Incident Radiation

We need to determine the solid angle subtended by the disk from the light source. Setting up a system of spherical polar coordinates with the light source as the origin and the polar axis in the negative $z$ -direction, we see that the ray of light emitted by the source in a direction determined by polar angles $\theta,\phi$ will only hit the disk if it strikes a chord of length $2R\cos\phi$ , for $-\tfrac12\pi \le \phi \le \tfrac12\pi$ . Thus the ray of light will strike the disk provided that $\tan\theta \le 2\cos\phi \hspace{2cm} -\tfrac12\pi \le \phi \le \tfrac12\pi$ Thus the solid angle subtended by the disk is $\begin{aligned} \Omega & = \int_{-\frac12\pi}^{\frac12\pi} \,d\phi \int_0^{\tan^{-1}(2\cos\phi)} \sin\theta\,d\theta \; = \; \int_{-\frac12\pi}^{\frac12\pi} d\phi\,\Big[-\cos\theta\Big]_{\theta=0}^{\tan^{-1}(2\cos\phi)} \\ & = \int_{-\frac12\pi}^{\frac12\pi} \Big(1 - \frac{1}{\sqrt{1 + 4\cos^2\phi}}\Big)\,d\phi\; = \; \pi - 2\int_0^{\frac12\pi} \frac{d\phi}{\sqrt{5 - 4\sin^2\phi}} \\ & = \pi - \frac{2}{\sqrt{5}}K\big(\sqrt{\tfrac45}\big) \end{aligned}$ using the complete elliptic integral $K$ of the first kind. Thus the percentage of emitted light that strikes the disk is $100 \times \frac{\Omega}{4\pi} \; = \; 25 - \frac{10\sqrt{5}}{\pi}K\big(\sqrt{\tfrac45}\big) \; \approx \; \boxed{8.93406}\%$

The most familiar scenario is that of an isotropic radiator (like a star) with an imaginary spherical shell surrounding it, through which the energy passes. If the radius of the shell is $d$ , the power density is the total radiant power divided by the surface area.

$P_d = \frac{P_{total}}{4 \pi d^2}$

For the sphere, we can simply integrate the scalar power density over the imaginary surface to calculate the total power, since the surface normal is everywhere parallel to the radiation flux. However, for arbitrary surfaces, this nice relationship does not hold.

For example, suppose there was an isotropic radiator in the middle of a right circular cylinder. In this case, integrating the simple scalar power density expression over the surface gives an absorbed power greater than that of the source. To fix this, we must introduce an additional term, which is the magnitude of the dot product between the unit flux directional vector and the unit surface normal. The differential absorbed power is:

$dP_{abs} = P_d \, {dA} \, |\hat{n} \cdot \hat{d}|$

where $P_d$ is the scalar power density, $dA$ is the differential surface area, $\hat{n}$ is the unit surface normal vector, and $\hat{d}$ is the unit displacement vector from the source to the surface. This is very similar to the solid angle calculation.

For this problem, begin by defining coordinates for the source:

$(S_x,S_y,S_z) = (R, 0, R)$

Use polar coordinates to define the coordinates of a point on the disk (notice that $r$ and $R$ are different):

$(x,y,z) = (r \,cos \theta, r \,sin \theta, 0)$

Differential disk area:

$dA = r \, dr \, d\theta$

Displacement vector:

$\vec{d} = (r \,cos \theta - S_x, r \,sin \theta - S_y, 0 - S_z) = (\Delta x, \Delta y, \Delta z)$

Unit displacement vector:

$\hat{d} = \frac{\vec{d}}{\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}}$

Unit surface normal ( $\hat{k}$ is a unit-vector in the z-direction):

$\hat{n} = \hat{k}$

Scalar power density:

$P_d = \frac{P_{total}}{4 \pi |\vec{d}|^2}$

Differential absorbed / incident power:

$dP_{abs} = P_d \, {dA} \, |\hat{n} \cdot \hat{d}|$

Putting all of these elements together in a double integral and solving numerically yields the result that approximately $8.9 \%$ of the total radiated power / energy is incident on the disk.

$P_{abs} = \int_0^{2 \pi} \int_0^{R} dP_{abs} = \int_0^{2 \pi} \int_0^{R} P_d \, {dA} \, |\hat{n} \cdot \hat{d}|$