Perception
Psych 293
Spring 2015
Exam 2 Answers
There were 135 points on the exam as a whole — 9 questions worth 15 points each. You earn points on each question starting at zero. A score of 7 or 8 means you got the question about half right. Those are positive points. A score of 10 means you got a lot of it right. A score of 12 – 14 means you missed just a little bit. And 15 is full credit on a question. On each page, the scores for each question are added, so you will see a total for the page. Those are added up on the back page.The average number of points, out of the 135, on this exam, was 80.11.
1. The graph of rhodopsin — Show wavelengths from 400 to 720 nm on the x axis and “Proportion of light absorbed” on the y axis. The y axis goes from 0 at the bottom to .1 at the top. The graph has one peak in the middle, at 510 nm.
2. The Hurvich & Jameson graph is the graph of the APPEARANCE of each wavelength of light in the visible spectrum. The graph consists of two lines going across the spectrum. One line is the blue-yellow line; the other is the red-green line. You can pick any number between 400 and 720 nm. Your graph should show a plausible color appearance at that wavelength. At 410 nm, the appearance is a mixture of red and blue, so the red and blue should each be large in their direction at that point. Moving to the right on the x axis, there continues to be both red and blue but the amounts get closer to zero. When the red-green line crosses zero, its interpretation changes from red to green. That means that colors in the middle of the spectrum contain green, but no red. Red comes back in the mid 500s. After the blue-yellow line crosses zero the interpretation changes from blue to yellow, showing that nearly all of the remaining colors on the spectrum contain some yellow, but no blue. The points on the graph are produced through the Hurvich & Jameson cancellation technique. For purplish or violet appearances (a mixture of red and blue) the given wavelengths have to plot a certain amount of blue and a certain amount of red. To determine the amount of blue, yellow is added to cancel the blue. To determine the amount of red, green must be added. When the blue is cancelled, the light will look like unique red (no blue). For each point on the x axis (a given wavelength), two points on the y axis must be determined.
3. The controversy was resolved by declaring both 3 color theory and 4 color theory correct. Hurvich & Jameson maintained that there are indeed 3 sets of cones in the retina and recognize that lights from 3 parts of the spectrum can be mixed to make any color of the rainbow. However, they also maintain that color appearance comes from another level. To explain color appearances requires 4 color, opponent process, theory. Here the primaries include yellow. They argue that red-green is one component of this system and blue-yellow is another pair. Their wiring diagram with 3 receptors linked to the two opponent pairs (plus black and white) shows the 3 color system and 4 color system on one diagram. Note that 3 color theory is based on the observation that all of the colors of the rainbow can be generated from combinations of 3 properly chosen lights. Yellow is not a primary because it can be made by mixing red and green lights. Which is what happens on your computer. In 4 color theory, there is a different definition of primaries. Here, the primaries are the component color experiences across the visible spectrum. Every wavelength has an appearance that is a mixture of red and blue, blue and green, green and yellow, and yellow and red. Yellow is a primary because it is a component of a range of appearances.
4. This is what one does in the Hurvich & James cancellation technique. If the appearance of a light is purplish, then adding yellow is what one does to cancel blue and estimate the amount of blue in the appearance of the purplish light. When enough yellow is added to cancel the blue, the light looks red.
5. The graph of rhodopsin shows that activation of rhodopsin cannot determine color (appearance based on wavelength alone) because photons can be absorbed at nearly any wavelength. Once enough photons are absorbed by the rhodopsin, there is no way to tell what wavelength of light was involved. Thus, a single receptor type in the eye cannot ever, alone, determine color. This is true for ANY type of visual pigment, including the 3 types in the cones. The fact that rhodopsin (the rod pigment) was used in the example is irrelevant to the point. The point holds for any pigment taken by itself because all pigments absorb light over a wide range of wavelengths. This could initially make color vision puzzling because it says that a long wavelength sensitive pigment, taken by itself, cannot be the basis for sensing long wavelength light as such. That is, just knowing that so-called long wavelength sensitive cones are activated is not, by itself, an indicator of the presence of long wavelength light.
