In the book AGI (Advanced Global Illumination), the Phong BRDF is



In Mitsuba and Tungsten, the Phong BRDF is

$f_r(x,\Psi\leftrightarrow\Theta)=[\cfrac{ratio_{diffuse}}\pi+\cfrac{(R\cdot\Theta)^n(n+2)(1-ratio_{diffuse})}{2\pi}]\cdot albedo$

Why there is a $\cfrac{n+2}{2\pi}$ in the specular part?

What's the relation between the two BRDFs?


1 Answer 1


It's the normalization factor to make sure the BRDF always reflects the same amount of energy regardless of the value of the specular exponent n. Without that factor, changing the specular exponent changes the overall reflectance of the material.

Further reading: The Blinn-Phong Normalization Zoo | The Tenth Planet Blog

  • $\begingroup$ Yes, I thought of this after posting the question. And I'm trying to calculate $k_s$. $\endgroup$
    – chaosink
    Aug 16, 2017 at 9:20
  • $\begingroup$ Wow, the link you give is amazing! $\endgroup$
    – chaosink
    Aug 16, 2017 at 9:22

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