Problem Set for Module 3
1. Disparity (20 points)
In lecture, we derived the formula for determining the distance of a point (from one’s eyes)
through binocular vision:
D = fb/( + )
***Numerator is constant
F= fo
...
Problem Set for Module 3
1. Disparity (20 points)
In lecture, we derived the formula for determining the distance of a point (from one’s eyes)
through binocular vision:
D = fb/( + )
***Numerator is constant
F= focal length of the eye (constant)
B= distance between two eyes (constant)
***denominator is changing
Alpha: distance from retina of left eye to object
Beta: distance from retina of right eye to object
Alpha plus beta= disparity
Inverse relationship
Large disparity= short distance
Small disparity= large distance
(The meaning of the symbols in this formula were all explained in lecture.) The denominator
here is called the disparity.
a) Start with an an object / point that is very far away (say the Distance is 1,000 meters).
a. Is the disparity large or small?
Small disparity
b. If the object moves a relatively small distance (say 1 meter), what kind of
change will this make to the disparity?
If the object starts at a very far away point and moves a small distance, this will
have a small effect on the disparity.
b) Since the retina is not, in fact, a smooth continuous surface, but is rather composed of
neurons (which have a finite size), it has a limited resolution. Explain how this fact
affects our ability to determine the distance of points that are very far from us (and
thus have a tiny disparity).
Because the retina is composed of neurons that have a limited resolution, when an
object reaches a certain distance away we are unable to detect a small change in that
distance. Therefore, when objects get really far away we are unable to use binocular
vision to determine disparity because the algorithm essentially becomes useless.
c) Start with an an object / point that is very close (say the Distance is 5 centimeters).
a. Is the disparity large or small?
Large disparity
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