6. The situation described is a light consisting of 3 component wavelengths compared to a light with a single wavelength. The question is always: “Do the two lights look exactly alike (yes or no)?” You need to notice that in this kind of an experiment, the two lights nearly always look different. We are varying the wavelengths in the adjustable light to try to find at least one combination of settings that make the two lights look identical. Therefore, we cannot tell anything until we have done a sweep across the entire range of adjustments possible. What we are looking for are situations where intensity adjustments of the component wavelengths NEVER yield a match. That is when we enter an N (no, there is never a match) into the table. The Y in the table means “yes, somewhere there is a setting that yields a match.” But if we have a set of wavelengths that can be varied to get a match to the other light, we need to keep in mind that the two lights looked different most of the time. If you missed the fact that the question was about only one setting and answered as if the question were about the relation between number of underlying pigments and number of adjustable wavelengths in the variable light, then you should answer “4 or more underlying pigments.” To fully distinguish lights on the basis of wavelength, a person has to have more underlying pigments than the number of adjustable wavelengths in the variable light.
7. This is the Hurvich & Jameson situation. They were measuring the appearance of every wavelength. Their graph shows that neary EVERY wavelength LOOKS LIKE a mixture of colors — at the short wavelengths the appearances are mixtures of red and blue. In the middle of the spectrum the appearances are mixtures of blue and green, then green and yellow, and finally yellow and red. Out of all the wavelengths, only 3 (the unique colors) do not look like mixtures. Thus, not only CAN a single wavelength look like a mixture, but nearly all of them do. From the standpoint of appearance, a “pure” color would be a unique one, one that does not appear to be a mix of colors. Because most single wavelengths look like mixtures, this is the opposite of metamers where the “purity” of a light is defined by the wavelengths. Here, the “purity” is defined by the appearance.
8. Draw the table that shows the relation between “number of underlying visual pigments” and “number of component wavelengths in a variable light” for a light matching experiment. Where the number of underlying pigments is more than the number of adjustable wavelengths, the two lights can be discriminated on the basis of wavelength and there never will be a match (enter an N in the table. N means “No there is NEVER a match.”) A “Y” means that somewhere in the set of adjustments a match CAN be found — Yes a match is possible.Think about Case 1 — a person with one underlying pigment looking at two lights of different wavelengths, but one wavelength in each light. Varying the intensity of the wavelength of the variable light eventually leads to a setting where the two lights look the same. A reason for this is found in what we learned from the Rhodopsin graph. A given number of photons can be absorbed by an receptor at ANY wavelength if we have enough light at that wavelength. That means that a match between the two lights is possible and the first entry in the table is a Y for “yes, there can be a match.” Remember that I stressed that the lessons of the Rhodopsin graph are the foundation for this whole section. What we learned in the first class covered on this exam contains the lessons you have to hang on to, then apply and extend as we go through the cases represented on the table.
The next case is when we go to a new person, who has 2 underlying visual pigments. For that person, the two lights, each with a single wavelength, but different wavelengths, can never be made to match using only intensity adjustments. In the table, where the number of underlying pigments is 2 and the number of adjustable wavelengths is 1, enter an N. If, however, we can add a second adjustable wavelength to our variable light, then the two lights can be made to match — at some setting. We enter a Y where the number of underlying pigments is 2 and the number of adjustable wavelengths is two. Notice that the number of underlying pigments is NOT variable within a person. That would take a major eye operation. If we were collecting data to fill in the table, each row (number of underlying pigments) would have to be a different person. The only row that would be easy to find people for would be the row with 3 underlying pigments.
9. The fantasy opponent system would allow people to see combinations of red and green, and blue and yellow. What could NOT be seen would be combinations of red and blue (thus, no magenta) and combinations of green and yellow (thus, no green yellows). One could see red-yellow, red-green, blue-yellow, and blue-green. A full answer would mention both the color combinations that could not be seen as well as the combinations that could be seen.