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Let us consider a needle of length 1, with one (fixed) end at the coordinate $(x,y,z)=(0,0,0)$. If we make the other end (wich is free) of the needle move, it will describe the unit sphere. We assume the distribution of the free end of the needle is uniform on the sphere, which means it is given by: $$\cos(\phi) d\phi d\theta$$

There is a light above the vertical axis of the sphere. What is the distribution of the length of the shadow of the needle?

So far, I have considered the hemisphere corresponding to $\phi \in [0,\frac{\pi}{2}[$. Let $t \in ]0,1[$. $$P(L<t)=P(\cos(\phi)<t)=P(\phi>\arccos(t))$$ $$P(L<t)=\int_{\arccos(t)}^{\frac{\pi}{2}}\cos(\phi)d\phi$$ $$P(L<t)=\int_{0}^{t}\frac{l}{\sqrt{1-l^2}}dl$$ $$P(L<t)= 1 - \sqrt{1-t^2}$$ Is it right? Is there an intuition of the result?

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Your notation is inconsistent with the usual spherical coordinates, by interchanging $\phi$ and $\theta$ and measuring $\phi$ up from the $(x,y)$ plane, but your area element is consistent with this by using $\cos \phi$ instead of $\sin \theta$. It is consistent with the usual latitude and longitude. If the light is very high, so the rays are parallel to the $z$ axis, the shadow is just the projection of the needle onto the $(x,y)$ plane and its length is $\cos \phi$. The intuitive result is that the probability that the shadow is less that $t$ is the solid angle of the sphere with $\cos \phi \lt t$.

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I used the French convention, I did not know it would be different in English. You are right about the unconventional way I measured $\phi$. Could you elaborate on the link between probability and solid angle? –  Wok Mar 29 '11 at 19:05

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