HOME >HOMEWORK #2 SOLUTION
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.
Light at 720nm is absorbed by red (L) cones with a relative absorbance that is over two orders of magnitude greater than that of the rods.† If dim red (720 nm) light is shone on the retina, then the rods will not see a bright light, and will not lose their light adaptation while the L cones will see the red light.†
This is one of the reasons why darkroom lights & ship navigator lights are deep red--so that operators donít lose their dark adaptation.
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?
In order to see white, you have to have equal relative absorbance in all of the cones.† A rainbow splits full spectrum light in such a way that each point along a transverse section of the rainbow represents only one wavelength of light.† Looking at the spectral sensitivity curves of the cones, it is clear that no one wavelength of light excites all three types equally.† Hence, there are no white points (bands) in the rainbow.† [If, however, you were to unevenly overlap two different rainbows, there would be white at points of intersection between different bands where two different wavelengths combined exited all three equally.]††
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?
Yes, you would see white where the two profiles overlap (L/S ~480nm, M/S ~470nm, L/M ~580nm)
What if you had four (or more) cones?
Still no white points.†
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?
1 thermal activation per 80 seconds = 1/80 thermal activation per second = 1/800 thermal activation per 100 ms.† Since thermal activations are random & independent, the probability that we get 2 activations within 100 ms of each other is (1/800)2.† If a photon response or thermal event lasts 100 ms, the probability that we have a thermal event occurring in 1 rod at any given time is 1/800.† The average # of thermal activations occurring at any instant in a pool of 250 rods is 250 x (1/800) = 250/800 = 0.3125.
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?
From Beer's law, we have the relationship between probability of absorbing a photon at a given wavelength:
where f is the fraction of light transmitted (not absorbed) of wavelength lambda.† l is the length of the photoreceptor and c is the concentration.† The alpha of lambda is a constant reflecting the probability of a photon being absorbed of wavelength lambda.† As we see, both concentration of pigment and the photoreceptor l affect the probability that light is absorbed by the rod.
To increase the fraction of absorbed light from 25% to 50% would require either the lengths to be:
and since Beer's law is symmetric in c and l, we could achieve the same increase in absorbed light by increasing the concentration by the same factor.
If we had to chose between increasing the length of the photoreceptor or increasing the concentration of pigment, then there should be no difference in noise between the two since if the photoreceptor is a cylinder of a fixed diameter, the # of pigment molecules should be the same in both cases.† However, we are given that the photoreceptor response slows down in the case of higher pigment concentration.† This would cause the photon response to last longer.† If we redo problem #4 with a longer duration of a singe photon response, we see that the mean # of thermal activations at any time is larger, and therefore the noise is higher.† Therefore, increasing the concentration will cause more noise than increasing the photoreceptor length.