# Homework Help: Light scattering (rayleigh scattering)

1. Nov 5, 2008

### spaghed87

1. The problem statement, all variables and given/known data
For what wavelength of light is the scattering only 2.00% that of light with a visible wavelength of 520nm?

NOTE: The intensity of scattered light is inversely proportional to the fourth power of the wavelength.

2. Relevant equations
I(light)/I(IR)=(wavelength of IR / wavelength of light)^4=200

3. The attempt at a solution
I(520)/I(IR)=(Lamda/520nm)^4=200

Where, Lamda is equal to the wavelength of the IR light.

Lamda is ~1960nm (this answer seems to be wrong, why? Do you see an error anywhere? This is going to be how you solve it so the error shouldn't be that big or that a different equation is needed. I know I'm on the right track.)

2. Nov 5, 2008

### borgwal

An elementary mistake: 2% has nothing to do with 200.

3. Nov 5, 2008

### spaghed87

well, according to this solution.

http://www.mscd.edu/~physics/denn/solutions/Ch23/EOC_Solution_23_25.pdf [Broken]

They used 100 for 1%?

This is the same problem except they are working with a scattering of 1.00% and a wavelength of 500nm.

100 is what the ratio of the intensities is equal to. However, I cannot see where the number 100 is coming from in the solution to the similair problem. They did not specify in the solution. Does the number 100 relate to 1% at all?

Last edited by a moderator: May 3, 2017
4. Nov 5, 2008

### borgwal

Yes, as in 1%=1/100.

5. Nov 6, 2008

### spaghed87

Thanks, I see now. I've passed differential equations even... I have always got stumped on simple mathmatical things like percentages though. For some reason my brain was thinking just to double the value of 100 for 2%. I do stupid stuff like that all the time. =(

6. Nov 6, 2008

### borgwal

You're welcome :-)