Of interest is the minimum displacement, Sx, of a black spot centered over the eye that will produce a contrast change of 0.01. The intensity contrast change, Q, is defined by
a. Find an expression for Ie(8x = 0) in terms of the spot half-width, a, and the receptor half-intensity angle, 0m/2. Recall that Ie = ko V2X S(u)F(u) du, where u is the spatial frequency in r/°.
b. Now shift the spot to the right by 8x. Find an expression for Ie(Sx). Note that AIe = Ie(Sx) - Ie(Sx = 0) > 0.
c. Find a numerical value for the 8x required to make C(8x) = 0.01. Let a = 1.5°, 0m/2 = 1.5°; x is in degrees. The answer will require the trial-and-error solution of a transcendental equation.
f(x) |
' -- -— (Motion to right) | ||
(Black spot object) |
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