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CS602 Assignment No 03 Fall 2019 Solution & Discussion Due Date: 22-01-2020

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sir plz cs602 3 assignment soultion jalde bana dyn

sir plz help kar dyn cs 602 3 assignment soultion plz

Question No:1                                                                                                                           Marks=15  

Suppose a Movie maker wants to create a night vision effect in a scene. The scene has light grey colored chair having RGB components (0.5,0.7,0.8). He has used four light sources. In terms of RGB the intensities of light sources are l1(0.4,0.7,0), l2(0.2,0.1,0.5), l3(-0.5,-0.6,0.5) and l4(-0.1,-0.2,0 ) ; Light source 3 and light source 4 are dark lights having negative colors.

 

  1. What will be the final light intensity? (5 marks)
  2. Calculate the overall color contribution of light in relation to the object surface? (5 marks)
  3. What color component(s) will remain absent from the scene? (3 marks)
  4. Which color the object will look like in presence of the light sources? (2 marks)

Hint: (You may check final color intensities given in color.pdf file)

 

Question No 2:                                                                                                                        Marks=05

Fiber Optics are widely used as optic cables to transmit light signals over long distances with minimal loss of data. It works on the principle of total internal reflection.

In order to simulate the fiber optic in computer graphics, calculate the critical angle where the inner core of fiber has index of refraction 1.5 while the cladding has index of 1.3.

ksi ko solution mila?

CS602 Assignment No. 3 Solution FALL 2019

Question no. 1:
Solution:
a. Final intensity of light is (0.4, 0.7, 0) + (0.2, 0.1, 0.5) + (-0.5, -0.6, 0.5) + (-0.1, -
0.2, 0) = (0, 0, 1)
b. Overall Contribution: id ⨂ cs = (0, 0, 1) ⨂ (0.5, 0.7, 0.8)
= (0, 0, 0.8)

c. Red and Green components will remain absent from the scene
d. The object will look blue in the presence of light sources.

Question no. 2:
Solution:
n1=index of refraction=1.5
n2=index of cladding=1.3
n1sinθc = n2sin90o
θc =
θc =
θc =

CS602 Solution 

link below link 

CS602_Assignment%20No.3_Fall%202019.doc

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