# Surface Brightness at any angle

## 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.

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SammyS
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## 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.

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

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

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