values of 0 or 1].
