Lecture 17:Achromatic System
Opponency vs. Tricromacy
-trichromacy is correct at the leve of the retina (3 cone types)
-experimental prediction: humanobserversshould beableto matchthe appearanceof anycolor
just by mixing in R,G,B
-oponency theory is correct at the level LGN and V1
-cell exhibit color antagonism
-R to G
-B to Y
“First trichromacy then opponency”
-building opponent axes from the 3 cone types
-take the sum of L and M cone responses
-red-green channel what should we do with the three responses to tell us where we are along
the red-green spectrum? If I looked only at S and M cones, it would only tell me if I had more
red or green compare the strength of the response of L and M cones (talk about ratio or
-key point is just compare (take the difference) of L cone responses and the M cone responses
(suppose the response difference is 0/equal, it would mean that it is achromatic/gray)
THE CHROMATIC SYSTEM
-The “Red-green” channel
-take the difference between L and M cone responses. -if the L and M cones are responding
equally gray patch (achromatic) L
L - M
-if the L - M cones are positive M
-if the L– M cones are negative
-if it is M-Lit is going to be opposite
-that comparison is going to show which part of the axes the color is.
Rather than simply combining L and M responses, we are comparing them and saying which
one is bigger.
The “Blue-yellow” channel/system
L (red) M (blue) S (green)
-we want to know if there’s more blue or more yellow.
Step1: Pull the L+M response (not caring about what’s red or green)
Step 2: compare it to the S response
-take the difference between the L and M response and the S response. (L+M-S)
-by comparing L+M vs. S (the minus is simply a way of asking is L+M bigger or is S bigger –
responding more strongly)
-if it is more blue the S response is larger
-if more yellow L+M is larger
-when we talk about yellow, we’re not concerned about red and green it’s ok to pull L+M
responses because you care about the overall spectrum (we don’t need to know the stronger L
response or stronger M response)
-you can very easily build up the three opponent color axes by combining the three cone
responses in difference ways.
Constancyin perception: your perception remains the same even though the image of the retina
is getting smaller/bigger. 3D shape seems constant even if our point of view changes
Color Constancy: changing light conditions change the composition of light reflected from a
-the retina sees color and shadow different colors, but we perceive it the same color. -when the light levels change, when the color composition of light changes, we still perceive
red as red and blue as blue.
-comparing how red color looks with different wavelength
composition is different (sunlight, fluorescent light, neon light,
-perceived surface color remains constant.
-in order to have constancy, it has to compare
-the reason why you see the dress in two different ways
-blue and black dress bright, slightly yellow light
-white and yellow under a shadow (dark, slightly blue)
Wavelength depend on colorof surface+wavelength composition of light thatis illuminating
on the scene.
-the blue and yellow is the same on the retina.
-RGB on the image RGB on the retina
Adelson’s checkboard illusion (the same concept as the dress) – this is light constancy.
The Color Constancy
The light Lthat is hitting our retina - reachingour eyes is a combination of two different things: 1. Spectral reflectance of the surface (S) – it is the physical property; for each wavelength, we
ask, what proportion of light in the wavelength does it balance out.
-for each wavelength, how much light is reflected back by the surface.
-graph what color would this it be? purplish (high wavelength in blue and red) type of
-the surface reflects back certain wavelengths more than it does with other wavelengths
-but the light that hits our eye also depends on:
2. Spectral composition of the
illuminant (the light source) (I)
-graph corresponds to sunlight-
it has approximately equal
wavelength (except for the blue)
-the reflectant of gray surface
let’s say that the reflectants of gray
surface is 60% (0.6) let’s say the light has 200 photons hitting the surface. How