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Light through a tube - Solid Angle (Oblique Cone)

  1. Jul 7, 2015 #1
    For a physics problem, I need to calculate the solid angle subtended by an oblique cone (cone where the apex does not lie on the line perpendicular to the cone's base from the center of the base).

    Consider the following problem:
    I have a 2D disk which emits light in an ever growing hemisphere (spanning a solid angle of ##2\pi##). Let the initial intensity of light be ##I##. Then I put a cardboard cylinder perpendicular to the disk with radius equal to the radius of the disk. This cylinder absorbs light when it hits the walls. The question is then what is the fraction of light that comes out of the cylinder.

    I am only interested in the fraction and not in the absolute number so this is what I did:

    Consider a diameter on the disk. Pick an arbitrary point at a distance ##x## from the top having a thickness ##dx##.
    The fraction of light that comes out from this element will simply be the ratio of the solid angle subtended by the cone (whose base is the other circular base of the cylinder) and the total angle of emission ##2\pi##.

    I now need to calculate the solid angle due to this oblique cone which I cannot seem to do. So since it is given that the length of the cylinder is much longer compared to the radius of its base, I thought I could brake the oblique cone into 2 half right cones and add up the solid angles due to them. I'm not sure if this is correct.

    (Check attachment for diagram)

    Also, the reason I only consider a single diameter is because of my interest in the fraction of light that gets through and not the absolute number (of photons) so if I calculate the fraction from one diameter, it will be the same for every other diameter and hence for the whole surface.

    Here is how I proceeded:

    Solid angle subtended by a right cone is given by:
    \begin{equation}
    \Omega = 2\pi (1-\cos \theta)
    \end{equation}
    Solid angle subtended by the upper half right cone:
    \begin{equation}
    \Omega_1 = \pi \left( 1 - \frac{l}{\sqrt{l^2+x^2}}\right)
    \end{equation}
    Solid angle subtended by the lower half right cone:
    \begin{equation}
    \Omega_2 = \pi \left( 1 - \frac{l}{\sqrt{l^2+(d-x)^2}}\right)
    \end{equation}
    Total solid angle:
    \begin{equation}
    \Omega_T=\Omega_1 + \Omega_2 = \pi \left(2 - \frac{l}{\sqrt{l^2+x^2}} - \frac{l}{\sqrt{l^2+(d-x)^2}}\right)
    \end{equation}
    Fraction $fr_x$ of useful atoms is the ratio of $\Omega_T$ and the total solid angle $2\pi$.
    \begin{equation}
    fr_x=\frac{1}{2\pi}\cdot \pi \left(2 - \frac{l}{\sqrt{l^2+x^2}} - \frac{l}{\sqrt{l^2+(d-x)^2}}\right)
    \end{equation}

    If the emissions per unit length is $\kappa$, the number of photons that will come out from the element ##dx## is ##fr_x\cdot \kappa dx##.

    On integrating, this gives the total number of photons that come out from the whole diameter ##d## in 3 dimensions.
    \begin{equation}
    \int_0^d \frac{1}{2}\left(2 - \frac{l}{\sqrt{l^2+x^2}} - \frac{l}{\sqrt{l^2+(d-x)^2}}\right) \kappa dx
    \end{equation}

    The total number of photons emitted by the diameter ##d## is ##\kappa d##.

    Therefore, the fraction of photons (light) emitted by the whole length ##d## will be:
    \begin{equation}
    F(l,d)=\frac{\int_0^d \frac{1}{2}\left(2 - \frac{l}{\sqrt{l^2+x^2}} - \frac{l}{\sqrt{l^2+(d-x)^2}}\right) \kappa dx}{\kappa d}
    \end{equation}
    On integration, this gives:
    \begin{equation}
    \boxed{F(l,d)=1-\left( \frac{l}{d}\right) \sinh^{-1}\left( \frac{d}{l}\right)}
    \end{equation}

    I'm not sure this line of thought is completely correct.
     

    Attached Files:

  2. jcsd
  3. Jul 7, 2015 #2

    Andy Resnick

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    Science Advisor
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    2016 Award

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