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Homework #1 Solutions

Phys 428-Summer 2001

1. Imagine that I show you two flashes of slightly different intensity. The first flash, on average, causes 100 more photons to be emitted than the second. Explain why it may be physically impossible to determine which flash was brighter if each flash is presented only once.

If a flash emits n photons on average, it will have a standard fluctuation of n^.5 due to the Poisson distribution of its emission.  Therefore, on any given trial, the fluctuations of each may be such that they have the same amount of emitted photons.  For instance, if the flashes are each around 10,000 photons, the average variation of each will be a hundred photons, and clearly you wouldn’t be able to tell the physical difference between the two on any given comparison.

2. You have a light source that produces flashes of an average intensity of 100 photons/flash. You shine this light source into your eye such that the flash illuminates 250 rods. You can see this flash 60% of the times that the flash is presented. You hypothesize that in order to see the flash, at least one rod must absorb 2 or more photons from the flash. Are your observations consistent with this hypothesis? What percentage of the photons must be absorbed by the rods? How does this number compare with the case that at least one rod must absorb 1 or more photons per flash in order to see the flash)?

So, the probability that more than one (two or more) photon is absorbed is:

Solving this numerically gives a value for alpha of .22 (22%)

Solving this numerically gives an alpha less than .01.

Thus, if we double the number of photons (from 1 to 2) required for a rod to detect a light, we decrease its sensitivity by more than an order of magnitude!

3. Imagine the retina being illuminated by a steady light. The number of absorbed photons in any given time interval is Poisson distributed. Most of the time, we do not perceive steady light as flickering. Some experiments have shown that the dimmest transient light increment (a dim flash superimposed on the steady background light) that we can detect scales with the square root of the background (steady) light intensity. Does this result make sense? Why or why not?

The Background light is Poisson distributed.  This means that the background light source has a standard fluctuation that goes as the square root of the background (steady) light intensity.  Any light superimposed on this background must be greater than the existing noise to be detected as something other than noise.  So the result presented in the question definitely makes sense.

4. Your retina is not perfect. There are noise sources due to spontaneous activity in the retina. This noise causes your retina to effectively always "see" a dim background light (that also follows Poisson statistics) even in complete darkness. Imagine we extend the experiment described in question 3 to very dim background light intensities and get the result shown below. Do the results (shown below) make sense? Why or why not?

We know that the standard variation of a Poisson-distributed background goes as the square root of the mean intensity of the background.  Any detectable dim flash must have magnitude above this variance. 

(Dimmest detectable flash) ~ (Background light level)^1/2

On a log-log plot, this appears as a straight line with slope 1/2 (A).

However, the constant (slope zero) thermal noise causes a constant noise effect, and the detectable dim flash must be above the noise due to fluctuations in this as well (B).  Adding these gives the plot showin in the homework.

(A)                                                                        (B)