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.


1 Answer 1


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.

  • $\begingroup$ 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 $L$. $\endgroup$ Commented May 23, 2019 at 3:59
  • $\begingroup$ 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 $L \cos(\theta)$. $\endgroup$ Commented May 23, 2019 at 6:34
  • $\begingroup$ 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 $E=E$. $\endgroup$ Commented May 23, 2019 at 8:23

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