Power of radiation received by a surface

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Using Stefan Boltzmann Law ,

The power emitted per unit area by the small sphere is P/A = eσT⁴ .

The power absorbed per unit area by the surface is given as P/A = eσT₀⁴ .

How should I use the distance between the two surfaces ?

 

[Bandersnatch]

It looks like 'a small sphere' can be taken to mean 'a point-like source'.
How does the density of radiation emitted from a point source change with distance?

 

[Jahnavi]

How does the density of radiation emitted from a point source change with distance?

Thanks for replying .

I am not sure but is Power inversely proportional to the square of the distance form the point source ?

I haven't come across term like density of radiation from a point source . Could you explain ?

 

[Bandersnatch]

Sure. So you have a point source (also works for any spherically-symmetric source) that radiates some amount of energy per unit time - i.e., has some power.

Now, this radiation spreads in every direction (imagine a single pulse going out). In 2 dimensions that'd mean a circle of growing radius, in 3D it's a sphere of growing radius.

We can then define radiation density at some distance from the source as the amount of emitted radiation (power) spread over the entire area of a sphere $\rho_r = P/A(d)$. The farther you go, more spread out the initial pulse of radiation, so there's less of it at any given point.

If the power is constant (the source keeps radiating the same amount of energy each instant), the density varies only with the inverse of area of a sphere, which means $\rho_r \propto 1/d^2$.
I.e., it's the inverse square law, as you have correctly guessed.

To get the power received by some surface at distance d from the source, you just multiply the local radiation density by the receiver area. ($P_{rec}=A_{rec}\rho_r$)

Now, the power is received by the same receiving surface both before and after, so how big it is won't affect the result (hence it's not given in the question).

So you just need to combine the source power growing as per the S-B law with the received power decrease as per the inverse square law.

 

[Jahnavi]

We can then define radiation density at some distance from the source as the amount of emitted radiation (power) spread over the entire area of a sphere ρr=P/A(d)ρr=P/A(d)\rho_r = P/A(d). The farther you go, more spread out the initial pulse of radiation, so there's less of it at any given point.

So , at any point at a distance r from the point source , Power received per unit area would be P/(4πr²) ?

 

[Bandersnatch]

So , at any point at a distance r from the point source , Power received per unit area would be P/(4πr²) ?

Yes.

 

[Jahnavi]

To get the power received by some surface at distance d from the source, you just multiply the local radiation density by the receiver area. (Prec=ArecρrPrec=ArecρrP_{rec}=A_{rec}\rho_r)

Now, the power is received by the same receiving surface both before and after, so how big it is won't affect the result (hence it's not given in the question).

I don't understand why the shape of the surface doesn't matter .

If we consider a plane surface at some distance from a point source P something like a man standing in front of a vertical wall . Just like the wall is moved away , if the plane surface is moved by a distance D away from the point source , only the point where the perpendicular from the point P meets the plane moves away by a distance D from the point source .

The distance of the point source from any other point doesn't increase by D , so how can we apply inverse square proportionality for an arbitrary shape ?

 

[Bandersnatch]

I don't understand why the shape of the surface doesn't matter .

I'm pretty sure you're overthinking it, considering what the question gives you. Yes, the shape would matter in the case you describe. But since the question doesn't include any detailed information on how the receiver is shaped or sized, nor how close to the source it is exactly, it's safe to assume that it's one of the 'spherical cow' type of questions.
It only wants of you to show that you understand the proportionality relationships involved. Just assume the receiver is negligibly small as compared to the distances involved, so that local radiation density is approximately constant across its entire surface.
(note the 'approximately' qualifier in the question)

 

[Jahnavi]

Ok. I get your point .

it's safe to assume that it's one of the 'spherical cow' type of questions

What's that ?

 

[Bandersnatch]

It's a running joke, that physics problems and physicists in general are sometimes using absurdly-sounding approximations, because that's the only feasible way to calculate something.
There's even a wikipedia page on it:
https://en.wikipedia.org/wiki/Spherical_cow

 

[Jahnavi]

Thanks !