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Given a Lambertian emitter (or reflector) of area $A$ that is emitting a total power (resp. flux) of $\Phi$ (units $W$). To calculate the radiance $L$ (units $\frac{W}{m^2 sr}$) the solid angle $\Omega$ is needed:

$$ L = \frac{\Phi}{A \cdot \Omega}$$

Question: Is the solid angle $\Omega = \pi$ or $\Omega = 2 \pi$?

Approach 1: My initial feeling was $2 \pi$, because $L$ is constant in all directions of the hemisphere for Lambertian emitters, and the hemisphere has the solid angle $2 \pi$.

Approach 2: The radiance of a Lambertian emitter is $L = \frac{I(\theta)}{A \cdot cos(\theta)}$ (which is constant as $I(\theta) = I_{max} \cdot cos(\theta)$ for Lambertian emitters). Thus, $I(\theta) = L \cdot A \cdot cos(\theta)$ (units $\frac{W}{sr}$), which gives the total flux or power $\Phi$ (units $W$) when integrated over the hemisphere: $$\Phi =\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} I(\theta) \cdot sin(\theta) d\theta d\varphi $$ $$ = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} L \cdot A \cdot cos(\theta) \cdot sin(\theta) d\theta d\varphi$$ $$ = 2\pi \cdot L \cdot A \int_{0}^{\frac{\pi}{2}} cos(\theta) \cdot sin(\theta) d\theta$$ $$ = 2\pi \cdot L \cdot A \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{2}} sin(2\theta) d\theta$$ $$ = \pi \cdot L \cdot A $$

And thus $L = \frac{\Phi}{\pi \cdot A}$ and $\Omega = \pi$, which is contradiction to the initial approach of $2\pi$.

So which one is correct and why? (Calculation says $\pi$, but $2 \pi$ is more intuitive...).

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Here's a more detailed explanation of the source of the apparent discrepancy. From SPIE

enter image description here

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  • $\begingroup$ Thanks, I understand now that the projected solid angle of the hemisphere is $\pi$. But is there a way to see why a projected angle is needed at all? Maybe it's naive, but as there is a solid angle in the definition of the radiance I would have just used the real solid angle, not a projected one. $\endgroup$ Jun 19 '21 at 8:29
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    $\begingroup$ For completeness (found answer in the provided link): The radiance is the emitted power divided by projected area in the direction of the angle of view and solid angle. However, this is for small solid angles. If the "detector" sees the area under many angles of view, because it is so large (and that's the case when we are interested in the emission into the whole hemisphere), then the radiance is calculated as power divided by (true) emission area and projected solid angle (which is pi instead of 2 pi). $\endgroup$ Jun 19 '21 at 19:26
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The projected solid angle of the hemisphere is $\pi$ from its definition $\Omega=A/r^2$, where $A$ is the corresponding surface area.

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