# Radiation Intensity (heat)

1. Sep 9, 2014

### Chacabucogod

So I understand that a black body is a diffuse emitter, and that it radiates energy at a rate of σT^4.
I also understand the correspondence between a solid angle and the superficial area of a sphere.

$dS=r^2sin(θ)dθd\phi=\frac{ds}{r^2}=d\omega=\frac{dAcos(\alpha)}{r^2}$

What I don't understand is how they get to the formulation for intensity of radiation.

Why is the intensity of radiation proportional to the cos of the angle between the two areas? Isn't the intensity the same for every part of the half sphere since it's a black body?

Last edited: Sep 9, 2014
2. Sep 12, 2014

### elegysix

intensity is a vague word when it comes to radiation.
Are you talking about irradiance (W/m^2)?

A sketch of the areas and angles you're talking about might help clarify the problem

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