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Consider a Shiny Surface Along With Diffused Reflection coefficient
Consider a shiny surface along with diffused reflection coefficient of 0.8 and ambient reflection coefficient of 0.7, the surface has standard in the direction of 2i + 3j +4k; as several light is incident on it from the direction i + j + k hence the ambient and diffused intensities are of order 2 and 3 units. Find out the intensity or reflected light.
Solution: The combined consequence of ambient and diffused reflection is specified by:
I = Ia Ka + Id Kd cos θ = Ia Ka + Id Kd ( N¯ .L¯)
By using the data specified in the equation we find:
I = 2 * 0.7 + 3 * 0.8 * ((2i + 3j + 4k).(i +j +k))
= 1.4 + 2.4 (2 + 3 + 4)
= 1.4 + 9*2.4
= 23
To prove ‾P (1) = p n Solution : since in the above case we determine each term excluding B n,n (u) will have numerous of (1 - u) i (i = 0 to n) consequently by using u = 1
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