I'm approaching to radiometry and I'm struggling with one of the properties of Lambertian surfaces. We know that the radiance is the radiant flux emitted, reflected, transmitted or received by a surface, per unit solid angle per unit project area: $$L_{e,\Omega}=\frac{\partial^2 \phi_{e}}{\partial^{\Omega}\partial({Acos\theta})}$$ where $\Omega$ is the solid angle, $\phi_{e}$ is the radiant power and $Acos\theta$ is the projected area.

Since the radiant intensity $I_{e,\Omega}$ is the radiant flux emitted, reflected, transmitted or received, per unit solid angle, we can simplify the equation as $$L_{e,\Omega}=\frac{\partial{I_{e,\Omega}}}{\partial(Acos\theta)}$$ Now, I would like to introduce the hypotesis of Lambertian surface. We know that for a Lambertian surface the radiant intensity is proportional to the cosine of the angle between the direction of the viewer and the normal to the surface. If $I_0$ is the incident radiant intensity, the Lambert's cosine law states that radiant intensity at a certain angle $\theta$ is $$I=I_0cos\theta$$ The problem arises now. How can I mix up these information to demonstrate that the radiance is indipendent from the angle of view? If I just plug in the Lambert's cosine law, I get $$L_{e,\Omega}=\frac{\partial({I_0cos\theta})}{\partial({Acos\theta})}$$ How can I solve this derivative? Thank you in advance!


1 Answer 1


We have the function $f(\theta)=I_0\cos\theta$ for which we wish to find $\partial_{A\cos\theta}f(\theta)$. First, rewrite $A\cos\theta$ as $u$, and then we will express $f$ in terms of $u$:

$$\begin{align*}u&=A\cos\theta\\\implies\theta&=\cos^{-1}\frac{u}{A}\\\implies f(u)&=I_0\cos\left(\cos^{-1}\frac{u}{A}\right)\\&=\frac{I_0}{A}u\\\implies\partial_uf(u)&=\boxed{\frac{I_0}{A}}\end{align*}$$ which is indeed independent of $\theta$.


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