# Urr, this doesn't seem like it should be very hard but well, good 'ol me

1. Jan 30, 2006

### schattenjaeger

For this problem assume the "EM spectrum of astrophysical relevance" is between 1m and 10^-3nm. Calculate the fraction of the visible light by assuming a) a linear scale and b)a logarithmic scale. Explain the difference..

hur? I'm guessing this is one of those math skills I've either blanked on or never knew but should've, but as far as calculating the fraction goes I can only think to like divide the range of visible light(which is like 3 x 10^-7, right?)by the total range, which would be, pretty much, 1? (1-10^-12 to be exact)and then I spose that'd be pretty linear, or whatever.

So assuming that out of my ass speculation is right, I still dunno what to do with a logarithmic scale. I could take the log of that 3x10^-7 but that's a negative answer. I could piddle with the units so it wasn't negative, but that seems awfully...wrong. So primarily my question is a "what the hell am I doing?" problem.

And one quickie, it wants me to derive Wien's displacement law by setting (we'll make y be my lambda) dB/dy=0, where B is that 2hc^2etc./e^hc/kt -1 with the formatting kept poor just so you get the point. The problem only wants us to work it out to where we get something only solvable by numerical methods, then stop, so if I take the derivative of that and set it equal to 0, my next stop would be to solve for lambda*T, I guess, and it'd equal what it equals in the actual formula, would that be the step to stop at?

2. Jan 30, 2006

### SpaceTiger

Staff Emeritus
Yep.

The logarithm of a number less than one is negative -- this is perfectly alright. Imagine I express the entire range of wavelengths in "log meter" units, meaning, for example, 1 cm is expressed as "-2" and 10-7 m is "-7". In this unit system, what is the range spanned by visible light? What is the range of the EM spectrum mentioned in the question?

Give it a try and see what you get. I suspect you won't be able to solve for $\lambda T$ or it wouldn't need to be done numerically.

3. Jan 30, 2006

### schattenjaeger

Ah, ok, thanks!

4. Jan 31, 2006

### schattenjaeger

So doing it like that for the logarithmic scale I got like 2 percent which sounds decent enough, as for the linear scale, since the length of the whole thing is just 1meter(unless I wanna be just THAT accurate)is the percent 3x10^-5?(3x10^-7 *100 percent)?

5. Jan 31, 2006

### SpaceTiger

Staff Emeritus
Sounds good.