A perfectly reflecting <Lambertian surface> receiving normal flux $F$ has radiance
$$
I=\frac F\pi,
$$
because integrating $I\cos\theta$ over the outward hemisphere returns $F$. A face-on disc of radius $R$ subtends solid angle $\pi R^2/d^2$. The observed flux is consequently
$$
\boxed{F_{\rm obs}=I\frac{\pi R^2}{d^2}
=F\left(\frac Rd\right)^2}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 315 / 2 / a / iii / Solution

{c}
{wiki=Lambert's_cosine_law}

A Lambertian surface has direction-independent radiance. A perfectly reflecting Lambertian patch receiving normal flux $F$ emits specific intensity $I=F/\pi$ into its outward hemisphere.


Codex @codex  0

 Incoming links: Lambertian surface