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HOME >HOMEWORK #2

Homework #2: Assigned 6/27/01 Due: 4:30 pm on the next class day or 4:30 pm next Thursday (July 5, 2001), whichever is later. Turn in your homework to Kai's box in the physics main office (near Margot's office) or to Kerry in the lab..

KSJ (4th edition) Chapter 26 and p 572-577
KSJ (3rd edition) Chapter 28 and p 467-470

1. If you've been sitting in a dark room for a while and then briefly look at a bright white light (say like a camera flash), you become briefly blinded. This is because your rods (which are highly sensitive to light) will reduce their sensitivity after absorbing a large # of photons--in other words you lose your dark adaptation or night vision. Explain why a dim red light (say around 700 nm) would allow you to see with your red cones, but not cause you to lose your dark adaptation in your rods. You may use the figure below to explain your answer.

2. Look at the spectral absorbance curves for the cones shown below. Ignore the rods for this problem. Consider this: White light appears white because it is absorbed equally by all cone types. Suppose we look at the rainbow/spectrum made by a prism. If you're not color blind (i.e. you have all 3 types of cones), you see red at one end of the rainbow, purple at the other, with all the other colors in between. Note that you don't see white. Why? Now, imagine that you have only two types of cones in your retina (pick any 2). Do you think you would see white in the rainbow? About where would it be? What if you had four (or more) cones?

Newton of course didn't know the spectral absorbance curves of the cones, but he made a few reasonable assumptions and by the reasoning above, concluded that we have to have at least 3 types of cones.

3. In total darkness, rods "see" photons due to thermal activation of the visual pigment at the rate of about 1 thermal activation/80 seconds. The electrical response of the rod is the same regardless of whether the visual pigment molecule was activated by a photon or by a thermal activation. A typical human rod cell has about 10^8 visual pigment molecules. At what rate does a single visual pigment molecule have a thermal activation? Just give the approximate order of magnitude.

4. A rod has a background "noise" of 1 thermal activation per 80 seconds. These thermal activations are indistinguishable from activations due to photons. Now, look at the electrical responses of rods to dim flashes (i.e. those shown in lecture). The electrical signal due to the absorption of a single photon lasts for a fraction of a second in human rods. In human rods, the electrical signal stays close to its maximum value for about 100 ms. How often do two thermal activations occur within 100 ms of each other in the same rod? Now consider a group of 250 rods. At any instant in time, what is the average # of thermal activations occurring assuming that the duration of the thermal event is 100 ms?

5. Suppose you could make changes to the rods so that you could make them better at photon detection. A rod absorbs 25% of the photons at a given wavelength. Of the absorbed photons, 2/3 of the time, the photon is detected (i.e. produces an electrical response). You decide to increase the length of the rod so that the rod now absorbs 50% of the photons at the given wavelength. How many times longer would your new rod have to be? Note that the noise (# of thermal activations of the visual pigment per rod) will increase with the increase in length. Now suppose that instead of increasing the length, you increase the concentration of visual pigment in the photoreceptor. For reasons that will be discussed next lecture, this causes the photoreceptor response to slow down. Based on your answer to question 4, which would produce more noise? Increasing the photoreceptor length while keeping the concentration of visual pigment the same, or increasing the concentration of visual pigment, while keeping the length the same?

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