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jerrypiglet

Why would $dE(\omega_i) = dL(\omega_i)cos\theta_i$? Two sides of this equation seem to have different units?

kayvonf

Recall from the radiometry lecture that $\frac{dE}{dw} = L$. So $dE = Ldw$. Here, I'm writing dL as a shorthand for Ldw. And don't forget that the cosine comes from radiance L being defined in terms of differential area normal to the ray direction (as defined by $\omega$), and irradiance being an energy density with respect to differential area on the surface. This conversation is discussed in greater detail here.