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forestmine
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Homework Statement
Suppose you have a disk of radius r at x=y=z=0 with its normal pointing up along the z-axis. The disk radiates with specific intensity [itex]I(\theta)[/itex] from its upper surface. Imagine the observer plane is at z=Z, where Z is much greater than r. Let [itex]I(\theta)[/itex] = I = constant. Calculate the flux crossing the observer plane as a function of position. Integrate the flux over the entire plane to show that it is equal to the flux emitted by the source.
Homework Equations
[tex]F = \int{I(\theta) cos\theta d\Omega}[/tex]
[tex]d\Omega = sin\theta d\theta d\phi[/tex]
The Attempt at a Solution
I feel like I'm either complicating this way more than need be, or I'm oversimplifying. I think I'm really just caught up in the coordinates for some reason. Anyway, here's what I'm thinking.
First off, because z >> R, we can think of a series of parallel rays coming off of the disk and hitting the observer plane some very large distance away. If we were to project the disk onto the observer plane, the flux should be the same everywhere within that projected region and zero outside of it.
From our above expression for the flux:
[tex]F = \int{I(\theta) cos\theta d\Omega} = I\int\int cos\theta sin\theta d\theta d\phi[/tex]
since we're treating I as a constant.
At this point, I think I'm getting hung up on the units of integration. [itex]\phi[/itex] should go from 0 to 2pi, and it makes sense that it shouldn't depend on position in [itex]\phi[/itex]. Now, does [itex]\theta[/itex] range from 0 to pi/2 (since we're dealing with the top side of a disk)?
If that's the case, we get [itex]F = I \pi[/itex].
I think that's showing the flux emitted by the source and NOT integrating the flux over the entire observing plane, which should evidently be equal. Not sure how to go about that part...
Now, I'm also confused about the first part of the question, which asks to calculate the flux crossing the plane as a function of position. The only difference then would be that our theta limits go from 0 to theta, right?
[tex]F = I \pi sin^{2}\theta[/tex]
Now I need to get theta in terms of x, y, and z, right? In this case, z is fixed, so the flux won't depend on z, while it certainly would if we cared about the flux at different points along Z. On the other hand, if we go to some x and y that are outside of the radius r, there should be no flux. But beyond that, I guess I'm getting lost in coordinates. I would think that [itex]\sin\theta = r^{2}/\sqrt{y^{2}+z^{2}}[/itex] but that's contrary to my idea that it shouldn't depend on z for our plane that is located at z=Z.
Hopefully I'm at least somewhat on the right track. I feel like I'm headed in the right direction, but I could definitely use some guidance. Thanks!