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akannan

If we have the assumptions for directional lighting on the previous slide, then $L^{in}(\omega_{in}) = s \delta (\omega = \omega_0)$ for some constant $s$. But we also know that $\cos \theta_{in} = \frac{\overrightarrow{\ell} \cdot \widehat{n}}{|| \overrightarrow{\ell} ||}$ .

Going backwards from the final equation $I = a(\widehat{n} \cdot \overrightarrow{\ell})$, does this necessarily mean that $s$ is the magnitude of $\ell$?

mpotoole

Yup, exactly! Let's derive this explicitly here.

Suppose that we have a diffuse BRDF, $f(\omega_{in},\omega_{out}) = \frac{\rho}{\pi}$ for $\rho \in [0,1]$. And suppose that we have a directional light source $L^{in}(\omega_{in}) = s\delta(\omega_{in}-\omega_0)$.

Then, plugging this into our reflectance equation, we get

$L^{out}(\omega) = \int_{\Omega_{in}} f(\omega_{in},\omega_{out}) L^{in}(\omega_{in}) \cos(\omega_{in}) d\omega_{in}$

$= \int_{\Omega_{in}} \frac{\rho}{\pi} s \delta(\omega_{in}-\omega_0) (\omega_{in} \cdot n) d\omega_{in}$

$=\frac{\rho}{\pi} (s\omega_0 \cdot n)$.

There are some minor differences between this and the n-dot-l equation written on this slide. Specifically, the albedo $a = \frac{\rho}{\pi}$, and the lighting vector $l = s\omega_0$.