I'm a little bit confused with the transition from just setting solid colors to object to involving light and its different components. For instance, the following formula $R = I K_d max(0, n.l)$ that computes the involvement of the diffuse component of light in the object's color. The part where we compute the dot product makes sense because indeed visually points in the object where the normal is orthogonal on the light vector should be dark. However, I can't get to understand what's the purpose of $K_d$. Is there any example of an object in real life that would for instance have low $K_d$ (hardly react to the existence of a light source)? Why would changing $K_d$'s value ever make sense?

Thank you.

  • 1
    $\begingroup$ You can think of $k_d$ as a descriptor for how much light the object reflects. Also $k_d$ is typically a 3 component vector in order to rrprrsent color (RGB). $\endgroup$
    – lightxbulb
    May 21, 2022 at 22:46

1 Answer 1


Let's ignore $K_d$'s dependence on wavelength for a moment (which @lightxbulb mentioned in the comments). Consider a bright white surface, and a grey one. The difference between them is precisely that the latter reflects less light; this, has lower value of $K_d$.

In general, one might loosely think of $K_d(\lambda)$ as the colour of the material (usually represented as a triple of values for $\lambda$ equal to reed, green, and blue lights wavelength).

  • $\begingroup$ My problem with the notion of "color of the material" is that it then get's multiplied by light's color and color multiplication makes little sense. $\endgroup$
    – Essam
    May 23, 2022 at 6:56
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    $\begingroup$ @Essam Color multiplication is reflection. Take a material that reflects $k_{d,r}=80\%$ of incoming red light (in a trichromatic model). And let the arriving red light be $L_r$, then the reflected light (ignoring Lambert's cosine law) is $k_{d,r} L_r = 0.8L_r$. This simply means that $20\%$ was absorbed. $\endgroup$
    – lightxbulb
    May 23, 2022 at 9:26

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