Surface Brightness at any angle

[dykuma]

Homework Statement

Here is a picture of the problem
https://www.dropbox.com/s/2bps6ga2o4hjpgw/hw4.png?dl=0
For those who don't want to click the link, basically the problem wants me to calculate the surface brightness of a flat surface at any angle if the source function S = B(T), which is a constant.

It then wants me to consider limits at optical depths τ<<1 and τ >>1, and to know why brightness is not infinite at θ = π/2,

Homework Equations

Non were provided. My guess for the first part is that I should use something similar to
F=∫ I⋅Cos(θ) dΩ. I have no idea for the second part.

The Attempt at a Solution

I found that for a flat source Ω=2π, so the result of this integral may be F= I⋅2π⋅Cos(θ), however, then the
second question makes no sense, as Cos(π/2)=0 (unless that's the point). Further, I don't know if that is even the right equation, as I am not sure I want flux for this.

I am also unsure how to handle the optical depth portion of the question.

 

[SammyS]

Homework Statement

Here is a picture of the problem
https://www.dropbox.com/s/2bps6ga2o4hjpgw/hw4.png?dl=0
For those who don't want to click the link, basically the problem wants me to calculate the surface brightness of a flat surface at any angle if the source function S = B(T), which is a constant.

It then wants me to consider limits at optical depths τ<<1 and τ >>1, and to know why brightness is not infinite at θ = π/2,

Homework Equations

Non were provided. My guess for the first part is that I should use something similar to
F=∫ I⋅Cos(θ) dΩ. I have no idea for the second part.

The Attempt at a Solution

I found that for a flat source Ω=2π, so the result of this integral may be F= I⋅2π⋅Cos(θ), however, then the
second question makes no sense, as Cos(π/2)=0 (unless that's the point). Further, I don't know if that is even the right equation, as I am not sure I want flux for this.

I am also unsure how to handle the optical depth portion of the question.

You may get more response if that image is posted directly.

 

[dykuma]

You may get more response if that image is posted directly.

Thank you! I was unable to do so, I apologize.

 

[dykuma]

I'm going to bump this question one more time. I can literally find nothing about this problem online, or in two different textbooks.