I am trying to understand the equation of the radiance, but there is one thing i don't understand: $L_r = \frac{d^2\Phi}{d\omega \space dA\cos{\theta}}$

Why is that second exponent there in the numerator?

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d/dw (dphi/dA) = d^2phi/(dwdA) –  Chris Feb 3 at 10:24
Normally radiance is also per solid angle, but I'll just assume your interpretation in my answer. –  Chris Feb 3 at 14:36
Just in case it wasn't clear from his answer, $d^2$ does not mean something to the second power. It means you are taking the second derivative of the variable. –  tpg2114 Feb 3 at 16:15

$L_r = \frac{d^2\Phi}{d\omega \space dA\cos{\theta}}$

I can separate out the derivatives in the above equation to get:

$L_r = \frac{d}{d\omega}(\frac{d\Phi}{ \space dA\cos{\theta}})$ [note, alternatively I could have done $L_r = \frac{d}{dA}(\frac{d\Phi}{ \space d\omega\cos{\theta}})$, but I believe the former will allow for a more intuitive explanation of the formula.]

Let us examine the portion in brackets, that is, $\frac{d\Phi}{ \space dA\cos{\theta}}$. Firstly, note that $\theta$ represents the angle between the surface normal and the specified direction from which $L_r$ is measured, so we can think of $A cos\theta$ as being the area perpendicular to the direction L is measured.

This term is thus calculating how much power there is in the light, per the component of unit area that is perpendicular to the light, at a particular point.

The reason for the derivative, rather than simply a "divide by" is because we must consider each "infintisimally small" surface element separately, because each surface element could have a different orientation, and thus a different value of $\cos{\theta}$, since $\theta$ depends upon the orientation of the surface element. A simple "divide by" over a large area would thus not work, as there would be no value of $\theta$ applicable to the entire area.

Let us now consider the $\frac{d}{d\omega}$ component. This operator acts on the quantity "Power per perpendicular unit area", and thus evaluates to "Power per perpendicular unit area, per unit frequency". The power per unit area could be due to a contribution from a variety of frequencies, and this quantity $L_r$ can thus tell us how the power contributions change as frequency changes. Ultimately the quantity could be used to evaluate the power contributions in a certain frequency range, by computing the integral $\int_{\omega_0}^{\omega_0 + \delta} L_r d\omega$.

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Thanks, you made this clear to me :) –  user408141 Feb 5 at 17:40