For a Lambertian surface, the radiance at a point is

$$L = \frac{\alpha}{\pi}\cos\theta_i E_0$$

If we have an infinite flat surface, albedo = 1, illuminated directly normal, then we get

$$L = \frac{E_0}{\pi}$$

But now suppose we are sitting very close to the surface. The surface occupies $$2\pi$$ sr of our view, so the irradiance we measure is

$$E_r = 2\pi L = 2E_0$$

Is this correct? I was expecting to get $$E_0$$ back again. $$2E_0$$ makes it feel like we're violating the conservation of energy or something.

I don't really understand what are you doing. I think your first equation shouldn't have a cos factor in it.

We have the relation,

$$BRDF = dL_r / dE_i$$

That is the brdf is the ratio of reflected radiance to incoming "irradiance". Re-arranging this gives us,

$$dL_r = BRDF * dE_i$$

For diffuse surfaces we know,

$$BRDF = \alpha/\pi$$

Substituting in above equation we have,

$$dL_r = \alpha/\pi * dE_i$$

Irrespective of what angle we view, the radiance is the same. Now for a diffuse surface with albedo = 1 we have what you said,

$$dL_r = dE_i/\pi$$

Now in order to measure Irradiance, we can use the relation,

$$E = \int_{2\pi} L \cos(\theta) d\omega$$

$$E = \displaystyle\int_{\phi = 0}^{2\pi} \displaystyle\int_{\theta = 0}^{\pi/2} L \cos(\theta) \sin(\theta) d\theta d\phi$$

$$E = E/\pi * \pi$$

$$E = E$$

So you got back the same irradiance.

• The first cosine is for the direction of incoming light, to get the "true" irradiance or whatever. E.g. $E_0$ coming in at a $90^\circ$ angle doesn't light up the surface at all. I think my discrepancy comes from the fact you integrated $L\cos(\theta)$ instead of just integrating $L$. May 23 '19 at 3:59
• 1). I know. But if you are putting in the cosine then you have to do it properly. $dE$ will change into $dL_f$. Because as shown below $dE = dL \cos(\theta) d\omega$ 2) Why would you just integrate $L$. You get irradiance by integrating $L \cos(\theta)$. May 23 '19 at 6:34
• Ah I thought I had already taken care of the $\cos\theta$ when I was calculating solid angle for my actual problem (not a flat plane). But that's just it, I needed $\cos^2\theta$ in total, and now I get $E=E$. May 23 '19 at 8:23