Inverse Square Law and Absorption

We are going to explore two important features of how waves (light, sound, x-rays, etc) propagate through things. The first of these is the spreading out of waves due to the inverse square law. This is what we explored in lab and in class. The calculations are confusing but not particularly difficult once you get them written down correctly. The second process is absorption which is an exponential process. As we did in class I will ask you to figure out the important constants in the exponential equation in a variety of settings.

Useful numbers: In physics we usually think of watt/m² as the best way of describing how much energy is arriving at some location like the earth of my study desk. On the other hand, designers and architects tend to describe the illumination that exists in various human settings in terms of lumens. This has always been confusing to me as a physicist so I spent some time sorting this out. The relationship is complicated but for our purposes the best model is that 93 lumens of sunlight is equivalent to 1 watt. Because incandescent lights and CFL's are a different mix of colors there is a slightly different conversion factor for them but we are going to ignore that difference and assume that they work much like sunlight. You also need to know that 1 lux = 1 lumen/m and that 1 lux is roughly the same the illumination of a full moon at night, 1 mlux is starlight (no moon) on a clear winter night, and 1^.10⁵ lux is bright sunlight in the summer.

1) As we have seen in class the energy output of the sun is 3.8^.10²⁶ watts. Calculate the brightness of the sunlight in watt/m² when it gets to the earth (top of the atmosphere) and when it gets to Neptune. Earth is 1.5^.10¹¹ m from the sum and Neptune is 4.5^.10¹² m from the sun. Convert the brightness of the sun at the top of Neptune's atmosphere to lux and tell me if the sun looks brighter or dimmer than our moon (seen from earth) when you're visiting Neptune?

2) A 100 W lightbulb (26 W CFL) produces 1600 lumens of light. How far away do I have to be, in meters, for the brightness of that lightbulb is roughly the same (1 mlux) as starlight? Why does this explain our "dark sky" ordinance in Deschutes county that says any outdoor lights must be screened so their light does NOT travel outward to our neighbor's property? The actual ordinance only applies to lights greater than 1800 lumens and says:

"must limit direct line-of-sight of the fixture’s lamp to the property on which the fixture is installed"

3) A standard green laser pointer produces 4 lumens and will make a spot that is only 5 cm in radius at 100 m. (This is a lot like your floodlight calculation from lab) What is the intensity of laser beam, in lux, at this distance? Given that 100 lux will make you see spots what would happen if you go this laser in your eyes? The concern for airline pilots is that one can purchase laser pointers that produce several hundred lumens and can produce temporary blindness at several kilometers which is why it is seriously illegal to direct laser pointers into the sky near airports.

Now for some exponential calculations......

4) In the deep ocean far away from land 1% of the light makes it through 100 m of water (that's very clear water!). The intensity of light at the surface is 1000 watt/m². After you determine N₀ and λ in the expression N=N₀e^{λ x} then calculate the depth (in m) where the light level is 1 mwatt/m². This is equivalent to full darkness. Given that ocean is mostly 3 km deep does this explain why it is pitch dark at the bottom of the ocean?

5) There is a rule of thumb in business and banking that 72/(interest rate %) tells you how many years it will take to double your money. Check this by doing the following calculations just like we did in class. Assume you have an interest rate of 5%. Start with some amount of money and calculate how much you will have after each of the first three years. Then determine N₀ and λ in the expression N=N₀e^{λ t}. Then set N = 2N₀ and solve for the time at which this happens (you will need to use natural logs (ln) on your calculator to do this). Compare the time you get to 72/5. We'll talk in class.

6) In the expression N=N₀e^{λ x} the units of λ are 1/distance. λ is called the coefficient of linear attenuation. For many metals that we might want to x-ray (airplane parts for instance) λ is roughly 100 1/cm. If my part is 15 cm thick what percent of my x-rays are getting through the sample?

7) When we x-ray the human body the same ideas apply. For bone the coefficient of linear attenuation (for a particular x-ray energy) is roughly 0.86 1/cm while for muscle tissue it is only 7.55^.10^-2 1/cm. In an x-ray of the leg there is about 14 cm of muscle tissue most places and about 4 cm of bone in certain places. Assuming an initial x-ray "brightness" of 100 "lumens" determine the brightness of the x-rays passing through each part of the leg.