# Homework Help: Power radiated through a hole in small wavelength distribution

1. Feb 5, 2014

### PsychonautQQ

1. The problem statement, all variables and given/known data
A cavity at Temperature 6000k has an energy distribution corresponding to a blackbody. We make a small hole in it 1mm in diameter.

Calculate the power radiated through the hole of wavelength interval between 550nm and 551nm.
HINT: when dλ is small (such in this case), certain assumptions can be made regarding λ and a product can be used to approximate an integral

2. Cool Equation that's probably useful:
u(λ)= 8*pi*h*c*(λ^-5)*(e^(hc/λkt)-1)^(-1)

λT = 2.898*10^-3 maybe this will be useful? (wiens shiznit)

R=σT^4

3. The attempt at a solution
Hope someone is smart out there :X hah.

Alright so first of all i'm having some conceptual problems with this... the power radiated THROUGH the hole isn't dependent on the overall surface area of the cavity? It seems it's not as it gave us no information about the size or shape of the cavity.. so that's a bit offputting from the start...

So the total power radiated from the cavity is σT^4, because the cavity approximates a blackbody. But how much is radiated through the HOLE at that wavelength? any guidance?

2. Feb 5, 2014

### BOYLANATOR

You can imagine the hole as being the sole source of radiation. Looking at a hole in a 5800K cavity is no different than looking at the 5800K Sun. It looks like your cool equation is for power per unit surface area per unit of wavelength. How can you find power per unit surface area?

3. Feb 6, 2014

### PsychonautQQ

ahh I was thinking about the radiation being radiated OUT of the hole.. into it!! duh.. thanks man :D.

By sureface area you just mean the area of the hole? No information is given on the surface area of the cavity.. thanks again dood

4. Feb 6, 2014

### BOYLANATOR

The radiation is coming out of the hole. It doesn't matter how big the cavity is, the radiation can only escape through the hole. So yes, I mean the surface area of the hole.

5. Feb 7, 2014

### PsychonautQQ

can radiation not strike the atoms making the cavity of the wall, increasing their temperature and then radiate away NOT through the hole as thermal power?

So what I have done so far is I have
u(λ)dλ = (8*pi*hc*λ^(-5) * dλ ) / (e^(hc/λkT)-1)
where I put λ = 550.5 since that's between 550 and 551.
I put λ as 1 nm and the rest are constants or given information.
So I can get a numerical value for this equation, which gives me the energy density inside the cavity for the wavelength's between 550 and 551.

Any tips as what do to from here?

6. Feb 7, 2014

### PsychonautQQ

To find the power radiated would I take through the hole R(lambda) = c/4 * u(lambda)? I feel like I would have to add information about the radius of the hole somewhere right? Would R(Lambda) = c/4 * u(lambda) the total power?