Cool Visual Illusions: Craik-O'Brien-Cornsweet Effect

This week's visual illusion is related to Mach bands, and similar in some ways to the watercolor effect. It's called the Craik-O'Brien-Cornsweet effect (or just the Cornsweet effect)1. This is the best example I've ever seen (from here):

i-40e191c517865f75cd38969eac76ed29-purvescornsweet.jpg

What you should see is a dark square over a light square (almost white). Now take your finger and place it over the boundary between the two squares. What do you see? Two squares of the same color!

What's going on here? Well, at the boundary between the two squares, the top one really is dark, and the bottom one really is white. The visual system seems to be taking the dark gray and white at the edges of the squares, and spreading it across them (sort of like the visual system spreads the boundary colors throughout the center of the watercolor effect figures), so that otherwise identical squares look very different. Here is a more traditional version2:

i-a03d7160c760126951159f918b894695-purvescornsweet2.jpg

How this happens, exactly, is still a matter of contention, but most explanations do include our old friend the center-surround receptive field, working much as they do in perceiving Mach bands (see the post linked above). As with Mach bands, Dale Purves and his colleagues have argued that the Cornsweet effect is instead a result of the way our visual system is wired. Since the visual system is wired the way it is because of the normal properties of the visual environment it was built to interpret, it can be fooled when some properties of a stimulus mimic properties commonly found in the visual environment, while others don't. If this explanation is true, then making Cornsweet effect stimuli look even more like ordinary visual scenes should enhance the Cornsweet effect, while making the stimuli look less like ordinary scenes should decrease the effect. Purves et al. (reference in footnote 2) tested this prediction in several ways, one of which was to change the orientation of the stimuli. I'll let Purves et al. describe their reasoning to you:

Because humans evolved in an environment in which the primary source of illumination is usually from above (i.e., from the sun), the spatial arrangement of the same objects can look quite different when they are turned upside down. Thus, if the Cornsweet stimulus is rotated from its usual horizontal presentation such that the dark gradient is above and the light gradient below... the stimulus is more likely to have been generated by light from above (because the direction of the gradients is consistent with a doubly curved surface arranged in this way). If, on the other hand, the same stimulus is rotated 180°... this likelihood is diminished.

Accordingly, when the equiluminant territory adjoining the dark gradient is uppermost its surface is likely to be less reflective than that of the lower territory. The reason is that the possible sources of the stimulus include at some higher level of probability an object whose uppermost surface is better lit than the lower surface (as indicated in the cutaway view to the right of the stimulus). When two surfaces return the same amount of light to the eye and one is better lit than the other, the better lit surface will always have been the less reflective. Because the visual system, according to our theory, constructs percepts based on the relative probabilities of the possible sources of the stimulus, the statistical influence of this increased probability causes the uppermost of the two equiluminant adjoining surfaces to appear darker than the lower one. (p. 8547)

So, the Cornsweet effect should be much stronger when, at the boundary between the two squares, the dark color is on top and the lighter color on the bottom, than when the light color is on the top and the dark on the bottom. To see that this is the case, check out Purvis et al.'s stimuli:

i-17850aa96cac68a38977c4c9b414a0cc-purvescornsweet3.jpg

Consistent with Purves et al.'s prediction, the effect should be stronger for the figure on the left than the one on the right. As with Mach bands then, it may turn out that our visual system is just perceiving these figures the way it is "designed" to, based on the statistical properties of its environment. Or, it could still turn out to be the case that the receptive fields on the retina and directly connected to those on the retina are being tricked by overlapping areas of differing brightness. Only further research will tell.

One thing is certain, though. No matter how many times we look at these figures knowing that they are the same color, some part of our visual system is going to be tricked, and we're going to perceive one as being lighter than the other. When you think about it, that's a little disturbing. No matter how much we know, we still can't see the figures the way they actually are. Reality matters to the visual system only statistically.


1Cornsweet, T.N. (1970). Visual Perception. New York, NY: Academic Press.
2From Purves, D., Shimpi, A., Lotto, R.B. (1999). An empirical explanation of the Cornsweet effect. Journal of Neuroscience, 19, 8542-8551.

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Very facinating stuff. Though I have a problem with the two pictures in the bottom. Or actually with the right most ones. My brain tells me that the sheets are the same color and I cannot produce a feeling that they would be different. Just doesn't happen. The colors on the left look very different, on the other hand. The same happens with the text book picture (the second from the top) - I don't see a difference in the colors, but I am very much fooled with the colors of the first multicolored picture in the top.

I think that the role of the Mach effect in enhancing volume illusion thus solar lighting illusion is better shown in a figure like this:

Figure

In the right figure, we really have the feeling that the upper surface is darker than the lower. This feeling disappears in the left figure - though it looks like there is a very faint reflected light on the bottom of the lower surface. Of course, all the surfaces are the same value of grey.