The figure below shows three different fields of view in the form of circular cones with different spreading angles $\theta$, lengths $R$ and diameters $D$.
The cones are thin enough so that $R \approx H$, where H is central length of the cone, and
$$sin \theta = \text{opposite/hypotenuse} \approx \theta = \frac{D/2}{R}$$where D is the diameter of the cone.
Note that cones a) and b) have the same spreading angle $\theta1$ but different lengths $R1$ and $R2$, and that cones a) and c) have the same length $R1$ but different spreading angles $\theta1$ and $\theta 2$
Suppose the red line is a wall with uniform temperature emitting blackbody irradiance $E^*$. Find:
1) The irradiance E reaching a), b) and c) (the cone tips) assuming the power is spreading out into a hemisphere of radius $R$ in each case.
2) The radiance L = E/$\Delta \omega$ at a), b) and c).
If you do it right, you should find that the three radiances are identical, i.e. that for small angles and uniform emitters, L is independent of distance to the target.
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from IPython.display import Image
Image('figures/three_cones.png')
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a) power at D1 = $E^* \times \pi (D_1/2)^2$ (Watts) $E_a$ at tip of cone = $$\frac{E^* \pi (D_1/2)^2}{2 \pi R_1^2}$$
a) solid angle = $\Delta \omega = A/R^2 = (\pi (D_1/2)^2)/R_1^2$
a) Radiance = $L_a = E_a/\Delta \omega = E^*/(2 \pi)$
b) power at D2 = $E^* \times \pi (D_2/2)^2$ (Watts) $E_b$ at tip of cone = $$\frac{E^* \pi (D_2/2)^2}{2 \pi R_2^2}$$
b) solid angle = $\Delta \omega = A/R^2 = (\pi (D_2/2)^2)/R_2^2$
b) Radiance = $L_b = E_b/\Delta \omega = E^*/(2 \pi)$
c) power at D2 = $E^* \times \pi (D_2/2)^2$ (Watts) $E_c$ at tip of cone = $$\frac{E^* \pi (D_2/2)^2}{2 \pi R_1^2}$$
c) solid angle = $\Delta \omega = A/R^2_1 = (\pi (D_2/2)^2)/R_1^2$
c) Radiance = $L_c=E_c/\Delta \omega = E^*/(2 \pi)$
It looks like $L_a = L_b = L_c$ which shows that L is independent of distance for these examples. Note that this isn't the reason we know that L is conserved -- it's because Maxwell's equations show that conservation of L is a property of electromagnetic radiation. Given this, we can check the approximations we made in estimating the flux for this problem. How close is our solution to what we get using the correct formula for flux?
We saw in the flux_to_radiance notebook that the exact formula to convert from radiance to flux for a hemisphere was:
$$ E = \int_0^{2\pi} \int_0^{\pi/2} L \cos \theta \sin \theta \, d\theta \, d \phi $$The only difference for a cone is the limited zenith angle, if the half-angle (that is the angle from the centerline to the side) of the cone is $\theta$, the we've got:
$$ E = \int_0^{2\pi} \int_0^{\theta} L \cos \theta \sin \theta \, d\theta \, d \phi $$(note that I've changed my definition of $\theta$ from the figure above by dividing by 2. Here's the new setup for a cone of radius $r$ and length $l$.
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Image("figures/full_triangle.png")
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So what is $\theta$? It has to be $\theta = \arctan(r/l)$. Wikipedia says that:
$$\cos \theta = \cos\left ( \arctan(r/l) \right ) = \frac{1}{\sqrt{1 + (r/l)^2}}$$So try a change of variables to do the integral: let $\mu = \cos \theta$ so $d\mu = - \sin \theta \, \cos \theta\, d\theta$
Since $L$ is independent of angle, we can integrate out $\phi$ and do the transformation to get:
$$ E = 2 \pi \int_0^{\theta} L \cos \theta \sin \theta \, d\theta = 2 \pi \int_{\cos\theta}^1 L \mu d\mu$$since $\cos 0 = 1$.
Solving this yields:
$$E = \pi L \left [ \frac{\left ( r^2/l^2 \right )}{1 + \left( r^2/l^2 \right )} \right ]$$Note that the limits work correctly -- as we move the cone tip towards the surface, $r/l \to \infty$ and we get the fat cone limit
$$E = \pi L \frac{\infty}{\infty} = \pi L$$In agreement with the equation in the flux_to_radiance notebook
And as we zoom out, $r/l \to 0$ and the thin cone limit becomes:
$$ E = \pi L \left ( \frac{r^2}{l^2} \right )$$or rearranging:
$$ E = L \left ( \frac{\pi r^2}{l^2} \right ) = L \left ( \frac{\Delta A}{l^2} \right ) \approx L \Delta \omega$$as expected.
How does this limit compare with the answers we got to the question assuming spreading into a hemisphere? Let's assume that L doesn't depend on distance, so that $L = L^* = E^*/\pi$. If we insert that into our last equation we get:
$$E = L \left ( \frac{\Delta A}{l^2} \right ) = \frac{E^*}{\pi} \left ( \frac{\Delta A}{l^2} \right ) = \left ( \frac{E^* \Delta A}{\pi l^2} \right )$$So we're off by a factor of 2 in the denominator. That's because the power isn't being evenly distributed into a hemisphere from the circular patch on the surface, so our assumption above that the power spreading out evenly into $2 \pi l^2$ above isn't accurate.
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