I have got to calculate total luminous flux incident on a plane section by calculating the solid angle subtended by the slotted (plane) section, having four identical rectangular slots each having width $5$cm & length $15$cm measured from the center $O$ (As shown in the diagram below) at the point $P$ lying at a distance $OP=6$cm on the axis passing through the center $O$ & normal to the plane of paper.
If the slotted section has inner radius $r=5cm$ & outer radius $R=20 cm$ then find out the solid angle subtended by this section at the point $P$ (lying perpendicularly outwards to the plane of paper)
Let $a = 6$. Let $X$ be the slotted section and $Y$ be the slot.
Change coordinates so that $X$ and $Y$ are living on the plane $\mathcal{P} = \{ (x,y,z) : z = a \}$ and the reference point for computing the solid angle is located at the origin. The geometric setup is basically the same as in this answer. Please read the method 3 in that answer first, in particular the Lemma there. We are going to need it later.
For any geometric shape $Z \subset \mathcal{P}$, let $\Omega_Z$ be the solid angle subtended by $Z$. Consider the projection $\eta : \mathcal{P} \to S^2$ as in above answer. $$\mathcal{P} \ni (x,y,a) \quad\mapsto\quad \frac{1}{\sqrt{x^2+y^2+a^2}}(x,y,a) \in \mathcal{S}^2$$ It is clear $\Omega_Z$ is simply the area of $\eta(Z)$ on $S^2$.
The key observation is under $\eta$, straight lines on $\mathcal{P}$ get mapped to geodesics on $S^2$. So the boundary of $Y$ get mapped to a polygon of 16 sides on $S^2$ and $12$ of them are geodesics.
We will use Gauss-Bonnet theorem to evaluate $\Omega_X$ and $\Omega_Y$. In particular, we will use following version of the theorem on $S^2$:
If $D_\rho$ is a disk on $\mathcal{P}$ centered at $O = (0,0,a)$ with radius $\rho$, one can show that $$\int_{\eta(\partial D_\rho)} k_g ds = 2\pi\frac{1}{\sqrt{1+(\rho/a)^2}}$$
It is clear $X \cup Y = D_R$ with $R = 20$, we find
$$ \Omega_X + \Omega_Y = \Omega_{D_R} = 2\pi \left( 1 - \frac{a}{\sqrt{a^2+R^2}} \right) = 2\pi\left(1 - \frac{3}{\sqrt{109}}\right) $$
What's remain is to compute $\Omega_Y$. Since the inner radius $r = 5$ is the same as the width of the beam, the four curved sections of $\partial Y$ contribute one-third of that of a full circle. Since $k_g = 0$ on geodesics, we find
$$\int_{\partial Y} k_g ds = \frac{2\pi}{3}\frac{a}{\sqrt{a^2+r^2}} = 2\pi\left(\frac{2}{\sqrt{61}}\right)$$
To compute the $16$ external angles, let us walk around $\partial Y$ counter-clockwisely.
Eight of the external angles look like what happen at vertex $A$ (please refer to the figure below).
The vertex $A$ is a distance $\rho_A = 5$ from $O$. The incoming curve is perpendicular to the radial direction while the outgoing straight edge is making an angle $\frac{\pi}{6}$. Using the Lemma in method 3 of the aforementioned answer, the external angle at $A$ is equal to $$ \theta_A = \tan^{-1}\left(\sqrt{1 + (\rho_A/a)^2}\tan\frac{\pi}{3}\right) - \frac{\pi}{2} = \tan^{-1}\left(\frac{\sqrt{183}}{18}\right) - \frac{\pi}{2} $$
The remaining eight external angles look like what happen at vertex $B$.
The vertex $B$ is at a distance $\rho_B = \sqrt{15^2 + (5/2)^2} = \frac52\sqrt{37}$ from $O$. The incoming edge is making an angle $\tan^{-1}\frac16$ to the radial direction while the outgoing edge is making an angle $\tan^{-1}6$ to the radial direction. The external angle at $B$ is equal to $$\begin{align} \theta_B &= \left( \pi - \tan^{-1}\left(6\sqrt{1+ (\rho_B/a)^2}\right) \right) - \tan^{-1}\left(\frac16\sqrt{1+ (\rho_B/a)^2}\right)\\ &= \pi - \tan^{-1}\left(\frac{\sqrt{1069}}{2}\right) - \tan^{-1}\left(\frac{\sqrt{1069}}{72}\right) = \tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right) \end{align} $$
Combine all these, we find
$$\begin{align} \Omega_Y & = 2\pi\left(1 - \frac{2}{\sqrt{61}}\right) - 8\left( \tan^{-1}\left(\frac{\sqrt{183}}{18}\right) + \tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right) - \frac{\pi}{2} \right)\\ \implies \Omega_X & = 8\left( \tan^{-1}\left(\frac{2\sqrt{1069}}{25}\right) + \tan^{-1}\left(\frac{\sqrt{183}}{18}\right) \right) - 2\pi\left( 2 + \frac{3}{\sqrt{109}} - \frac{2}{\sqrt{61}} \right)\\ & \approx 2.038049336500276 \end{align} $$