Suppose $P_1$ and $P_2$ two different points on surface. And let

  • $l$: be the light source direction,
  • $n$: normal vector
  • $r$: specular reflection direction
  • $v$: viewing direction .

In case 1 (at $P_1$) the angel between $l$ and $n$ is $\theta_1.$ And specular reflection $r$ coincide with $v.$ So we get maximum specular intensity $I_1$,say. Like the image below:

enter image description here

In case 2 (at $P_2$) $l$ is positioned as case 1. But $n$ is different on $P_2$. The angle between $l$ and $n$ is $\theta_2.$ And in this case also specular reflection $r$ coincide with $v.$ So --let's say that-- the maximum specular intensity is $I_2$. In this case we have different angle between $n$ and $l,$ different $r,$ different $v$ from case1.

Like the image below :

enter image description here

My question is specular intensity $I_1$ has greater intensity than $I_2$ or $I_1$ and $I_2$ are equal ?

  • $\begingroup$ It depends on the brdf. $\endgroup$
    – lightxbulb
    Commented Jan 4, 2022 at 8:57
  • $\begingroup$ @lightxbulb it's not possible to say if there is greater angle between n and l then intensity is less? $\endgroup$
    – S. M.
    Commented Jan 4, 2022 at 9:47
  • $\begingroup$ That's the case for a Lambertian where you get a $\frac{k_d}{\pi}\cos\theta$ ($\cos\theta = \vec{n}\cdot\vec{l}$) modulating factor in the rendering equation. However if I pick a different brdf this doesn't have to be the case anymore. $\endgroup$
    – lightxbulb
    Commented Jan 4, 2022 at 11:02
  • $\begingroup$ @lightxbulb I don't know brdf. I know basic specular reflection. If there is greater angle between n and l then angle between r and n also increase. Then intensity also less. Am I correct? $\endgroup$
    – S. M.
    Commented Jan 4, 2022 at 11:14
  • $\begingroup$ See en.m.wikipedia.org/wiki/… Idk what brdf you mean by "basic specular reflection". If you mean ideal mirrors then the angle is irrelevant. If you mean (Blinn-)Phong then the angle is again irrelevant if $v$ and $r$ coincide. If you mean modified (Blinn-)Phong such that it is energy conserving (i.e. doesn't cancel $\cos\theta$ from the rendering equation), then the modulating factor in case 2 is smaller. $\endgroup$
    – lightxbulb
    Commented Jan 4, 2022 at 12:15

1 Answer 1


The rendering equation is the standard equation that people use to approximate geometric optics in computer graphics. The equation reads:

$$L(x, \omega_o) = L_e(x, \omega_o) + \int_{\Omega_x}f(\omega_o, x, \omega_i) L_i(x, \omega_i)\cos\theta_i\,d\omega_i,$$

where $L$ is the radiance function (we want to find this function to compute the radiance arriving at the virtual camera sensors in the scene corresponding to specific pixels). $L_e$ is the emitted radiance from surface point $x$ in direction $\omega_o$ and it is non-zero only for light-sources - it basically encodes the strength/colour of light sources in the scene. $\Omega_x$ is the upper hemisphere around the normal $N_x$ at point $x$. The Lambertian factor is $\cos\theta_i = \omega_i \cdot N_x$ where $\omega_i$ is a direction from $\Omega_x$. It accounts for light getting spread out over a larger area for larger angles with the normal. $L_i(x, \omega_i)$ is the radiance arriving at $x$ from direction $\omega_i$. In vaccum we have $L(\gamma(x,\omega_i), -\omega_i) = L_i(x,\omega_i)$, where $\gamma$ is the ray-tracing function. Finally $f$ is the brdf: it describes the light scattering properties of the surfaces in the scene - e.g. wood scatters light differently than metal or an ideal mirror.

As I understand it, you are only interested in one bounce reflected light, this is accounted by changing $L_i$ to $L_e$ and ignoring the direct emission:

$$L(x, \omega_o) = \int_{\Omega_x}f(\omega_o, x, \omega_i) L_e(r(x,\omega_i), -\omega_i)\cos\theta_i\,d\omega_i$$,

and from now on I'll set $y=r(x,\omega_i)$ for simplicity. You also seem to be considering a single directional light, which amounts to setting a Dirac delta light source: $L_e(y,-\omega_i) = \delta(l - \omega_i)$. Setting $v=\omega_o, \, n=N_x$ and plugging into the equation yields:

$$L(x, v) = f(v, x, l)(n \cdot l)$$

The behaviour wrt the angle between $n$ and $l$ in the above depends on $f$. Let us consider some examples. The specular part of the Phong brdf is: $f(v,x,l) = C\frac{(r\cdot v)^{\alpha}}{n\cdot l}$ which results in:

$$L(x, v) = C(r\cdot v)^{\alpha}$$

and thus if $r=v$ then the angle between $n$ and $l$ doesn't affect the radiance. Now consider the energy-conserving modification of the Phong brdf: $f(v,x,l) = C(r\cdot v)^{\alpha}$, which results in:

$$L(x, v) = C(r\cdot v)^{\alpha}(n\cdot l)$$

If $r=v$ and $\theta = \angle (n, l)$ then if $\theta_1<\theta_2$ it follows that $L_1>L_2$. Now consider some imaginary brdf $f(v,x,l) = C\frac{(r\cdot v)^{\alpha}}{(n\cdot l)^2}$, then you have:

$$L(x, v) = C\frac{(r\cdot v)^{\alpha}}{n\cdot l}$$

and if $r=v$ it follows that $\theta_1<\theta_2 \implies L_1<L_2$. From this you can conclude that you can get all possible cases based on your choice of brdf. My guess is that for most physically plausible brdfs however the following would hold: $\theta_1<\theta_2 \implies L_1>L_2$ due to Lambert's cosine law.

  • $\begingroup$ Please don't use comments for elongated discussions; this conversation has been moved to chat. $\endgroup$
    – wychmaster
    Commented Jan 6, 2022 at 22:17

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