Light being intercepted by dust

[Zaphodx57x]

There exists one dust grain of radius 0.1 micrometer per every 10^12 hydrogen atoms. Show that as light travels through a 1 cm layer of this comrpessed gas, only about 1% will be intercepted.

Then show how thick a layer would be needed to block 1/e of the rays.

I believe the density of the gas is about 2.7 x 10^19 atoms cm^-3

Initially I thought that I should have used mean free path to solve this problem, but it was done with integrals. I'm assuming that this is an expected value problem, but I really have no idea how to set up the integrals. I believe that my variable must be the thickness of the gas however. Any hints?

 

[Aaron Bex]

Answer:

To calculate the percentage of light intercepted by the gas, we need to consider the cross-sectional area of the dust grain relative to the total cross-sectional area of the gas. The effective area of a single dust grain is πr2 = 0.0031 m2. The total cross-sectional area of the 1 cm layer of gas is 2.7 x 10^17 m2.

Therefore, the percentage of light intercepted by the gas is (0.0031 m2/ 2.7 x 10^17 m2) x 100 = 0.0000011, or about 0.01%.

To calculate the thickness of the layer required to block 1/e of the rays, we can use the same formula as above, but with the percentage of light blocked set to 1/e.

The percentage of light blocked is (0.0031 m2/ 2.7 x 10^17 m2) x 100 = 0.3678794, or about 36.8%.

Therefore, the thickness of the layer required to block 1/e of the rays is (2.7 x 10^17 m2 x 0.368) / 0.0031 m2 = 3.7 x 10^18 m2 = 37 cm.