# Lambertian surface and the luminous intensity

I am currently studying the basics of photometry to better understand the rendering equation of Kajiya. One thing I'm currently struggling with is Lambert's cosine law.

Let's go over the premises:

• A lambertian surface scatters light evenly in all directions.
• The projected surface area decreases with the cosine of the viewing angle.

For the luminous intensity from a viewing angle $$I_\theta$$ I have the following equation: $$I_\theta = I_0 \cdot cos(\theta).$$ My question is: why should the luminous intensity be decreasing with the angle if I can still see the whole surface? If the light is scattering evenly in every direction, as was the premise for a lambertian surface, why should the projected surface area matter at all. Shouldn't I still recieve the same amount of light, but only in a more concentrated beam?

Thank's in advance to anyone who can help me get out of my confision, which I've been stuck in for the whole last week.

• I think there might be some confusion in what is meant by luminous intensity. Are you talking about the luminous intensity emitted by the Lambertian surface, or the luminous intensity received by a distant observer/detector? The cosine dependence applies to the former, while your claims are correct for the latter.
– Puk
Commented Dec 6, 2023 at 6:17

## 1 Answer

Assuming we are talking about the luminous intensity emitted/scattered by the Lambertian surface, maybe this will help. A piece of ordinary, non-glossy paper is a pretty good diffuse reflector. Each piece of it looks equally bright, regardless of viewing angle.

Now suppose you are looking at a sheet of paper from far away. The luminous intensity of the paper in a given direction $$\theta$$ (where $$\theta$$ is your viewing angle) is proportional to the luminous flux your eye receives. Each solid angle of the paper looks equally bright to you regardless of $$\theta$$. However, for large $$\theta$$, the entire sheet of paper looks smaller to you (i.e. it subtends a smaller solid angle in your field of view), so you receive less luminous flux, hence the $$\cos\theta$$ factor.

If you were thinking of the luminous intensity perceived by an observer in the direction of the source, while viewing the source at angle $$\theta$$, then you would be correct (precisely because the paper's brightness per unit solid angle is independent of viewing angle). But this is not normally what is meant by the luminous intensity of a source/scatterer.

• My confusion came from the fact that I was focussing too much on the luminous flux, which stays constant for each direction, whereas it's the solid angle which is responsible for the cosine dependancy. Thank you for your thorough explanation! Commented Dec 11, 2023 at 0:34