## Problems

5.1. A "one-dimensional" photoreceptor is modeled by a Gaussian DSF, s(x):

x = 9m/2 is the acceptance half-angle; the ct parameter is chosen so s(±8m/2)

= 0.5. For this to happen, ct = 8m/2/ ^ln(4) . When the black spot is directly over the receptor (centered at x = 0), the object has a one-dimensional intensity distribution, f(x) = Io[1 - U(x + a) + U(x - a)] (see Figure 5.2-4). a. Derive an expression for the absorbed intensity contrast, Cie = (Iemax -Iemin)/(Iemax + Iemin). Ie is the absorbed intensity in the rhabdom of a retinula cell. In general, Ie is given by k P™

Clearly, Iemin occurs when the spot is at x = 0, and Iemax occurs when the spot is at x.

b. Plot log(Cie) vs. a/8m/2 for 0 3 a/8m/2 3 100. (Use semilog paper.) 5.2. A "one-dimensional" photoreceptor is modeled by a hyperbolic DSF, s(x).

This receptor views a one-dimensional, square-wave object as shown in

Figure P5.2. The square wave has period A and peak intensity, Io.

a. Find a general expression for Cie when the object is shifted by A/2 (one-half spatial wavelength. Assume limiting resolution similar to the development in the second example in Section 5.2.2.

b. Plot and dimension Cie vs. 8m/2/A.

c. Now consider the small movement sensitivity of the receptor. The stripes are now shifted by x = A/4 so that the falling edge of the square-wave intensity is at the origin. A small displacement of the square wave to the right by x = +8x will cause a AIe > 0. A small displacement of the object to the left by x = -8x will cause a AIe < 0. Now assume

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