|
⇤ ← Revision 1 as of 2012-05-27 03:20:11
Size: 1072
Comment:
|
Size: 1736
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 6: | Line 6: |
| If we map angular space around the sphere in spherical coordinates, with θ as the angle of the reflection from the light source axis, and φ being the angle around the axis, then the light leaving at angle $\theta$ is bounced off the sphere at angle $\theta/2$ The steradian area of the an angle element $ d\theta ~ d\phi $ is $ \sin \theta ~ d\theta ~ d\phi $. This is associated with an intercepted light area on the sphere of $ 0.5 ~ sin( \theta ) ~ \sin( \theta / 2 ) ~ \cos( \theta / 2 ) ~ d\theta ~ d\phi $ or $ { 1 \over 4 } ( 1 - cos( \theta ) ) ~ sin( \theta ) ~ d\theta ~ d\phi $ | If we map angular space around the sphere in spherical coordinates, with $\theta$ as the angle of the reflection from the sphere, then the place were it reflects on that sphere is a ring with a radius $2\pi~\sin(\theta/2)$ and a width $d\theta/2$. Projected towards the light source, that makes an annular ring with width $\cos(\theta/2)d\theta/2$. So the total light collected is light in = $2\pi~\sin(\theta/2)\cos(\theta/2)d\theta/2~=~2\pi\left(1-\cos(\theta)\right)d\theta/4$. That is projected onto the "sky sphere", a ring with diameter $2\pi~\sin(\theta)$ and a width of $d\theta$, for an angular area of $2\pi~\sin(\theta)d\theta$. The intensity in that ring is: |
| Line 8: | Line 12: |
| The light per steradian is thus $ \left( { 1 \over 4 } ( 1 - cos( \theta ) ) ~ sin( \theta ) ~ d\theta ~ d\phi \right) / \left( sin( \theta ) ~ d\theta ~ d\phi ~\right) $ |
intensity( θ ) = $\left(2\pi\left(1-\cos(\theta)\right)d\theta/4\right)~/~\left(2\pi~\sin(\theta)d\theta\right)$ |
| Line 12: | Line 14: |
| or | intensity( θ ) = $\left(1-\cos(\theta)\right)~/~\left(4~\sin(\theta)\right)$ |
| Line 14: | Line 16: |
| $ { 1 \over 4 } ( 1 - \cos( \theta ) ) $ | Dividing by $\sin(\theta)$ is awkward, it blows up at θ=0 . So let's multiply top and bottom by $1+\cos(\theta)$ : intensity( θ ) = $\left(1-\cos(\theta)\right)\left(1+\cos(\theta)\right)~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)$ intensity( θ ) = $\left(1-\cos(\theta)^2\right)~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)$ intensity( θ ) = $\sin(\theta)^2~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)$ intensity( θ ) = $\sin(\theta)~/~(4\left(1+\cos(\theta)\right)$ SOMETHING IS WRONG - Retroreflection cannot be zero! |
Reflection From a Sphere
Assume a distant source uniformly illuminates a mirror sphere with a radius of 1 unit. What is the brightness per steradian of the reflection at angle \theta from the light source?
If we map angular space around the sphere in spherical coordinates, with \theta as the angle of the reflection from the sphere, then the place were it reflects on that sphere is a ring with a radius 2\pi~\sin(\theta/2) and a width d\theta/2. Projected towards the light source, that makes an annular ring with width \cos(\theta/2)d\theta/2. So the total light collected is
light in = 2\pi~\sin(\theta/2)\cos(\theta/2)d\theta/2~=~2\pi\left(1-\cos(\theta)\right)d\theta/4.
That is projected onto the "sky sphere", a ring with diameter 2\pi~\sin(\theta) and a width of d\theta, for an angular area of 2\pi~\sin(\theta)d\theta. The intensity in that ring is:
intensity( θ ) = \left(2\pi\left(1-\cos(\theta)\right)d\theta/4\right)~/~\left(2\pi~\sin(\theta)d\theta\right)
intensity( θ ) = \left(1-\cos(\theta)\right)~/~\left(4~\sin(\theta)\right)
Dividing by \sin(\theta) is awkward, it blows up at θ=0 . So let's multiply top and bottom by 1+\cos(\theta) :
intensity( θ ) = \left(1-\cos(\theta)\right)\left(1+\cos(\theta)\right)~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)
intensity( θ ) = \left(1-\cos(\theta)^2\right)~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)
intensity( θ ) = \sin(\theta)^2~/~\left(4~\sin(\theta)\right)\left(1+\cos(\theta)\right)
intensity( θ ) = \sin(\theta)~/~(4\left(1+\cos(\theta)\right)
SOMETHING IS WRONG - Retroreflection cannot be zero!